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.

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Updated on: 10-Oct-2022

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