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Show that $3 + \sqrt{2}$ is an irrational number.
Given: $3\ +\ \sqrt{2}$
To do: Here we have to prove that $3\ +\ \sqrt{2}$ is an irrational number.
Solution:
Let us assume, to the contrary, that $3\ +\ \sqrt{2}$ is rational.
So, we can find integers a and b ($≠$ 0) such that $3\ +\ \sqrt{2}\ =\ \frac{a}{b}$.
Where a and b are co-prime.
Now,
$3\ +\ \sqrt{2}\ =\ \frac{a}{b}$
$\sqrt{2}\ =\ \frac{a}{b}\ -\ 3$
$\sqrt{2}\ =\ \frac{a\ -\ 3b}{b}$
Here, $\frac{a\ -\ 3b}{b}$ is a rational number but $\sqrt{2}$ is irrational number.
But, Rational number $≠$ Irrational number.
This contradiction has arisen because of our incorrect assumption that $3\ +\ \sqrt{2}$ is rational.
So, this proves that $3\ +\ \sqrt{2}$ is an irrational number.
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