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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.
$p:$ If $x$ is an integer and $x^2$ is even then $x$ is also even.
Solution:
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Let $\mathrm{x}$ is not even -i.e. $x=2 n+1$
$\therefore x^2=(2 n+1)^2=4 n^2+4 n+1=4\left(n^2+\mathrm{n}\right)+1$ $4\left(x^2+x\right)+1$ is odd. i.e. "If $q$ is not true then $p$ is not true" is proved
Hence, the given statement is true.
$\therefore x^2=(2 n+1)^2=4 n^2+4 n+1=4\left(n^2+\mathrm{n}\right)+1$ $4\left(x^2+x\right)+1$ is odd. i.e. "If $q$ is not true then $p$ is not true" is proved
Hence, the given statement is true.
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