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Proof Methods: Part 2 Sections 3.1-3.6 More Number Theory Definitions Divisibility n is divisible by d iff kZ | 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

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proof methods part 2

Proof Methods: Part 2

Sections 3.1-3.6

more number theory definitions
More Number Theory Definitions
  • Divisibility
    • n is divisible by d iff
      • kZ | 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
transitivity proof

P(x)

Q(x)

Transitivity Proof

Prove 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 rZ defn of div

b|c  c = b*s for some sZ 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
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….

xZ | (a|b and b|a)  (ab)

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

Show: ab

a|b  b = a*k for some kZ defn of div

b|a  a = b*l for some lZ 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 ab

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 ab

 The proposed divisibility property is False!

more number theory definitions cont

2

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,rZ | 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
Even/Odd is a Special Case of Divisibility
  • We say that n is divisible by d if kZ | n=d*k
    • n is divisible by 2 if kZ| n = 2k (even)
      • The other case is n = 2k+1 (odd, remainder of 1)
    • n is divisible by 3 if kZ| n = 3k
      • The other cases are n = 3k+1 and n = 3k+2
    • n is divisible by 4 if kZ| n = 4k
      • The other cases are n = 4k+1, n = 4k+2, n = 4k+3
    • n is divisible by 5 if kZ| n = 5k
      • The other cases are …….
parity of integers
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
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
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 kZ | 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

n 2 2 proof cont
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 kZ

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 kZ

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

n 2 2 proof cont11
n2-2 Proof (cont)

Case 3: n = 3k+2 for kZ

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
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
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
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 counterexample15
Proof by Counterexample

Is the following True or False?

x,yR,  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
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