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Let p : If x is an integer and x2 is even, then x is even,

Question:

Let $p$ : If $x$ is an integer and $x^{2}$ is even, then $x$ is even,

Using the method of contrapositive, prove that p is true.

Solution:

Let $p: x$ is an integer and $x^{2}$ is even.

q: x is even

For contrapositive,

$\sim p=x$ is an integer and $x^{2}$ is not even.

$\sim \mathrm{q}=\mathrm{x}$ is not even.

The contrapositive statement is: If $x$ is an integer and $x^{2}$ is not even, then $x$ is not even.

Proof;

Let $x$ be an odd/not even integer

x = 2n + 1

$\{2 n$ must be an even integer as when an integer is multiplied with an even integer, the answer is always even. Adding one ensures that the integer is odd after the multiplication. $\}$

$\rightarrow \mathrm{x}^{2}=(2 \mathrm{n}+1)^{2}$

$\rightarrow \mathrm{x}^{2}=4 \mathrm{n}^{2}+4 \mathrm{n}+1$

$\left\{4 n^{2}\right.$ and $4 n$ are even irrespective of integer $n$ 's value. Adding 1 makes the not even/odd $\}$

Thus, if $x$ is an integer and $x^{2}$ is not even, then $x$ is not even.