Question
Prove that $2-3\sqrt{5}$ is an irrational number.
✓
Answer
Let us assume that $2-3\sqrt{5}$ is rational.
Then, there exist positive co primes a and b such that,
$2-3\sqrt{5}=\frac{\text{a}}{\text{b}}$
$2-\frac{\text{a}}{\text{b}}=3\sqrt{5}$
$\frac{2\text{b}-\text{a}}{\text{b}}=3\sqrt{5}$
$\frac{2\text{b}-\text{a}}{3\text{b}}=\sqrt{5}$
This contradicts the fact that $\sqrt{5}$ is an irrational number.
Hence $2-3\sqrt{5}$ is irrational.
Then, there exist positive co primes a and b such that,
$2-3\sqrt{5}=\frac{\text{a}}{\text{b}}$
$2-\frac{\text{a}}{\text{b}}=3\sqrt{5}$
$\frac{2\text{b}-\text{a}}{\text{b}}=3\sqrt{5}$
$\frac{2\text{b}-\text{a}}{3\text{b}}=\sqrt{5}$
This contradicts the fact that $\sqrt{5}$ is an irrational number.
Hence $2-3\sqrt{5}$ is irrational.
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