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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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