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**Quotient-Remainder Theory, Div and Mod**

If ๐ and ๐ are integers and ๐>0, then ๐ ๐๐๐ฃ ๐=๐ and ๐ ๐๐๐ ๐=๐ โบ ๐=๐๐+๐ where ๐ and ๐ are integers and 0โค๐<๐. Quotient: q=๐ ๐๐๐ฃ ๐= ๐ ๐ Reminder: r=๐ ๐๐๐ ๐=๐โ๐๐

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Exercise Prove that for all integers a and b, if a mod 7 = 5 and b mod 7 = 6 then ab mod 7 = 2. What values are ๐, ๐, and ๐?

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Exercise Prove that for all integers a and b, if a mod 7 = 5 and b mod 7 = 6 then ab mod 7 = 2. Hint: ๐ = 7 ๐ = 7๐+5, b = 7๐+6 ๐๐ = 7๐+5 7๐+6 =49๐๐+42๐+35๐+30=7 7๐๐+6๐+3๐+4 +2

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**Floor & Ceiling Floor: If ๐ฅ is a real number and ๐ is an integer, then**

Definition: Floor: If ๐ฅ is a real number and ๐ is an integer, then ๐ฅ =๐ โบ ๐โค๐ฅ<๐+1 Ceiling: if ๐ฅ is a real number and ๐ is an integer, then ๐ฅ =๐ โบ ๐โ1<๐ฅโค๐ ๐ฅ ๐ ๐+1 floor of ๐ฅ= ๐ฅ ๐ฅ ๐โ1 ๐ ceiling of ๐ฅ= ๐ฅ

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**Relations between Proof by Contradiction and Proof by Contraposition**

To prove a statement โ๐ฅ in ๐ท, if ๐ ๐ฅ then ๐(๐ฅ) Proof by contraposition proves the statement by giving a direct proof of the equivalent statement โ๐ฅ in ๐ท, if ๐ ๐ฅ is false then ๐ ๐ฅ is false Proof by contradiction proves the statement by showing that the negation of the statement leads logically to a contradiction. Suppose ๐ฅ is an arbitrary element of ๐ท such that ~๐(๐ฅ) ~๐(๐ฅ) Sequence of steps Suppose โ๐ฅ in ๐ท such that ๐(๐ฅ) and ~๐(๐ฅ) Same sequence of steps Contradiction: ๐(๐ฅ) and ~๐(๐ฅ)

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**Summary of Chapter 4 Number theories: Proofs:**

Even, odd, prime, and composite Rational, divisibility, and quotient-remainder theorem Floor and ceiling The irrationality of 2 and gcd Proofs: Direct proof and counterexample Indirect proof by contradiction and contraposition

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**๐ is rational โ โ integers ๐ and ๐ such that ๐ = ๐ ๐ and ๐ โ 0.**

The Irrationality of 2 How to proof: Direct proof? Proof by contradiction? Proof by contraposition? If ๐ is a real number, then ๐ is rational โ โ integers ๐ and ๐ such that ๐ = ๐ ๐ and ๐ โ 0. A real number that is not rational is irrational.

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**The Irrationality of 2 Proof by contradiction: Starting point:**

Negation: 2 is rational. To show: A contradiction. 2 = ๐ ๐ , where ๐ and ๐ are integers with no common factors and ๐โ 0, by definition of rational. ๐ 2 = 2 ๐ 2 , by squaring and multiplying both sides with ๐ 2 ๐ 2 is even, then ๐ is even. Let ๐ = 2๐ for some integer ๐. (2๐) 2 =4 ๐ 2 =2 ๐ 2 , by substituting ๐ = 2๐ into ๐ 2 = 2 ๐ 2 . ๐ 2 is even, and so ๐ is even. Hence both ๐ and ๐ have a common factor of 2.

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**Irrationality of 1+3 2 Proof by contradiction: Starting point:**

Negation: is rational. To show: A contradiction.

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**1+3 2 = ๐ ๐ for some integers ๐ and ๐ with ๐โ 0.**

Irrationality of Proof by contradiction: By definition of rational, = ๐ ๐ for some integers ๐ and ๐ with ๐โ 0. It follows that 3 2 = ๐ ๐ โ by subtracting 1 from both sides = ๐ ๐ โ ๐ ๐ by substitution = ๐โ๐ ๐ by the rule for subtracting fractions with a common denominator Hence, 2 = ๐โ๐ 3๐ by dividing both sides by 3. ๐ โ ๐ and 3๐ are integers and 3๐โ 0 by the zero product property. Hence 2 is quotient of the two integers ๐ โ ๐ and 3๐ with 3๐โ 0, so 2 is rational by the definition of rational. This contradicts the fact that 2 is irrational.

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**Property of a Prime Divisor**

Proposition 4.7.3 For any integer ๐ and any prime number ๐, if ๐|๐ then ๐ | (๐+1) If a prime number divides an integer, then it does not divide the next successive integer. Starting point: there exists an integer ๐ and a prime number ๐ such that ๐ | ๐ and ๐ | (๐ + 1). To show: a contradiction. ๐ = ๐๐ and ๐ + 1 = ๐๐ for some integers ๐ and ๐ by definition of divisibility. It follows that 1 = (๐ + 1) โ ๐ = ๐๐ โ ๐๐ = ๐(๐ โ ๐ ), ๐ โ ๐ =1/๐, by dividing both sides with ๐. ๐ > 1 because ๐ is prime, hence, 1/๐ is not an integer, thus ๐ โ๐ is not an integer, which is a contradict ๐ โ๐ is an integer since ๐ and ๐ are integers. if ๐ and ๐ are integers and ๐โ 0: ๐|๐ โบ โ an integer ๐ such that ๐=๐โ๐

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**Infinitude of the Primes**

Theorem Infinitude of the Primes The set of prime numbers is infinite. Proof by contradiction: Starting point: the set of prime number is finite. To show: a contradiction. Assume a prime number ๐ is the largest of all the prime numbers 2, 3, 5, 7, 11,. . . ,๐. Let ๐ be the product of all the prime numbers plus 1: ๐ = (2ยท3ยท5ยท7ยท11ยท ยท ยท๐) + 1 Then ๐ > 1, and so, by Theorem (any integer larger than 1 is divisible by a prime number) , ๐ is divisible by some prime number q. Because q is prime, q must equal one of the prime numbers 2, 3, 5, 7, 11, , ๐. Thus, by definition of divisibility, ๐ divides 2ยท3ยท5ยท7ยท11ยท ยท ยท๐, and so, by Proposition 4.7.3, ๐ does not divide (2ยท3ยท5ยท7ยท11ยท ยท ยท๐) + 1, which equals ๐. Hence N is divisible by q and N is not divisible by q, and we have reached a contradiction. [Therefore, the supposition is false and the theorem is true.]

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**Greatest Common Divisor (GCD)**

The greatest common divisor of two integers a and b is the largest integer that divides both ๐ and ๐. Exercise: gcd 72,63 =9, since 72=9โ8 and 63=9โ7 gcd , = 2 20 , since = 2 20 โ 5 20 and = 2 30 โ 3 30 Definition Let ๐ and ๐ be integers that are not both zero. The greatest common divisor of ๐ and ๐, denoted gcd(a, b), is that integer ๐ with the following properties: ๐ is common divisor of both a and b, in other words, ๐ | ๐, and ๐ | ๐. For all integers ๐, if ๐ is a common divisor of both ๐ and ๐, then ๐ is less than or equal to ๐. In other words, for all integers ๐, if ๐ | ๐, and ๐ |๐, then cโค๐.

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**Greatest Common Divisor (GCD)**

Lemma 4.8.1 If ๐ is a positive integer, then gcdโก(๐,0) = ๐. Proof: Suppose ๐ is a positive integer. [We must show that the greatest common divisor of both ๐ and 0 is ๐.] ๐ is a common divisor of both ๐ and 0 because r divides itself and also ๐ divides 0 (since every positive integer divides 0). No integer larger than ๐ can be a common divisor of ๐ and 0 (since no integer larger than ๐ can divide ๐). Hence ๐ is the greatest common divisor of ๐ and 0.

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**Greatest Common Divisor (GCD)**

Lemma 4.8.2 If ๐ and ๐ are any integers not both zero, and if ๐ and ๐ are any integers such that ๐=๐๐+๐, then gcdโก(๐,๐)=gcdโก(๐,๐) Proof: [The proof is divided into two sections: (1) proof that gcdโก(๐, ๐) โค gcdโก(๐, ๐ ), and (2) proof that gcdโก(๐, ๐ ) โค gcdโก(๐, ๐). Since each gcd is less than or equal to the other, the two must be equal.] gcdโก(๐, ๐) โค gcdโก(๐, ๐): [We will first show that any common divisor of ๐ and ๐ is also a common divisor of ๐ and ๐.] Let ๐ and ๐ be integers, not both zero, and let ๐ be a common divisor of ๐ and ๐. Then ๐ | ๐ and ๐ | ๐, and so, by definition of divisibility, ๐= ๐๐ and ๐ = ๐๐, for some integers ๐ and ๐. Now substitute into the equation ๐ = ๐๐ + ๐ to obtain ๐๐ = (๐๐)๐ + ๐.

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**Greatest Common Divisor (GCD)**

Lemma 4.8.2 If ๐ and ๐ are any integers not both zero, and if ๐ and ๐ are any integers such that ๐=๐๐+๐, then gcdโก(๐,๐)=gcdโก(๐,๐) Proof (contโ): gcdโก(๐, ๐) โค gcdโก(๐, ๐): [We will first show that any common divisor of ๐ and ๐ is also a common divisor of ๐ and ๐.] ๐๐ = (๐๐)๐ + ๐. Then solve for ๐ : ๐ = ๐๐ โ (๐๐)๐ = (๐ โ ๐๐)๐. But ๐ โ ๐๐ is an integer, and so, by definition of divisibility, ๐ | ๐ . Because we already know that ๐ | ๐, we can conclude that ๐ is a common divisor of ๐ and ๐ [as was to be shown].

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**Greatest Common Divisor (GCD)**

Lemma 4.8.2 If ๐ and ๐ are any integers not both zero, and if ๐ and ๐ are any integers such that ๐=๐๐+๐, then gcdโก(๐,๐)=gcdโก(๐,๐) Proof (contโ): gcdโก(๐, ๐) โค gcdโก(๐, ๐): [Next we show that gcdโก(๐, ๐) โค gcdโก(๐, ๐).] By part (a), every common divisor of ๐ and ๐ is a common divisor of ๐ and ๐ . It follows that the greatest common divisor of ๐ and ๐ is defined because ๐ and ๐ are not both zero, and it is a common divisor of ๐ and ๐ . But then gcdโก(๐, ๐) (being one of the common divisors of ๐ and ๐) is less than or equal to the greatest common divisor of ๐ and ๐ : gcdโก(๐, ๐) โค gcdโก(๐, ๐ ). gcdโก(๐, ๐) โค gcdโก(๐, ๐): The second part of the proof is very similar to the first part. It is left as an exercise.

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**The Euclidean Algorithm**

Problem: Given two integer A and B with ๐ด>๐ตโฅ0, find gcdโก(๐ด,๐ต) Idea: The Euclidean Algorithm uses the division algorithm repeatedly. If B=0, by Lemma we know gcdโก(๐ด,๐ต) = ๐ด. If B>0, division algorithm can be used to calculate a quotient ๐ and a remainder ๐: ๐ด=๐ต๐+๐ where 0โค๐<๐ต By Lemma 4.8.2, we have gcdโก(๐ด,๐ต)=gcdโก(๐ต,๐), where ๐ต and ๐ are smaller numbers than ๐ด and ๐ต. gcdโก(๐ด, ๐ต) = gcdโก(๐ต, ๐) = โฏ = gcdโก(๐ฅ, 0) = ๐ฅ ๐=๐ด ๐๐๐ ๐ต

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**The Euclidean Algorithm - Exercise**

Use the Euclidean algorithm to find gcd(330, 156).

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**The Euclidean Algorithm - Exercise**

Use the Euclidean algorithm to find gcd(330, 156). Solution: gcd(330,156) = gcd(156, 18) 330 mod 156 = 18 = gcd(18, 12) 156 mod 18 = 12 = gcd(12, 6) 18 mod 12 = 6 = gcd(6, 0) 12 mod 6 = 0 = 6

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**An alternative to Euclidean Algorithm**

If ๐โฅ ๐> 0, then gcdโก(๐,๐)=gcdโก(๐, ๐โ๐)

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**An alternative to Euclidean Algorithm**

If ๐โฅ ๐> 0, then gcdโก(๐,๐)=gcdโก(๐, ๐โ๐) Hint: Part 1: proof gcdโก(๐,๐)โคgcdโก(๐, ๐โ๐) every common divisor of a and b is a common divisor of b and a-b Part 2: proof gcdโก(๐,๐)โฅgcdโก(๐, ๐โ๐) every common divisor of b and a-b is a common divisor of a and b.

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**Homework #5 Problems Converting decimal to rational numbers.**

4.2.2: 4.2.7: โฆ

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Homework #5 Problems 1. Converting decimal to rational numbers : Solution: = โ10000= : โฆ Solution: let X = x = โฆ. 10x = โฆ x โ 10x = X = / 99990

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Exercise Prove that for any nonnegative integer ๐, if the sum of the digits of ๐ is divisible by 9, then ๐ is divisible by 9. Hint: by the definition of decimal representation ๐= ๐ ๐ 10 ๐ + ๐ ๐โ1 10 ๐โ1 +โฏ+ ๐ ๐ 0 where ๐ is nonnegative integer and all the ๐ ๐ are integers from 0 to 9 inclusive. = ๐ ๐ ( 9999โฏ999 ๐ 9 โฒ ๐ +1)+ ๐ ๐โ1 ( 9999โฏ999 ๐โ1 9 โฒ ๐ +1)+โฏ+ ๐ 1 (9+1)+ ๐ 0 =9 ๐ ๐ 11โฏ11 ๐ 1 โฒ ๐ + ๐ ๐โ1 11โฏ11 ๐โ1 1 โฒ ๐ +โฏ+ ๐ ๐ ๐ + ๐ ๐โ1 +โฏ+๐ 1 + ๐ 0 = an integer divisible by 9 + the sum of the digits of ๐

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Homework #5 Problems Theorem: The sum of any two even integers equals 4k for some integer k. โProof: Suppose m and n are any two even integers. By definition of even, m = 2k for some integer k and n = 2k for some integer k. By substitution, m + n = 2k + 2k = 4k. This is what was to be shown.โ Whatโs the mistakes in this proof?

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