Question

show that √2-3√5 is irrational​

Answer 1

ANSWER:

• (√2-3√5) is an irrational number.

GIVEN:

• Number = (√2-3√5)

TO PROVE:

• (√2-3√5) is an Irrational number.

SOLUTION:

Let (√2-3√5) b are rational number which can be expressed in the form of p/q where p and q have no other common factor than 1.

$\implies \: \sqrt{2} - 3 \sqrt{5} = \dfrac{p}{q} \\ \\ \implies \: \sqrt{2} - \dfrac{p}{q} = 3 \sqrt{5} \\ \\ \: \: squaring \: both \: sides \: we \: get. \\ \\ \implies \: ( \sqrt{2} - \dfrac{p}{q} ) {}^{2} = (3 \sqrt{5} ) {}^{2} \\ \\ \implies \: 2 + \frac{p {}^{2} }{q {}^{2} } - \frac{2 \sqrt{2}p }{q} = 45 \\ \\ \implies \: 2 - 45 + \frac{p {}^{2} }{q {}^{2} } = \dfrac{2 \sqrt{2} p}{q} \\ \\ \implies \: - 43 + \frac{p {}^{2} }{q {}^{2} } = \frac{2 \sqrt{2} p}{q} \\ \\ \implies \: \frac{ - 43q {}^{2} + p {}^{2} }{q {}^{2} } = \frac{2 \sqrt{2}p }{q} \\ \\ \implies \: \frac{p {}^{2} - 43q {}^{2} }{q {}^{2} } \times \frac{q}{2p} = \sqrt{2} \\ \\ \implies \: \frac{p {}^{2} - 43q {}^{2} }{2pq} = \sqrt{2}$

Here:

• (p²-43q²)/ 2pq is a rational number but √2 is an irrational number.
• Thus our contradiction is wrong.
• So (√2-3√5) is an irrational number.