# Real Numbers : Exercise 1.3 (Mathematices NCERT Class 10th)

Q.1     Prove that $\sqrt 5$ is irrational.
Sol.       Let us assume, to the contrary, that $\sqrt 5$ is rational.
Now, let $\sqrt 5 = {a \over b}$ , where a and b are coprime and $b \ne 0$. Squaring on both side, we get
$5 = {{{a^2}} \over {{b^2}}} \Rightarrow 5{b^2} = {a^2}$                                                          ...(1)
This shows that ${a^2}$ is divisible by 5
It follows that a is divisible by 5                                              ...(2)
$\Rightarrow a = 5m$ for some integer m.
Substituting a = 5m in (1), we get
$5{b^2} = {\left( {5m} \right)^2} = 25{m^2}$
or ${b^2} = 5{m^2}$
$\Rightarrow {b^2}$ is divisible by 5
and hence b is divisible by 5                                                   ...(3)
From (2) and (3), we can conclude that 5 is a common factor of both a and b.
But this contradicts our supposition that a and b are coprime.
Hence, $\sqrt 5$ is irrational.

Q.2     Prove that 3 + $2\sqrt 5$ is irrational.
Sol.       Let us assume, to the contrary, that $3 + 2\sqrt 5$, is a rational number.
Now, let  $3 + 2\sqrt 5 = {a \over b}$, where a and b are coprime and $b \ne 0$
$\Rightarrow 2\sqrt 5 = {a \over b} - 3$ or $\sqrt 5 = {a \over {2b}} - {3 \over 2}$
Since, a and b are integers.
Therefore, ${a \over {2b}} - {3 \over 2}$ is a rational number
$\Rightarrow$ $\sqrt 5$ is a rational number.
But $\sqrt 5$ is an irrational number.
This shows that our assumption is incorrect.
So, $3 + 2\sqrt 5$ is an irrational number .

Q.3     Prove that the following are irrationals :
(i) ${1 \over {\sqrt 2 }}$     (ii) $7\sqrt 5$     (iii) $6 + \sqrt 2$
Sol.       (i) Let us assume, to the contrary, that ${1 \over {\sqrt 2 }}$ is rational. That is, we can find co-prime integers p and $q( \ne 0)$ such that
${1 \over {\sqrt 2 }} = {p \over q} \Rightarrow {{1 \times \sqrt 2 } \over {\sqrt 2 \times \sqrt 2 }} = {p \over q} \Rightarrow {{\sqrt 2 } \over { 2 }} = {p \over q}$
$\Rightarrow \sqrt 2 = {{2p} \over q}$
Since p and q are integers, ${{2p} \over q}$ is rational, and so $\sqrt 2$ is rational.
But this contradicts the fact that $\sqrt 2$ is irrational.So, we conclude that ${1 \over {\sqrt 2 }}$ is irrational.

(ii) Let us assume, to the contrary, that $7\sqrt 5$ is rational.
That is, we can find co-prime integers p and $q( \ne 0)$ such that $7\sqrt 5 = {p \over q}$
Since p and q are integers, ${p \over {7q}}$ is rational and so is $\sqrt 5$
But this contradicts the fact that $\sqrt 5$ is irrational . So, we conclude that $7\sqrt 5$ is irrational.

(iii) Let us assume, to the contrary, that $\sqrt 2$ is rational. That is, we can find integers p and $q( \ne 0)$ such that
$6 + \sqrt 2 = {p \over q} \Rightarrow 6 - {p \over q} = \sqrt 2$
$\Rightarrow \sqrt 2 = 6 - {p \over q}$
Since p and q are integers, we get $6 - {p \over q}$ is rational, and so $\sqrt 2$ is rational.
But this contradicts the fact that $\sqrt 2$ is irrational.
So, we conclude that 6 + $\sqrt 2$ is irrational