CBSE BoardEnglish MediumSTD 11 ScienceMathsMathematical Reasoning1 Mark
Question
Show that the following statement is true "The integer $n$ is even if an only if $n^2$ is even"
✓
Answer
The given statement can be re$-$written as
$"$The necessary and sufficient condition that the integer $n$ is even is $n^2$ must be even$"$
Let $p$ and $q$ be the statements given by
$p:$ the integer $n$ is even.
$q: n^2 $ is even.
The given statement is
$"p$ if and only if $q"$
In order to check its validity, we have to check the validity of the following statements.
$"$If $p,$ then $q"$
$"$if $q,$ then $p"$
Checking the validity of $"$if $p,$ then $q":$
The statement $"$if $p,$ then $q"$ is given by:
"If the integer $n$ is even, then $n^2 $ is even$"$
Let us assume that $n$ is even. Then,
$n = 2m,$ where $m$ is an integer
$\Rightarrow n^2 = (2m)^2$
$\Rightarrow n^2 = 4m^2$^
$\Rightarrow n^2 $is an even integer
Thus, $n$ is even $\Rightarrow n^2 $ is even
$\therefore "$if $p$, then $q"$ is true.
Checking the validity of $"$if $q$, then $p":$
"if $n$ is an integer and $n^2 $ is even, 'then $n$ is even$"$
To check the validity of this statemens, we will use contrapositive method.
So, let $n$ be an odd integer. Then,
$n $ is odd
$\Rightarrow n = 2k + 1$ for some integer $k:$
$\Rightarrow n^2 = (2k + 1)^2$
$\Rightarrow n^2 = 4k^2 + 4k + 1$
$\Rightarrow n^2$ is not an even integer
Thus, $n$ is not even $\Rightarrow n^2$ is not even
$\therefore "$if $q,$ then $p"$ is true.
Hence, $"p$ if and only if $q"$ is true.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.