Floor and Ceiling CSE 20— Discrete Math Ceiling(x): ! x " � The least-integer greater than or equal to x � Formal definition: the unique integer n such that n -1 < x � n Floor(x): # x $ � The greatest-integer less than or equal to x � Formal definition: the unique integer n such that n � x < n+1 Summer, 2006 Graph: July 13 (Day 4) Methods of Proof Number Theory Instructor: Neil Rhodes 2 Floor and Ceiling Usage ! x " = x ↔ x is an integer ↔ # x $ = x You’re guessing a number between 1 and 1000 � Each answer is “too high”, “too low”, or “you got it” � What is the maximum number of guesses required to guess the number? For real x, integer n: # x $ = n ↔ x-1 < n � x � ! x " = n ↔ x � n < x+1 � ! x+n " = ! x " +n � # x $ = - ! - x " � ! x " = - # -x $ � 3 4
Proof of Ceiling/Floor Properties Disproof of Ceiling/Floor Properties ! x " = n ↔ x � n < x+1 # x+y $ = # x $ + # y $ # x $ = - ! - x " 5 6 Theorems with Ceiling Proof by Contradiction To Prove p � n if n is even � Suppose p is false � n � 2 � … = n + 1 if n is odd � Show a contradiction: q and ~q 2 2 � Therefore, our assumption (that p was false) is false, and hence that p is true. 7 8
Proof by Contradiction Proof by Contradiction Every integer is either odd or even, but not both The product of an irrational and a non-zero rational is irrational. 9 10 Proof by Contraposition Differences between the Two To prove ∀ x ∈ D, P(x) � Q(x) Proof by contraposition � Any such proof can be rewritten as a proof by contradiction � Rewrite (mentally) as ∀ x ∈ D, ~Q(x) � ~P(x) – Assume P(x) and ~Q(x). Show ~P(x), a contradiction � Prove contrapositive directly (use generalizing from generic particular) – Suppose x is an element of D such that Q(x) is false – Show that P(x) is false Proof by contradiction � More powerful. Can’t necessarily be rewritten as a proof by contraposition Example: If the square of an integer, n, is even, then n is even � Can show any contradiction 11 12
Proof by Contradiction There are an Infinite Number of Primes � 2 is irrational. Proof: � Proof: � Suppose the set of primes is finite. � Suppose that � 2 is rational List all the primes in this finite set: P1, P2…,Pn. Construct the number m=p1•p2•…•Pn + 1. Since m is bigger than any pi, m is not prime remainder of m/pi is 1 (by QRT), so pi doesn’t divide m for any i. But, due to Prime Factorization theorem, m has prime factors (since m > 1). So m has some prime factor not equal to any pi, so the finite set of primes is missing a prime. � But that’s a contradiction. � But that is a contradiction. 13 14 gcd, lcm Computing GCD Greatest Common Divisor (GCD) Calculating GCD of 24 120 � GCD(m, n) is the largest integer k that divides integers m and n � Prime factorization – k | m and k | n � GCD(m, n) is a linear combination (with integer coefficients) of m and n – ∃ i, j ∈ Z: gcd(m, n) = im + jn � Euclid’s algorithm � To calculate GCD(m, n) 134 % 24 = 14 134 = 5•24 + 14 – Compute prime factorization of m and n – gcd(m, n) = common prime factors (and powers) of m and n 24 % 14 = 10 24 = 1•14 + 10 � Euclid’s algorithm – int gcd(m, n) 14 % 10 = 4 14 = 1•10 + 4 – if (n == 0) return m 10 % 4 = 2 10 = 2•4 + 2 else return gcd(n, m mod n) Least Common Multiple (LCM) 4 % 2 = 0 4 = 2•2 + 0 � LCM(m, n) is the smallest integer k such that integers m and n divide k gcd = 2 – m | k and n | k � To calculate LCM(m , n) – Compute prime factorizations of m and n – lcm(m, n) = union of prime factors (and powers) of m and n 15 16
GCD as Linear Combination Why does Euclid’s Algorithm Work? Given m, n, GCD(m, n), find integer i, j such that GCD(r, 0) = r (r a positive integer) � GCD(m, n) = im + jn � Use Euclid’s algorithm 134 % 24 = 14 134 = 5•24 + 14 24 % 14 = 10 24 = 1•14 + 10 14 % 10 = 4 14 = 1•10 + 4 10 % 4 = 2 10 = 2•4 + 2 4 % 2 = 0 4 = 2•2 + 0 � Go backwards 2 = 10-2•4 2 = 10-2•(14-10) = 3•10-2•14 2 = 3•(24-1•14) - 2•14 = -5•14 + 3•24 2 = -5(134-5•24) +3•24 = 28•24-5•134 17 18 Why does Euclid’s Algorithm Work? GCD(m, n) = GCD(n, m mod n) � Prove GCD(m, n) � GCD(n, m mod n) � Prove GCD(m, n) � GCD(n, m mod n) 19
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