Question

Prove that each of the following numbers is irrational:

(i) $\sqrt{6}$
(ii) $\left(2-\sqrt{3}\right)$
(iii) $\left(3+\sqrt{2}\right)$
(iv) $\left(2+\sqrt{5}\right)$
(v) $\left(5+3\sqrt{2}\right)$
(vi) $3\sqrt{7}$
(vii) $\frac{3}{\sqrt{5}}$
(viii) $\left(2-3\sqrt{5}\right)$
(ix) $\left(\sqrt{3}+\sqrt{5}\right)$

Answer 1

(i) Let $\sqrt{6}=\sqrt{2}\times \sqrt{3}$ be rational.
Hence, $\sqrt{2},\sqrt{3}$ are both rational.
This contradicts the fact that $\sqrt{2},\sqrt{3}$ are irrational.
The contradiction arises by assuming $\sqrt{6}$ is rational.
Hence, $\sqrt{6}$ is irrational.

(ii) Let $2-\sqrt{3}$ be rational.
Hence, $2$ and $2-\sqrt{3}$ are rational.
∴ $(2-2+\sqrt{3})=\sqrt{3}$ = rational [∵ Difference of two rational is rational]
This contradicts the fact that $\sqrt{3}$ is irrational.
The contradiction arises by assuming $2-\sqrt{3}$ is rational.
Hence, $2-\sqrt{3}$ is irrational.

(iii) Let $3+\sqrt{2}$ be rational.
Hence, $3$ and $3+\sqrt{2}$ are rational.
∴ $3+\sqrt{2}-3=\sqrt{2}$ = rational [∵ Difference of two rational is rational]
This contradicts the fact that $\sqrt{2}$ is irrational.
The contradiction arises by assuming $3+\sqrt{2}$ is rational.
Hence, $3+\sqrt{2}$ is irrational.

(iv) Let $2+\sqrt{5}$ be rational.
Hence, $2+\sqrt{5}$ and $\sqrt{5}$ are rational.
∴ $(2+\sqrt{5})-2=2+\sqrt{5}-2=\sqrt{5}$ = rational [∵ Difference of two rational is rational]
This contradicts the fact that $\sqrt{5}$ is irrational.
The contradiction arises by assuming $2-\sqrt{5}$ is rational.
Hence, $2-\sqrt{5}$ is irrational.

(v) Let, $5+3\sqrt{2}$ be rational.
Hence, $5$ and $5+3\sqrt{2}$ are rational.
∴ $(5+3\sqrt{2}-5)=3\sqrt{2}$ = rational [∵ Difference of two rational is rational]
∴ $\frac{1}{3}\times 3\sqrt{2}=\sqrt{2}$ = rational [∵ Product of two rational is rational]
This contradicts the fact that $\sqrt{2}$ is irrational.
The contradiction arises by assuming $5+3\sqrt{2}$ is rational.
Hence, $5+3\sqrt{2}$ is irrational.

(vi) Let $3\sqrt{7}$ be rational.
∴ $\frac{1}{3}\times 3\sqrt{7}=\sqrt{7}$ = rational [∵ Product of two rational is rational]
This contradicts the fact that $\sqrt{7}$ is irrational.
The contradiction arises by assuming $3\sqrt{7}$ is rational.
Hence, $3\sqrt{7}$ is irrational.

(vii) Let $\frac{3}{\sqrt{5}}$ be rational.
∴ $\frac{1}{3}\times \frac{3}{\sqrt{5}}=\frac{1}{\sqrt{5}}$ = rational [∵ Product of two rational is rational]
This contradicts the fact that $\frac{1}{\sqrt{5}}$ is irrational.
∴ $\frac{1\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}=\frac{1}{5}\sqrt{5}$
So, if $\frac{1}{\sqrt{5}}$ is rational, then $\frac{1}{5}\sqrt{5}$ is rational.
∴ $5\times \frac{1}{5}\sqrt{5}=\sqrt{5}$ = rational [∵ Product of two rational is rational]
Hence, $\frac{1}{\sqrt{5}}$ is irrational.
The contradiction arises by assuming $\frac{3}{\sqrt{5}}$ is rational.
Hence, $\frac{3}{\sqrt{5}}$ is irrational.

(viii) Let $2-3\sqrt{5}$ be rational.
Hence $2$ and $2-3\sqrt{5}$ are rational.
∴ $2-(2-3\sqrt{5})=2-2+3\sqrt{5}=3\sqrt{5}$ = rational [∵ Difference of two rational is rational]
∴ $\frac{1}{3}\times 3\sqrt{5}=\sqrt{5}$ = rational [∵ Product of two rational is rational]
This contradicts the fact that $\sqrt{5}$ is irrational.
The contradiction arises by assuming $2-3\sqrt{5}$ is rational.
Hence, $2-3\sqrt{5}$ is irrational.

(ix) Let $\sqrt{3}+\sqrt{5}$ be rational.
∴ $\sqrt{3}+\sqrt{5}=a$, where $a$ is rational
∴ $\sqrt{3}=a-\sqrt{5}$ ... (1)
On squaring both sides of equation (1), we get
$3=(a-\sqrt{5}{)}^2=a^2+5-2\sqrt{5}a$
⇒ $\sqrt{5}=\frac{a^2+2}{2a}$
This is impossible because right-hand side is rational, whereas the left-hand side is irrational.
This is a contradiction.
Hence, $\sqrt{3}+\sqrt{5}$ is irrational.