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## Proof Methods: Part 2

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### Proof Methods: Part 2

Sections 3.1-3.6

More Number Theory Definitions

- Divisibility
- n is divisible by d iff
- kZ | n=d*k
- d|n is read “d divides n” where n and d are integers and d 0 (note: d|n d/n)
- Other ways to say it:
- n is a multiple of d
- d is a factor of n
- d is a divisor of n
- d divides n
- Properties of Divisibility
- Thm 3.3.1 Transitivity
- a,b,c Z, If a divides b and b divides c, then a divides c
- Thm 3.3.2 Divisibility by a Prime
- Any integer n>1 is divisible by a prime number

Q(x)

Transitivity ProofProve that for all integers a, b, and c, If a divides b and b divides c, then a divides c

a,b,c Z, If a|b and b|c then a|c

Suppose: a, b, and c are ints such that a|b and b|c

Show: a|c or c = a*(int) [remember: n=d*k by defn]

a|b b = a*r for some rZ defn of div

b|c c = b*s for some sZ defn of div

c = (a*r)*s substitution

c = a*(r*s) associative

Let k=r*s be an int mult of ints

So, c = a*k

a divides c or a|c defn of div

Proof by Counterexample

Is it true or false that for all ints a and b, if a|b and b|a then a=b?

a,b Z, If a|b and b|a then a=b

Negating gives….

xZ | (a|b and b|a) (ab)

Suppose: a and b are ints such that a|b and b|a

Show: ab

a|b b = a*k for some kZ defn of div

b|a a = b*l for some lZ defn of div

b = (b*l)*k = b*(l*k) substitution & assoc

Since b 0, cancel b’s giving 1=l*k (l and k are divisors of 1)

Thus, k and l are both either 1 or -1

If k = l = 1, then b=a

If k = l = -1, then b = -a and so ab

This means you can find a counterexample by taking b = -a

Example, a = -2 and b = 2; a|b = 2|-2 and b|a = -2|2 but ab

The proposed divisibility property is False!

4 11

8

3

q

r

3

9 32

27

5

32 div 9

32 mod 9

More Number Theory Definitions (cont.)- Quotient-Remainder Theorem
- Given any integer n and positive int d

unique q,rZ | n=d*q + r and 0 ≤ r < d

- Example: 11/4
- 11 = 2*4 + 3
- Div/Mod
- n div d = int quotient when n is divisible by d
- n mod d = int remainder when n is divisible by d

n div d = q n mod d = r

n = d*q + r

Even/Odd is a Special Case of Divisibility

- We say that n is divisible by d if kZ | n=d*k
- n is divisible by 2 if kZ| n = 2k (even)
- The other case is n = 2k+1 (odd, remainder of 1)
- n is divisible by 3 if kZ| n = 3k
- The other cases are n = 3k+1 and n = 3k+2
- n is divisible by 4 if kZ| n = 4k
- The other cases are n = 4k+1, n = 4k+2, n = 4k+3
- n is divisible by 5 if kZ| n = 5k
- The other cases are …….

Parity of Integers

- How can we prove whether every integer is either even or odd?
- By Q-R Theorem we know that n = d*q + r and 0 ≤ r < d
- if d = 2, then there exists integers q and r such that

n = 2q + r and 0 ≤ r < 2

- Evaluating the cases gives….

n = 2q + 0 n = 2q + 1

even parity (n=2k) odd parity (n=2k+1)

- Theorem 3.4.2: Any two consecutive integers have opposite parity

Applying the Q-R Theorem

Given any integer n, apply the Q-R Theorem to n with d = 4

- This implies that there exist an integer quotient q and remainder r such that

n = 4q + r and 0 ≤ r < 4

- Hence,

n = 4q, n = 4q+1, n = 4q+2, n = 4q+3

- Look at Theorem 3.4.3: The square of any odd integer has the form 8m+1

Divisibility Proof

Prove n2 – 2 is never divisible by 3 if n is an integer

Discussion: What does it mean for a number to be divisible by 3? If a is divisible by 3 then kZ | a=3k and the remainder is 0. Other options are a remainder of 1 and 2. So, we need to show that the remainder when n2 – 2 is divided by 3 is always 1 or 2.

There are 3 possible cases:

Case 1: n = 3k

Case 2: n = 3k + 1

Case 3: n = 3k + 2

n2-2 Proof (cont)

Suppose: n is a particular but arbitrarily chosen integer

Show: When n2 – 2 is divided by 3 the remainder is always 1 or 2

Case 1: n = 3k for kZ

n2-2 = (3k)2 - 2 substitution

9k2 - 2 = 3(3k2) – 2 mult & factoring

3(3k2 - 1) + 1 rearranging

The remainder when dividing by 3 is 1

Case 2: n = 3k+1 for kZ

n2-2 = (3k+1)2 - 2 substitution

9k2 + 6k + 1 - 2 = 3(3k2 + 2k) – 1 mult & factoring

3(3k2 + 2k - 1) + 2 rearranging

The remainder when dividing by 3 is 2

n2-2 Proof (cont)

Case 3: n = 3k+2 for kZ

n2-2 = (3k+2)2 - 2 substitution

9k2 + 12k + 4 - 2 multiplying

3(3k2 + 4k) + 2 rearranging

The remainder when dividing by 3 is 2

In each case the remainder when dividing n2-2 by 3 is nonzero. Thus proving the theorem.

Unique Factorization Theorem

- Theorem 3.3.3: Any integer n > 1 is either prime or can be written as a product of prime numbers in a way that is unique (Fundamental Theorem of Arithmetic)

1) prime number

2) product of prime numbers

- Example
- n = 4 = 2*2, where 2 is a prime number
- n = 7 = 7*1, where 7 is a prime number
- n = 100 = 10*10 = 2*2*5*5, where 2 and 5 are prime numbers

Standard Factor Form

- Because of UFT, any integer n > 1 can be written in ascending order from left to right

n = p1e1 p2e2 p3e3 …. Pkek

- Where

k is a positive integer

p1 pk are prime numbers

e1 ek are positive integers

p1 < p2 < p3 < … < pk

- Example

k = 100 = 10*10 = 2*2*5*5 = 2252

n = 3300 = 100*33 = 4*25*3*11 = 2*2*5*5*3*11 = 22 3 52 11

Even More Number Theory Definitions (cont.)

- Floor/Ceiling
- Floor of x x
- unique integer n such that n ≤ x < n+1
- Ceiling of x x
- unique integer n such that n-1 < x ≤ n
- Example
- X = 37 / 4 = 9 ¼
- x = 9 and 9 ≤ 9 ¼ < 10
- x = 10 and 9 < 9 ¼ ≤ 10
- Note: Floor or Ceiling of an integer is itself!

Proof by Counterexample

Is the following True or False?

x,yR, x + y = x + y

Method 1:

Suppose: x and y are particular but arbitrarily chosen real numbers such that x = y = ½

Show: The statement x + y = x + y is False

½ + ½ = 1 = 1 substitution

½ + ½ = 0 + 0 = 0 substitution

x + y x + y

Method 2:

Rewrite as a negation x + y x + y

Prove negation is True

Other Floor/Ceiling Theorems

Theorem 3.5.1: For all Real numbers x and all integers m, x + m = x + m

- Intuitive since m is an integer and its floor is always itself

Theorem 3.5.2: The Floor of n/2

n/2 if n is even

n/2 =

(n-1)/2 if n is odd

- Intuitive since when n is even, n/2 is an integer and the floor of an integer is itself or n/2

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