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The Book ofIntegers
Iulia Gugoiu
Copyright © 2007 by La Citadelle
www.la-citadelle.com
FOR HIGH SCHOOL STUDENTSFOR HIGH SCHOOL STUDENTS
MATHEMATICSMATHEMATICS
Teodoru GugoiuB.Sc. Hons., M.Sc., B.Ed.Mathematics TeacherWilliam Lyon Mackenzie CIToronto District School Board
B.Sc., B.Eng.Mathematics & Science Teacher
Iulia & Teodoru Gugoiu
The Book of Integers - Mathematics for High School Students
© 2007 by La Citadelle4950 Albina Way, Unit 160Mississauga, OntarioL4Z 4J6, [email protected]
All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writting from the publisher.
ISBN 0-9781703-1-8
Edited by Rob Couvillon
“Simplicity is complexity resolved.”
“Don't look for obscure formulae or mysteries.It is pure joy that I am giving you.”
Constantin Brancusi, romanian sculptor
(see on the left side a drawing of “The Endless Column”)
Acknowledgements
We want to thank to all our students. Their feedback, comments, criticism, ideas, and requests help us to become better teachers.
Iulia & Teodoru Gugoiu
everyday
Topic
1. Understanding integers2. Absolute value, sign, and opposite3. Number line4. Comparing integers5. Applications of integers6. Addition of integers. Axioms7. Additions of integers using the number line8. Addition of integers using rules9. Subtraction of integers10. Order of operations (I)11. Equalities and inequalities12. Equivalent equalities and inequalities13. Equations14. Inequations15. Multiplication of integers (I)16. Multiplication of integers (II)17. Order of operations (II)18. Division of integers (I)19. Division of integers (II)20. Order of operations (III)21. Equalities and equations22. Proportions and equations23. Inequalities and inequations24. Powers25. Exponent rules26. Order of operations (IV)27. Divisors28. Divisibility rules29. Prime factorization30. Set of divisors31. Highest Common Factor (HCF)32. Least Common Multiple (LCM)33. Square roots34. Cubic roots35. Roots of superior order36. Roots rules37. Equations with powers and radicals38. Link between radicals and powers39. Order of operations (V)40. Substitution
Answers
Page
56789
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45
Content
© La Citadelle www.la-citadelle.com
The Book of Integers
1. Understanding Integers
5
Iulia & Teodoru Gugoiu
1.1 (Definition) Integers are signed numbers made of:a) a sign, + for positive numbers and - for negative numbers b) an absolute value, that is a natural number Example: -5.The sign is - . The absolute value is 5.
1. Identify whether the following numbers are integers or not:
1.2 (No Sign Rule) If an integer does not have a sign, by default the sign is considered positive. So, the numbers 5 and +5 are identical.In general:
1.3 (Number Zero) 0 is considered neither positive nor negative. However, some generalizations require us to consider the following numbers: +0 and -0. Still, these two instances of the number zero are equal to 0. So:
1.4 (The Set of Integers) The set of all integers can be represented by using the set notation as:
1.5 (Boundless) The set of integers is unlimited or boundless. There is neither a biggest integer nor a smallest integer. The number of elements in this set (the number of integers) is infinite.
1.6 (Infinity) Although a largest integer does not exist as a regular number, the symbol (called plus infinity) is used to express an unlimited large positive number.Similarly, the symbol (called minus infinity) is used to express an unlimited large (as absolute value) negative number.These two symbols are not considered numbers because the arithmetic of these numbers is completely different from the arithmetic of regular numbers. For example:
¥+
¥-
2. For each of the following integers, identify the sign and the absolute value:
3. Use the short notation (by dropping the sign) to rewrite the following integers:
4. Rewrite the following numbers as integers (including the sign):
5. Classify the following numbers as positive, negative or neither:
6. Find a way to represent the set of negative integers (see 1.4 for the set of positive integers).
101)22)3)4)10)11)2)0)2)3) -----+ jihgfedcba
3)0)1)2)5) +++++ edcba
55)41)300)20)1) edcba
1000)123)11)7)5)0)10)2)5)1) -----+ jihgfedcba
10. Find the value of the following expressions that contain the infinity symbol:
aa +=
000 -=+=
,...}3,2,1,0,1,2,3{..., +++---=Z
,...}3,2,1{ +++=+Z
definednotis¥-¥
¥=+¥ 5¥
´¥¥
¥´¥
¥
¥¥¥´¥+¥¥+¥-¥
2)
1)10)
1000))
/))10))1)
3 jihgf
edcba
25,13,1,3 -+-+
)10)01.0)3)3
2)1)5)5.7)0)2)1) kjihgfedcba ---+ ¥-
7. Use the set notation (see 1.4) to represent all odd positive integers greater than 0 and less than 10.
8. Find the logical value (true or false) of each of the following sentences:
a) All integers are positive.b) There is a largest positive integer.c) The smallest integer is 0.d) Any natural number is also an integer.e) There are more positive integers than negative integers.f) is a positive integer.¥+
9. For each case, identify the integers that satisfy the given properties:
a) is positive and has an absolute value equal to 7b) is negative and has an absolute value equal to 11c) has an absolute value equal to 5d) has an absolute value equal to 0e) is positive and has an absolute value equal to ¥
The set of positive integers is:
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The Book of Integers
2. Absolute value, sign, and opposite
6
Iulia & Teodoru Gugoiu
1. Use absolute value notation (see 2.1), and find the absolute value of the following integers:
2.5 (Opposite Numbers) Two integers are called opposite if they have the same absolute value but opposite signs.So, +5 and -5 are opposite.+5 is the opposite of -5-5 is the opposite of +5
2.6 (General Definition) The opposite of any number a is -a. So, to get the opposite of a number, just place the - sign in front of the number. Examples:The opposite of 5 is -5.The opposite of -5 is -(-5) that is equal to 5.
2.7 (The Opposite of an Opposite) The opposite of an opposite of a number is equal to the number itself:
Example. Let’s consider the number 3. The opposite of this number is -3. The opposite of the number -3 is +3, which is equal to the original number 3.
2.8 (The Opposite of Zero) The opposite of 0 is 0:
2.1 (Absolute Value) If you drop the sign of an integer, you get its absolute value. The absolute value of an integer is a natural number. Use the function | | to express the absolute value. Examples:
(Read: the absolute value of -4 is 4)
2.2 (Number Zero) The absolute value of 0 is 0.
2.3 (Sign) If you drop the absolute value of an integer you get its sign. Use the function sign() to express the sign of an integer. Examples:
2.4 (Number Zero) The sign of 0 is not defined. 0 is neither a positive nor a negative number.
© La Citadelle
4|4| =-
7|7||7|
6|6|
=+=
=+
0|0| =
+=+=
-=-
+=+
)10()10(
)7(
)5(
signsign
sign
sign
aa =-- )(
000 =+=-
2. For each of the following absolute values, find the possible value(s) of the integer(s):
3. For each case, find the sign of the integer (see 2.3):
4. For each case, find the opposite of the given integer:
5. Find the value of each expression. One case is solved for you as an example:
6. Find the value of each expression that contains the absolute value function. One case is solved for you as an example:
7)7)12)100)10)0)3)2)2)3) -----+ jihgfedcba
123)5)10)3)2)1)0) gfedcba
7)1)1)10)100)10)0)4)5)3) jihgfedcba -+---
77)7)2)11)0)20)3)3)20)12) -+---- jihgfedcba
)))32((()))35(())745())811())21()
)))20(())))4(())3())5())2()
+-+--+--+---+-
+-+----++---
jihgf
edcba
|2||8|)|3||5|)|2|2)|2||5|)|1||3|)
|)7(|)|3|)|)4(|)|25|)|)41(|)
-¸--´--´----+-
-----+--+-
jihgf
edcba
Example:
5)5())5(()))5((()))32((() =--=-+-=-+-=+-+-j
7. Find the value of each expression that contains the absolute value function. One case is solved for you as an example:
||)2|1(|)||8||)||55|5|)|)2(|)|)47(|)
||3||7||)|3||5|)3|2|1)2|3|))1|3(|2)
|2|5)|5||13|))1|5(|)|5|
|10|)|3||2||1|)
-+-----------
------+-++-+-´
---´-+---
--+-+-
onmlk
jihgf
edcba
Example:
325|2||5|) =-=---g
Example:
4|4||37|||3||7||) ==-=---j
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The Book of Integers
3. Number Line
7
Iulia & Teodoru Gugoiu
1. For each point represented on the following number line, find the corresponding integer:
3.1 (The Number Line) An intuitive representation of integers is based on the number line. There is one to one correspondence between the set of integers and a set of equidistant points on a line. This correspondence is illustrated in the following figure:
3.2 (Positive and Negative Integers) The positive integers are situated to the right side of the origin O. The negative integers are situated to the left side of the origin O.
3.3 (Number Zero) Number zero corresponds to the origin O.
© La Citadelle
0 +1 +2 +3-1-2-3-4 +4
O
3.4 (Absolute value) The absolute value of an integer is equal to the distance between the origin O and the corresponding point on the number line. The greater the absolute value of an integer, the greater the distance between the corresponding point on the number line and the origin.
0 +1 +2 +3-1-2-3-4 +4
O M
3|3| =+
3.5 (Opposites) Two integers that are opposite to each other correspond to points on the number line that are symmetrically positioned relative to the origin O.
0 +1 +2 +3-1-2-3-4 +4
O NM
Example. The absolute value of the number +3 is 3. M is the point corresponding with +3 on the number line. The distance between M and O is equal to 3.
Example. The numbers -3 and +3 are opposite to each other. -3 corresponds to point M on the number line and +3 corresponds to N. The points M and N are symmetrically positioned relative to the origin O. The distances MO and NO are both equal to 3.
2. Plot the following points on the number line below:
3. A part of a number line is represented in the figure below. All the points are equidistant. Find the integers that correspond to the given points:
4. Find the corresponding integer and its absolute value, by calculating the distance between each point and the origin O.
5. For each of the points below (A, B, C, D, E, F), find the symmetrical point relative to the origin O (see 3.5):
+2
A
O
BC DE
+2+2+1 +3 +4 +5 +6 +7 +8 +9-8 -7 -6 -5 -4 -3 -2 -1 0
)3())4())5())7())2() --+- EeDdCcBbAa
-15
D EA
-35
B C
D EA B C
+10
+2+2+1 +3 +4 +5 +6 +7 +8 +9-8 -7 -6 -5 -4 -3 -2 -1 0
A B C D E
6. For each case, use the number line to identify an integer that is:
a) 2 units left of +7b) 3 units right of +2c) 2 units left of -5d) 4 units right of -2e) 3 units from +5f) 2 units from -6g) 3 units from the origin
+2+2+1 +3 +4 +5 +6 +7 +8 +9-8 -7 -6 -5 -4 -3 -2 -1 0
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The Book of Integers
4. Comparing Integers
8
Iulia & Teodoru Gugoiu
1. For each case identify the true statement:4.1 (Less, Equal, Greater) Any two integers a and b can be compared. The result of this comparison is that only one of the following statements is true:
4.4 (Comparing negative integers, 0, and positive integers) Any negative integer is less than 0 and 0 is less than any positive integer. Examples:
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4.2 (The Number Line) On the number line a smaller integer is placed at the left side of a bigger integer. So:
4.5 (Transitivity) The order (comparison) relation between two integers has the transitivity property:
bthatgreaterisaba
btoequaltisaba
bthatlessisaba
>
=
<
...3210123... <+<+<+<<-<-<-<
4.3 (Order) As you can see from the previous relation, the set of integers is an ordered set. Integers can be sorted (ordered) from smaller integers to bigger integers. There is neither a smallest integer nor a largest integer.
cathencbandbaif <<<
Example:
323002 +<-+<<- thenand
So, any negative integer is less than any positive integer.
4.6 (Comparing positive integers) When you compare two positive integers, the greatest integer is the one with the greatest absolute value. Example:
3553 +>++<+ or
4.7 (Comparing negative integers) When you compare two negative integers, the greatest integer is the one with the smallest absolute value. Example:
7557 ->--<- or
Indeed, -7 is at the left side of -5 and therefore is smaller.
70
02
+<
<-
07
20
>+
->
...3210123... >->->->>+>+>+
2. Sort the integers from the smallest to the greatest (use the < symbol):
3. Sort the integers from the greatest to the smallest (use the > symbol):
5. Place the symbols <, =, or > between each pair of integers to obtain true statements:
4. Use the transitivity property to combine the two true statements and get a third true statement (see 4.4):
6. Find the logical value (true or false) of each statement:
9. Find the set of values of the integer x so each statement is true:
8. Find the logical value (true or false) of each statement (Note: a<b<c is true if both a<b and b<c are true):
00;00;00)01;10;01)33;33;33)
50;50;50)35;35;35)21;21;21)
>=<>=<>=<
>=<>=<>=<
fed
cba
3;2;1;0;2;3)3;3;2;2;1;1)2;0;2;1;3)
0;2;1;3)2;1;0)1;2)
---++-+-+---+
---++
fed
cba
5;1;1;2;3)5;5;4;4;2;2)4;0;2;2;3)
0;1;2;3)2;1;0)1;2)
---++-+-+--++
------
fed
cba
1223)3110)4225)
1223)2002)2110)
-<--<-->-->+<<-
>><<-<<
andfandeandd
andcandbanda
57)55)42)24)73)
00)50)11)01)21)
+-+++----
---
jihgf
edcba
33)21)53)24)75)
11)50)11)22)23)
+>-->+<+->--<
+>-<-=+>->
jihgf
edcba
7. Two more comparison operators are (greater or equal operator) and (less or equal operator). Find the logical value (true or false) of each statement:
£³
03)20)033)45)32)
11)22)21)10)11)
³--³+³--£-+£
+£-³+-³-£³
jihgf
edcba
203)310)135)135)011)
113)021)321)101)210)
->>+->->-<-<-->->->->
+>->->>+>->-<<-<<
jihgf
edcba
22)30)40)14)31)
51)01)31)12)31)
->³->>+££->>->>-
£<³³->>+<<-<<
xjxixhxgxf
xexdxcxbxa
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The Book of Integers
5. Applications of Integers
9
Iulia & Teodoru Gugoiu
1. Use integers to describe the position of an object. The object is positioned:
A) 5 units right of the origin B) 4 units left of the originC) in the origin D) 1 unit rightE) 2 units left
5.1 (Position) Let’s suppose that an object is placed at point A on the number line.
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5.2 (Displacement) A change in position of the object is called displacement. The displacement of the object from the point A to the point B can be described:a) in words as: 5 units to the rightb) using an integer as: +5
The position of the object can be described:a) in words as: 2 units left of the origin Ob) using an integer as: -2If the object is moved from point A to point B, its new position can be described:a) in words: 3 units right of the origin Ob) using an integer: +3
0 +1 +2 +3-1-2-3-4 +4
OA B
0 +1 +2 +3-1-2-3-4 +4
OA B
+5
The displacement of the object from point B to point C can be described:a) in words: 2 units to the leftb) using an integer: -2
0 +1 +2 +3-1-2-3-4 +4
O C B
-2
5.3 (Temperature) Temperature can be expressed using integers. The temperature of 0 temperature of 20
0
0
°C corresponds to 0. A °C above °C is +20. A temperature 30°C below 0°C is -30. The smallest possible temperature is approximately 273°C below
°C and corresponds to -273.An increase in temperature of 10°C corresponds to +10 and a decrease in temperature of 40°C corresponds to -40.
5.4 (Altitude) Altitude can be expressed using integers. Sea level corresponds to 0. An altitude of 100m above the sea level corresponds to +100. A depth of 50m below sea level corresponds to -50. Rising 200m corresponds to +200 and falling 500m corresponds to -500.
5.5 (Bank Account) Bank account can be expressed using integers. A debit of $100 corresponds to +100. A credit of $25 corresponds to -25. A deposit of $50 corresponds to +50 and a withdraw of $75 corresponds to -75.
2. Use integers to describe the displacement of an object. The object is:
A) moved 2 units to the right B) moved 2 units to the leftC) fixed in the origin D) moved 3 units rightE) moved 5 units left
3. Use integers to describe the outside temperature. The temperature:
A) is 10 degrees above 0 B) is 15 degrees below 0C) for melting ice D) for boiling waterE) the lowest possible
4. Use integers to describe the change in the temperature. The temperature is:
A) increasing by 10 degrees B) decreasing by 5 degreesC) the same D) 3 degrees moreE) 10 degrees less
5. Use integers to describe the position of an object relative to sea level. The object is:
A) 1000m above B) 2m belowC) 20m under sea level D) at sea levelE) 7m higher than sea level
6. Use integers to describe the change in the altitude of an object. The object is:
A) raised 2m B) lowered 5mC) moved 7m upward D) moved 3m downwardE) not moved at all
7. Use integers to describe the balance of a bank account. The balance is:
A) a debit of $500 B) a credit of $150C) the initial value D) a debt of $31E) $125 cash
8. Use integers to describe the transaction in a bank account. The transaction is:
A) a deposit of $50 cash B) a deposit of a $550 chequeC) a withdraw of $40 cash D) a payment of a $45 billE) just updating the address
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The Book of Integers
6. Addition of Integers. Axioms
10
Iulia & Teodoru Gugoiu
1. For each case, identify the addends and the sum:6.1 (Addition) Addition is a binary operation (represented by the symbol +) between two operands (called terms or addends) to obtain a result called the sum. Example:
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The terms (addends) are 3 and 2.The sum is 5.
6.2 (Commutative Property) The order in which the addends are added is not important. This is the commutative property of addition. This property can be written as:
5)2()3( +=+++
abba +=+
Example:
6)2()4()4()2( +=+++=+++
6.3 (The Additive Identity) 0 is the additive identity of addition satisfying the property:
aaa =+=+ 00
Example:
5)5(00)5( -=-+=+-
6.4 (Opposites Cancellation) The sum of a number (integer) and its opposite is 0:
0)()( =+-=-+ aaaa
Example:0)3()3( =-++
6.5 (The Associative Property) If an expression contains more than 2 terms then the order in which the additions are done is not important. For 3 terms this property states that:
)()( cbacba ++=++
Example:
6333)21(
651)32(1
=+=++
=+=++
6.6 (Application) By combining the previous properties the sum of a complex addition (involving more terms) might be calculated very easily.Example:
7)7(00
)7()]4()4[()]2()2[(
)4()7()2()4()2(
-=-++
=-+-+++++-
=-+-+++++-
426)10)6(4)4)3(7)
4)2()2()1)1()2()321)
+==++=-+
+=++++=-++=+
fed
cba
2. Use the commutative property (see 6.2) to rewrite these additions:
)5()4()4)3()0)1()
05)32)10)
-+-+-+-
+++
fed
cba
3. Use the additive identity property (see 6.3) to find each sum:
)5(0)0)4()0)1()
05)10)00)
-++++-
+++
fed
cba
4. Use the opposites cancellation property (see 6.4) to find each sum:
)11(0)11(0))3(03)50)5()
)7()7())0()0())10()10()
)5(5))2()2())1(1)
+++-+-+++++-
-++-++++-
-++++--+
ihg
fed
cba
5. Use the associative property (see 6.5) to find each sum easily. One case is solved for you as an example. You might also need to apply the commutative property.
)11(511))7()7(3)5)5()3()
)37)17((17))316(14))75()5()
1)195())2())2(5()7)35()
+++-++-++-+-
+-+++++-
++++-+++
ihg
fed
cba
6. Group the terms conveniently (see 6.6) to find the sum. One case is solved for you as an example:
54)5(3)9(2)1313)4(17)19)3(11)
8)6(42)8642))1(39)3(1)
52)5(3)6)10(4))2()1(321)
)2()4()2()4()35)3())1()2(1)
++-++-++++-++-+
+-+++++-+++-+
++-++-+-+-+++
-+++++-++--+-+
lkj
ihg
fed
cba
Example:
000)9)9(()55(
))54()9(())5()32((54)5(3)9(2)
=+=+-+-=
=++-+-++=++-++-+l
Example:
550)5())11()11((
)511()11()11()511()11(511)
=+=++-++=
=+-++=+++-=+++-i
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The Book of Integers
7. Addition of integers using the number line
11
Iulia & Teodoru Gugoiu
1. Use the number line to find each sum:7.1 (Addition Redefined) The addition operation can be interpreted as:
old value + change = new value
Example. Let’s find the sum of:
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using this new definition of addition together with the number line:
+3 is the old value-5 is the change-2 is the new valueSo:
7.2 (More addends) If more terms (addends) are to be added, then the meaning of the addition operation is:
old+change+change+...+change=new
Example. Let’s find the sum of:
So:
)5()3( -++
0 +1 +2 +3-1-2-3-4 +4
O
-5
)1()5()3()6()4( -+++-+++-
0 +1 +2 +3-1-2-3-4 +4
+6-3
+5-1
7.3 (The hidden addition sign) When adding more numbers, usually the addition operator (+) between the numbers can be dropped without affecting the clarity of the expression. So, an expression as:
14532 -+-+-
is in fact equivalent of:
)1()4()5()3()2( -+++-+++-
Let’s use number line to find the sum:
0 +1 +2 +3-1-2-3-4 +4
+3-5
+4
-1
So:
2)5()3( -=-++
3)1()5()3()6()4( +=-+++-+++-
114532 -=-+-+-
2. Find the addition operation that corresponds to each arrow on the number line:
3. Use the number line to find each sum:
4. Find the addition operation that corresponds to the sequence of arrows on the number line:
5. Simplify the following expressions by hiding the addition operators between two consecutive terms. Do not calculate the sum.
6. Rewrite each expression by revealing the hidden addition operators. Do not calculate the sum.
7. Use the number line to calculate each sum:
)5()3()0)2())6()2())5()3()
)1()2())1()4())3()2())3()4()
)2()1())2()3())3(0))2(0)
-+++-++--++
-+-++--+--++
++-+++-+++
lkji
hgfe
dcba
0 +1 +2 +3-1-2-3-4 +4
a)
b)c)
d) e)
)3()3()2()2())3()1()2()5())1()2()1()
)6()4()2())3()2()1(0))3()2()1()
)3()5()2())3()1()3())3()2()1()
-+-++++-+-+-++-+-+-
-+++--+++-+++-++
-+-++-+++--+++
ihg
fed
cba
0 +1 +2 +3-1-2-3-4 +4
)5()4()2()1())1()3()5())3()2()2()
)5()2())2()3())2()1()
-+++++--+-++++-+-
-+-++--++
fed
cba
212345)3322)4321)
53)32)21)
---+-++--+-+-
--+--
fed
cba
642531)10631)8421)
7654321)5342)5432)
5420)531)222)
2345)321)531)
+++---+-+--+-
+-+-+-++---+-
-+-+-----
-+----+-
lkj
ihg
fed
cba
8. Find the value of each expression. One case is solved for you as an example:
|2||6|5)|5||2|)|3|2)|52|) ---------- dcba
Example:
321265|2||6|5) -=--=--=----d
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The Book of Integers
8. Addition of integers using rules
12
Iulia & Teodoru Gugoiu
1. For each case, calculate the sum of the positive integers (see 8.2):8.1 (Addition Rules) The method of using the number line to add two or more integers is not useful for adding large numbers (in absolute value). In these cases the addition rules presented bellow must be used.
© La Citadelle
Examples:
8.2 (The Addition of Positive Integers) If you add two or more positive integers:a) the sign of the result is positiveb) the absolute value of the result is the sum of the absolute values of each of the terms
...)(...)()()( ++++=++++++ cbacba
9)423()4()2()3(
7)52()5()2(
+=+++=+++++
+=++=+++
8.3 (The Addition of Negative Integers) If you add two or more negative integers:a) the sign of the result is negativeb) the absolute value of the result is the sum of the absolute value of each of the terms
...)(..)()()( +++-=+-+-+- cbacba
Examples:
8.4 (The Addition of Integers with Opposite Signs) If you add two integers with opposite signs:a) the sign of the sum is the sign of the integer with the biggest absolute valueb) the absolute value is the difference between the greatest absolute and the smallest absolute valueCase 1. If the absolute value of the positive number is greater than the absolute value of the negative number:
10)4321(4321
8)413()4()1()3(
4)31()3()1(
-=+++-=----
-=++-=-+-+-
-=+-=-+-
)()()( bAbA -+=-++
Case 2. If the absolute value of the negative number is greater than the absolute value of the positive number:
)()()( aBBa --=-++
Examples:
Examples:
3)36(63
2)35()3()5(
+=-+=+-
+=-+=-++
3)14(14
5)27()7()2(
-=--=+-
-=--=-++
2. For each case, calculate the sum of the negative integers (see 8.3):
3. In each of the following cases, the absolute value of the positive integer is greater than the absolute value of the negative integer. Find the sum:
4. In each of the following cases, the absolute value of the negative integer is greater than the absolute value of the positive integer. Find the sum:
5. For each case, calculate the sum:
333221)300020010)302510)321)
5432112345)3525550)3251250)1351000)
)375()25())35(200)123321)24101)
3321))17(5))7()3()31)
++++++++
++++
++++
+++++++
ponm
lkji
hgfe
dcba
5150015050200)552510)432)
1001234)33332222)35001000)
)350(450))125()125())25(200)
)3()3())3()2()21)
-----------
------
-+--+--+-
-+--+---
lkj
ihg
fed
cba
)50()150())3333()1111()300250)12502500)
)125()550())300()100()12525)50100)
)15()55())55()20()2015)510)
75))6()3())2()4()35)
-++++-+--
-++++-+--
-++++-+--
+-++--++-
ponm
lkji
hgfe
dcba
)350()125())2200()4400()4005500)2000150)
)250()150())150()600()125150)20050)
)40()15())25()30()1025)103)
49)85))3()7())4()3()
-++++-+--
-++++-+--
-++++-+--
+--++--++
ponm
lkji
hgfe
dcba
)2()3()1()5()3()1()1550)0325125)
)300()125()125()1505)220200)
1001000100)100101)502010)
655)420)4321)
++++-+-+-++----
-+++--+++
--------
+-+-----
lkj
ihg
fed
cba
6. Group the terms conveniently to find more easily each sum. One case is solved for you as an example:
252015105)15662534)1523105)
46545)7283)16743)
-+-+---+-+-+--
-+-++--++-+-
fed
cba
Example:
90)10100(10100)1525()6634(15662534) -=--=+-=-+--=--+-e
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The Book of Integers
9. Subtraction of integers
13
Iulia & Teodoru Gugoiu
9.1 (Subtraction) Subtraction is a binary operation between an operand called the minuend and an operand called the subtrahend to get a result called the difference:
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differencesubtrahendminuend =-
9.2 (Number Line) You can use the number line to calculate the difference between two integers. To do that use this rule:
changevalueoldvaluenew =-
Example:
0 +1 +2 +3-1-2-3-4 +4
O
+5
)2()3( --+
where:-2 is the old value+3 is the new value+5 is the change from the old value to the new valueSo:
5)2()3( +=--+
9.3 (Subtraction Redefined) Subtraction between a minuend and a subtrahend can be redefined as an addition between the minuend and the opposite of the subtrahend:
)( baba -+=-
Example:
9)5()4()5()4( -=-+-=+--
9.4 (Anti-commutative Property) The subtraction operation has the anti-comutative property:
)( abba --=-
Example:
2)3()5(
2)5()3()5()3(
+=+-+
-=-++=+-+
9.5 (Sign Rules) The following sign rules can be useful when dealing with addition and subtraction of integers:
aaa
aa
aa
aaa
=+=--
-=+-
-=-+
=+=++
)(
)(
)(
)(
Examples:
23553)5()3(
945)4()5(
3)3())3((
=-=+-=--+-
-=--=-++-
=--=+--
1. Use the number line to find each difference (see 9.2):
2. Find the subtraction operation that corresponds to each arrow on the number line (see 9.2):
)5()7())5(2)52))5()3()
)2(0))1()4())3()2())1()5()
)2()2())2()3())1()3())1()2()
--------+
+-+-+---+--
---+-++----+
lkji
hgfe
dcba
0 +1 +2 +3-1-2-3-4 +4
a) b)c)
d) e)
3. Rewrite each subtraction as an addition with the opposite, then find the result:
)7(5))2()7())3()2())5()3()
)1(3))1()4())4()2())2()3()
)2(5)35)40)51)
+----------+
----++-++-+
-----
lkji
hgfe
dcba
4. Use the anti-commutative property to calculate the differences (see 9.3):
143123))5(0))12()6()107)
144))8()3()40)73)
-+-+-+-
-+-+--
hgfe
dcba
5. Use the sign rules to simplify the following expressions (see 9.4):
)))5((()))5(()))6(())4()
))2(())3())3())5()
-+---+------
+--+--+++
hgfe
dcba
6. Find the value of each expression. One case is solved for you as an example:
))1(())1(()))3(())5(()))1(())4(()
))6(())2(())))4((())6(()))6(())4(()
))5(()3()))2(()3())3()5()
+--+-+---+---+----
-+--+--+-+-+-------
+--+---+--+-++
ihg
fed
cba
+2+2+1 +3 +4 +5 +6 +7 +8 +9-8 -7 -6 -5 -4 -3 -2 -1 0
Example:
1046)4(()6()))4((())6(() =+=--+--=-+-+-+-e
7. Find the value of each expression. One case is solved for you as an example:
|3||2||1|)|7|5||)|3||4|)|5||2|) -+---+--------+ dcba
Example:
2|2||75||7|5||) =+=+-=+--c
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The Book of Integers
10. Order of operations (I)
14
Iulia & Teodoru Gugoiu
1. Use the “left to right” rule (see 10.1) to find the result of each sequence of operations:
10.1 (Priority) The addition and the subtraction are considered operations of the same priority. Therefore, if more operations appear within an expression then the operations must be done in the order they occur (from left to right). Example:
© La Citadelle
330344
3413
34152
-=-=--=
=--++=
=--++-
10.2 (Grouping) Sometimes it is useful to group all the positive terms and negative terms separately:
51512
)753()246(
752436
-=-=
=++-++=
=--++-
10.3 (Brackets) To change the order of operations, you can use brackets. If brackets appear within an expression then the operations inside of the brackets must be done first. Example:
)()( dbecaedcba +-++=+-+-
Example:
51311)3(1
)43()52(1
=++=+--=
=+-+--
10.4 (Nested Brackets) In complex expressions the brackets can be nested one inside of another. In this case start with the inner-most operations . This rule might be applied more than once if necessary. Example:
4)1(12
)232()12(2
))2(32())1(2(2
))64(32()3)21(2(2
-=---=
=+--+-+-=
=-------+-=
=----+---+-
10.5 (Standard Grouping Symbols) To differentiate nested grouping symbols, three types of standard grouping symbols are used: ( ) called parentheses, [ ] called square brackets, and { } called braces. By convention the order of nesting is:
{ [ ( ) ] }Example:
10}12{2}2]5[5{2
}2]1)0()1(3[5{2
}2]1)165()32(3[5{2
=+-=+--+-=
=+-----+-=
=+-+--+----+-
52015105)65432)28765)
2345)5432)4321)
432)143)321)
-+---+-+-+-+-
-+-++--+-+-
-+----+
ihg
fed
cba
2. Use grouping of positive and negative terms (see 10.2) to find the result of each sequence of operations:
5030252015)246810)654321)
3145)3251)4143)
432)142)523)
-+--+-+--+-+-
---++--+---
++-+---+
ihg
fed
cba
3. For each case, first do the operations inside of the brackets, then find the result (see 10.3):
2)54()32(1))75()12(0))52()25()31()
)74()53(1))43()52())42()42()
)52(0))53(2))42(1)
--+--+--+-+-----
----------+
---+-+--
ihg
fed
cba
4. For each case, complete the operations starting with the inner-most brackets (see
4)3)2)1)65(4(3(2(1))1)24()75(()43()
)))14((()))14((()))85(3(2)
)1)1)3((())4)53(()25()
2)1)53(2()))21(3(2)
----------+------
--+--+--+---+-
-+-----+---+-
+------
hg
fe
dc
ba
5. For each case, find the result using the order of operation for “standard grouping symbols” (see 10.5):
1}2]1)97(2[3{2))]}13()95[()]53()21[({1)
)]}32(3[2{1))]21(3[)}32()]21(1{[)
)]}26(5[)75({2))]43(3[)]43(3[)
]4)29(3[)75()]1)74(2[1)
----+-+-+---------
+-------+-----
+------++---
+---+-------
hg
fe
dc
ba
6. Find the value of each expression. One case is solved for you as an example:
|4)54(3|)|73|)73()|)74(3|)
|)3||2(||2|)|)3|2(|)7|5()|)5|3(2)
----------
--------------
fed
cba
Example:
112)1(2)32(2|)3||2(||2|) -=+-=---=---=------c
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The Book of Integers
11. Equalities and Inequalities
15
Iulia & Teodoru Gugoiu
1. Find the logical value (true or false) of each statement (see 11.1):11.1 (True/False Statements) In mathematics ideas are expressed as statements. A statement can only be true or false.Example: “3<4” is a true statement.Example: “3 is a divisor of 10” is a false statement.
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11.2 (Relational Operators) Some of the most common types of statements are binary relations involving the following relational operators:
Examples:
11.4 (Symmetry) Equality has the symmetry property that states:
bababa
bababa
¹³>
=£<
truetrue
falsetrue
22;11
20;03
-³--¹
+=<-
11.3 (Equality) If the relational operator is “=” then the binary relation is called equality. Examples:
false
true
4532
00
-=-
=
or:
“a=b“ is true if and only if “b=a“ is true.“a=b“ is false if and only if “b=a“ is false.Example:
523352 +-=+Û+=+-
abba =Û=
abifonlyandifba ==
11.5 (Inequality) If the relational operator is not “=” then the binary relation is called inequality. Examples of inequalities:
truetrue
truefalse
375;221
321;321
-³--<--
=+¹+
11.6 (Mirroring) For any inequality there is an equivalent inequality obtained by mirroring of the original:
abba
abbaabba
abbaabba
¹Û¹
£Û³<Û>
³Û£>Û<
;
;
The original inequality is true if and only if the mirrored inequality is true. The original inequality is false if and only if the mirrored inequality is false. Examples:
2222
533353
1001
-¹+Û+¹-
+->Û<+-
<Û>
121)2)93)
215)10)21)
=--
>=
fintegeraniseofdivisoraisd
bydivisibleiscba
2. Find the logical value (true or false) of each statement containing operational operators (see 11.2):
22)2515)52)22)
11)30)10)22)
10)12)11)11)
->-->+-³-+£
+>-->+<-¹
->>-+=-=+
lkji
hgfe
dcba
3. Do the required operations on both sides and find the logical value (true or false) of each of the following equalities (see 11.3):
50155)143532)123321)
3652)862)213)
542)121)321)
-=-+-=-++-=-+
-=--+=---=-
-=--=-=+
ihg
fed
cba
4. Use the symmetry property (see 11.4) to rewrite each equality and then find the value (true or false) of each statement:
)74(30)317531)2453)
753)742)231)
---=--=-+--=+-
-=---=--=
fed
cba
5. Do the required operations on both sides and find the logical value (true or false) of each of the following inequalities (see 11.5):
0)52()74(1))62(153)
)2010(15)2015(10))83(31)23(2)
228642))41(2)73(1)
2251)5162)
>--+---->-
--£--+--³+--
--¹-+----£--
--³---<--
hg
fe
dc
ba
6. Use mirroring (see 11.6) to rewrite the following inequalities and then find the value (true or false) of each statement:
432321))22()85(1)364532)
22)21)10)
+-¹+----£--+-³+-
+<-->->
fed
cba
7. Find the logical value (true or false) of each statement containing operational operators and absolute value function. One case is solved for you as an example:
|1||1|)|3|0)|1|0)|5|5)
|1|0)|1||2|)|3||4|)|3||3|)
+>-->+<-¹
-<-<-+>++=-
hgfe
dcba
Example:
)(12|1||2|) falsec <Û-<-
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The Book of Integers
12. Equivalent Equalities and Inequalities
16
Iulia & Teodoru Gugoiu
1. Do the required operation, then simplify, and finally find the logical value (true or false) of each statement:
12.1 (Adding or Subtracting) By adding (or subtracting) the same number to (from) the left side and to (from) the right side of an equality (inequality) you get an equivalent equality (inequality):
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Examples:
.etc
cbcaba
cbcaba
+<+Û<
+=+Û=
)(
42442242
falseareboth
++=+-+Û+=-+
)(
2422314231
trueareboth
-+<-+Û+<+
12.2 (Moving Terms) Moving a term from one side of an equality (inequality) to the other side requires changing the term to its opposite in order to get an equivalent equality (inequality):
bcacba
bcacba
+=Û=-
-=Û=+
Examples (true statements):
12532513
22312321
-+-¹-Û+-¹+-
+=+Û=+-
12.3 (Empty side) After removing the last term from one side of an equality (inequality) to the other side replace the empty side with 0. Indeed:
abbaba -=Û=+Û= 00
03213210
321202321
¹---Û++¹
+-+-=Û-=-+-
12.4 (Positive Terms) By moving terms from one side of an equality (inequality) to the other side, it is always possible to have an equivalent equality (inequality) with positive terms on both sides. Example:
522144142542 ++=++Û--=-+-
12.5 (Simplifying) If both sides of an equality (inequality) contain the same term then this term can be cancelled out:
.etc
bacbca
bacbca
<Û-<-
=Û+=+
Examples (true statements):
42134321
765765
-¹-Û+-¹+-
<Û-<+-
Examples (true statements):
6;263)5;532)1;217)
4;042)3;132)2;22)
addfaddeaddd
addcaddbadda
-£--³----<+-
->+-+¹-+-=+
2. By moving terms from one side to the other side (see 12.2), find 3 equivalent equalities (inequalities):
101551510)23312)1235)
164)432)132)
-+-£-+-³+--¹+-
->+--<---=-
fed
cba
3. Find an equivalent equality (inequality) by moving all terms from the right side to the left side:
151051510)3221)223)
210)413)422)
+-£---³--+¹
+->--<--=-
fed
cba
4. Find an equivalent equality (inequality) by moving all terms from the left side to the right side:
23573)1510105)311)
2121)231)374)
-+£+-+--³--¹+
->+-<---=-
fed
cba
5. Find an equivalent equality (inequality) by moving terms from one side to the other side so finally each side to have only positive terms (see 12.4):
321432)52421)43)
23)12)121)
---£----³+--¹-
->--<--=-
fed
cba
6. Find an equivalent equality (inequality) by canceling out the identical terms on both sides (see 12.5):
5323312)114421)155)
2232)133)11121)
+-+-£-+-+-³-+-+-¹-
+->-+-<-+-=+-
fed
cba
7. Find an equivalent equality (inequality) so one side is 0 and the other side is an integer. One case is solved for you as an example:
2)85()53()62)1)84(2())23()32(1)
2135)2142)213)
+£-----³+-----¹--
-->+--<---=-
fed
cba
Example:
)(010878762)1)84(2() truee ³Û³+-Û-³-Û--³+---
8. Find the logical value (true or false) of each statement. One case is solved for you as an example:
21|2||1|)|3||1|)|8||8|)|7||7|)
0|1|1)|2||5|)|3||0|)|2||2|)
-->-+-+£-+³-+¹-
=-+->--<+=-
hgfe
dcba
Example:
)(3332121|2||1|) trueh ->Û->+Û-->-+-
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The Book of Integers
13. Equations
17
Iulia & Teodoru Gugoiu
1. Find if the given value of x is a solution of the following equations (use substitution as explained in 13.2).
13.1 (Equation) An equation is an equality containing an unknown number represented by a letter and called variable. Example:
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521 -=+- x
In this equation x is the unknown number or variable.
13.2 (Solution) A solution of an equation is the value (number) of the variable that makes the equation a true statement. If an equation has more solutions, these solutions together form a set solution. Example:
31 -=+x
A solution of this equation is x = -4. Indeed, if you replace x by -4 in the equation, you get a true statement:
314 -=+-
13.3 (Solving) To solve an equation means to write a sequence of equivalent equations until you isolate the variable. Example:
6432 -=-+ xa) regroup the left side (LS) and simplify the right side (RS): 2)32( -=-+x
b) simplify LS: 21 -=-x
c) move -1 from LS to RS: 12 +-=x
d) simplify RS: 1-=x
13.4 (Variable on RS) If the variable appears on the RS of an equation, then use the symmetry property of an equality. Example:
5523
233152
=Û=Û=+Û
Û-=Û-+=+-
xxx
xx
13.5 (Opposite Variable) If -x appears instead of x in the equation, then move -x to the other side. Example:
11
212121
-=Û=-Û
Û=-Û+=Û=-
xx
xxx
13.6 (Brackets) If the equation contains brackets, then simplify and expand (remove) the brackets until you succeed in isolating the variable. Example:
4:4:
51:
51:
512
5)1(2:
5)23(2
5))2(3(2
-==-
=-
=-
=--
=+-
=+--
=---
xmirrorxsimplify
xRStoxmove
xsimplify
xexpand:
xsimplify
xexpand:
x
4;5)2(1)2;11)2;132)
0;32)1;21)0;0)
-=-=---=+=--=+=-
=+=---==+==-
xxfxxxexxd
xxcxxbxxa
2. Solve for x by isolating the variable on the left side and moving all the other terms to the right side (see 13.3):
530)2153)12)
40302010)3132)53)
51)13)21)
+--=--=--=-
-=-+-+=-+++=+
+=+--=-=+
xixhxg
xfxexd
xcxbxa
3. Solve for x by isolating the variable on the right side and moving all the other terms to the left side (see 13.4):
15105)321)32)
201015)322)211)
55)32)21)
-=---=----=-
-+=-+--=+-+=-
+=-+-=-=-
xixhxg
xfxexd
xcxbxa
4. Use the “simplify and expand” method explained in 13.6 to solve the following equations for x:
)]15(5[105))32(1)]3(4[2)
0)3()23())2(53)
1)2(2)2)1(1)
+---=---=---
=--++-=-
-=--+=--
xfxe
xdxc
xbxa
5. Solve for x using the “working backward” method (the first case is solved for you as an example):
))1)(4(53(2)74(1)52)1)3(2(1)
1)1)5(2(1)01)5()52()
)15(510)3)1(2)
------=--=-----
-=+---=+---
--=--=--
xfxe
xdxc
xbxa
4
51155)1(23)1(3)1(2)
-=Û
Û-=Û-=Û-=--Û--=--Û-=--
x
xxxxxa
Example:
6. Solve for x. One case is solved for you as an example:
|3||2||1|||)|2||1|||)12||)
|5|||)6||1)5||)
02||)0||)1||)
-=-+---=-+-=-
-==+=-
=-==
xixhxg
xfxexd
xcxbxa
Example:
22||213||321|||3||2||1|||) ±=Û=Û-+=Û=+-Û-=-+-- xxxxxi
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The Book of Integers
14. Inequations
18
Iulia & Teodoru Gugoiu
14.1 (Inequation) An inequation is an inequality containing an unknown number represented by a letter and called variable. Example:
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52 -³+x
14.2 (Solution) A specific value of the variable that makes the inequation a true statement is called a solution of the inequation. Example:
51 <- x
x = 1 is a solution of this inequation. Indeed, by replacing x with 1, the previous inequation become a true statement:
50511 <Û<-
14.3 (Set Solution) In general, an inequation has more than one solutions. The set solution is the set of all solutions of a given inequation. Example:
Zxx Î-³ ;1
The set solution is:
,...}4,3,2,1,0,1{ ++++-=x
and has the following graphical representation:
0 +1 +2 +3-1-2-3-4 +4
14.4 (Solving) Solving an inequation implies finding a sequence of equivalent inequations until the variable is isolated. Example:
2
2
123
;132
£
³
³+-
Î-³+--
x
x
x
Zxx
0 +1 +2 +3-1-2-3-4 +4
14.5 (Brackets) If an inequation contains brackets, then simplify and expand (remove) the brackets until the variable is isolated. Example:
0
0
415
415
4]1[5
4]23[5
4)]2(3[5
>
<
<--
<--
<+-
<+--
<---
x
x
x
x
x
x
x
0 +1 +2 +3-1-2-3-4 +4
1. Find if the given value of x is a solution of the given inequation (use substitution as explained in 14.2).
1);2(42)3;25)4;03)
2;23)2;12)2;01)
+=--->-+=-³-+=£+-
-=-<+-+=->--=¹+
xxfxxexxd
xxcxxbxxa
2. On a number line, graphically represent the set solution of each inequality:
12)32)0)1)2)
33)31)1)3)2)
>-£-³+£³+£-³
+<<-+<->-<->-¹
xandxjxandxixhxgxf
xexandxdxcxbxa
3. Solve for x by isolating the variable on the left side:
15102030)5321)5231)
23)12)02)
-³-++--£---³-+
-<-+>+¹-
xfxexd
xcxbxa
4. Solve for x by mirroring the inequation:
xhxgxfxe
xdxcxbxa
³-³£+³-
<<->-¹-
3)0)5)7)
0)5)1)1)
5. Solve for x by isolating the variable on the right side, simplifying, and then mirroring the inequation (the first case is solved for you as an example):
10515105)5385)3274)
321)53)40)
++-³+-+-£+--+³-
+-<-->-+¹
xfxexd
xcxbxa
444040) -¹Û¹-Û¹-Û+¹ xxxxaExample:
6. Solve for x by moving the variable to the other side, simplifying and eventually mirroring the inequality:
1510105)4219)1351)
531)415)30)
+--³+-+-£+---³-
+-<--->+---¹
xfxexd
xcxbxa
7. Solve for x using the “expand the brackets and simplify” method (see 14.5):
)]25(10[155))25(3)]1(2[1)
2)5()23())5(55)
3)1(5)5)2(3)
---->-+---³-----
-£--+--+-<-
->---¹--
xfxe
xdxc
xbxa
8. Solve for x using the “working backward” method (the first case is solved for you as an example):
]1)43([)]1(3[)]52(1[)0]2)3([)75()
123)2())52(41)
72)3(5))3(23)
+------>---³+-----
+-£+---+--<-
->----¹-
xfxe
xdxc
xbxa
835
53)3(5)3(23)3(23)
¹Û+¹Û
Û¹-Û--¹-Û--¹--Û--¹-
xx
xxxxa
Example:
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The Book of Integers
15. Multiplication of Integers (I)
19
Iulia & Teodoru Gugoiu
1. Use the short notation for the following additions:15.1 (Shortcut) Multiplication was initially designed as a shortcut for addition of like terms:
© La Citadelle
anaaaatimesn
´=++++ 44 344 21 ...
n is called the multiplier, a is called the multiplicand, and the result of the multiplication operation is called the product. Example:
1025222225
=´=++++ 44 344 21times
15.2 (Commutativity) The multiplication operation is commutative (the multiplier and the multiplicand can be interchanged without affecting the product):
abba ´=´
Example:
1555553
153333335
3
5
=++=´
=++++=´
43421
44 344 21
times
times
15.4 (Multiplying integers with opposite signs) The product of two integers with opposite signs is negative:
)()()( baabba ´-=´-=-´
Example:
)32(6)3(2:
)3(2)3()3(
6)3()3(
´-=-=-´
-´=-+-
-=-+-
So
15.5 (Multiplying negative integers) The product of two negative integers is positive:
baabba ´=-´-=-´- )()()()(
Example: 12)43()4()3( +=´+=-´-
15.6 (Multiplying by 0) The product of any number and 0 is 0:
000 =´=´ aa
Example: 0)7(00)7( =-´=´-
15.7 (Multiplicative Identity) The product between a number and 1 is equal to that number (1 is called the multiplicative identity):
aaa =´=´ 11
Example: 3)3(11)3( +=+´=´+
15.3 (Multiplying positive) The product of two positive integers is positive:
)()()()()( baabba ´+=+´+=+´+
7777777)111111))2()2()2()
33333)555)111)
--------------+-+-
-----++++
fed
cba
2. Use the expanded notation for the following multiplications:
)18(3))1(1))10(5))4(3))3(5)
03))5(5))1(2))5(4)23)
+´-´-´-´-´
´+´-´-´´
jihgf
edcba
3. Use the commutativity property of multiplication to expand the following multiplications in two different ways (see 15.2):
20)11)33)43)25)
35)26)24)32)51)
´´´´´
´´´´´
jihgf
edcba
5. Multiply the following integers with opposite signs (see 15.4):
)6()5()2)10()3)4())3(3))1(1)
)2(6))2()2()1)3())5()1())3(2)
+´-´-´--´-´
-´+´-´--´+-´
jihgf
edcba
6. Multiply the following negative integers (see 15.5):
4. Multiply the following positive integers (see 15.3):
)5()6())10()5()1010))2()3()
)10(3)5)3())5(2)11)
+´++´+´+´+
+´´++´´
hgfe
dcba
)5()5())2()10())9()1())4()3())4(5)
)4()2())2()5())1()3())3()2())1(1)
-´--´--´--´--´-
-´--´--´--´--´-
jihgf
edcba
7. Multiply the following integers (see 15.6 and 15.7):
)10(1)01)110))1(1)
11)00))5(0))1(0)
-´´´--´
´´+´-´
hgfe
dcba
8. Multiply the following integers:
1)111())2468(0))1()531())1()123()
)50()6())50()6())20()6()66)
)25()5())40()5())2()30())20()5()
)10()10()10)10())10(10)1010)
)4()5()4)5())4(5)45)
´--´-´+-´-
-´+-´-´-´
+´--´++´--´-
-´-´--´´
-´-´--´´
tsrq
ponm
lkji
hgfe
dcba
9. Find the value of each expression:
|0|)5432()0|5|)|3||3|)|2|1)
|)1()333(|)|)5(0|)|)4()2(|)|)3(2|)
´-´--´--´
-´--´+´--´
hgfe
dcba
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The Book of Integers
16. Multiplication of Integers (II)
20
Iulia & Teodoru Gugoiu
1. Complete the following multiplication statements:16.1 (Divisors) The operation of multiplication has a special meaning when the multiplicand and the multiplier are integers:
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cba =´
a and b are called factors or divisors of the product c.c is a multiple of a and b.Example:
16.2 (Associativity) The operation of multiplication has the property of associativity (the order in which 3 numbers are multiplied is not important):
15)5()3( -=+´-
-3 and +5 are factors or divisors of -15.-15 is a multiple of -3 and +5.
cbacbacba ´´=´´=´´ )()(
By convention, if more than one multiplication operations appear, the operations are done in the order they occur, from left to right. Example:
24)4()6()4()2()3( +=-´-=-´+´-
16.3 (Distributivity) The multiplication operation is distributive over the addition operation:
cabacba ´+´=+´ )(Example. The expression:
)]4()5[()3( -++´-
can be evaluated in two different ways:
îíì
-=+-=-´-++´-
-=+´-
31215)4()3()5()3(
3]1[)3(
16.4 (Positive Factor) If the number in front of the bracket is positive, the signs of the final terms are coincident with the signs of the terms inside of the brackets:
eadacabaedcba ´-´+´-´=-+-´ )(
16.5 (Negative Factor) If the number in front of the bracket is negative, the signs of the final terms are opposite to the signs of the terms inside of the brackets:
dacabadcba ´+´+´-=--´- )(
îíì
=++-=´+´+´-
=-´-=
=--´-
93126134323
9)3(3
)142(3Example:
îíì
=+-=´+´-´
=´=
=+-´
81086524232
8)4(2
)543(2Example:
)()12(36))()15(60))()10(100)
0)()1())(11)12)()3()
)()5(10))()4(16))(420)
´-=-´-=+´-=-
=´-´=--=´-
´-=+´-=´=-
ihg
fed
cba
2. Use the associativity property (see 16.2) to calculate easily the product:
2)15()75()4())4()5(6)25())4()5(15)12()
4)5(3)25())5()10(2)11()2)15(4)2()
5)2(9))5()7()4()5)3(2)
´-´-´--´-´´--´-´´-
´-´´--´-´´-´-´´-
´-´-´+´-´-´
ihg
fed
cba
3. Use the distributivity property (see 16.3) to rewrite the following expressions by expanding the brackets. Then evaluate the final expression.
)152()4())53()2())25(1)
)423(5))52()3())43(2)
)]4()1()3[()2())]5()2[()3())43(2)
+--´--´-++´-
-+-´++´+--´
-++--´+-++´-+´
ihg
fed
cba
4. Use the distributivity property (see 16.3) to factor out the common factors, then evaluate the final expression. One case is solved for you as an example:
42532522525)315215)410310210)
113112511111)243221)223212)
275717)105310)5515)
5443)5232)3121)
´+´-´+-´-´-´+´-´
´-´+´-´´-´+´-´+´-´-
´+´-´´-´´+´-
´-´-´-´´+´
lkj
ihg
fed
cba
Example:
4)2(2)431(2423212243221) -=-´=-+-´=´-´+´-=´-´+´-h
5. Evaluate each of the following expressions in two different ways (see 16.3, 16.4 and 16.5):
)321()5())152()3())51(2)
)213(2))31()2())43(2)
)]3()2()1[()4())]3()2[()1())32(1)
+--´-+-´-++´-
-+-´+-´--´-
--++-´+-++´-+´
ihg
fed
cba
6. Rewrite one of the factors of multiplication to calculate the product more easily. One case is solved for you as an example:
)51()19())21()19())4(51)
)6(102)5)39()993)
+´--´--´
-´´-´
fed
cba
Example:
969311000
5120100051)20(1)20(5051)20()150(
151)20(51)120(5151)120()51()19()
-=+-=
=+--=+-´+-´=+-´+=
=´+-´=+-´=´+-=+´-f
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The Book of Integers
17. Order of Operations (II)
21
Iulia & Teodoru Gugoiu
1. For each of the following expressions, analyze each operator and decide if it is an unary or binary operator:
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17.1 (Unary & Binary Operators) In mathematics, the symbols + and - are used as unary operators to express the sign (of a number or of opposite) and as binary operators to express the addition and the subtraction operations. Example:
)4()3()2(1 --+´-+-
First, second and forth - operator are unary. Third - operator is binary.First + operator is binary and second + operator is unary.The x operator is always a binary operator.The brackets are used here to express the sign of numbers, not to change the order of operations.
17.2 (Default Order of Operations) If an expression contains additions, subtractions and multiplications but not grouping symbols then the default order of operations is:a) do all the multiplication operations in the order they appear (from left to right)b) do all the additions and subtractions in the order they appear (from left to right)Example:
347
)4()7()4()6(1
)4()3()2(1
-=+-=
=---=---+-=
=--+´-+-
17.3 (Order of Operations Algorithm) If an expression contains grouping symbols then use the following algorithm to find the value of the expression:1. Identify the innermost bracket(s)2. Perform the operations inside the innermost bracket(s) using the default order of operations3. Replace the innermost bracket(s) with the result you got on step 24. If there are still more grouping symbols then go back to step 15. If there are no more grouping symbols, then perform the operations of the remaining expression using the default order of operations.Example:
75935)1(]9[3
5)1(]9162[3
5)1(]9)4(42[3
5)43(]9)51(42[3
-=+--=+-´+-=
=+-´-++-=
=+-´--´-+-=
=+-´--´-+-
3655)1)6()5()5(6)
67))5(())5()3()2()1()
+´--+-´+--´
´-+--+--´++-
dc
ba
2. Use the “left to right” rule (see 17.2) to calculate the value of the following expressions containing more than one multiplications:
)4()3(5)2(10)2)1()2(4)6()5)4(3)2(1)
)10(0)5(10)1003)2(0)0)4()5(1)
)2()2()2()2())2()3()1(2))3()2()1()
)3()2(1)3)2(1)321)
-´-´´-´´-´-´´-´-´´-´
-´´-´´´-´´-´-´
-´-´-´--´-´-´-´-´-
-´-´´-´´´
lkj
ihg
fed
cba
3. Use the default order of operations (see 17.2) to calculate the value of the following expressions (brackets are used here to express the sign of integers):
254321)2)4(3)2()1())4()3(21)
4321)43)2(1)4321)
4321)4321)4321)
3)2(1)321)321)
321)321)321)
´+´-´´-´--´--´-´+-
´-+--+-´+´´-
+-´-´´-´-´
+-´-+´-´--
-´+´´+
onm
lkj
ihg
fed
cba
4. Use the order of operations algorithm (see 17.3) to calculate the value of the following expressions:
2)54)32((1)2)4())32()1(())43()21()
4)32(1))4()32(1))43()21()
4)32(1)43)21()4)32(1)
)3)2(1())32(1)3)21()
)32(1))32(1)3)21()
´+´-´´-+--´---´+-
´-+--´+-´+-´--
+-´-´´-´-´
+-´-+´-´--
-´+´´+
onm
lkj
ihg
fed
cba
)))223(53(23(32)4)5()2)1()2)5)3(2(((()
1)1(}3]13)212()1()2[(4{]2)1()34[(2)
)1()3)2(2()24(2)21()2()121()
42)12()4)4(3)3(2()
´-´--´-´-´--´--´-+-´-
--´+-´´-´-´-´---´+-´-
-´+-´´+-´´+-´-´´-
-´--´+-´--´
s
r
q
p
)1()2(3)2(3212)1()2(4)2()1(3)4(5)
)5(2)2(2)2()4(3)2()1()2(321)
)4(21)5(4)4(3)3(2)
-´-´--´´´+´-´-´-+´-´+-´-
-´+-´´---´´-+-´-´´-
-´´--´+-´--´
r
q
p
5. Calculate the value of the following expressions:
|)2(3||655|)|3|2|)2()5(|))53(|53|)
|5||3||2||1|)1|)6()5(||)5(6|)|32|2)
-´-´---´--´+--´-
+--´++-+-´+--´-´
fed
cba
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The Book of Integers
18. Division of Integers (I)
22
Iulia & Teodoru Gugoiu
1. Change each of the multiplication statements to an equivalent division statement:
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18.1 (Division) The division operation was initially designed as the opposite operation of multiplication. So, a multiplication statement like:
cba =´
is considered equivalent to any one of the following division statements:
a
cbor
b
ca ==
Example:
ïî
ïí
ì
-
-=+
+
-=-
Û-=+´-
3
124
4
123
12)4()3(
18.2 (Definitions) The division operation is an operation between a dividend and a divisor to obtain a result called quotient:
quotientdivisor
divident=
18.3 (Sign Rules) The sign rules for division are similar to the sign rules for multiplication:
b
a
b
a
b
a
b
a
b
a
b
a
b
a
b
a
+=-
--=
+
-
-=-
++=
+
+
Examples:
76
42
6
424
5
20
5
20
53
15
3
152
5
10
5
10
=+=-
--=-=
+
-
-=-=-
+=+=
+
+
18.4 (Division Operators) There are four operators used to express the division operation:
bababab
a:/ ¸
Examples:
18.5 (Division by 1) Any number divided by 1 is equal to the initial number:
aa
=1
18.6 (Division by 0) Division by 0 is not defined.
definednota
=0
6)3(:)18(5)2()10(
3)2/()6(24
8
-=-++=-¸-
-=+--=+
-
000)00)1()5)3(15)1025)
11)1()4)2(8))3()3(9))1(12)
=´=´-´-=--=´-
-=´-´-=--´-=-´=-
hgfe
dcba
2. Change the division statements to equivalent multiplication statements:
1010
100)0
2
0)2
3
6)
5
102)2
4
8)
45
20)10
1
10)0
1
0)4
3
12)1
1
2)
=-
-=
-=
-
--=--=
-
-=-
-=-
=-=-
-=-
jihgf
edcba
3. Find the value of each expression (see 18.3):
5
25)
6
12)
5
15)
3
9)
4
16)
3
9)
6
18)
5
20)
4
4)
2
0)
5
10)
2
2)
-
-
-
-
-
-
-
-
-
----
-
lkjihg
fedcba
4. Find the value of each of the following expressions containing different types of division operators (see 18.4):
)5(:25))2(8)10/)20()3
9)
)5(:5))4()8())2/(6)1
3)
--¸-+
-
-+¸---
+
hgfe
dcba
5. Find the value of the following special cases of integer division (see 18.5 and 18.6):
0
0)
7
0)
1
6)
0
2)
1
5)
11
11)
10
10)
1
1)
5
5)
5
5)
1
3)
3
3)
lkjihg
fedcba
--
+
-
+
-
----
-
-
++
6. Find the value of each expression:
25
625)
125
1000)
150
600)
40
200)
50
250)
11
121)
50
100)
25
75)
10
50)
5
45)
12
36)
7
28)
---
+
-
+
-
----
-
-
+
-
+
-
lkjihg
fedcba
7. Find the value of each expression containing the absolute value function. One case is solved for you as an example:
5
15)
11
0)
|5|
|0|)
5
|10|)
4
20)
|3|
18)
4
|12|)
|5|
|15|)
|5|
5)
3
|3|)
----
---
-
-
-
--
-
-
-
--
jihgf
edcba
Example:
225
10
5
|10|) =-=
-=
-
-g
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The Book of Integers
19. Division of Integers (II)
23
Iulia & Teodoru Gugoiu
1. Write the following ratios of integers as rational numbers in lowest terms:
© La Citadelle
19.1 (Rational Numbers) The set of integers Z is not closed under the operation of division. This means that the quotient of two integers might not be an integer. In these cases the quotients of two integers define more complex numbers in mathematics than integers called rational numbers. Example:
3
7
+
-
19.2 (Lowest Terms) The quotient of two integers can be simplified by dividing both the dividend and the divisor by common factors until you get the quotient in lowest terms. Example:
3
4
2)6(
2)8(
6
8
3)18(
3)24(
18
24
+
-=
¸+
¸-=
+
-=
¸+
¸-=
+
-
19.3 (Signed Decimals) If the quotient of two integers is not an integer, you can express the quotient using a signed terminating or non-terminating decimal.
6.2...666.23
56.2
5
13-=-=
+
--=
-
+
19.4 (Multiples) Given an integer, you can find its multiples by multiplying the integer by +1, -1, +2, -2, +3, -3, ....So, the multiples of -3 are:
,...12,12,9,9,6,6,3,3 +-+-+-+-
19.5 (Exact Division) If the dividend is a multiple of the divisor, the quotient is an integer. In this case the dividend is divided exactly by the divisor. Example:
43
12-=
-
+
19.6 (Remainder) If the dividend is not a multiple of the divisor, the division has a quotient and a remainder. The quotient is an integer and the remainder is a natural number less than the absolute value of the divisor. Here is the division statement and some examples:
||0; drrdqDord
rq
d
D<<+´=+=
1)2(59152
9
12)5(9152
9
1)2()4(9142
9
1249142
9
+-´=-=-
-
+´-=--=-
+-´-=-=-
+´==
becauseR
becauseR
becauseR
becauseR
16
18)
16
24)
10
25)
12
30)
12
20)
6
8)
18
32)
15
20)
4
6)
6
16)
4
14)
4
3)
-
-
-
-
-
-
-
-----
-
lkjihg
fedcba
2. Express the following ratios of integers as signed terminating decimals (see 19.3):
20
5)
16
24)
10
35)
24
6)
25
10)
15
3)
10
2)
10
7)
8
5)
12
3)
5
2)
4
1)
-
--
-
-
--
-
-
-
-
--
-
lkjihg
fedcba
3. Express the following ratios of integers as signed non-terminating repeating decimals (see 19.3):
90
15)
24
8)
15
5)
6
2)
44
4)
12
2)
15
20)
12
4)
6
4)
3
1)
---
-
---
-
--
-
-
-
jihgf
edcba
4. For each given integer, find 5 positive and 5 negative multiples:
10)8)4)5)2)
7)1)6)5)2)
+++-+
---+-
jihgf
edcba
8. Find the quotient and the remainder for each case (see 19.6):
4
15)
4
15)
4
15)
4
15)
3
7)
3
7)
3
7)
3
7)
-
--
-
-
--
-
hgfe
dcba
7. Rewrite each of the following division statements (see 19.6):
345
17)34
5
17)23
5
17)23
5
17)
334
9)33
4
9)12
4
9)12
4
9)
RhRgRfRe
RdRcRbRa
=-
--=
--=
-=
=-
--=
--=
-=
6. Find the quotient and the remainder for each case:
2
15)
3
15)
4
15)
7
15)
3
11)
5
9)
4
7)
2
5)
hgfe
dcba
5. Rewrite each of the following division statements:
1102
21)43
5
19)25
3
17)13
3
10)
324
11)03
3
9)12
4
9)12
3
7)
RhRgRfRe
RdRcRbRa
====
====
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The Book of Integers
20. Order of Operations (III)
24
Iulia & Teodoru Gugoiu
1. Use the “left to right” order of operations to find the value of each of the following expressions (see 20.1):
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20.1 (Consecutive Divisions) If more divisions occur consecutively within an expression, the divisions must be done one by one in the order that they occur (from left to right). Example:
20.2 (Brackets) Use brackets to specify a particular order of operations. Example:
2)2()4()2()6()24( -=-¸+=-¸-¸-
8)3()24()]2()6[()24( -=+¸-=-¸-¸-
20.3 (Consecutive Multiplications and Divisions) Multiplications and divisions are considered operations of equal priority. If multiplications and divisions appear consecutively, perform the operations in the order that they occur (from left to right). Brackets can be used to specify a different order. Examples:
6323)12()24(
3)12()4()6(
=´=´-¸-=
=´-¸+´-
4222)3()6(
2)]4()12[()6(
=´=´-¸-=
=´-¸+¸-
20.4 (Default Order of Operations, Updated) Multiplication and division are considered to have higher priority than addition and subtraction. If an expression contains addition, subtraction, multiplication and division but not grouping symbols, then the default order of operations is:a) do all the multiplication and division operations in the order that they appear (from left to right)b) do all the addition and subtraction operations in the order that they appear (from left to right). Example:
20.5 (Order of Operations Algorithm) If an expression contains grouping symbols, use the Order of Operations Algorithm presented in 17.3 and the Default Order of Operations presented above (20.4). Example:
2)6()2(6
)2(33)6(32
-=---+-=
=-´-¸-+´-
244284]14[2
59]12)14(3[2
)5())3(6(
]1)2()2()7(3[2
2)10())4(1232(
]1)3(6)24()321(3[2
-=+-=+´-=
=-+----´-=
=-+--+
+--+-´-´-=
=¸-+-¸-´+
+--¸+¸-´´+-´-
)1()2()3(236)5)1()3(75))5()2()1(40)
)1()2()2(32)2)2(264)2)3(224)
2)5(50)5)2(20)2)2(12)
-¸-¸-¸¸-¸-¸-¸-¸-¸-¸
-¸-¸-¸¸-¸¸¸-¸¸-
¸-¸-¸-¸-¸-¸
ihg
fed
cba
2. Use the “left to right” order of operations to find the value of each of the following expressions (see 20.3):
)2()8()6()4(1)3)2()3(2)2)3()6(3)
)1()1()1(5)2)2(216)3)1(212)
2)5()1(5)2)5(10))2()3(4)
-´-¸-´-´´-¸-´¸-¸-´-
-¸-´-¸¸-´¸´-¸¸-
´-¸-´´-¸--¸-´
ihg
fed
cba
3. Start with the inner-most brackets and find the value of each expression (see 20.2):
)5(25
)5(10
)12(24
)3()12()
)33(94
)2(2)3()
)]2(5[20
6)4()3()
4
)4(16
)2(2
)12()1()
)2(4
1612))2(
4
82)
)]2()4[()]3(8[)]4)5[(20))]2()8[(16)
-¸
-¸¸
-¸
-¸-
´-¸´
-´´-
-´¸
¸-´--
-¸¸
-´
-´-
-´
-¸--¸
-
´-
-´-¸-´´-¸--¸-¸
ihg
fed
cba
4. Use the “default order of operations” to find the value of each of the following expressions (see 20.4):
1)1()1()1(1))3(24)8(2))1(3)12(24)
34)5(2054)5)2(3)5(15)13)2(101)
5)1()3(210))2(3412))2(4321)
--¸-+-´--´-¸-´-´+-¸
´+-¸+´-+-´+-¸+--¸´-
+-¸-´--´+¸--¸-´+
ihg
fed
cba
5. Use the “order of operations algorithm” (see 20.5) to find the value of each of the following expressions:
)2(23
)4(163
1)2(2
)6(22)
1)2()1(
1)3(122))2(
5
4231)
)1)2(2(]4)2(3[)2)333(12)1)5()321()
-´+
-¸+-
--´
-´+-
+-´-
+-¸+--´
-
´--
--´¸--´--´+-¸-+-¸´-
fed
cba
)2(232
)5(204
53
8))2104()]2(45[)]27(15[)
32)5(15
6)3(21)2()]5)2)6(2[()53()]43(2[1)
-´+´-
-¸+-+
-
-¸-´-´+¸+-¸--
-÷÷ø
öççè
æ
+-¸
¸-´+-´-¸´-+¸+-´---
ji
hg
)]3(3[5
15
3
152
244
)2(23
3)5(3
2
10
)2(2
153
45
2102
24)3(
)
3)5(2
2852
21
3)1315(
)3(91
3)554(
)3()9(
1)1322(
2
254)
5)254)5(3()1)3(9()22210()4832(2)
-´¸-
´+´-
-´-+
-´-
+-´
--
¸
-´
-´-
´-´
¸--
+´-
+-´
¸+´¸
--
´+¸-
-¸+-
--´´
-¸-
-+´+--+
-
¸´
¸¸´--´++-¸´´+¸-+¸+---
m
l
k
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The Book of Integers
21. Equalities and equations
25
Iulia & Teodoru Gugoiu
1. All the following equalities are true statements. Multiply or divide both sides of each equality by the given integer and check if the statement remains true (see 21.1).
© La Citadelle
21.1 (Multiplying and Dividing) By multiplying (or dividing) the left side and the right side of an equality by the same number, you get an equivalent equality:
0;
0;
¹=Û=
¹´=´Û=
cc
b
c
aba
ccbcaba
true
false
;113
69
3
3693
;126
)3()4()3()2(42
-=-Û-
-=
-Û-=
+=-Û
Û-´-=-´+Û-=+Examples:
21.2 (Unknown Factor) Solution 1. Rewrite the multiplication as an equivalent division:
a
bxbxa =Û=´
42
88)2(82 +=
-
-=Û-=´-Û-=´- xxx
21.3 (Unknown Dividend) Solution 1. Rewrite the division as an equivalent multiplication:
baxba
x´=Û=
12)3()4(34
-=+´-=Û+=-
xx
21.4 (Unknown Divisor) Solution 1. Rewrite the division as an equivalent multiplication and then back as an equivalent division:
b
axxbab
x
a=Û´=Û=
46
24)6(246
24-=
+
-=Û´+=-Û+=
-xx
x
Solution 2. Divide the equality by the second factor:
a
bx
a
b
a
xabxa =Û=
´Û=´
Solution 2. Multiply the equation by the divisor:
baxbaa
xab
a
x´=Û´=´Û=
Solution 2. Multiply the equation by the divisor and divide the equation by the initial quotient:
63
18;
3
18
)3(
)3(;18)3( -=
-
+=
-
+=
-
´-+=´- x
xx
18)9()2(
);9()2()2(
)2(;92
-=+´-=
+´-=-
´-+=-
x
xx
b
ax
b
xb
b
axbaxbx
x
ab
x
a=
´=´=´=´= ;;;;
2;4
4
4
8;48;4
8;4
8=
´=´=´=´= x
xxxx
xx
6;3)3(5)3(4)1;3)5()105(4)
2;4)3(22)3;321)
-+-´=+´--+-¸-=
-+-´=---=-
bydividedbymultiplyc
bydividebbymultiplya
2. Solve each of the following equations for x by replacing each multiplication statement with an equivalent division statement (see 21.2):
)3(0)520))3(9)210)
147)124)8)4()153)
-´=´-=--´=´-=-
-=´=´-=-´-=´
xhxgxfxe
xdxcxbxa
3. Solve each of the following equations for x by dividing both sides by a convenient integer (see 21.2):
)5(0)525))3(18)321)
16)4()153)12)2()41)
-´=´=--´=-´-=-
-=-´-=´--=-´-=´-
xhxgxfxe
xdxcxbxa
4. Solve each of the following equations for x by replacing the division statement with an equivalent multiplication statement (see 21.3):
11)
37)
71)
65)
010
)44
)45
)43
)
-
-=-
-=
+=-
-=
=+=-
-=-=-
xh
xg
xf
xe
xd
xc
xb
xa
5. Solve each of the following equations for x by multiplying both sides by a convenient integer (see 21.3):
33)
25)
34)
23)
03
)62
)43
)31
)
-
-=-
-
-=-
-=
-=-
=-
-=-
-=-
+=-
xh
xg
xf
xe
xd
xc
xb
xa
6. In order to solve each of the following equations for x, rewrite the division statement as a multiplication statement and then back as a division statement (see 21.4):
xl
xk
xj
xi
xh
xg
xf
xe
xd
xc
xb
xa
-==-
-=
-=-
-=-
-
-=+
-=-=-
+=-
-=-
--=+=
-
287)
255)
82)
11)
369)
505)
217)
183)
624
)525
)315
)15
)
7. In order to solve each of the following equations for x, multiply and then divide both sides by convenient integers (see 21.4):
xl
xk
xj
xi
xh
xg
xf
xe
xd
xc
xb
xa
-=-
-==-
-=-
-=-
-
-=-
-=-=-
-=-
+=-
-+=
-+=
-
131)
3003)
369)
306)
408)
217)
3311)
162)
530
)20100
)735
)11
)
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The Book of Integers
22. Proportions and Equations
26
Iulia & Teodoru Gugoiu
1. Use cross-multiplication and division to solve for x the following equations (see 22.2):
© La Citadelle
22.1 (Proportions) A proportion is an equation with a ratio on each side:
d
c
b
a=
22.2 (Solving a proportion) To solve a proportion means to find one term of the proportion when three other terms are given. Use cross-multiplication and division to isolate the unknown term:
b
cax
b
xb
b
caxbca
c
b
x
a ´=
´=
´´=´= ;;;
Example:
14;5
90
5
5;905;
30
3
5-=
-=
´-=´=
-x
xx
x
22.3 (Shortcut for a proportion) To find a term of a proportion multiply the adjacent terms and divide by the opposite term:
a
cbd
b
dac
c
dab
d
cba
d
c
b
a
´=
´=
´=
´=
Û=
;;;
Example:
203
60
3
5)12(
53
12-=
-=
´-=Û=
-x
x
22.4 (Other equations) The main idea in solving an equation is to isolate the variable using equivalent equations.Example 1:
95445
2
858)5(2
311)5(211)5(23
=+=Û=-Û
-
-=-Û-=-´-Û
+-=-´-Û-=-´--
xx
xx
xx
Example 2:
52
10
1
210210
2
4104)2(10
354)2(10
43)2(105
-=-
=Û=-
Û=¸-
Û-
-=¸-Û-=-´¸-
Û--+=-´¸-
Û+=+-´¸-
xx
x
xx
x
x
Example 3:
42
8
1
28
248
3
122
8
1
3
28
12
3
9
28
153
-=-
=Û=-
Û-=-
Û=+-
Û=
+-
Û-
-=
+-
+-
xx
xx
xx
210
0)
812
3)
4
25)
8
4
2)
-=
--=
--=
-
-=
-
xd
xc
xb
xa
2. Use cross-multiplication and division (the shortcut method explained at 22.3) to solve for x the following equations:
51
0)
63
2)
2
315)
5
5
1)
+=
--=
--=
-
-=
-
xd
xc
xb
xa
3. Use equivalent equations to solve for x the following equations:
1)32(410
1)5(3
)2(102
2)4(3)
3)2313(2]2412[)]3()32(5[)
)32(1
18
1
3)
10
5
6
)2(25)
35)3(2)21)1(3)
25
8
5
20)
3
6
16
4)
9
6
3
13)
5
53
10
6)
4
2
42
5)
2
4
2
3)
)5(25)146)222)
2512)534)932)
++-´--
--´-=
-¸--
¸-´-
´+´---=+¸-¸-´+´-
-¸-
-=
-
-=
-
-´-
-=-+¸-+=--´-
+¸
-=
--=
+¸
-
-
-=
-¸-
´-=
--=
+´
-
-
-=
-
-¸-=+-=-¸--=+¸
+´-=-=´--=-´
x
r
xq
xp
xo
xnxm
xl
xk
xj
xi
xh
xg
xfxexd
xcxbxa
4. Solve for x using the “working backward” method. One case is solved for you as an example.
3
3
3510
2)1)2()2(36)
06)7()318()3()32)4(2)1()
0415)4(5)054512)
5)4(5)8)4(10)
-=
+¸¸
--=-¸¸-¸
=+-¸¸-´¸-=-´´´-
=-¸´-´=+¸´¸-
-=-¸´-+=-´¸
xhxg
xfxe
xdxc
xbxa
Example:
52
110
1
210210)4(8108)4(10)
-=-
´=Û
Û-
=Û-=¸Û-¸+=¸Û+=-´¸
x
xxxxa
xh
xg
xf
xe
1
1
1)
82
4)
6
93)
8
10
4)
-=
+
-=
++
-=
--=
-
-
xh
xg
xf
xe
3
10
15)
204
5)
3
28)
14
4
7)
-=
+
-=
+
-
+
-=
-
-
-=
+
-
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The Book of Integers
23. Inequalities and Inequations (II)
27
Iulia & Teodoru Gugoiu
© La Citadelle
23.1 (Positive Numbers) By multiplying (or dividing) an inequality (or inequation) by a positive number you get an equivalent inequality (or inequation).Example 1 (all statements are false):
215
10
5
5105 +>-Û
+
+>
+
-Û+>-
Example 2 (all statements are true):
86);4()2()3()2(;43 -³--´+³-´+-³-
23.2 (Negative Numbers) By multiplying (or dividing) an inequality (or inequation) by a negative number you get an equivalent inequality (or inequation) only if you change its sense. Examples (all are true):
433
12
3
9129
632)3(1)3(21
£Û-
-£
-
-Û-³-
->-Û´->´-Û<
23.3 (Inequations) The main idea in solving an inequation is to write a sequence of equivalent inequations until you can isolate the variable. Examples:
23.4 (Inequations with Ratios) If you multiply an inequation by an expression containing the variable, you have to separately analyze the cases when the expression is positive or negative.
15);15()1()()1(;15
);3()5(5
)5(;35
.2
42
8
2
282.1
-³+´-³-´-+£-
-´-£-
-´--³
-
-
+<Û-
-<
-
´-Û->´-
xxx
xx
xx
x
1
62.2
+-
+>-
xEx
101.1 <Û>+- xxCase
solutionnoxandx
xxx
xxx
Þ><
>-<+-+>-´+-
+-
+´+->-´+-
41
4;31;6)2()1(
;1
6)1()2()1(
101.2 >Û<+- xxCase
}3,2{41
4;31;6)2()1(
;1
6)1()2()1(
++=Þ<>
<->+-+<-´+-
+-
+´+-<-´+-
xxandx
xxx
xxx
}1,2,3,4,5,6{60
;6;60
)(6;60
16
;3
318
3
1;3
18
;9
3)9(
2)9(;
9
32.1.
------=Þ-³<
-³£-<
-£³->
³-
-
-³´
--£
-
-´-£
-´-
-
-³
-
xxandx
xxthenxIf
solutionnoxxthenxIf
xxx
xxEx
1. All of the following inequalities are true statements. Multiply or divide both sides of each inequality by the given positive integer and check if the statement remains true (see 23.1).
2;4)6()3(4)3;3)4(120)
3;5)1(26)2;321)
bydividedbymultiplyc
bydividebbymultiplya
-³-¸-´-+-¸-£
+-´>++-<-
2. All of the following inequalities are true statements. Multiply or divide both sides of each inequality by the given negative integer and change the sense of the inequality. Check if the statement remains true (see 23.2).
2;2)3()3(2)3;1)5(100)
5;15)10(25)2;432)
--³-¸-´-+-¸-£
-+-´>+--<-
bydividedbymultiplyc
bydividebbymultiplya
3. Solve the following inequations for x by multiplying or dividing both sides by a convenient integer (see 23.3). Look only for integral solutions of x (that is , solutions where x is an integer).
110)
51)
10)
11)
30)
25)2
5)1
2)
48)36)102)93)
-£
->-<
-³
-³
->--£
--<
-
´<-´-£-³´-->´
xl
xk
xj
xi
xh
xg
xf
xe
xdxcxbxa
4. Solve the following inequations for x. Find only the integral solutions of x (x is an integer). An analysis of the sign of x might be necessary (see 23.4).
xl
xk
xj
xi
xh
xg
xf
xe
xd
xc
xb
xa
10025)
71)
62)
11)
105)
62)1
2)2
6)
51)2
2)0
2)0
1)
-³-
-³
-³-
-³-
-³-
->--£-<
-
-
->--³³
-<
5. Solve the following inequations for x by isolating the variable (see 23.3). Look only for integral solutions of x (x is an integer).
)62(3
10)
13
102)
3
82)
3
32)1(5)4
3
)1(2)2)1(38)
)2(510)9)32(3)8)2(2)
-´´-
->
+´->-
-
->-
-
+¸->--£
+´--+´<-
+-´-£-³-´´-->-´-
xi
xh
xg
xf
xexd
xcxbxa
6. Solve the following inequations for x. Look only for integral solutions of x (x is an integer). One case is solved for you as an example:
3|12
|)3|1|3)51||)0||)
0|2|)10|2|)0|1|)2||)
2|2|)10|2|)0|1|)2||)
<--->£+<
³-³´>-<
=-=´=-=
xlxkxjxi
xhxgxfxe
xdxcxbxa
Example:
002222202.2
42222202.1
2|2|)
=Û=Û=-Û=-Þ<Û<-
=Û+=Û=-Þ>Û>-
=-
xxxxxxCase
xxxxxCase
xd
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The Book of Integers
24. Powers
28
Iulia & Teodoru Gugoiu
1. Change the following repetitive multiplications to powers (see 24.1):
© La Citadelle
24.1 (Shortcut) Powers were introduced as shortcuts to repetitive multiplications of a number by itself:
44 344 21timesn
n aaaaa ´´´´= ...
aaaa ´´´´ ...
Example: 3)2()2()2()2( -=-´-´-
24.2 (Expanded and exponential notations) The expanded notation is:The exponential notation is:where a is called the base and n is called the exponent.The full expression represents a power.Examples:
na
na
2
4
)5()5()5(.2
)3()3()3()3()3(.1
-=-´-
-´-´-´-=-
24.3 (Negative exponents) A power with a negative exponent is defined as:
44 344 21timesn
n
n
aaaaaa
´´´´==-
...
11
Example:
8
1
)2()2()2(
1
)2(
1)2(
3
3
-=
-´-´-=
-=- -
As you can see from the previous example, an integer raised to a negative integral exponent is a rational number.
24.4 (Base 0 and 1) 0 cannot be a base for powers having integral negative exponents because division by 0 is not defined. Also 1 can be a base for any integral exponents. This operation doesn’t have an inverse operation (see sections about Roots).
11111
000000
3
4
=´´=
=´´´=
24.5 (Power Operator) The power is a binary operation between a base and an exponent. Usually the base is written normally and the exponent is written as superscript. Sometimes the ^ operator is used to denote a power operation, especially on calculators and in programing languages:
naan ^Û
Example:
9
1
)3()3(
1)2()^3( =
-´-=--
)1()1()1()1()1()1())2()2()2()
)10()10()10()10()10())4()
)5()5()5()5()3333333)
-´-´-´-´-´-+´+´+
-´-´-´-´-+
-´-´-´-´´´´´´
fe
dc
ba
2. Use expanded notation to express the following powers (see 24.2):
325323
243234
)6())7()1))5())10()10)
)8())2())3())1())2()3)
----
----+
lkjihg
fedcba
3. Use expanded notation to express the following powers (see 24.3):
45232
23312
)5()1))4())10()10)
)3()5))1())2()4)
-----
-----
---
--+
jihgf
edcba
4. Use expanded notation to find the value of the following expressions. One case is solved for you as an example.
023232
023232
124533
)10()5))2())2()0)0)
7)10))10())10())10())10()
)5()10))1())1()2)2)
---
----
---
---
---
--
srponm
lkjihg
fedcba
Example:
100
1
)10()10(
1
)10(
1)10()
2
2 =-´-
=-
=- -i
5. Use the ^ operator to express the following powers (see 24.4):
324312 )10()10))5())2()2)3) --- --- fedcba
6. Convert the ^ notation to the regular exponential notation (see 24.4):
)1()^10())2()^5()2)^5())2(^5)2^5) ------ edcba
7. Use the expanded notation to find the value of the following expressions. One case is solved for you as an example.
)3()^2())3()^10()2)^5())2(^10)2^3) ------ edcba
Example:
8
1
)2()2()2(
1
)2(
1)2()3()^2()
3
3
-=
-´-´-=
-=-=-- -e
8. Find the value of each expression containing powers and absolute value functions. One case is solved for you as an example.
|2||2||3|12 |2|)|5|)10)|10|)|2|) ------ ---- edcba
Example:
4
1
2
12|2|)
2
2|2| ===- ---e
24.5 (Exponent 0 and 1) Any number (excluding 0) raised to power of 0 is 1. Any number raised to power of 1 is equal to itself:
aaa == 10 1
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The Book of Integers
25. Exponent Rules
29
Iulia & Teodoru Gugoiu
1. Simplify the following expressions by adding the exponents (see 25.1):
© La Citadelle
25.1 (Adding Exponents) The product of two or more powers having a common base is a power with the same base and an exponent equal to the sum of exponents:
nmnm aaa +=´
Example:85353 )2()2()2()2( -=-=-´- +
25.2 (Subtracting Exponents) The division of two powers having a common base is a power with the same base and an exponent equal to the difference of exponents:
nm
n
m
aa
a -=
Example:235
3
5
)3()3()3(
)3(-=-=
-
- -
25.3 (Multiplying Exponents) A power raised to another power is a power having the base of the first power and the exponent equal to the product of exponents:
nmnm aa ´=)(
Example:63232 )5()5())5(( -=-=- ´
25.4 (Multiplying Bases) Multiplying two or more numbers and then raising the product to a power is equivalent to raising each number to that power and then multiplying all those powers:
nnnn xbaxba ´´´=´´´ ...)...(
Example:444 )3()2()]3()2[( +´-=+´-
25.5 (Dividing Bases) Dividing two numbers and then raising the ratio to a power is equivalent to raising each number to that power and then dividing the powers:
n
nn
b
a
b
a=÷
ø
öçè
æ
Example:
5
55
)3(
)2(
3
2
+
-=÷
ø
öçè
æ
+
-
25.6 (Simplification) Use exponent rules to simplify an expression containing powers. Example:
2)4(6
4
4
4
6
4
)2(2)2(3
4
223
)2()2()3(
)3(
)2(
)2(
)]3()2[(
)3()2(
6
])3()2[(
----
-
-
-
-
-
-´-´
-
-
-=-=-
-´
-
-=
=-´-
-´-=
-´-
325315223 00)55))10()10())3()3()22) ´´-´--´-´ edcba
2. Rewrite the following expressions as a product of two powers:
132213522123 )1()0)4))10())2()4) ++++++ --- fedcba
3. Simplify the following expressions by subtracting the exponents (see 25.2):
2
2
5
3
4
4
3
5
1
4
2
5
3
3)
)4(
)4()
)1(
)1()
)3(
)3()
)10(
)10()
3
3)
2
2) gfedcba
-
-
-
-
-
-
-
-
4. Rewrite the following expressions as a division of two powers:
122153634132 )1()4)1))10())5()5) ---+--- --- fedcba
6. Expand the brackets (write the expression as a product of two powers as explained at 25.4):
23234 )02())]3(2[))]2()5[())14())32() ´--´-´-´-´ edcba
7. Simplify (write each expression as a single power):
21524322 )5()5()00))10()10())1()1()22) -´-´-´--´-´ edcba
8. Expand the brackets (write the expression as a ratio of two powers, as explained at 25.5):
323524
3
5)
2
2)
3
0)
2
1)
3
2)
3
2) ÷
ø
öçè
æ
-÷ø
öçè
æ
+
-÷ø
öçè
æ÷ø
öçè
æ
-
-÷ø
öçè
æ -÷ø
öçè
æfedcba
9. Simplify (write each expression as a single power):
5
5
4
4
7
7
2
2
5
5
3
3
)3(
2)
)1(
0)
)2(
)6()
)4(
)1()
4
)2()
4
2)
---
-
-
--fedcba
10. Use exponents rules to simplify, then evaluate:
55
7
25
2
23
4
532
2
32
4
32
)5()2(
)10()
)5()5(
)5()
55
5)
)10()10()10())3()3(
)3()3()
2
22)
-´-
-
-´-
-
´
-´-´--´-
-´-´
-
-
fed
cba
5. Simplify (write the expression as one single power as explained at 25.3):
211123232323 ))7(())3()])2[())2())2())2() ------- -- fedcba
32
232
4
23
322
222
)2(16
)8()4()
)2(
)2()2()
])2()6[(
])3()4[()
-´
-´-
úúû
ù
êêë
é
-
-´-
-´-
-´--
--
-
ihg
12
22
222
3242
12
23
324
32
))4((
)2(2
])2()8[(
])4()4[()
)2()2(
)2()2(
))2(()2(
)2()2()2()
----
-
--
-
-
-´´
+¸-
-¸-÷÷
ø
ö
çç
è
æ
-´-
-¸-¸
-´-
-´-´-kj
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The Book of Integers
26. Order of Operations (IV)
30
Iulia & Teodoru Gugoiu
1. Evaluate the following expressions (see 16.1):
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26.1 (Negation versus Power) The operation of raising a number to a power has a higher priority than the negation operation (changing a number to its opposite). Examples:
16)2(
16)2(2
4
44
+=-
-=-=-
26.2 (Right to Left Order for Exponents) If an expression contains consecutive raising operations to a power, perform the raising operations from right to left:
cbcbbb aabutaacc ´=Û )()(
Examples:
6422)2(
512222
62323
9)3(3 22
===
===
´
26.3 (Bases or exponents as expressions) If the base or the exponent of a power is an expression itself, first evaluate the base and the exponent separately, then evaluate the power:
)()^( baab ÛExample:
16)2()46()432( 462232 =-=+-=+´- +-´+-
26.4 (Default Order of Operations Updated) If an expression contains additions, subtractions, multiplications, divisions and powers but not grouping symbols, then the default order of operations is:a) do all the powers operations from right to leftb) do all the multiplication and division operations in the order they appear (from left to right)c) do all the addition and subtraction operations in the order they appear (from left to right). Example:
26.5 (Order of Operations Algorithm) If an expression contains grouping symbols, use the Order of Operations Algorithm (17.3) and the Default Order of Operations (26.4). Example:
213618)3()3(218
9)27()3(4892
3)3()3(2832 2322
-=+--=---´+-=
=¸---´¸+´-=
=¸---´¸+´-
4)11(
)11616()116)2(8(
)116)24(8(
)116)484(8(
)1)4()284(2(
2
22
2
)1()2(
)1()53(223
=+=
=+¸=+¸-´-=
=+¸+-´-=
=+¸¸+-´-=
=+-¸¸+-´-
-´-
-´-
111021
004422
10)10))2())123())2())2()
)10()10))2()2))2()2)
----
--
-------
------
lkjihg
fedcba
2. Evaluate the following expressions (see 16.2):
22222)3(3)3(
333)3(233
)2())2()2))2())2()2)
2))2()2)2))2()2)2222222
22222
lkjihg
fedcba
---
-
---
---
3. Evaluate the following expressions (see 16.3):
2
2
2
2
39212
331
4322
2]4)32[(2232132
)24(
)48()510()
221
432))2()
])2()31[())228())321()
-
¸-´-÷ø
öçè
æ
-
´-
-´-
¸´--+´-+-
¸-
¸´¸-÷ø
öçè
æ
´--
´+-
-¸--+¸--´+
fed
cba
4. Use the default order of operations (see 16.4) to evaluate the following expressions:
2323422222
2222323222
2)5()2(3)3()241642)3321)
)6()3()2()32432)321)
¸-´--¸-¸-¸´-¸-´-
-¸-´-+¸---+-
fed
cba
6. Use the order of operations algorithm (see 16.4) to evaluate the following expressions:
2
2
23
32
22
3262
332
)22()42(2
2
23
224
323222)3()23(
23222232
32)33(
1)322()43(
222
123)
)2()2(22
2)2()2(2)
])22[()12)4(2
)12()12(2)
3]1)2()2[()]21()53(1[))423(2)
)2323()2510()523()5)3431()
2)32(
÷÷
ø
ö
çç
è
æ
+¸-
-´-´+-´
÷÷
ø
ö
çç
è
æ
+-
--
-´-¸-
+-´--
-÷÷
ø
ö
çç
è
æ
-+-¸
-´+-
--¸¸+-´+-+-+´--
-´+-´+¸-¸+--+´--
-
-
--
-´+-
-
hg
fe
dc
ba
2)13(1
22
4
12)3(
21
53
)2(
)2(2
12
3)
2)223()3()5225()23(2)2()
2
2
422
00
2
3
2
4
221
524
2
2
524322252332223
22
22
-÷÷
ø
ö
çç
è
æ
--
-´
-
-+-+
+
+-
÷÷
ø
ö
çç
è
æ
-
-¸-
-÷÷
ø
ö
çç
è
æ
-
-
-+--+-¸´+---+-
-÷÷
ø
ö
çç
è
æ
--
´-
--
j
i
5. Use the order of operations algorithm (see 16.4) to evaluate the following expressions:
222)22()2)222(2)222)
222)222))222(22)
22222)22222)22222)
22222)22222)22222)
22222
2222222
222
222
´+¸--+´¸-+´
¸-´-¸´++
´+¸+-+´¸--+¸-´
¸´+-´-+¸¸´-+
¸-
´++¸
lkj
ihg
fed
cba
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The Book of Integers
27. Divisors
31
Iulia & Teodoru Gugoiu
1. Find the logical value (true or false) of each statement:
© La Citadelle
27.1 (Divisors) An integer a is a divisor of the integer b if dividing b by a leaves no remainder. In this case there is an integer c so:
cabca
b´=Û=
27.2 (Definitions) Both a and c are called divisors or factors of a.Both a and c divide a.b is called a multiplier of both a and c.b is divisible by both a and c. For the example above:Both -3 and -4 are divisors of 12Both -3 and -4 divide 1212 is divisible by both -3 and -4
27.3 (Improper Divisors) Given any integer a (except 0 and 1) there are at least 4 divisors of this number that are improper divisors: 1, -1, a, -a. Indeed:
1;1;1
;1
-=-
=-=-
=a
a
a
aa
aa
a
27.4 (Proper Divisors) Any divisor of an integer that is not an improper divisor is called a proper divisor.Example: The proper divisors of -6 are:+2, -2, +3, and -3
27.5 (Prime Numbers) A prime number is a positive integer (a natural number) with no proper divisors. The only divisors of a prime number are 1 and the number itself (if you ignore -1 and the opposite of that number).
Example: 7 is a prime number. The only natural numbers that are divisors of 7 are the improper divisors 1 and 7.
27.6 (Composite Numbers) A natural number that is not a prime number is called composite number. A composite number has proper divisors (different from 1 and the number itself).Example: 12 is composite number. The improper divisors are 1 and 12. The proper divisors are 2, 3, 4, and 6. For this example the negative divisors were ignored.
Example: -3 is a divisor of 12 because:
Example: The improper divisors of -5 are:1, -1, -5, +5.
27.7 (Number 1) Conventionally, number 1 is neither a prime nor a composite number.
)4()3(1243
12-´-=Û-=
-
112)614)
205)83)
224)416)
103)205)
-+-
-++-
+--
-+--
divideshofmultipleaisg
dividesfofdivisoraise
bydivisibleisdofmultipleaisc
dividesbofdivisoraisa
2. Find the improper divisors of each integer:
3)11)20)5)8)10) ---+- fedcba
3. Find the proper divisors of each integer:
16)6)10)15)6)12) ----+- fedcba
4. Find all divisors of each integer:
21)30)20)24)18)9) ----+- fedcba
5. Find all prime numbers between 1 and 30.
6. Find all composite numbers between 31 and 50.
7. Classify each of the following numbers as either prime, composite or neither.
1)57)53)51)41)31)21)11) hgfedcba
8. Use the Sieve algorithm (below) to find the prime numbers between 1 and 100:
Eratosthenes
1. Create the list of all whole numbers between 2 and 100: list A (see the list below).2. Create an empty list and name it: list B (this is the list of prime numbers).3. Move the first number from list A to list B, then remove from list A all that number’s multiples.4. Repeat step 3 until no more numbers are left in list A.
The list of prime numbers is the list B.
100999897969594939291
90898887868584838281
80797877767574737271
70696867666564636261
60595857565554535251
50494847464544434241
40393837363534333231
30292827262524232221
20191817161514131211
1098765432List A:
List B:
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The Book of Integers
28. Divisibility Rules
32
Iulia & Teodoru Gugoiu
1. For each of the following numbers, find whether or not they are divisible by 2:
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28.1 (Divisibility by 2) An integer is divisible by 2 if the last digit is 2, 4, 6, 8, or 0. A number divisible by 2 is called an even number. Examples: 24, -12, -30.
28.2 (Divisibility by 3) An integer is divisible by 3 if the sum of all digits is divisible by 3. Examples: -111, 102, 750.
28.3 (Divisibility by 4) An integer is divisible by 4 if the number defined by the last two digits is divisible by 4. Examples: 1020, -132, -144..
28.4 (Divisibility by 5) An integer is divisible by 5 if the last digit is 5 or 0.
28.5 (Divisibility by 6) An integer is divisible by 6 if the number is divisible by 2 and 3. Examples: 102, -222, 1014.
28.6 (Divisibility by 7) Use the following algorithm:a) Drop the last digit and double it b) Subtract the result from the remaining number. c) If the final result is divisible by 7 then the original number is also divisible by 7. Example:28.7 (Divisibility by 8) An integer is divisible by 8 if the number defined by the last three digits is divisible by 8. Examples: 816, -8192, -2000.
28.8 (Divisibility by 9) An integer is divisible by 9 if the sum of all digits is divisible by 9. Examples: 369, -252, 126.
28.9 (Divisibility by 10) An integer is divisible by 10 if the last digit is 0.
28.10 (Divisibility by 11) Alternately add and subtract the digits from left to right. If the number you get is divisible by 11 then the original number also is. Example: 1353.
28.11 (Divisibility by 12) An integer is divisible by 12 if the number is divisible by 3 and 4. Examples: 22344, -132, -8652.
28.12 (Divisibility by 13) Use the following algorithm:a) Drop the last digit and multiply it by 9 b) Subtract the result from the remaining number. c) If the final result is divisible by 13 then the original number also is. Example: 3003.
771279;791 =´-
011
0;03531 ==-+-
013
0;03927
27339300
==´-
=´-
2. For each of the following numbers, find whether or not they are divisible by 3:
3. For each of the following numbers, find whether or not they are divisible by 4:
4. For each of the following numbers, find whether or not they are divisible by 5:
5. For each of the following numbers, find whether or not they are divisible by 6:
6. For each of the following numbers, find whether or not they are divisible by 7:
7. For each of the following numbers, find whether or not they are divisible by 8:
8. For each of the following numbers, find whether or not they are divisible by 9:
9. For each of the following numbers, find whether or not they are divisible by 10:
10. For each of the following numbers, find whether or not they are divisible by 11:
11. For each of the following numbers, find whether or not they are divisible by 12:
12. For each of the following numbers, find whether or not they are divisible by 13:
13. Use a calculator and the algorithms in 28.1 to 28.12 to find if:
36)31)112)21)0)14)55) gfedcba +--
12345)130)765)201)1111)222)1234) --+-- gfedcba
12386)148)966)701)510)312)120) --+-- gfedcba
12777)444)465)700)1225)115)220) --+-- gfedcba
9654)888)465)900)1158)135)246) --+-- gfedcba
23345)885)465)861)1152)635)231) --+-- gfedcba
9870)896)890)556)1152)620)248) --+-- gfedcba
5553)666)981)777)108)6588)1234) --+-- gfedcba
999)7890)1000)5555)1230) -+-- edcba
23432)642)22)1111)308)6171)1232) --+-- gfedcba
66)12345)111)222)324)600)240) --+-- gfedcba
74191)100)11)1313)182)3900)130) --+-- gfedcba
012345678998765432109)
12655555666661223334444)
1234513579864206)
11665554449998887776)
765432112345678983)
divided
ofmultipleaisd
ofdivisoraisc
ofmultipleaisb
ofdivisoraisa
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The Book of Integers
29. Prime Factorization
33
Iulia & Teodoru Gugoiu
1. Use the following factor trees and write down for each case the prime factorization of the number placed at the root of the tree:
© La Citadelle
29.1 (Fundamental Theorem of Arithmetic) Any natural number except 1 can be written as a product of prime numbers in a unique way called the prime factorization of that number. Use exponents to make the expression simpler. Example:
312 5325553221500 ´´=´´´´´=
29.2 (Factor Tree Method) The factor tree method is a method to find the prime factorization of a natural number. The root of the tree is the number itself. For example:
24
Using divisibility rules try to find a divisor of the given number. In our case 2 is a divisor. By dividing 24 by 2 you get 12. So:
2 is a prime factor and so is a leaf of the tree. Let’s continue the process with 12 until we get only prime factors (leaves):
24
2 12
24
2 12
2 6
2 3
All leaves of the tree generate the prime factorization of the given number:
13 32322224 ´=´´´=
29.3 (Uniqueness) Although for a natural number more than one factor tree is possible, the prime factorization of that number is unique. Example:
24
4 6
2 32 2
24
3 8
2 4
2 2
29.4 (Integers) Prime factorization can be extended to integers (except 0, 1, and -1) if only positive factors are considered. Example: -36
36
4 9
3 32 2
22 3236 ´-=-
So:
210
7 30
2 15
3 5
225
25 9
3 35 5
150
5 30
3 10
5 2
330
10 33
11 35 2
A) B) C) D)
3. Build a factor tree for each number and write down its prime factorization:
83)350)1024)250)100)80)64)40) hgfedcba
2. Complete the following factor trees and write down the prime factorization:
2
3
5 7
2 311 5
7
11
5 2
3 33 3
A) B) C) D)
4. Build at least 4 different factor trees for the number 600.
5. Build a factor tree for each of the following negative integers and write down its prime factorization:
12345)17)4096)300)88)12) ------ fedcba
6. Find the value of each number given by its prime factorization and then build a factor tree:
111112558
2222412123
117532)52)3)2)
532)52)753)32)
´´´´´
´´´´´-´
hgfe
dcba
7. Complete the following factor trees and write down the prime factorization:
24
A) B) C)
200 360
8. Three numbers a, b, and c are given by their prime factorization:
13122341223 75327532532 ´´´=´´´=´´= cba
Use the exponents rules to find the prime factorization for:
2
22 ))))/))
c
bafcbae
c
badacabcbbaa
´´´
´´
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The Book of Integers
30. Set of Divisors
34
Iulia & Teodoru Gugoiu
1. Use the set notation to express the following sets:
A) the set of prime numbers less than 10B) the set of divisors of 10C) the set of first 5 natural numbers starting with 0D) the set of first positive 5 multiples of 3E) the set of odd numbers between 10 and 20
© La Citadelle
30.1 (Set) A set is a collection of elements. One way to define a set is to enumerate all the elements that belong to the set.. Example: }5,2,1{
Note that the elements are separated by commas and written inside braces {}. This is called the set notation.
30.2 (Multiplying Sets) The product of two sets is a set of products of each element of the first set by each element of the second set. Example:
}21,15,14,10,7,5{
}73,53,72,52,71,51{
}7,5{}3,2,1(
=
=´´´´´´=
=´
30.3 (The Set of Divisors) To find the set of all divisors of a natural number:a) find the prime factorizationb) for each prime factor, build a set having elements the prime factor raised to all possible powers starting 0 and ending at the maximum power as found in the prime factorizationc) multiply the sets you get in step b) Example:
}90,18,30,6,10,2,45,9,15,3,5,1{
}518,118,56
,16,52,12,59,19,53,13,51,11{
}5,1{},18,6,2,9,3,1{
}5,1{}92,32,12,91,31,11{
}5,1{}9,3,1{}2,1{
}5,1{5
}9,3,1{3
}2,1{2
53290
1
2
1
121
=
=´´´
´´´´´´´´´=
=´=
=´´´´´´´=
=´´
Þ
Þ
Þ
´´=
30.4 (Integral Divisors) Given an integer you can use the algorithm presented above to find all integral divisors of that number. Example:
}24,8,12,4,6,2,3,1{
}38,18,34,14,32,12,31,11{
}3,1{}8,4,2,1{
}3,1{3
}8,4,2,1{2
3224
1
3
13
=
=´´´´´´´´=
=´
Þ
Þ
´-=-
So, the set of all integral divisors is:
}24,8,12,4,6,2,3,1{ ±±±±±±±±
2. Use the rule of “multiplying sets” presented in 30.2 and find the “product” of the following sets:
}3,2,1{}2,1{}1{)}2,1{}2,1{)}4,3{}2,1{)}5{}1{)
}2,1{}2,1{}2,1{)}4,3{}5,2{)}3,2{}3,2,1{)}2,1{}2,1{)
´´-´-´´
´´´´´
hgfe
dcba
3. For each of the following powers, build the set of all its divisors (see 30.3):
31032
4321
5)2)3)3)
2)2)2)2)
hgfe
dcba
Example:
}16,8,4,2,1{}2,2,2,2,2{2) 432104 =Þd
4. For each of the following natural numbers, find the set of positive divisors (see 30.3):
2048)625)100)36)20)16)10)6) hgfedcba
5. For each of the following integers, find the set of all divisors (positive and negative) (see 30.3):
512)1000)360)200)160)40)8) ----- gfedcba
6. For each natural number, find the number of divisors (see the example below):
5120)2500)6000)640)200)64)40)12) hgfedcba
Example:
.
40425)13()11()14(:
5326000:
6000)
314
divisorspositiveallofnumbertheisThis
themmultiplyandexponentsallto1Add
ionfactorizatprimethePerform
f
=´´=+´+´+
´´=
7. Compute the following operations with sets of numbers (see 30.2 for multiplication). One case is solved for you as an example:
}2,1{}6,4,2{)}2,1{)}5,2{)}3,2{}3,2,1{)}2,1{}2,1{) 32 ¸-+ edcba
Example:
}25,10,4{}25,10,10,4{}55,25,52,22{}5,2{}5,2{}5,2{) 2 º=´´´´=´=c
Note: In a set any element is unique (appears only once).
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The Book of Integers
31. Highest Common Factor (HCF)
35
Iulia & Teodoru Gugoiu
1. For each case: find the set of the positive divisors of all the given numbers then choose the highest common divisor (factor):
© La Citadelle
31.1 (HCF) The Highest Common Factor (HCF) of two or more natural numbers is the highest common natural number into which you can divide these numbers evenly (without a remainder). Example:
54
20;3
4
12;2
4
8
4
20,12,8:
===
=HCF
andGiven
31.2 (The Algorithm for finding HCF) The following algorithm allows you to find the HCF for any set of natural numbers:a) perform the prime factorization of each numberb) complete the missing prime factors with prime factors raised to power 0c) HCF is the number with all the prime factors raised to the lowest exponents.Example:
12327532)
7532180;7532120
;753284)
532180
;532120;73284)
180,120,84:
120012
10120113
1012
122
113112
=´=´´´=
´´´=´´´=
´´´=
´´=
´´=´´=
HCFc
b
a
andGiven
31.3 (HCF for integers) To calculate the HCF for a set of integers, consider only the positive divisors of these numbers. In this way, the HCF will always be a natural number. Example:
20532)
532180;532200)
532180;52200)
180200:
102
122203
12223
=´´=
´´=´´-=-
´´=´-=-
+-
HCFc
b
a
andGiven
31.4 (Factoring) Given an expression with two or more terms you can always factor the HCF of these terms. Example:
)752(6423012 -+-´=-+-
31.5 (HCF Function) The function HCF has as input a list of natural or integral numbers separated by commas, and as output the HCF of these numbers. Examples:
6)60,30,12(
5)15,10(
=-
=
HCF
HCF
60;18)100;250)90;36;81)24;64)48;36;8)20;12) fedcba
Example:
6
}60,30,20,15,12,10,6,5,4,3,2,1{}60,12,20,4,30,6,10,2,15,3,5,1{
}5,1{}12,4,6,2,3,1{}5,1{}3,1{}4,2,1{53260
}18,9,6,3,2,1{}18,6,2,9,3,1{}9,3,1{}2,1{3218
60;18)
112
21
=
==
=´=´´Þ´´=
==´Þ´=
HCF
f
3. For each case use the algorithm presented in 31.2 to find the HCF:
300;225)100;40;64)81;128)72;60;18)30;24) edcba
4. For each case use the algorithm presented in 31.2 to find the HCF (see also 31.3):
625;225;125)40;21)60;36;30)250;400) ------- dcba
5. Rewrite the following expressions by factoring out the HCF:
907260)25664)500250200)183012)
350100240)225125)8064)3624)
--+-+----
+-----
hgfe
dcba
7. Simplify the following expressions by canceling out the HCF.
52
37)
13065
5226)
150100
1257550)
9648
806432)
115150
906045)
8472
603624)
5664
4024)
4221
3514)
+-
-
+-
-
+-
+-
+-
+-+-
+-
+-
+-
--
--
+-
-
hgfe
dcba
Example: 22
4
64
532
)64(25
)532(25
150100
1257550) ==
+-
+-=
+-´
+-´=
+-
+-f
2. For each case, find the HCF by factoring out the common factors one by one (one case is solved for you as an example):
105;70)48;40)20;45;35)32;24)36;20;6)12;8) fedcba
Example:
3575
}3;2{75}21;14{5}105;70{)
=´
´´=´=
isHCFThe
f
6. Simplify the following expressions by canceling out one by one the common factors (one case is solved for you as an example):
10234
5117)
150125
5075)
10836
1207260)
9040
806030)
9680
642440)
5442
361830)
2520
3015)
1215
69)
+-
-
+-
-
+-
+-
+-
+-+-
+-
+-
+-
--
--
+-
-
hgfe
dcba
Example:
42
8
97
635
)97(3
)635(3
2721
18915
)2721(2
)18915(2
5442
361830) ==
+-
+-=
+-´
+-´=
+-
+-=
+-´
+-´=
+-
+-c
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The Book of Integers
32. Least Common Multiple (LCM)
36
Iulia & Teodoru Gugoiu
1. For each set of numbers, list the multiples until you can identify the LCM (see 32.2):
© La Citadelle
32.1 (LCM) The Least Common Multiple (LCM) of two or more natural numbers is the least common natural number which can be divided evenly (without a remainder) by each of these numbers. Example:
32.3 (The Algorithm for finding LCM. Solution 2) The following algorithm allows you to find the LCM for any set of natural numbers:a) perform the prime factorization of each numberb) complete the missing prime factors with prime factors raised to power 0c) LCM is the number having all the prime factors raised to the greatest exponents.Example:
32.5 (LCM for integers) To calculate the LCM for a set of integers, consider only the positive multiples of these numbers. In this way, the LCM will always be a natural number. Example:
32.6 (LCM Function) The function LCM has as input a list of natural or integral numbers separated by commas, and as output the LCM of these numbers. Examples:
120)15,8(;30)15,10( =+-= LCMLCM
26
12;3
4
12
12;64
==
=LCMandGiven
32.2 (The Algorithm for finding LCM. Solution 1) List all the multiples of each given number until you find a first common multiple. That is the Least Common Multiple. Example:
24
,...30,24,18,12,66
,...28,24,20,16,12,8,44
,...30,27,24,21,18,15,12,9,6,33
6,4,3
=
Þ
Þ
Þ
LCM
andGiven
600532
53225;53212
5328;25,12,8
213
200012
003
=´´=
´´=´´=
´´=
LCM
andGiven
30532;53215
5326;15,6
111110
011
=´´=´´=
´´-=-+-
LCM
andGiven
32.4 (LCM & HCF) The product of any two natural numbers (except 1) is equal to the product of LCM and HCF of these numbers. Example:
HCFLCM
HCFLCM
andGiven
´==´
==
1801512
3;60
1512
20;12)35;25)15;6;4)12;5)10;8;3)15;6) fedcba
2. For each set of numbers, use the algorithm presented in 32.3 to find the LCM:
32;24)25;30)8;12;5)20;16)18;15;10)10;4) fedcba
3. For each pair of numbers, prove that the product of the numbers is equal to the product between the HCF and the LCM of these numbers:
16;14)25;10)8;6)30;12)15;10)12;10) fedcba
4. For each set of integers find the LCM:
25;20)5;12;18)16;12)30;25;20)8;3) -------- edcba
5. Find the value of the following expressions containing HCF and LCM functions (see 31.5 and 32.6):
)25,15,10()25,15,10(
251510)
)5,10,16(
)50,100,160()
)80,100,120())35,25()35,25(
3525)
)20,15(
)32,48())8,6()24,20()
)16,12())8,6()
HCFLCMh
LCM
HCFg
HCFfHCFLCM
e
LCM
HCFdLCMHCFc
LCMbHCFa
´
´´´
´
-
Example:
2052)80,100,120(
5280;52100;532120
)80,100,120()
12
1422113
=´=
´=´=´´=
HCF
HCFf
6. LCM is used to add (or subtract) two or more fractions with unlike denominators. The Least Common Denominator (LCD) is the LCM of all denominators. To add the following fractions, convert them into equivalent fractions having a denominator equal to LCD and then add (or subtract) numerators. One case is solved for you as an example:
10
1
8
3)
16
1
12
1)
15
1
10
1)
6
5
4
3)
6
1
3
2)
3
1
2
1) ---+-+ fedcba
Example:12
19
12
109
12
10
12
9
6
5
2
2
4
3
3
3
6
5
4
3) =
+=+=´+´=+c
7. LCM is used to add (or subtract) two or more rational numbers with unlike denominators. The Least Common Denominator (LCD) is the LCM of all denominators. To add the following rational numbers, convert them into equivalent numbers having a denominator equal to LCD and then add (or subtract) numerators. One case is solved for you as an example:
16
1
12
5)
15
1
10
1)
6
5
4
1)
6
5
3
2)
3
1
2
1)
-+
-
--
--+
-
-+
-
--
-edcba
Example:
30
1
30
23
30
2
30
3
15
1
2
2
10
1
3
3
15
1
10
1
15
1
10
1)
-=
+-=+
-=´+
-´=+
-=
--
-d
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The Book of Integers
33. Square Roots
37
Iulia & Teodoru Gugoiu
1. Find the squares of the following numbers:
© La Citadelle
33.1 (Square Roots) The square root operation is the inverse operation to raising a number to the power 2:
axsoxthatisa =2
33.2 (Radicand) Because any real number (including an integer) squared is positive or 0, the radicand (the expressions inside the radical sign) is positive or 0:
Example: 42;24 2 ==
0³aifrealisa
Example:
01616 <-- becauserealnotis
33.3 (Principal Root) If more numbers squared are equal to the radicand, then by convention the square root is represented by the positive number called the principal root:
0³a
Example: 39;93;9)3( 22 ===-
33.6 (Square root of a square) The square root of a negative number squared gives you the opposite of the original number:
33.5 (Square Root of a Square) The square root of a positive number squared gives you back the original number:
33.7 (Algorithm for finding the square root) To find the square root of a number:a) do the prime factorizationb) the square root has the same factors as the original number, but all exponents are divided by 2Example:
33.8 (Real Numbers) If the radicand of a square root is not a perfect square, then the square root of that number is a real number. You can use a calculator to find a decimal approximation of the real number. Example:
3;39)3( 2 -==- not
847327056
7327056?;7056
112
224
=´´=
´´==
3932 ==
141.12 @
33.4 (Basic Square Roots) Here is a list of some basic roots (square roots):
;525255;416164
3993;2442
1111;0000
22
22
22
=Þ==Þ=
=Þ==Þ=
=Þ==Þ=
200)100)50)20)15)12)11)10)9)
8)7)6)5)4)3)2)1)0)
rqponmlkj
ihgfedcba
--
--
2. Find the following square roots (use the results you have got at Exercise 1):
16)144)16)49)40000)2500)
10000)100)400)121)64)9)
36)81)25)0)4)1)
rqponm
lkjihg
fedcba
3. Find the value of the following expressions containing square roots of squares (see 33.5):
222222
222222
99999)123)100)20)13)10)
9)5)7)0)3)1)
lkjihg
fedcba
4. Find the value of the following expressions containing square roots of squares (see 33.6):
22222
22222
)125())100())30())10())11()
)3())8())7())1())5()
-----
-----
jihgf
edcba
5. Find the value of the following expressions (see 33.7):
64422442
624222422
)5()10()3)20())3()5()5)7()
310)332)53)32)
-´-´--´-´-
´´´´´
hgfe
dcba
6. Use the prime factorization algorithm to find the following square roots (see 33.7):
Example:
12500125100510510510)5()10() 322/62/46464 =´=´=´=´=-´-h
202500)7056)15625)2500)2916)1089) fedcba
7. Use a calculator to find the following roots. Round the answer to the nearest hundredth (see 33.8):
99999)1234)50)10)7)3)2) gfedcba
8. Use a calculator to find the following roots. Round the answer to the nearest thousandth (see 33.8):
12345)777)200)30)10)3)2) gfedcba
Example:
23.31699999) @f
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The Book of Integers
34. Cubic Roots
38
Iulia & Teodoru Gugoiu
© La Citadelle
34.2 (Useful Cubic Roots) Here is a list of some basic cubic roots:
51251255;464644
327273;2882
1111;0000
3333
3333
3333
=Þ==Þ=
=Þ==Þ=
=Þ==Þ=
34.4 (Sign) The cubic root of a positive number is positive (because a positive number cubed is positive). The cubic root of a negative number is negative (because a negative number cubed is negative). Examples:
34.3 (The Cubic Root of a Cube) The cubic root of a number cubed is the original number:
aa =3 3
34.5 (The Algorithm for finding the cubic root) To find the cubic root of a number:a) perform the prime factorizationb) the cubic root has the same factors as the original number, but all exponents are divided by 3Example:
12321728
321728
?1728
123
36
3
=´=
´=
=
34.6 (Real Numbers) If the radicand of a cubic root is not a perfect cube, the cubic root of that number is a real number. You can use a calculator to find a decimal approximation of that real number. Example:
154.2103 @
32727)3(
288)2(
111)1(
33
33
33
-=-Þ-=-
-=-Þ-=-
-=-Þ-=-
Examples:
28)2(;3273 33 333 3 -=-=-==
5125;5125 33 -=-=
34.1 (Cubic Roots) The cubic root operation is the inverse operation to raising a number to the power of 3:
axsoxthatisa =33
Example:
82;28 33 ==
1. Find the cubes of the following numbers:
12)11)200)100)50)20)10)
6)5)4)3)2)1)0)
nmlkjih
gfedcba
2. Find the following cubic roots (use the results you have found at Exercise 1):
333333
333333
1331)125000)8000)27)1000)125)
64)8)216)1)1728)0)
lkjihg
fedcba
3. Find the cubes of the following numbers:
12)300)100)40)20)10)
6)5)4)3)2)1)
------
------
lkjihg
fedcba
4. Find the following cubic roots (use the results you found in Exercise 3):
33333
33333
1728)125)8000)64)1000)
27)216)8)216)1)
-----
----
jihgf
edcba
5. Find the value of the following expressions containing cubic roots of cubes (see 34.3):
3 33 33 33 33 33 3 12345)12)10)5)3)1) fedcba
6. Find the value of the following expressions containing cubic roots of cubes (see 34.3):
3 33 33 33 33 3 )123())10())4())2())1() ----- edcba
7. Find the value of the following expressions:
3 363 633 363 63
3 3693 363 633 33
)10()2()3)10())4()5()5)4()
532)310)52)32)
-´-´--´-´-
´´´´´
hgfe
dcba
8. Use the prime factorization algorithm to find the following cubic roots (see 34.5):
Example:
40)10(4)10()2()10()2()10()2() 123/33/63 36 -=-´=-´-=-´-=-´-h
33333 1728000)32768)13824)91125)5832) edcba
9. Use a calculator to find the following roots. Round the answer to the nearest thousandth (see 34.6):
333333 99999)1234)100)80)20)7) fedcba --
Example:
416.4699999) 3 @f
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The Book of Integers
35. Roots of superior order
39
Iulia & Teodoru Gugoiu
© La Citadelle
35.1 (n-th Order Roots) The n-th order root operation is the inverse operation to raising a number to the power n:
35.4 (Basic n-th order roots) Here is a list of some basic n-th order roots:
212818)2(
381813;216162
1111;0000
77
4444
5544
-=-Þ-=-
=Þ==Þ=
=Þ==Þ=
35.5 (The n-th order root of a positive number raised to power n) The n-th order root of a positive number raised to power n is the original number:
0; >= aaan n
35.7 (The Algorithm for finding the n-th root) To find the n-th root of a number:a) perform the prime factorizationb) the n-th root has the same factors as the original number, but all exponents are divided by nExample:
12323220736 124 484 =´=´=
Examples:
1010;536 65 5 ==
35.6 (The n-th order root of a negative number raised to power n) The n-th order root of a negative number raised to power n is the original number if n is odd and the opposite of the original number if n is even. Examples:
22256)2(;2)2(8 888 85 5 ===--=-
axsoxthatisa nn =Example:
32)2(;232
162;216
55
44
-=--=-
==
35.2 (Radicand) Because any real number (including an integer) raised to an even power is positive or 0 the radicand must be positive or 0:
evenisnifaan 0; ³
35.3 (The Principal Root) If n is even, more numbers raised to the power of n are equal to the radicand. By convention the n-th root is represented by the positive number called the principal root:
Example:
081814 <-- becauserealnotis
evenisnifan 0³
Example:
381;81)3(;813 444 ==-=
1. Find the values of the following powers:
5445465
4645445
)10())10())3())2())2())1())1()
100)10)5)4)3)2)0)
------- nmlkjih
gfedcba
2. Find the value of the following roots (use the results you found in Exercise 1):
954557
464455
512)243)1)100000)32)1)
81)1000000)625)16)1)0)
------
-
lkjihg
fedcba
3. Find the value of the following expressions containing roots of powers (see 35.5):
11 117 79 97 74 45 5 54321)12)10)2)4)1) fedcba
10 105 57 75 56 6 )123())10())4())1())1() ----- edcba
4. Find the value of the following expressions containing roots of powers (see 35.6):
5. Find the value of the following expressions (see 35.7):
5 10157 1475 1054 48
5 510104 484 1245 105
)10()2()3)10())4()5()5)4()
332)310)52)32)
-´-´--´-´-
´´´´´
hgfe
dcba
6. Use the prime factorization algorithm to find the following roots (see 35.7):
76654 78125)4096)46656)59049)20736) edcba
7. Use a calculator to find the following roots. Round the answer to the nearest hundredth:
745654 12345)100)100)125)100)10) -- fedcba
Example:
100000)10(10100000) 55 -=-Û-=-i
Example:
5432154321) 11 11 =f
Example:
123123)123() 10 1010 10 ==-e
Example:
8001008)10()2()10()2()10()2() 235/105/155 1015 -=´-=--=-´-=-´-h
Example:
84.312345) 7 -@-f
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The Book of Integers
36. Roots Rules
40
Iulia & Teodoru Gugoiu
1. Use the product rule (see 36.1) to simplify the following expressions:
© La Citadelle
36.1 (Product Rule) The product of two or more like roots is equal to the root of the product of the radicands:
nnn baba ...... ´´=´´
Examples:
101010101010
284242
2
2333
==´=´
==´=´
36.2 (Simplifying Radicals) The product rule allows you to simplify radicals. Example:
252522522550 =´=´=´=
36.3 (Quotient Rule) The ratio of two like roots is equal to the root of the ratio of the radicands:
nn
n
b
a
b
a=
Example:
4642
128
2
128 333
3
===
36.4 (Simplifying Radicals) The quotient rule allows you to simplify radicals. Example:
2
5
4
25
4
25==
36.5 (Power Rule) A root raised to a power is equal to the root of the radicand raised to the same power:
n mmn aa =)(
Example:33 223 255)5( ==
36.6 (Canceling Out) The n-th order root raised to the power of n is equal to the radicand:
aa nn =)(Examples:
32)32(;3)3( 552 -=-=
36.7 (Simplifying Expressions) You may use all the above rules to simplify expressions containing radicals. Example:
65
152
)5(
)15()2(
5
152
:;6)6(5
30
5
30
5
152
5
152
2
222
2
2
222
=´
=´
=÷÷
ø
ö
çç
è
æ ´
==÷÷
ø
ö
çç
è
æ=
=÷÷
ø
ö
çç
è
æ=
÷÷
ø
ö
çç
è
æ ´=
÷÷
ø
ö
çç
è
æ ´
or
82)1010)555)44)
162)42)444)22)
17 47 333344
5533333
´´--´-´-´
´-´´´´
-hgfe
dcba
2. Use the product rule to simplify the following radicals (see 36.2):
335 63 4
4 6335
500)32)128)10)10)200)
10)32)54)128)18)8)
lkjihg
fedcba
--
-
3. Use the quotient rule (see 36.3) to simplify the following expressions:
3
3
4
4
5
5
5
5
3
3
3
3 4
3
3
3
3
500
4)
120
15)
64
4)
3
96)
4
128)
24
3)
75
3)
10
10)
4
500)
2
250)
2
16)
2
50)
-
-
-
--
lkjihg
fedcba
4. Use the quotient rule to simplify the following expressions (see 36.4):
43353
81
625)
27
1000)
64
125)
243
32)
27
8)
25
9) fedcba
-
-
-
-
5. Use the power rule (see 36.5) to rewrite (simplify) the following expressions:
2324435233 )5())10())5())3())5())2() -fedcba
6. Use the canceling out rule (see 36.6) to simplify the following expressions:
5544332332 )5())10())10())20())5())2() -- fedcba
Example:
333 33 33 443 5555555)5() ´=´=´==d
7. Simplify the following expressions with radicals:
2
3
2
3)
54
32)
5
32)
5
2)
2
3)
3232
162)
52
108)
5
202)
1030
3)
2
63)
3
33
3323
3
32
55
55
33
33
3
33
´÷÷
ø
ö
çç
è
æ
´
´÷÷
ø
ö
çç
è
æ ´÷÷
ø
ö
çç
è
æ
-
-÷÷
ø
ö
çç
è
æ
´
´
´
´
-
-´-
´
´
jihgf
edcba
2
3
32
333
33
516
23)
654
321)
20
5
10
3
60
2)
62540
32)
62150
105)
86
322)
÷÷
ø
ö
çç
è
æ
´
´÷÷
ø
ö
çç
è
æ
´´
´´´´
´´
-´
´´
´
´
´´
pon
mlk
Example:
2
1
4
1
4
1
48
12
48
12
86
322
86
322) =====
´
´´=
´
´´k
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The Book of Integers
37. Equations with powers and radicals
41
Iulia & Teodoru Gugoiu
1. Check by substitution if the given x is a solution of the given equation:
© La Citadelle
37.1 (Roots) Consider the equation:
axn =
Any number x satisfying the equation is called a root of the equation. Example:
42 =x
4)2(2 2 =±±= becausex
The two roots of this equations are:
37.2 (n is even) In the case that n is even (e.g. 2, 4, 6, ...) the equation:
0; ³= aaxn
has two real roots given by:
n ax ±=
There are no real roots if a<0. Examples:
rootsrealnox
xx
Þ-=
±=±=Þ=
1
38181
2
44
37.3 (n is odd) In the case that n is odd (e.g. 3, 5, 7, ...) the equation:
has always a real solution given by:
axn =
n ax =
Examples:
23232
32727
55
33
==Þ=
-=-=Þ-=
xx
xx
37.4 (Unknown Radicand) To find the solution of the equation:
axn =
raise left and right sides to power n:
nnnn axax =Þ=)(
Examples:
8)2(2
2555
33
2
-=-=Þ-=
==Þ=
xx
xx
37.5 (More Steps Solution) You may need to do more steps in order to solve an equation with powers and radicals. Example:
28
8816
1648
4138
32718
;27)18(
3
3
23
3
33
33
==
=-=
==+
=+=+
==-+
=-+
x
x
x
x
x
x
1024;2)10000;10)125;5)
216;6)125;5)1;1)
64;4)16;4)9;3)
1045
337
322
-=-==-=-=-=
=====-=
=-==-==-=
xxixxhxxg
xxfxxexxd
xxcxxbxxa
2. Solve the following equations (see 37.2):
1024)625)64)81)256)
10000)64)625)4)25)
104648
42422
=====
===-==
xjxixhxgxf
xexdxcxbxa
3. Solve the following equations (see 37.3):
100000)216)128)64)32)
1000)128)125)32)64)
53735
37353
=-==-==
-=-==-==
xjxixhxgxf
xexdxcxbxa
4. Solve for x (see 37.4):
10)5)1)2)3)
1)2)3)2)2)
43333
43
=-=-===
===-==
xjxixhxgxf
xexdxcxbxa
5. Solve for x (see 37.4):
327)464)21024)5125)
264)11)28)525)
=-=-=-=-
=-=-==
xxxx
xxxx
hgfe
dcba
Example:
3)4()4(64)4(464) 3 =Û-=-Û-=-Û-=- xg xxx
7. Solve for x (see 37.5):
43
343
33
3
335 34 2
223333
3 3
3
27
3
3
103)027)14()1()
5
1
8
5)
8
64
425
4)83)1(5)222)
)5()3()813
)16)5(2)
3)2(3)242)21)
=´
´=-+-´+=
--
=-
--=++´=-´
-=+-=+=+´
=-´-=-´=-
xlxk
xj
xixhxg
xfx
exd
xcxbxa
6. Find x that satisfies the following equations:
xexdxcxbxa xxxxx ===== 46656)3125)256)27)4)
Example:
64665664665646656) 6 =Û=Û=Û= xxxe xx
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38. Link between Radicals and Powers
42
Iulia & Teodoru Gugoiu
1. Convert these radical notations into exponential notations:
© La Citadelle
38.1 (The Link between Radicals and Powers) If the radicand is positive, a radical expression can be converted to a power according to the following formula:
Example:0;
1
>= aaa nn
2/133 =
38.4 (Multiplying Unlike Radicals) Using the link between radicals and powers, we can now multiply unlike radicals:
nm nmnm
nm
nm
nmnmnm
aaa
aaaaaa
´ +
´
++
=´
Þ==´=´
1111
38.2 (Generalization) The precedent relation can be generalized to:
0; >= aaa n
mn m
Example:3/23 2 55 =
Example:
66 532 323 322222 ===´´ +
38.5 (Dividing Unlike Radicals) Using the link between radicals and powers, we can now divide unlike radicals:
mn mnmn
mn
nm
n
m
n
m
aaa
a
a
a
a ´ -´
--
====
11
1
1
Example:
55 215 6
15 253 35
5
3
5
3
5
3
422
888
8
8
8
8
8
===
====-
-=
-
- ´ -
38.6 (Radical of Radicals) If the radicand of a radical is another radical, you can use the following formula to simplify the expression:
nmm n aa ´=
Example:
2646464 6233 === ´
Indeed:
nmnmmnmnm n aaaaa ´´ ====
1111
)()(
38.3 (Simplifying Radicals) The following formula allows you to simplify radicals:
n kn
k
nm
kmnm km aaaa === ´
´´ ´
Example:
555 26 3 ==
5104543 2)1024)64)2)5)27)2) -gfedcba
2. Convert these exponential notations into radical notations:
4/12/13/12/13/12/12/1 625)100)27)10)8)3)2) gfedcba
3. Convert these radical notations into exponential notations:
3 23 67 36 55 23 23 2)3)7)10)5)7)2) gfedcba
4. Convert these exponential notations into radical notations:
3/55/43/43/24/52/35/3 5)20)7)9)10)5)3) gfedcba
5. Simplify the following radicals (see 38.3):
6 310 55 103 94 23 94 27)2)2)2)25)2)3) gfedcba
6. Write the following products of radicals as unique radicals (see 38.4):
543644354
453433
3333)222)101010)22)
22)55)1010)33)
´´´´´´´´
´´´´
hgfe
dcba
7. Write the following ratios of radicals as unique radicals (see 38.5):
43
6
6
4
4
53
4
3
3 3
3)
10
10)
5
5)
2
2)
5
5)
10
10)
2
2) gfedcba
8. Write the following expressions with radicals as unique radicals (see 38.6):
65 33 34 333 64)4)10)5)10)2)2) gfedcba
9. Write the following expressions with radicals as unique radicals (one case is solved for you as an example):
4
33
33
3 24
53
34
5
4
34
3
5
5)55)
5
5)10)2)
3
3)
2
2
2
2)
33
33)
10
1010)2
2
2)
jihgf
edcba
´
´´
´´´
Example:
1212
11
12
13
12
3
12
4
12
6
4
1
3
1
2
1
4
1
3
1
2
143
101010
101010101010101010)
´==
====´´=´´
+
++++f
Example:
12 512
5
12
3
12
2
4
1
6
1
4
1
6
1
4
1
3
1
2
1
4
1
3
1
2
1
4
3222222222
2
22
2
2) ====´=´=´=´
++-a
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39. Order of Operations (V)
43
Iulia & Teodoru Gugoiu
1. Find the value of each expression:
© La Citadelle
39.1 (Radical Symbol) The radical symbol is also considered a grouping symbol:
)( )(nn aa Û
So, to evaluate an expression containing a radical symbol, follow these steps:a) evaluate the order of the radical (n). This must be a positive integer.b) evaluate the radicand (a).c) evaluate the value of the radical.Example:
416652652 2432 ==+´=+´-´
39.2 (Nested Radicals) If the radicand of a radical contains other radical symbols, start the evaluation of the expression with the innermost radical. Example:
39)4(5645 3 ==--=--
39.3 (Order of Operations) If an expression contains all kind of operations including radicals, then consider radicals as independent units. Evaluate the radicals and substitute their values in the original expression. All previously explained rules for the order of operations still apply. Examples:
5)3(2274.1 3 =--=--
72314))2(5(2
2
)2()2()3225(4
8.2 5
3
-=´--=´-+--
=
=-´-´-+--
1394))2(1()24(
)81()216(.3
22
232
=+=--+-=
=--+-
1323662
332322
22)32()32(.4
-=-=--+=
=´-´-´+
+´=+´-
1616)1()59(43
45
)125)3((6427
1625.5
2
3 634
33
-=´-=-´-
-=
=-+-´-+
-
8
27)27(
8
1)27(
2
1
)3(2)27()4(.6
3
339383
-=-´=-´=
=-´=-´ --
42
31)32( 33)4()8(
)2(2
5
452 2)3()1( 3
)1(
)2(1)6])2()3[(2))2()8()
)2()7)3()54)2()
25
22
-
----+-´--¸-
+-+´-
--¸-
-´-´-
fed
cba
2. Find the value of each expression (see 39.2):
33 278 3933
3 34 223 44
6464))2527()64)2()
)84(2)7)41()41()1231)
-----´-
--´++´----
fed
cba
Example:
39811
81
)1(
)2(1) 24
2
3
==+=-
-=-
--f
Example:
022426446464) 32278 3 33
=-=-=-=-f
3. Use the order of operations to find the value of each expression:
úú
û
ù
êê
ë
é
--+-
-----´
-¸
-´-¸-´-
---++-
+--÷øö
çèæ
--÷÷
ø
ö
çç
è
æ-´¸´-
-+-+---+
--
´÷÷
ø
ö
çç
è
æ+´+´-
--+-´-¸+
5
3
43
323255
27433
5
3333
22
2227
3
3 2
2
3
3
234325
53
3
3
2
33 2333 233
3
)64()32(
)4()3(64)81
3216
8898)
)321()321()16
256
1
1)
)21()21()100)
)3(38
16))54516()
)27()8()32()3264
2725)
5
4
125
64))4455()45()
8226)25162184)
nm
lk
ji
hg
fe
dc
ba
4. Use a calculator to evaluate the following expressions. Round the answer to the nearest thousandth.
Example:
242212211122122111
)21()21()21()21()21()21() 22
´-=´-´-´-´-´+´-´-´=
=+´+--´-=+--j
53 32
1
3
)21(5233
2
31))25())32(10)211)
-+
÷÷
ø
ö
çç
è
æ+---+- dcba
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40. Substitution
44
Iulia & Teodoru Gugoiu
1. For each case find whether or not the expression is well-formed:
© La Citadelle
40.1 (Algebraic Expressions) An algebraic expression contains a well-formed sequence of operations between numbers or/and variables. Examples:
formedwellisyx
formedwellnotisyx
-´+-
-´+-
2
/2 -
+
+--+-´--
-++-´-
+-- xj
zy
xixhxgxf
yxexdyxcxbxa
y 2)
2))5)2)3()
/)))5)3)
4(2
2
40.2 (Substitution) Substitution is the process of replacing the variables of an expressions with numbers in order to evaluate the expression. Example: The value of the following expression:
2-- yxif:
5
3
+=
-=
y
x
is: 10282)5()3( -=--=-+--
2. If x = -2, find the value of each expression:
)2()1())1
2)
2
4))1()
2)32))4())2()3)
3 5
3
+´-+
´
+-+
´-+´¸-´-´+
xxjxix
xh
xgxf
xxexdxxcbxxa
x
x
3. If x = -2 and y = +3, find the value of each expression:
)2()1())()3
2))
1
1)
))2(2)))2)
2222
233
-´++¸´´-
´÷ø
öçè
æ
+
-
+´-¸-¸´
yxjyxyxiy
xhyg
x
yf
yxeyxdxcxbyxa
x
yy
40.3 (Named Expressions or Formulas) You can give a name to an expression to make it available for further reference. Example:
ïî
ïíì
´=
=
2/
/
2tas
mFa
If F = 10; m = 5; t = 2 then:
ïî
ïíì
=´=´=
===
42/222/
25/10/
22tas
mFa
4. For each case, use substitution to find the value of the named expression:
3
2
3
:
3
4
:)
uv
zu
yxz
Find
y
x
Givena
=
-=
+-=
-=
-=
2
3
:
3
4
2
:)
edf
cbe
cbd
Find
c
b
a
Givenb
a
aa
´=
-´=
+=
-=
-=
=
222
)(2
:
3
5
2
:)
HWLD
HWLV
HWHLWLA
Find
H
W
L
Givenc
++=
´´=
´+´+´´=
=
=
=
40.4 (Function Notation) You can use an expression to define a function. A function has a name followed (in parentheses) by a list of parameters separated by commas. Example:
22
2
),(
3)(
yxyxg
xxxf
+=
´-=
You can use substitution to find the value of a function for the given values of parameters. Example:
5916)3()4()3,4(
431)1(3)1()1(
22
2
=+=++-=+-
=+=-´--=-
g
f
5. The functions f, g and h are defined below. Find the values of these functions for each case:
)2)1,3(())1),1(),0,0(()))1(),0(()))1(()
)0,1())2,4,4())2,1())1()
/
2),,(;),(;)(
22223
+
------
+
´+=--+=-+=
gfhfghgffgfffe
gdhcgbfa
zxy
yxzyxhyxyxyxgxxxxf
40.5 (Order of Operations) An expression can also contain functions. To find the value of the expression, replace the functions with their values. Example:
3201
222)2(;011)1(
?)]2([)1(1;)(
2
22
22
-=-+=
=-==-=
=-+=-=
E
ff
ffExxxf
6. The functions f(x) and g(x,y) are defined below. Find the value of each expression. One case is solved for you as an example:
)0,0()0(1))]1,2([)))1(1),0(1()))0(1()
)1,0()4(2))0,1(
)2())0,0()1,1())0()1()
)1(
1),(;)(
)2(
2
2
gfhggffgfffe
gfdg
fcggbffa
y
xyxgxxxf
f ´++-+
´---
--
+
-=-=
164)4(4
1
24222)2(;4
1
)11(
12)1,2(;)]1,2([)
2212
2
2
)2(
===÷ø
öçè
æ
-=-=-==+
-=
---
fggg f
Example:
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The Book of Integers
Answers
45
Iulia & Teodoru Gugoiu
1. Understanding integers (Page 5)
nokyesjnoiyeshnog
yesfyesenodyescnobyesa
)))))
)))))).1
101,)22,)3,)4,)10,)
11,)2,)0,)2,)3,).2
-+--+
-+-+
jihgf
edundefinedcba
3)0)1)2)5).3 edcba
55)41)300)20)1).4 +++++ edcba
negativejpositivei
negativehnegativegpositivefneithere
negativedpositivecnegativebpositivea
))
))))
)))).5
,...}4,3,2,1{.6 ----=-Z
}9,7,5,3,1{.7
falsef
falseetruedfalsecfalsebfalsea
)
))))).8
integerannotedorcba )0)55)11)7).9 -+-+
definednotjihgf
definednotedcba
)0))))
))))).10
¥¥¥
¥¥¥¥
2. Absolute value, sign, and opposite (Page 6)
7)7)12)
100)10)0)3)2)2)3).1
jih
gfedcba
123)5)10)3)2)1)0).2 ±±±±±± gfedcba
++-
+-+--+
)))
))))))).3
jih
gfedefinednotdcba
77)7)2)
11)0)20)3)3)20)12).4
+-+
+--++-
jih
gfedcba
5)2)2)
3)3)20)4)3)5)2).5
+--
------+
jih
gfedcba
4)15)4)
3)4)7)3)4)3)5).6
jih
gfedcba --
3)08)5)2)3)4)2)6)
5)8)3)10)6)2)6).7
--
-
nmlkjih
gfedcba
3. Number line (Page 7)
8)4)5)1)2).1 -+-+- EDCBA
+2+2+1 +3 +4 +5 +6 +7-5 -4 -3 -2 -1 0
A BC DE
.2
0)10)20)30)40).3 EDCBA ----
7)3)2)5)8).4 ++--- EDCBA
)6(')3(')1(')4(')6('.5 --+++ EDCBA
33)48)
82)2)7)5)5).6
+---
+++-++
orgorf
oredcba
4. Comparing integers (Page 8)
00)
01)33)50)35)21).1
=
>=<><
f
edcba
320123)321123)
32012)1023)
201)21).2
<<<-<-<-+<+<+<-<-<-
+<<<-<-<<-<-
<<-+<+
fe
dc
ba
52113)
542245)20234)
3210)201)21).3
->->->>+
->->->+>+>->>+>+>
->->->->>->-
f
ed
cba
13)
30)45)13)22)20).4
-<-
->+<-><-<
f
edcba
57)
55)42)24)73)
00)50)11)01)21).5
+<-
+=+<+-<-->-
=->-><-<
j
ihgf
edcba
falsejtrueitruehfalsegfalsef
falseefalsedfalsecfalsebtruea
)))))
))))).6
falsejtrueifalsehtruegtruef
trueetruedtruectruebtruea
)))))
))))).7
truejtrueitruehfalsegfalsef
falseefalsedfalsectruebtruea
)))))
))))).8
}2,1,0,1{)
}1,2{)}4,3,2,1,0{)}3,2,1,0{)}2{)
}5,4,3,2{)}1,0{)}0,1,2{)}0,1{)}2{).9
-
---
---
j
ihgf
edcba
7. Addition of integers using the number line (Page 11)
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The Book of Integers
Answers
46
Iulia & Teodoru Gugoiu
5. Applications of integers (Page 9)
2)1)0)4)5).1 -+-+ edcba
6. Addition of integers. Axioms (Page 10)
sumaddendsfsumaddendse
sumaddendsdsumaddendsc
sumaddendsbsumaddendsa
6;4,2)10;6,4)
4;3,7)4;2,2)
1;1,2)3;2,1).1
+
-+++
+-+
)4()5()
)3(4))1(0)50)23)01).2
-+-
-+-++++
f
edcba
5)4)1)5)1)0).3 -+- fedcba 8. Addition of integers using rules (Page 12)
5)3)0)2)2).2 -+-+ edcba
273)100)0)15)10).3 -+-+ edcba
10)3)0)5)10).4 -+-+ edcba
7)0)20)2)1000).5 +--+ edcba
0)3)7)5)2).6 edcba -+-+
125)31)0)150)500).7 +--+ edcba
0)45)40)550)50).8 edcba --++
casesallfor0.4
5)3)
3)37)33)7)25)5)15).5
ih
gfedcba -
0)30)27)8)20)
9)5)0)3)0)5)2).6
lkjih
gfedcba -
2)2)4)2)3)
3)5)1)1)5)3)2).1
--+--
--+++-
lkjih
gfedcba
)2(0))3()4()
)4()4())2()1())5()2().2
++++-
-++-+--++
ed
cba
2)1)
4)4)2)2)6)5)0).3
--
-----
ih
gfedcba
)5()2()3()5()1(.4 ++-+++-++
5421)
135)322)52)23)21).5
-++-
--+----+--
f
edcba
)2()1()2()3()4()5()
)3()3()2()2())4()3()2()1()
)5()3())3()2())2()1().6
-+-+-+++-++
++++-+-++-+++-
-+-++--++
f
ed
cba
3)6)5)4)
2)2)3)1)6)2)6)3).7
++-
----
lkji
hgfedcba
3)3)1)3).8 --- dcba
356)3210)65)
6)66666)4075)1575)1135)400)
235)444)125)54)22)10)4).1
pon
mlkjih
gfedcba
1905)90)9)1334)5555)4500)
800)250)225)6)5)3).2
------
------
lkjihg
fedcba
100)2222)
50)1250)425)200)100)50)
40)35)5)5)2)3)2)2).3
po
nmlkji
hgfedcba
225)2200)5100)1850)
100)450)25)150)25)5)
15)7)5)3)4)1).4
----
------
------
ponm
lkjihg
fedcba
3)20)200)300)10)222)
1000)111)80)6)2)10).5
-----
----
lkjihg
fedcba
15)90)1)6)0)10).6 -- fedcba
Page 15 - continue
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The Book of Integers
Answers
47
Iulia & Teodoru Gugoiu
9. Subtraction of integers (Page 13)
10. Order of operations (I) (Page 14)
12. Equivalent equalities and inequalities (Page 16)
falsef
falseetruedfalsecfalsebfalsea
)
))))).1
falseltruektruejtrueifalsehtrueg
trueftrueetruedfalsecfalsebtruea
))))))
)))))).2
trueftrueetrued
truectruebfalsea
;43);00);37)
;42);42);04).1
£³-<-
->-¹=
2)7)3)8)2)
3)1)6)0)1)4)3).1
---
--
lkjih
gfedcba
2)4()2()3)2()1()
5)3()2()4)4(0)2)4()2().2
-=+-++=--+
+=--+-=+-+=---
ed
cba
12)5)1)8)4)
5)2)1)7)2)4)4).3
--
---
lkjih
gfedcba
20)
5)6)3)10)5)4)4).4
-
-------
h
gfedcba
5)5)6)4)2)3)3)5).5 ---- hgfedcba
0)2)
3)8)10)10)2)5)2).6
ih
gfedcba --
2)2)7)3).7 dcba --
5)4)
0)2)4)2)3)2)0).1
--
--
ih
gfedcba
50)6)
3)3)1)4)5)5)0).2
-
-----
ih
gfedcba
1)5)
8)6)4)8)7)4)3).3
--
-
ih
gfedcba
5)4)8)4)5)3)5)2).4 ---- hgfedcba
9)
4)1)1)9)6)2)5).5
-
-----
h
gfedcba
0)0)6)1)1)4).6 fedcba --
falseitruehfalseg
falseftrueetruedfalsecfalsebtruea
)))
)))))).3
trueftrueetruedfalsecfalsebtruea )))))).4
truehfalseg
trueftrueefalsedtruectruebtruea
))
)))))).5
trueftrueetruedtruectruebfalsea )))))).6
falsehfalseg
trueffalseetruedfalsectruebtruea
))
)))))).7
1510101550
;155101510;101551510)
23312;312230;023312)
3215;35120;01235)
614;6410;0164)
342;3240;0432)
3210;0132;312)
..2
+--+-£
+-£+--£+-
³++--+-+-³³-++-
-¹+--+-¹¹+-+-
-->--+->>++-
+-<-++-<<+--
+--==+-+-=
f
e
d
c
b
a
varymayanswerThe
0151051510)
03221)0223)
021)0413)0422).3
£-+--
³++--¹--
>-<++-=+--
f
ed
cba
573230)
10515100)1130)
21210)3120)7430).4
-+--£
++--³--¹
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15. Multiplication of integers (I) (Page 19)
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Answers
48
Iulia & Teodoru Gugoiu
13. Equations (Page 17)
14. Inequations (Page 18)
yesfnoeyesdnocnobyesa )))))).1
2)1)3)0)1)2)6)2)1).2 ihgfedcba -
10)0)
5)5)7)0)10)5)1).3
ih
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25)1)2)6)1)0).4 fedcba ---
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2)1)
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0 +1 +2 +3-1-2-3-4 +4
0 +1 +2 +3-1-2-3-4 +4
0 +1 +2 +3-1-2-3-4 +4
0 +1 +2 +3-1-2-3-4 +4
0 +1 +2 +3-1-2-3-4 +4
0 +1 +2 +3-1-2-3-4 +4
0 +1 +2 +3-1-2-3-4 +4
0 +1 +2 +3-1-2-3-4 +416. Multiplication of integers (II) (Page 20)
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55)3251(11)4)431(2)
4)231(2)14)251(7)
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19. Division of Integers (I) (Page 23)
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The Book of Integers
Answers
49
Iulia & Teodoru Gugoiu
18. Division of integers (I) (Page 22)
20. Order of operations (III) (Page 24)
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17. Order of operations (II) (Page 21)
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23. Inequalities and inequations (Page 27)
Page 27 - continue
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The Book of Integers
Answers
50
Iulia & Teodoru Gugoiu
21. Equalities and equations (Page 25)
22. Proportions and equations (Page 26)
24. Powers (Page 28)
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27. Divisors (Page 31)
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The Book of Integers
Answers
51
Iulia & Teodoru Gugoiu
25. Exponents rules (Page 29)
26. Order of operations (Page 30)
28. Divisibility rules (Page 32)
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30. Set of divisors (Page 34)
© La Citadelle www.la-citadelle.com
The Book of Integers
Answers
52
Iulia & Teodoru Gugoiu
29. Prime factorization (Page 33)
31. Highest Common Factor (HCF) (Page 35)
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f
e
d
c
ba
22)15)40)16)12)7)8)6).6 hgfedcba
}6,4,3,2,1{)
}8,4,2,1{)}25,10,4{)}1,0,2,1{)}4,3,2{).7
e
dcba --
35)8)5)8)2)4).2 fedcba
75)4)1)6)6).3 edcba
25)1)6)50).4 dcba
3
4)
5
2)2)1)
7
15)4)
15
8)1).7 hgfedcba ---
34. Cubic roots (Page 38)
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The Book of Integers
Answers
53
Iulia & Teodoru Gugoiu
32. Least Common Multiple (LCM) (Page 36)
35. Roots of superior order (Page 39)
60)175)60)60)120)30).1 fedcba
800)90)
80)80)108)300)250)18).5
--
-
hg
fedcba
96)150)120)80)90)20).2 fedcba
1122224)505250)24248)
606360)305150)602120).3
´=´=´=
´=´=´=
fed
cba
100)180)48)300)24).4 edcba
5)8
1)20)1)
15
4)20)48)2).5 hgfedcba -
40
11)
48
1)
30
1)
12
19)
2
1)
6
5).6 fedcba
48
23)
30
1)
12
13)
2
3)
6
1).7 ----- edcba
33. Square roots (Page 37)
40000)10000)2500)
400)225)144)121)100)81)64)
49)36)25)16)9)4)1)0).1
rqp
onmlkji
hgfedcba
4)
12)4)7)200)50)100)10)20)
11)8)3)6)9)5)0)2)1).2
r
qponmlkj
ihgfedcba
99999)123)100)20)
13)10)9)5)7)0)3)1).3
lkji
hgfedcba
125)100)
30)10)11)3)8)7)1)5).4
ji
hgfedcba
12500)
180)75)175)270)54)45)6).5
h
gfedcba
450)84)125)50)54)33).6 fedcba
23.316)
13.35)07.7)16.3)65.2)73.1)41.1).7
g
fedcba
108.111)875.27)
142.14)477.5)162.3)732.1)414.1).8
gf
edcba
1728)1331)
8000000)1000000)125000)8000)1000)
216)125)64)27)8)1)0).1
nm
lkjih
gfedcba
11)50)20)3)
10)5)4)2)6)1)12)0).2
lkji
hgfedcba
1728)27000000)
1000000)64000)8000)1000)
216)125)64)27)8)1).3
--
----
------
lk
jihg
fedcba
12)5)20)4)
10)3)6)2)6)1).4
----
-----
jihg
fedcba
12345)12)10)5)3)1).5 fedcba
123)10)4)2)1).6 ----- edcba
40)90)
100)100)360)300)50)6).7
--
--
hg
fedcba
120)32)24)45)18).8 edcba
416.46)
726.10)642.4)309.4)714.2)913.1).9
f
edcba --
100000)10000)
81)32)16)1)1)100000000)
1000000)625)1024)81)16)0).1
-
--
nm
lkjihg
fedcba
2)3))10)2)
1)3)10)5)2)1)0).2
----
--
lkdefinednotjih
gfedcba
54321)12)10)2)4)1).3 fedcba
123)10)4)1)1).4 edcba ---
5)4)6)9)12).6 edcba
84.3)16.3)51.2)24.2)51.2)78.1).7 -- fedcba
38. The link between radicals and powers(Page 42)
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The Book of Integers
Answers
54
Iulia & Teodoru Gugoiu
36. Roots rules (Page 40)
39. Order of operations (V) (Page 43)
4
3)3))32(4)
2)24)1000)2)1000)
3)2)5
4)1)2)15).3
nml
kjihg
fedcba
+´
´---
-
2)10)5)2)2)2)4)2).1 hgfedcba ---
5
1)
22
1)
2
1)2)2)
2
1)
5
1)10)5)55)2)5).3
--
-
lkjihg
fedcba
3
5)
3
10)
4
5)
3
2)
3
2)
5
3).4 fedcba
-
-
37. Equations with powers and radicals (Page 41)
noiyeshyesg
yesfyesenodnocyesbyesa
)))
)))))).1
2)5)2)3)2)
10)8)5))5).2
±±±±±
±±±±
jihgf
edcsolutionsnoba
10)6)
2)4)2)10)2)5)2)4).3
ji
hgfedcba
-
----
10000)125)1)
8)27)1)16)9)8)4).4
jih
gfedcba
--
-
6)5)4)3)2).6 edcba
10)4)27)64)2)3)
8512)3)3)5)2)5).7
lkjihg
andfedcba
-±
---
4
33
625)
100)27)10)8)3)2).2
g
fedcba
33)2)4)8)5)8)9).5 gfedcba
3)4)2)4)2)3).1 fedcba -
0)8)2)2)2)1).2 fedcba --
148.3)070.2)310.2)503.0).4 dcba --
33
534
335
45)42)28)
1010)1010)210)10010)
42)23)42)23)22).2
lkj
ihgf
edcba
-
-
-
3433 25)100)55)39)25)32).5 fedcba
5)10)10)20)5)2).6 -- fedcba
20
3)
20
1)
20
1)
10
1)
6
1)
2
1)
2
3)
10
3)
5
6)
5
2)
2
3)
2
1)2)2)
10
1)3).7
ponmlkji
hgfedcba
-
-
3)3)10)3)
6))3)2).5
hgfe
dnumbernaturaloddanycba
definednotg
fedcba
)
1024)64)2)5)27)2).1 10
1
4
1
5
1
4
1
3
1
2
1
3
227
3
6
5
5
2
3
2
2
3
2)3)7)10)5)7)2).3 gfedcba
35 4
33 245 3
255)20)
77)9)1010)55)3).4
´
´´´
gf
edcba
60 1712 1112
20 94 315 812 76 5
33)2)1010)
2)2)5)10)3).6
´´ hgf
edcba
4
6
12
206
126
3)
10
1)5)
2
1)
5
1)10)2).7
g
fedcba
12
15941266
64)
4)10)5)10)2)2).8
g
fedcba
243 2
3
12
8
660 13
60 712 712 5
5)5)5
1)10)
2)3
1)
2
1)3)10)2).9
jihg
fedcba
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The Book of Integers
Answers
55
Iulia & Teodoru Gugoiu
40. Substitution (Page 44)
nojyesinohyesgnof
noeyesdyescyesbnoa
)))))
))))).1
0)2
1)2)1)1)
4)2)4)4
1)8).2
jihgf
edcba
-
---
38,30,62)
225,3,25)2,8,11).4
===
======
DVAc
fedbvuza
54)2)0)1)2)6)8)3).5 -- hgfedcba
1)36)3
2)
9
1)4)
1)2
1)2)8)3).3
-
---
jihgf
edcba
1)16)0)0)3)1)1)0).6 hgfedcba -