Prove that $2\sqrt3 − 1$ is an irrational number.


Given: $2\sqrt3\ −\ 1$

To do: Here we have to prove that $2\sqrt3\ −\ 1$ is an irrational number.

Solution:

Let us assume, to the contrary, that $2\sqrt3\ −\ 1$ is rational.

So, we can find integers a and b ($≠$ 0) such that  $2\sqrt3\ −\ 1\ =\ \frac{a}{b}$.

Where a and b are co-prime.

Now,

$2\sqrt3\ −\ 1\ =\ \frac{a}{b}$

$2\sqrt3\ =\ \frac{a}{b}\ +\ 1$

$2\sqrt3\ =\ \frac{a\ +\ b}{b}$

$\sqrt3\ =\ \frac{a\ +\ b}{2b}$

Here, $\frac{a\ +\ b}{2b}$ 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 $2\sqrt3\ −\ 1$ is rational.



So, this proves that $2\sqrt3\ −\ 1$ is an irrational number.

Updated on: 10-Oct-2022

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