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

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Answer:

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

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.

User Alexis Diel
by
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