Question:
Prove that $2 \sqrt{3}-1$ is an irrational number.
Solution:
Let us assume that $2 \sqrt{3}-1$ is rational .Then, there exist positive co primes $a$ and $b$ such that
$2 \sqrt{3}-1=\frac{a}{b}$
$2 \sqrt{3}=\frac{a}{b}+1$
$\sqrt{3}=\frac{\frac{a}{b}+1}{2}$
$\sqrt{3}=\frac{a+b}{2 b}$
This contradicts the fact that $\sqrt{3}$ is an irrational
Hence $2 \sqrt{3}-1$ is irrational
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