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Math 135: Lecture 3: Forwards and Backwards

Denition 3.1. An integer m divides an integer n , and we write m | n , if there exists aninteger k so that n = km .

Example 3.2. Divisibility

(a) 3 | 6

(b) 5 6

(c) a | 0

(d) For all non-zero integers a , 0 a since

Proposition 3.3 (Transitivity of Divisibility TD ). Let a,b,c Z. If a | b and b | c, then a | c.

Proof of TD:

1. Since a | b, there exists an integer r so that b = ra = b.

2. Since b | c, there exists an integer s so that c = sb = c.3. Substituting ra for b in the previous equation, we get c = sb = s (ra ) = ( sr )a .

4. Since sr is an integer, a | c.

Reading a Proof

1. Explicitly identify the and the .

2. Explicitly identify the .

3. Record any , usually denitions or previous knowl-edge.

4. each step with reference to the denitions, previous knowledge or techniquesused.

5. where necessary and justify these steps with reference to thedenitions, previous knowledge or techniques used.

Analysis of Transitivity of Divisibility (TD)Let a,b,c Z. If a | b and b | c, then a | c.

Hypothesis:

Conclusion:

Core Proof Technique:

Preliminary Material:

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Math 135: Lecture 3: Forwards and Backwards

Example 3.4. Analysis of Proof of TD .Proof of TD:

1. Since a | b, there exists an integer r so that ra = b.

2. Since b | c, there exists an integer s so that sb = c.

3. Substituting ra for b in the previous equation, we get c = sb = s (ra ) = ( sr )a .

4. Since sr is an integer, a | c.

Proposition 3.5 (Divisibility of Integer Combinations DIC ). Let a , b and c be integers. If a | b and a | c, then a | (bx + cy ) for any x, y Z.

Proof of DIC:

1. Since a | b, there exists an integer k such that b = ka .

2. Since a | c, there exists an integer l such that c = la .

3. Let x and y be any integers.

4. Now bx + cy = kax + lay = a (kx + ly ).

5. Since kx + ly Z, it follows that a | (bx + cy ).

Exercise 3.6. Analysis of Divisibility of Integer Combinations ( DIC )Let a , b and c be integers. If a | b and a | c, then a | (bx + cy ) for any x, y Z.

Hypothesis:

Conclusion:

Core Proof Technique:

Preliminary Material:

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Math 135: Lecture 3: Forwards and Backwards

Exercise 3.7. As in the previous example, justify each step of the proof of DIC .

1. Since a | b, there exists an integer k such that b = ka .

2. Since a | c, there exists an integer l such that c = la .

3. Let x and y be any integers.

4. Now bx + cy = kax + lay = a (kx + ly ).

5. Since kx + ly Z, it follows that a | (bx + cy ).

Proposition 3.8 (Bounds by Divisibility BBD ). If a | b and b = 0 then |a | | b| .

Proof of BBD: Since a |b, there exists an integer q so that b = qa .Since b = 0, q = 0.But since q = 0 and q is an integer, |q | 1.Thus |b| = |qa | = |q || a | | a | , that is |a | | b| .

Proposition 3.9 (Division Algorithm DA ). If a and b are integers, and b > 0, then there existunique integers q and r such that

a = qb + r where 0 r < b.

Example 3.10.a = q b + r

20 = 7 +21 = 7 +

20 = 7 +

Division Algorithm - RemarksIf a and b are integers, and b > 0, then there exist unique integers q and r such that

a = qb + r where 0 r < b.

q is called the and r is called the .

r is always .

r is always positive or zero.

a | b if and only if the

Though commonly known as the Division Algorithm , it not really an algorithm since itdoesnt provide a nite sequence of steps that will construct q and r .

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Math 135: Lecture 3: Forwards and Backwards

Denition 3.11 (Greatest Common Divisor) . Let a and b be integers, not both zero. An integerd > 0 is the greatest common divisor of a and b, written gcd( a, b ), if and only if

1. d | a and d | b , and

2. If c | a and c | b then c d .

Example 3.12. (a) gcd(24 , 30) =

(b) gcd(17 , 25) =

(c) gcd( 12, 0) =

(d) gcd( 12, 12) =

(e) gcd(0 , 0) =

Greatest Common Divisor - RemarksLet a and b be integers, not both zero. An integer d > 0 is the greatest common divisor of a and b, written gcd( a, b ), if and only if

1. d | a and d | b (common ), and

2. if c | a and c | b then c d (greatest ).

Every integer is a divisor of 0, since 0 = 0 m for every integer m .

For a = 0, the defn implies gcd( a, 0) = |a | and gcd( a, a ) = |a | .

We dene gcd(0 , 0) as 0.

Proposition 3.13 (GCD With Remainders GCD WR ). If a and b are integers not both zero,and q and r are integers such that a = qb + r , then gcd( a, b ) = gcd( b, r ).

Hypothesis:

Conclusion:

Key QuestionA key question is used when working backwards.

Example 3.14. Possible Question: How do we show that two integers are equal?

Better Question: How do we show that a number is a gcd?Answering a Key Question

1. First, give an abstract answer that contains no symbols from the specic problem.

2. Second, apply this abstract answer to the specic problem using appropriate notation.

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Math 135: Lecture 3: Forwards and Backwards

[How do we show that a number is a gcd?]

1. Abstract Answer:

2. Specic Answer:

Example 3.15 (Discovering A Proof of GCD WR ). If a and b are integers not both zero,and q and r are integers such that a = qb + r , then gcd(a, b ) = gcd( b, r ).Fill in the lines of the proof as we discover them.

A1 :

A2 :

B4 :

B3 :

B2 :

1.

2.

B1 :

Writing A Condensed Proof of GCD WRIf a and b are integers not both zero, and q and r are integers such that a = qb + r , thengcd(a, b ) = gcd( b, r ).

Proof of GCD WR: Let d = gcd( a, b ). We will use the denition of gcd to show that d = gcd( b, r ). Sinced = gcd( a, b ), d | b. Observe that r = a qb. Since d | a and d | b, d | (a qb) by the Divisibility of IntegerCombinations. Hence d | r , and d is a common divisor of b and r . Let c be a divisor of b and r . Since c | b andc | r , c | (qb + r ), again by the Divisibility of Integer Combinations. Now a = qb + r , so c | a . Because d = gcd( a, b )and c | a and c | b, c d .Remarks

1. If a = b = 0 this proposition is also true since the only possible choices for b and r areb = r = 0.

2. In general, there are many ways to work forwards and backwards.

3. Working backwards does not necessarily produce steps in the same order as the condensedproof.

4. Condensed proofs are short and usually omit the discovery process.

5. Be sure that you can identify where each of your hypotheses was used in the proof.

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