Proof Methods: Part 2

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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**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 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!**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,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