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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.
So, this proves that $2\ +\ 5\sqrt{3}$ is an irrational number.
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