Example Proofs: Biconditional
Proofs Workshop
Note: the proofs in this handout are not necessarily in the same form as they were presented at the workshop. In particular, any errors you spot here are entirely accidental, not deliberate. 1 Prove that n 2 is odd if and only if n is odd Proof of a biconditional: First we note that we have to prove that n 2 is odd if and only if n is odd. So we have to prove that n 2 is odd if n is odd, and that n 2 is odd, only if n is odd. Let P be the proposition that n 2 is odd and let Q be the proposition that n is odd (so the statement is in the form P ! Q ). Then we have to prove that Q ! P and P ! Q . Let us begin by proving Q ! P . Suppose Q is true, so that n is odd. Then, since n is odd, n may be written in the form n = 2k + 1 where k is an integer. 2
2
(
2
2
)
Then n = 2k +1 = 4k + 4k +1 = 2 2k + 2k +1 = 2m +1 where m = 2k 2 + 2k . Then n 2 is odd
(
)
2
because m = 2k + 2k is an integer. So, if Q is true, then P is true, and we have succeeded in proving half the result, Q ! P . To prove the converse, P ! Q , we prove instead the logically equivalent statement not-Q ⇒ not-P. Suppose not-Q; in other words, suppose that n is not odd, so n is even. Since n is even, n = 2k for some integer k. 2
( )
Then n = 2k
2
( )
= 4k 2 = 2 2k 2 = 2m, where m = 2k 2 is an integer.
Hence n 2 is even. This is the statement not-P. We've proved not-Q ⇒ not-P. Hence P ! Q which completes the proof. An alternative method: The truth table for P ! Q is: P
Q
P !Q
T
T
T
T
F
F
F
T
F
F
F
T
This shows that P ! Q is true only when P and Q are both true or both false. We use this idea in the following proof:
n 2 + n = n n +1 is even for all integers n, since either n or n + 1 is even.
(
)
Hence n 2 and n are both even or both odd, since the sum of an odd integer and an even integer is odd. Therefore n 2 is odd if and only if n is odd.
Shirleen Stibbe
http://www.shirleenstibbe.co.uk
n
2 Prove that 2 − 1 is a multiple of 3 if and only in n is an even integer. Proof of a biconditional Suppose n is an even integer. n
Then n = 2k for some integer k, and 2 − 1 = 2 k
k
2k
k
k
− 1 = (2 − 1)(2 + 1).
k
Since 2 − 1, 2 and 2 + 1 are 3 consecutive integers, one and only one of them is a multiple of 3. k
The only divisors of 2 are powers of 2, so 2 k
k
is not a multiple of 3.
k
Therefore one of 2 − 1 and 2 + 1 must be a multiple of 3, and 3 divides 22k − 1. Suppose n is not an even integer. n
Then n = 2k +1 for some integer k, and 2 − 1 = 2 We know that 3 divides 2
2k
2k+1
− 1 = 2(2
2k
− 1 ) + 1.
n
− 1, so 2 − 1 is one more than a multiple of 3, and so is not a multiple of 3.
n
Hence 2 − 1 is a multiple of 3 if and only n is an even integer. And a couple more: We also presented two incorrect biconditionals, where we claimed that ¬q ⇒ ¬p was the converse of p ⇒ q , and we had therefore proved it 'both ways'. For one, the statement was correct: Let x, y ∈ ℝ. Then |x| + |y| = |x + y| if and only if xy ≥ 0. And for the other, it was incorrect: Let a , b be integers > 0, where a is a multiple of b. Then for any integer c > 0, c and b are co-‐prime if and only if c and a are co-‐prime. You will find details of both of them under Common Errors on this page of my web site: http://shirleenstibbe.co.uk/proofs
Shirleen Stibbe
http://www.shirleenstibbe.co.uk