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Show that $5 − 2\sqrt{3}$ is an irrational number.
Given: $5\ −\ 2\sqrt{3}$
To do: Here we have to prove that $5\ −\ 2\sqrt{3}$ is an irrational number.
Solution:
Let us assume, to the contrary, that $5\ −\ 2\sqrt{3}$ is rational.
So, we can find integers a and b ($≠$ 0) such that $5\ −\ 2\sqrt{3}\ =\ \frac{a}{b}$.
Where a and b are co-prime.
Now,
$5\ −\ 2\sqrt{3}\ =\ \frac{a}{b}$
$5\ −\ \frac{a}{b}\ =\ 2\sqrt{3}$
$\frac{5b\ -\ a}{b}\ =\ 2\sqrt{3}$
$\frac{5b\ -\ a}{2b}\ =\ \sqrt{3}$
Here, $\frac{5b\ -\ a}{2b}$ is a rational number but $\sqrt{3}$ is irrational number.
But, Rational number $≠$ Irrational number.
This contradiction has arisen because of our incorrect assumption that $5\ −\ 2\sqrt{3}$ is rational.
So, this proves that $5\ −\ 2\sqrt{3}$ is an irrational number.
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