# 3.    Show that the following statement is true by the method of contrapositive.                p: If x is an integer and $x^2$ is even, then $x$ is also even.

Given, If x is an integer and $x^2$ is even, then $x$ is also even.

Let, p : x is an integer and $x^2$ is even

q: $x$ is even

In order to prove the statement “if p then q”

Contrapositive Method:  By assuming that q is false, prove that p must be false.

So,

q is false: x is not even $\impies$$\dpi{80} \implies$ x is odd $\dpi{80} \implies$ x = 2n+1 (n is a natural number)

$\\ \therefore x^2 = (2n+1)^2 \\ \implies x^2 = 4n^2 + 4n + 1 \\ \implies x^2 = 2.2(n^2 + n) + 1 = 2m + 1$

Hence $x^2$ is odd $\dpi{80} \implies$$x^2$ is not even

Hence p is false.

Hence the given statement is true.

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