Ch 1.5: Basic Proof Methods II

1. Ch 1.5: Basic Proof Methods II • Proof by Contraposition of P => Q • Suppose ~Q • …proof details here…. • Therefore, ~P (via a direct proof) • Thus, ~Q => ~P. • Therefore, P => Q 

2. Example 1 • Let m be an integer. Prove that if m^2 is even, then m is even. • Proof: Suppose that m is odd. • Then m = 2k +1 for some integer k. • It follows that m^2 = (2k+1)^2 = 4k^2 + 4k + 1 •  m^2 = 2(2k^2 + 2k) + 1 •  m^2 = 2n+1, where n = 2n+1 is an integer. • Thus m^2 is odd. • Therefore, m odd implies m^2 is odd. • Hence m^2 even implies m even, by contraposition. 

3. Example 2 • Prove Let x be a positive real number. • Prove that if x > 1, then x^2 – 1 > 0. • Proof: Suppose x is positive and x^2-1 <= 0. • Then (x-1)(x+1) <= 0. • It follows that 0 < x <= 1. • Therefore, x^2-1 <= 0 implies x <= 1. • Hence for x positive, x > 1 implies x^2 – 1 > 0, by contraposition. 

4. Contradiction • Proof by Contradiction. Want to prove P. • Suppose ~P • …proof details here…. • Therefore, Q. • …proof details here…. • Therefore, ~Q. • Thus, Q/\ ~Q, a contradiction. • Therefore, P. • Note: Q is not necessarily known at onset of proof.

5. Example 3 • Prove that sqrt(2) is irrational. • Proof: Assume false: sqrt(2) = p/q for integers p, q. • WLOG, assume p & q have no common factors. • Then 2 = p^2/q^2 => 2q^2 = p^2. • Thus p^2 is even, and hence p is even (by Example 1). • Since p & q have no common factors, q is odd. • Now p is even, so p = 2k for some integer k. • Thus p^2 = 4k^2, hence 2q^2 = 4k^2 (since 2q^2 = p^2). • Then q^2 = 2k^2, and therefore q is even (by Example 1). • This is a contradiction. Hence sqrt(2) is irrational. 

6. Example 4 • Prove that the set of primes is infinite. • Proof: Assume false: Let {p1,p2,…pk} be the set of primes, k in N. • Let n = (p1*p2*…*pk) + 1. • Since n is in N, it has a prime divisor q > 1. (See Ch 2.5.) • Thus q divides n. • Now q is in {p1,p2,…pk}, so it divides p1*p2*…*pk. • Therefore, q divides n - p1*p2*…*pk. (See Ch 1.4, p32.) • But n - p1*p2*…*pk = 1. • Thus q divides 1, which implies q = 1. • This is a contradiction. Hence the set of primes is infinite. 

7. Biconditional • Two-Part Proof of a Biconditional P Q. • Show P => Q by any method • Also, show Q => P by any method. • Therefore, P Q.

8. Example 5 • Let a be a prime; b, c in N. Prove that a|bc iff a|b or a|c. • Proof: Let a be a prime and b,c in N. • (=>) Suppose a|bc. Then bc = ka, for some k in N. • Also, b = p1p2…pm, c = q1q2…qn, pi and qi prime. • Then bc = (p1p2…pm)(q1q2…qn) • Since a is prime and a|bc, a = pi for some i or a = qj for some j. • Thus either b is an integer multiple of a or c is an integer multiple of a. • Therefore a|b or a|c. 

9. Example 5 • Let a be a prime; b, c in N. Prove that a|bc iff a|b or a|c. • Proof: Let a be a prime and b,c in N. • (<=) Suppose a|b or a|c. Then b = ma for some m in N or c = na, for some n in N. • Thus bc = (ma)c = (mc)a = ka, k = mc in N • or bc = b(na) = (bn)a = ja, j = bn in N • Therefore a|b or a|c. 

10. Example 6 • Sometimes it is possible to prove both case of a biconditional simultaneously. 

11. Basic Proof Methods Summary • In this section and the last one we learned how to prove P => Q using a direct proof and a proof by contraposition. • We learned how to prove P using contradiction. • We learned how to prove P  Q by proving the cases P => Q and Q => P separately or simultaneously. • Sometimes we can prove a proposition by more than one method. See example on page 41. • Most propositions can be proved true or false. Some cannot, and they are called undecidables. (This doesn’t happen in Math 240!).

12. Homework • Read Ch 1.5 • Do 43(3a-d,4a,b,6a-c,7a,b,12a-d)