Question
Show that $2-\sqrt{3}$ is an irrational number.
✓
Answer
Let us assume that $2-\sqrt{3}$ is rational.
Then, there exist positive co primes a and b such that,
$2-\sqrt{3}=\frac{\text{a}}{\text{b}}$
$\sqrt{3}=2-\frac{\text{a}}{\text{b}}$
This implies, $\sqrt{33}$ is a rational number, which is a contradiction.
Hence, $2-\sqrt{3}$ is irrational number
Then, there exist positive co primes a and b such that,
$2-\sqrt{3}=\frac{\text{a}}{\text{b}}$
$\sqrt{3}=2-\frac{\text{a}}{\text{b}}$
This implies, $\sqrt{33}$ is a rational number, which is a contradiction.
Hence, $2-\sqrt{3}$ is irrational number
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