Let \[p:\]If \[x\]is a number and \[{{x}^{2}}\]is even, then, at that point, \[x\]is likewise even

Let \[q:\text{ }x\]is a number and \[~{{x}^{2}}~\]is even

\[r:\text{ }x\]is even

By contrapositive technique, to demonstrate that \[p\]is valid, we expect that \[r\]is bogus and prove that \[q\]is too

Let \[x\]isn’t even

To refute that \[q\]is, it must be demonstrated that \[x\]isn’t a number or \[~{{x}^{2}}~\]isn’t even

\[x\]isn’t demonstrates that \[~{{x}^{2}}~\]is likewise not even.

Henceforth, explanation \[q\]is bogus.

In this way, the given assertion \[p\]is valid