1 / 16

Proof Methods: Part 2

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

issac
Download Presentation

Proof Methods: Part 2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Proof Methods: Part 2 Sections 3.1-3.6

  2. 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

  3. 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

  4. 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!

  5. 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

  6. 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 …….

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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.

  12. 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

  13. 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

  14. 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!

  15. 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

  16. 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

More Related