Shirleen Stibbe http://www.shirleenstibbe.co.uk

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 n2 is odd if and only if n is odd

Proof of a biconditional:

First we note that we have to prove that n2 is odd if and only if n is odd.

So we have to prove that n2 is odd if n is odd, and that n2 is odd, only if n is odd.

Let P be the proposition that n2 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.

Then n2 = 2k +1( )2= 4k 2 + 4k +1= 2 2k 2 + 2k( )+1= 2m+1 where m = 2k 2 + 2k . Then n2 is odd

because m = 2k 2 + 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.

Then n2 = 2k( )2= 4k 2 = 2 2k 2( ) = 2m, where m = 2k 2 is an integer.

Hence n2 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:

n2 + n = n n+1( ) is even for all integers n, since either n or n + 1 is even.

Hence n2 and n are both even or both odd, since the sum of an odd integer and an even integer is odd.

Therefore n2 is odd if and only if n is odd.

Shirleen Stibbe http://www.shirleenstibbe.co.uk

2 Prove that 2n − 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.

Then n = 2k for some integer k, and 2n − 1 = 22k − 1 = (2k − 1)(2k + 1).

Since 2k − 1, 2k and 2k + 1 are 3 consecutive integers, one and only one of them is a multiple of 3.

The only divisors of 2k are powers of 2, so 2k is not a multiple of 3.

Therefore one of 2k − 1 and 2k + 1 must be a multiple of 3, and 3 divides 22k − 1.

Suppose n is not an even integer.

Then n = 2k +1 for some integer k, and 2n − 1 = 22k+1 − 1 = 2(22k − 1 ) + 1.

We know that 3 divides 22k − 1, so 2n − 1 is one more than a multiple of 3, and so is not a multiple of 3.

Hence 2n − 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