Discrete CS103X: Structures Homework Assignment 2: Solutions
Due February 1, 2008
Exercise 1 (10 Points). Prove or give a counterexample for the following: Use the Fundamental Theorem of Arithmetic to prove that for n N, n is irrational unless n is a perfect square, that is, unless there exists a N for which n = a2 . Solution: We will prove the statement by contradiction. Assume n N is not a perfect square, yet its square root is a rational number p for coprime integers p, q, q where q = 0. So n = p or n = ( p )2 . Without loss of generality, we can assume both q q p and q are non-negative. If p = 0, then n = 0 which is a perfect square, contradicting our assumption. So we can assume both p and q are positive. By the Fundamental Theorem of Arithmetic, we can uniquely write both p and q as products of primes, say p = p1 p2 . . . pm and q = q1 q2 . . . qn . Since p and q are coprime, they have no common factors, so pi = qj for all 1 i m, 1 j n. We have: p2 = (p1 p2 . . . pm )2 = p2 p2 . . . p2 1 2 m and
2 2 2 q 2 = (q1 q2 . . . qn )2 = q1 q2 . . . qn
Now p2 andq 2 cannot have any common factors > 1 if they did have a common factor d > 1, any prime factor f of d (and there must be at least one such) must also be a common prime factor of p2 and q 2 (transitivity of divisibility). By the Fundamental Theorem of Arithmetic, p2 p2 . . . p2 is the unique prime factorization of p, so f 1 2 m must be one of the primes p1 , p2 , . . . , pm . Similarly, f must also be one of the primes q1 , q2 , . . . , qn . But this contradicts our statement that no pi = qj . So p2 and q 2 are coprime. A ratio of natural numbers in lowest terms is itself a natural number if and only if its denominator is 1. Since n N , we must have q 2 = 1, which implies q = 1. But then n must be the perfect square p2 , which contradicts our assumption. The statement is thus proved by contradiction. Exercise 2 (20 Points). Prove or disprove, for integers a, b, c and d: (a) If a|b and a|c, then a|(b + c). (b) If a|bc and gcd(a, b) = 1, then a|c. (c) If a and b are perfect squares and a|b, then 1 a| b.
(d) If ab|cd, then a|c or a|d. Solution: (a) If a|b then b = ma for some integer m and if a|c then c = na for some integer n. Thus, (b + c) = ma + na = (m + n)a. Since m + n is an integer, a|(b + c) (b) The proof is essentially identical to that of Theorem 5.1.1.a. Since gcd(a, b) = 1, there exist integers u, v with au + bv = 1. Multiply both sides by c to get c = auc + bcv (by the result of the first part of this exercise). We know that a|bc so a|bcv and of course a|auc, so a|(auc + bcv). Thus a|c. (c) Proof by contradiction: Assume b is not divisible by a. Consider the prime factorizations of a and b - there must be some prime p that appears m times in the prime factorization of a and n times in the prime factorization of b with m > n. The prime factorizations of perfect squares include every element of their square roots factorization twice, so p must occur 2m times in the prime factorization of a and 2n times in the prime factorization of b. But 2m > 2n, which implies that b is not divisible by a, a contradiction. Therefore a| b. (d) False. One possible counterexample is a = 10, b = 1, c = 4, d = 25. Exercise 3 (25 Points). On Euclids algorithm: (a) Write the algorithm in pseudo-code. (10 points) (b) Prove that Euclids Algorithm correctly finds the GCD of a and b in a finite number of steps. (10 points) (c) Use the algorithm to calculate gcd(1247, 899). Write out the complete sequence of derivations. (5 points) Solution: (a) Procedure GCD-Euclid Input: Integers a, b, not both 0. i. a = |a|, b = |b| ii. If b > a then swap a and b first iii. If b = 0, then return a iv. q = ba/bc (Quotient) v. r = a - q b (Remainder) vi. If r = 0, then return b vii. Return GCD-Euclid(b, r) 2
The first three lines are needed only for the recursive call and can be factored out with a second procedure. The following nonrecursive version also works: Procedure GCD-Euclid-Nonrecursive Input: Integers a, b, not both 0. i. a = |a|, b = |b| ii. If b > a then swap a and b iii. If b = 0, then return a iv. Do v. q = ba/bc (Quotient) vi. r = a - q b (Remainder) vii. If r = 0, then return b viii. a = b ix. b = r x. Loop (b) Theorem. Euclid's Algorithm correctly finds the GCD of a and b in a finite number of steps. Proof. We will prove the correctness of the algorithm in the context of the recursive listing above. The proof for the non-recursive version is identical except that it is slightly more difficult to phrase correctly. The first couple of lemmas help to justify the assumption a > b > 0 in the lecture notes. Lemma 1. d|a if and only if d | |a|. Proof. First we show that if d|x, then d| - x. By the Division Algorithm, x = qd for some integer q, so -x = -qd = (-q)d. Since -q is obviously integral since q is (and the Division Algorithm guarantees that this is an unique representation given x, d), d divides -x. So if d|a, then |a| is either a, which is trivially divisible by d, or -a, which by the above reasoning is also divisible by d. Similarly, if d | |a|, then a is either |a| or -|a|, both of which are divisible by d as above. This proves the result. Lemma 2. gcd(a, b) = gcd(|a|, |b|) = gcd(|b|, |a|) Proof. Let d = gcd(a, b). Since d = gcd(a, b), d|a and d|b, so by Lemma 1, d | |a| and d | |b|, i.e. d is a common divisor of --a-- and --b--. Now we prove by contradiction that d is the greatest such divisor. Assume there is some c > d such that c | |a| and c | |b|. Then by Lemma 1, c|a and c|b. So c is a common divisor of a and b strictly greater than the GCD of a and b, which contradicts the definition of the GCD. Therefore we must have gcd(a, b) = gcd(|a|, |b|). Also, the definition of gcd(x, y) is clearly symmetric in x and y, so gcd(|a|, |b|) = gcd(|b|, |a|). Hence proved. 3
Back to the original problem. Let P (n) be the following statement: "Euclids Algorithm finds the correct GCD of a and b in a finite number of steps, for all 0 a, b n (a and b not both 0)." will We prove P (n) holds for all positive integers n by induction. We assume a b: if not, the first couple of steps of the algorithm will take their absolute values and swap them if necessary so the relation holds, and by Lemma 2 the GCD of these two new values is precisely the same as the GCD of the original values. The base case, n = 1, has two possibilities: a = 1, b = 0, or a = 1, b = 1. In the first case, the third line of the algorithm returns the correct GCD 1, and in the second case r evaluates to 0 before any recursive calls, so the correct GCD b = 1 is returned, in a finite number of steps (no recursive calls, so at most 6 lines of pseudocode). Now assume P (n) is true and consider P (n + 1), where we allow the values of a and b to be at most n + 1. If b = 0, the third line returns the correct GCD, a. If b|a, gcd(a, b) = b, which is the value returned by the algorithm since r evaluates to 0. Otherwise, the algorithm recursively computes and returns gcd(b, r). Now by Lemma 4.3.1 in the lecture notes, gcd(a, b) = gcd(b, r). Also, since 0 r < b n + 1, r and b are both at most n (if b was n + 1, a would also be n + 1, which implies b|a, which is a case we have already handled) and not both zero. Hence by the inductive hypothesis, Euclids Algorithm correctly computes gcd(b, r) in a finite number of steps, which implies that it also correctly computes gcd(a, b) in a finite number of steps, since the sequence of steps before the recursive call adds only a finite overhead. Hence P (n + 1) is true. This proves the claim by induction. The truth of P (n) for all n N+ obviously implies the theorem. (c) The first step of the algorithm swaps 1247 and 899 so a = 1247, b = 899. We tabulate the values of a, b and r in each successive iteration until r = 0: Iteration a b r 1 1247 899 348 2 899 348 203 3 348 203 145 4 203 145 58 5 145 58 29 The value of b when r is zero is 29, so this is the GCD. Exercise 4 (20 Points) Some prime facts: (a) Prove that for every positive integer n, there exist at least n consecutive composite numbers. (10 points)
4
(b) Prove that if an integer n 2 is such that there is no prime p n, then n is a prime. (10 points) Solution:
n that divides
(a) Recall the definition n! = 1 2 3 n for any positive integer n. Consider the consecutive positive integers (n + 1)! + 2, (n + 1)! + 3, . . . (n + 1)! + (n + 1). By definition of factorial, all integers from 2 to (n+1) divide (n+1)!, so 2|((n+1)!+2), 3|((n + 1)! + 3), and so on up to (n + 1)|((n + 1)! + (n + 1)) (remember the property of divisibility proved in Exercise 3a.) Thus all members of this sequence are composite, making n consecutive composite numbers. This sequence can be generated for any n, so for all n there exists at least n consecutive composite numbers. (b) Proof by contradiction: Assume n is not prime. By Theorem 5.1.2, n = p1 p2 . . . pk for primes p1 p2 . . . pk , and since n is not prime k 2. Since no prime less than or equal to n divides n, n < p1 p2 . Then p1 p2 > n, so n = p1 p2 . . . pk > n, a contradiction. Thus our assumption was false and n must be prime. Exercise 5 (25 Points) A fun game: To start with, there is a chart with numbers 1211 and 1729 written on it. Now you and I take turns and you go first. On each players turn, he or she must write a new positive integer on the board that is the difference of two numbers that are already there. The first person who cannot create a new number loses the game. For example, your first move must be 1729 - 1211 = 518. Then I could play either 1211 - 518 = 693 or 1729 - 518 = 1211, and so forth. (a) Prove every...