Question

Prove that the following statement is true : If x, $y\in Z$ such that x and y are odd, then xy is odd.

Answer 1

Let p: $x,y\in Z$such that x and y are odd.

And q:xy is odd.

Then, we have to prove that xy is odd.

Direct Method

We assume that p is true and show that q is true p is true means x and y are odd integers. Then ,

$x=2m+1$ for some integer m

and $y=(2n+1)$ for some integer n,

$\therefore xy=(2m+1)(2n+1)$

$=(4mn+2m+2n+1)$

$=2(2mn+m+n)+1,$which is clearly odd.

Thus, $p\Rightarrow q.$

Hence , the given statement is true.

Contrapositive method

We will show that ~$q\Rightarrow$~ $p.$

Here ~q: It is false that both x and y are odd.

This means at least one of x and y is even.

Let x be even, Then x=2n for some integer n,

$\therefore$ xy=2ny for some integer n.

This shows that xy is even.

Thus , ~ p is true [ $\therefore$~p : zy is even]

$\therefore$~$q\Rightarrow p.$

So, the given statement is true.