Rewrite the following statement with if-then in five different ways conveying the same meaning. If a natural number is odd, then its square is also odd.

Solution: if p then q
$\to$ $p\Rightarrow q$ : a natural no is odd implies that its square is odd
$\to$ $nq\Rightarrow np$ : if square of natural no is not odd then all the natural no is not odd.
$\to$ q is necessary condition for p : the square of natural no being odd is necessary condition for the no to be odd
$\to$ p is sufficient condition for q : a no is odd is sufficient to say its square is odd
$\to$ p only if q : a natural no is odd only if its square is odd