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# Show that the following number is irrational.

$6+\sqrt{2}$

Given: $6\ +\ \sqrt{2}$

To do: Here we have to prove that $6\ +\ \sqrt{2}$ is an irrational number.

Solution:

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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