Question

If a square number is even then so is its square root.

Answer 1

Answer:

If $n^2$ is even, then so is n.

Step-by-step explanation:

We will prove the above theorem with the help of contrapositivty.

The contrapositive statement for given statement is, if n is odd then, $n^2$ is odd. Now, all we need to prove is for odd n we have odd $n^2$.

Let n be odd, then it can be written in the form $n = 2t + 1$, where t is an integer,

$n^2 = (2t + 1)^2 = 4t^2 + 1 + 4t = 2(2t^2 + 2t) + 1$

$n^2$ is odd as it can be expressed in the form $n^2 = 2s + 1, \text{ where s =}~ 2t^2 + 2t}$.

Hence, by contrapositivity, it can be said that if a number is even, then so is its square root.

But the given statement is true for $m > 2$.