Question.
Prove that the following are irrationals :
(i) $\frac{1}{\sqrt{2}}$
(ii) $7 \sqrt{5}$
(iii) $\mathbf{6}+\sqrt{\mathbf{2}}$
Prove that the following are irrationals :
(i) $\frac{1}{\sqrt{2}}$
(ii) $7 \sqrt{5}$
(iii) $\mathbf{6}+\sqrt{\mathbf{2}}$
Solution:
(i) Let us assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is rational. That is we can find coprime integers a and $b(b \neq 0)$ such that,
$\frac{1}{\sqrt{2}}=\frac{\mathbf{P}}{q}$
Therefore, $q=\sqrt{\mathbf{z}_{\mathbf{p}}}$
Squaring on both sides, we get
$\mathrm{q}^{2}=2 \mathrm{p}^{2}$ ...(1)
Therefore, 2 divides $\mathrm{q}^{2}$
so, 2 divides q
so we can write $\mathrm{q}=2 \mathrm{r}$ for some integer $\mathrm{r}$
squaring both sides, we get
$\mathrm{g}^{2}=4 \mathrm{r}^{2}$ ...(2)
From (i) \& (ii), we get
$2 p^{2}=4 r^{2}$
Therefore, 2 divides $\mathrm{p}^{2}$
So, 2 divides p
So, p & q have atleast 2 as a common factor.
But this contradict the fact that $\mathrm{p} \& \mathrm{q}$ have no common factor other than 1 .
This contradict our assumption that $\frac{1}{\sqrt{2}}$ is rational. So, we condude that $\frac{1}{\sqrt{2}}$ is irrational.
(ii) Let us assume, to the contrary, that $7 \sqrt{5}$ is rational.
That is, we can find coprime integers a and $b(b \neq 0)$ such that $7 \sqrt{\mathbf{5}}=\frac{\mathbf{a}}{\mathbf{b}}$
Therefore, $\frac{\mathbf{a}}{\mathbf{7 b}}=\sqrt{\mathbf{5}}$
Since $a$ and $b$ are integers, we get $\frac{\mathbf{a}}{\mathbf{7 b}}$ is rational, and so $\frac{\mathbf{a}}{\mathbf{7 b}}=\sqrt{\mathbf{5}}$ is rational.
But this contradicts the fact that $\sqrt{5}$ is irrational. This contradiction has arisen because of
our incorrect assumption that $7 \sqrt{5}$ is rational.
So, we conclude that $7 \sqrt{5}$ is irrational.
(iii) Let us assume, to the contrary, that $6+\sqrt{\mathbf{2}}$ is rational.
That is, we can find coprime integers a and $b(b \neq 0)$ such that $6+\sqrt{\mathbf{2}}=\frac{\mathbf{a}}{\mathbf{b}}$
Therefore, $\frac{\mathbf{a}}{\mathbf{b}}-6=\sqrt{\mathbf{2}}$
$\Rightarrow \frac{\mathbf{a}-\mathbf{6 b}}{\mathbf{b}}=\sqrt{\mathbf{2}}$
Since $\mathrm{a}$ and $\mathrm{b}$ are integers, we get $\frac{\mathbf{a}}{\mathbf{b}}-6$ is rational, and so $\frac{\mathbf{a}-\mathbf{6 b}}{\mathbf{b}}=\sqrt{\mathbf{2}}$ is rational.
But this contradicts the fact that $\sqrt{\mathbf{2}}$ is irrational. This contradiction has arisen because of our incorrect assumption that
$6+\sqrt{2}$ is rational.
So, we conclude that $6+\sqrt{2}$ is irrational.
(i) Let us assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is rational. That is we can find coprime integers a and $b(b \neq 0)$ such that,
$\frac{1}{\sqrt{2}}=\frac{\mathbf{P}}{q}$
Therefore, $q=\sqrt{\mathbf{z}_{\mathbf{p}}}$
Squaring on both sides, we get
$\mathrm{q}^{2}=2 \mathrm{p}^{2}$ ...(1)
Therefore, 2 divides $\mathrm{q}^{2}$
so, 2 divides q
so we can write $\mathrm{q}=2 \mathrm{r}$ for some integer $\mathrm{r}$
squaring both sides, we get
$\mathrm{g}^{2}=4 \mathrm{r}^{2}$ ...(2)
From (i) \& (ii), we get
$2 p^{2}=4 r^{2}$
Therefore, 2 divides $\mathrm{p}^{2}$
So, 2 divides p
So, p & q have atleast 2 as a common factor.
But this contradict the fact that $\mathrm{p} \& \mathrm{q}$ have no common factor other than 1 .
This contradict our assumption that $\frac{1}{\sqrt{2}}$ is rational. So, we condude that $\frac{1}{\sqrt{2}}$ is irrational.
(ii) Let us assume, to the contrary, that $7 \sqrt{5}$ is rational.
That is, we can find coprime integers a and $b(b \neq 0)$ such that $7 \sqrt{\mathbf{5}}=\frac{\mathbf{a}}{\mathbf{b}}$
Therefore, $\frac{\mathbf{a}}{\mathbf{7 b}}=\sqrt{\mathbf{5}}$
Since $a$ and $b$ are integers, we get $\frac{\mathbf{a}}{\mathbf{7 b}}$ is rational, and so $\frac{\mathbf{a}}{\mathbf{7 b}}=\sqrt{\mathbf{5}}$ is rational.
But this contradicts the fact that $\sqrt{5}$ is irrational. This contradiction has arisen because of
our incorrect assumption that $7 \sqrt{5}$ is rational.
So, we conclude that $7 \sqrt{5}$ is irrational.
(iii) Let us assume, to the contrary, that $6+\sqrt{\mathbf{2}}$ is rational.
That is, we can find coprime integers a and $b(b \neq 0)$ such that $6+\sqrt{\mathbf{2}}=\frac{\mathbf{a}}{\mathbf{b}}$
Therefore, $\frac{\mathbf{a}}{\mathbf{b}}-6=\sqrt{\mathbf{2}}$
$\Rightarrow \frac{\mathbf{a}-\mathbf{6 b}}{\mathbf{b}}=\sqrt{\mathbf{2}}$
Since $\mathrm{a}$ and $\mathrm{b}$ are integers, we get $\frac{\mathbf{a}}{\mathbf{b}}-6$ is rational, and so $\frac{\mathbf{a}-\mathbf{6 b}}{\mathbf{b}}=\sqrt{\mathbf{2}}$ is rational.
But this contradicts the fact that $\sqrt{\mathbf{2}}$ is irrational. This contradiction has arisen because of our incorrect assumption that
$6+\sqrt{2}$ is rational.
So, we conclude that $6+\sqrt{2}$ is irrational.