Question

prove that 7 + 3 root 5 is an irrational number​

Answer 1

Answer:

⇒ Assume that 7 - 3√5 is rational. So let x = 7 - 3√5, for a rational number x.

⇒ In x = 7 - 3√5, both sides are rational as assumed.

$.x = 7 - 3 \sqrt{5} \\ .3 \sqrt{5} = 7 - x \\ . \sqrt{5} = \frac{7 - x}{3}$

⇒ Here, as both sides of x = 7 - 3√5 are rational, then so should be √5 = (7 - x)/3. But this contradicts our earlier assumption that 7 - 3√5 is rational, because the RHS of √5 = (7 - x)/3 is rational while the LHS is irrational.

⇒ Hence proved