Prove that  $2\ +\ 5\sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.

Given: $2\ +\ 5\sqrt{3}$

To do: Here we have to prove that  $2\ +\ 5\sqrt{3}$  is an irrational number.

Solution:

Let us assume, to the contrary, that  $2\ +\ 5\sqrt{3}$  is rational.

So, we can find integers a and b ($≠$ 0) such that  $2\ +\ 5\sqrt{3}\ =\ \frac{a}{b}$.

Where a and b are co-prime.

Now,

$2\ +\ 5\sqrt{3}\ =\ \frac{a}{b}$

$5\sqrt{3}\ =\ \frac{a}{b}\ -\ 2$

$5\sqrt{3}\ =\ \frac{a\ -\ 2b}{b}$

$\sqrt{3}\ =\ \frac{a\ -\ 2b}{5b}$

Here,  $\frac{a\ -\ 2b}{5b}$  is a rational number but  $\sqrt{3}$  is irrational number. 

But, Irrational number  $≠$  Rational number.

This contradiction has arisen because of our incorrect assumption that  $2\ +\ 5\sqrt{3}$  is rational.

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Updated on: 10-Oct-2022

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