Question

p : If $x$ is an integer and $x_{2}$ is even then $x$ is also even

Open in App

Solution

Verified by Toppr

Let $q$ : $x$ is an integer and $x_{2}$ is even

$r$ : $x$ is even

To prove that $p$ is true by contrapositive method we assume that $r$ is false and prove that $q$ is also false.

Let $x$ is not even

To prove that $q$ is false, it has to be proved that $x$ is not an integer or $x_{2}$ is not even.

$x$ is not even implies that $x_{2}$ is also not even.

Therefore statement $q$ is false.

Thus the given statement $p$ is true.

Video Explanation

Was this answer helpful?

0

0