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Prove that the following are irrationals.
(i) $ \frac{1}{\sqrt{2}} $
(ii) $ 7 \sqrt{5} $
(iii) $ 6+\sqrt{2} $.
To prove:
Here we have to prove that, the given numbers are irrationals.
Solution:
(i) $\mathbf{\frac{1}{\sqrt{2}}}$
Let us assume that $\frac{1}{\sqrt{2}}$ is a rational number.
So, $\frac{1}{\sqrt{2}}$ can be written in the form $\frac{a}{b}$, where a and b are co prime and b is not equal to 0.
Hence,
$\frac{1}{\sqrt{2}}\ =\ \frac{a}{b}$
$\frac{b}{a}\ =\ \sqrt{2}$
Here, $\frac{b}{a}$ is rational but $\sqrt{2}$ is irrational.
Rational can't be equal to Irrational.
This contradicts our assumption, that $\frac{1}{\sqrt{2}}$ is a rational number.
Therefore, $\frac{1}{\sqrt{2}}$ is irrational number.
(ii) $\mathbf{7\sqrt{5}}$
Let us assume that $7\sqrt{5}$ is a rational number.
Hence, $7\sqrt{5}$ can be written in the form of $\frac{a}{b}$, where a and b are co prime and b is not equal to 0.
$7\sqrt{5}\ =\ \frac{a}{b}$
$\sqrt{5}\ =\ \frac{a}{7b}$
Here, $\sqrt{5}$ is irrational but, $\frac{a}{7b}$ is rational.
Rational can't be equal to Irrational.
This contradicts our assumption, that the number $7\sqrt{5}$ is a rational number.
Therefore, $7\sqrt{5}$ is irrational number.
(iii) $6\ +\ \sqrt{2}$
Let us assume, to the contrary, that $6\ +\ \sqrt{2}$ is rational.
So, we can find integers a and b ($≠$ 0) such that $6\ +\ \sqrt{2}\ =\ \frac{a}{b}$.
Where a and b are co-prime.
Now,
$6\ +\ \sqrt{2}\ =\ \frac{a}{b}$
$\sqrt{2}\ =\ \frac{a}{b}\ -\ 6$
$\sqrt{2}\ =\ \frac{a\ -\ 6b}{b}$
Here, $\frac{a\ -\ 6b}{b}$ is a rational number but $\sqrt{2}$ is irrational number.
But, Irrational number $≠$ Rational number.
This contradiction has arisen because of our incorrect assumption that $6\ +\ \sqrt{2}$ is rational.
So, this proves that $6\ +\ \sqrt{2}$ is an irrational number.
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