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Types of Proofs
Types of ProofsMathematical Definitions and
Theorems
v Definitions
• A math emat ic al d ef in it io n
describes a concept or an
object in math.
• Definitions must clearly state
what constitutes that object
and what does not.
• After defining a concept or anobject, mathematical
statements can be made about
them usually stating whether
or not they possess a certain
property.
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Types of Proofs
v Theorems
• Mathematical statements may
be true or not.
A theorem is a mathematicalstatement that has been
proven to be true.
Lemmas are similar to
theorems except that theyhave minor significance and
are often used in proving
theorems. In other words,
they are stepping stones in the
process of proving a theorem.
Corol lar ies are statements
that can be proven deductively
by applying other theorems.
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Types of Proofs
Types of Proof
v Direct Proof
• One of the popular methods of
proving the correctness of
statements is by using
established facts without
making any further
assumptions.
This is usually done by using
proven theorems or lemmas in
establishing the correctness of
the statement.
This type of proof is called the
d irec t p roo f or direct
argument .
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Types of Proofs
Examples:
Prove that the statement
“division is transitive”,
meaning, if x is divisible by y ,
and y is divisible by z , then x is
divisible by z .
Proof:
x is divisible by y means
x = (a)y
y is divisible by z means
y = (b)z
where a and b are
integers.
x = (a)y = (a)(b)z
x = (ab)z
Since ab is also an integer,
this means that x is
divisible by z .
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Types of Proofs
Prove that If x and y are odd,
then xy is also odd, where x and y are integers.
Proof:
x is an odd integer means:
x = (2a + 1)
y is an odd integer means:
y = (2b + 1)
where a and b are
integers.
xy = (2a + 1 )(2b + 1)
= 4ab + 2a +
2b + 1
xy = 2 (2ab + a + b)+ 1
Since (2ab + a + b) is an
integer, this means that xy
is also an odd integer.
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Types of Proofs
Prove that the sum of two
rational numbers is a rationalnumber.
Proof:
x is a rational number
means
x = a/b
y is a rational number
means
y = c/d
where a, b, c , and d are
integers.
x +y = a/b + c /d = (ad + cb)/(bd )
Since (ad + cb) and (bd )
are both integers, this
means that x + y is a
rational number.
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Types of Proofs
v Proof by Contradiction
• Proof b y c on trad ic t io n starts
by assuming the statement tobe proven is false and then
deriving consequences or
outcomes.
In the process, if acontradiction to what is known
to be true is reached, then it
implies that the initial
assumption that the statement
is false is incorrect.
Therefore, the statement must
be true.
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Types of Proofs
Examples:
Prove that if 3 x + 2 is odd,
then x is odd, where x is an
integer.
Proof:
Assume that 3 x + 2 is odd
but x is even.
x is even means
x = 2awhere a is an integer.
3 x + 2 = 3(2a) + 2
= 6a + 2
= 2(3a + 1)Since (3a +1) is an integer,
that means 3 x + 2 is even
(a contradiction).
Therefore the statement istrue.
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Types of Proofs
Prove that if x 2 is even, then x
is even.
Proof:
Assume that x 2 is even, but
x is odd.
x is odd means
x = 2a + 1
where a is an integer.
x 2 = (2a + 1)2
= 4a2 + 4a + 1
= 2(2a2 +2a) + 1
Since (2a2 + 2a) is an
integer, that means x 2 is
odd (a contradiction).
Therefore the statement is
true.
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Types of Proofs
Prove that the square root of
two is irrational.
Proof:
Assume that the square
root of two is rational.
If the square root of 2 is
rational, then it means that
where a and b are
integers.
Take note that a and bmust not have common
factors so that the fraction
will be in its simplest form.
This implies that either a or
b must be odd (or both).
b
a=2
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Types of Proofs
2 = a2/b2
2b2 = a2
This implies a2
is even. And if a2 is even, then a is
also even.
Since a is even, this
means that
a = 2 x
where x is an integer.
2b2 = a2
= (2 x )2
= 4 x 2
b2 = 2 x 2
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Types of Proofs
This again implies that b2
is even which makes beven also.
Since both a and b are
even, this produces a
contradiction to thestatement earlier that at
least one of them must be
odd.
This proves that the square
root of two is irrational.
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Types of Proofs
v Proof by Induction
• Proo f by induc t ion is useful
in proving that all elements of
an infinite set, such as the set
of natural numbers, possess a
certain property.
• Proof of induction starts off by
first proving that the property
in question is true for one of
the elements of the set,
usually the first one.
• The second step is to prove
that if the property is true for
one element of the set, it is
also true for the next element.
• Once these two proofs have
been established, then it can
be concluded that the property
is true for all members of the
set.
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Types of Proofs
• The first step is called the
bas is s tep while the second
step is called the i nduc t ion
step .
• Formal Definition:
Let P be the property to be
proven true for an infinite set.
Basis Step:
Prove that P (1) is true.
Induction Step:
Prove that if P (k ) is true,
then P (k + 1) is also true
for every positive integer k .
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Types of Proofs
Examples:
Prove that 1 + 2 + … + n =
½ n(n+ 1) for all n ≥ 1 (the sum
of the first n integers).
Basis Step: Prove P (1) is true.
Let n = 1:
1 = ½ (1) (1 + 1)
1 = 1
Induction Step: Prove that if
P (k ) is true, then P (k + 1) is
also true for all k .
Assume first that P (k ) is true.
That is,
( )12
1...21 +=+++ k k k
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Types of Proofs
Now prove that P (k + 1) istrue. That is,
( ) ( )( )212
11...21 ++=++++ k k k
( ) ( )( )212
11...21 ++=+++++ k k k k
( ) ( ) ( )( )212
111
2
1++=+++ k k k k k
( ) ( )[ ] ( )( )212
1121
2
1++=+++ k k k k k
[ ] ( )( )212
122
2
1 2++=+++ k k k k k
[ ] ( )( )212
123
2
1 2++=++ k k k k
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Types of Proofs
This proves that if P (k ) is true,
the P (k + 1) is also true for all
k .
Therefore, the property 1 + 2 +
… + n = ½ n(n+ 1) for all n ≥ 1
is true.
[ ] ( )( )212
123
2
1 2++=++ k k k k
( ) ( )( )212
1)2(1
2
1++=++ k k k k
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Types of Proofs
Prove that the sum of the first
n odd positive integers is n2 .
Basis Step: Prove P (1) is true.
Let n = 1:
1 = (1)2
1 = 1
Induction Step: Prove that if
P (k ) is true, then P (k + 1) is
also true for all k .
Assume first that P (k ) is true.
That is,
2)12(...531 k k =−++++
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Types of Proofs
Now prove that P (k + 1) istrue. That is,
This proves that if P (k ) is true,
then P (k + 1) is also true for allk .
Therefore, the statement that
the sum of the first n odd
positive integers is n2 is true.
( ) ( )2112...531 +=+++++ k k
( ) ( ) ( )211212...531 +=++−++++ k k k
( ) ( )22 112 +=++ k k k
( ) ( )22 11 +=+ k k
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Types of Proofs
Prove that n3 – n is divisible by
3 whenever n is a positive
integer.
Basis Step: Prove P (1) is true.
Let n = 1:
13 - 1 = 0
0 is divisible by 3.
Induction Step: Prove that if
P (k ) is true, then P (k + 1) is
also true for all k .
Assume first that P (k ) is true.
That is:
is divisible by 3.
k k −3
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Types of Proofs
Now prove that P (k + 1) is
true. That is:
is divisible by 3.
( ) ( )11 3 +−+ k k
( ) ( )1133 23 +−+++ k k k k
1133 23 −−+++ k k k k
k k k k −++ 33 23
k k k k 33 23 ++−
k k k k ++− 23 3
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Types of Proofs
Take note that it was
established earlier that k 3 – k
is divisible by 3. The term 3(k 2+ k ) is also divisible by 3. This
makes the entire term divisible
by 3.
This proves that if P (k ) is true,then P (k + 1) is also true for all
k .
Therefore, the statement that
n3 – n is divisible by 3
whenever n is a positive
integer is true.
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Types of Proofs
Prove that 7n – 1 is divisible by
6 for all n ≥ 1.
Basis Step: Prove P (1) is true.
Let n = 1:
71 - 1 = 6
6 is divisible by 6.
Induction Step: Prove that if
P (k ) is true, then P (k + 1) is
also true for all k .
Assume first that P (k ) is true.
That is:
is divisible by 6.
17 −k
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Types of Proofs
Now prove that P (k + 1) is
true. That is:
is divisible by 6.
17 )1( −+k
17 )1( −+k
177 −⋅ k
6777 +−⋅ k
6177 +−k
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Types of Proofs
Take note that it was
established earlier that 7 k – 1is divisible by 6. Therefore,
the term 7(7k – 1) is also
divisible by 6.
This proves that if P (k ) is true,then P (k + 1) is also true for all
k .
Therefore, the statement that
7n – 1 is divisible by 6 for all n
≥ 1 is true.
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Types of Proofs
Prove that 20 + 21 + 22 + … +
2n = 2(n+1) – 1 for all n ≥ 0.
Basis Step: Prove P (0) is true.
Let n = 0:
20 = 2(0+1) – 1
1 = 2 – 1
1 = 1
Induction Step: Prove that if
P (k ) is true, then P (k + 1) is
also true for all k .
Assume first that P (k ) is true.
That is,
( )122...222
1210−=++++
+k k
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Now prove that P (k + 1) is
true. That is,
( ) ( )122...222
111210−=++++
+++ k k
( ) ( )1222...222
21210−=+++++
++ k k k
( ) ( ) ( ) 12212 211 −=+− +++ k k k
( ) ( ) ( ) 12122 211 −=−+ +++ k k k
( ) ( ) 12122 21 −=−⋅ ++ k k
( ) ( ) 1212 22 −=− ++ k k
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