Question

Prove that the necessary and sufficient condition for an integer n to odd is that $n^2$ is odd.

Answer 1

Let p: The integer n is odd.

And $q:n^2$ is odd.

First we prove that $p\Rightarrow q.$

Let n be odd. Then, $n=(2k+1)$ for some integer k.

Then, $n^2(2k+1{)}^2=(4k^2+4k+1)=2(2k^2+2k)+1$

Thus, $n^2$ is 1 more than an even number and therefore, it is odd.

Thus, $p\Rightarrow q.$

Now, in order to prove that $q\Rightarrow p.$ it is sufficient to show that $p\Rightarrow q$ [contrapositive method]

Clearly, ~p: The integer n is even.

Let n=2k for some integer k.

Then , $n^2=4k^2,$which is even,

$\therefore$ n is even

$\Rightarrow n^2$ is even.

Consequently. ~p$\Rightarrow$~q and therefore , $q\Rightarrow p.$

Hence , $p\iff q,$i.e. the integer n is odd if and only if$n^2$ is odd.