Using the words necessary and sufficient rewrite the statement The integer `n` is odd if and only if `n^2` is odd Also check whether the statement is true.

Explanation

Let P: The interger n is odd.
And `q:n^2` is odd.
First n be odd: Then, n=(2k+1) for some interger K.
Then `n^2 (2k+1)^2=(4k^2+4k+1)=2(2k^2+2k)+1` This `n^2` is 1 more than an even number and therefore ,it is odd.
Thus `p rArr q`
Now ,in order to prove that `q rArr p` it is sufficient to show that ~`p rArr q `[Contrapositive method ]
Clearly ~ P The interge n is even
Let n = 2k for some interger k.
Then `n^2 = 4k^2` which is even
`therefore ` n is even `rArr n^2` is even .
Consequently ~ P `rArr ~` q and therefore `q rArr P.`
Hence `p iff ` q, i,e the interger n is odd if and only if `n^2` id odd .