Question

Prove that 5root2 - 2root3 is irratonal

Answer 1

Assume to reach our contradiction that 5√2 - 2√3 is rational, so that let x = 5√2 - 2√3, where x is a rational number which is the p/q form of 5√2 - 2√3.

Here, in x = 5√2 - 2√3, both sides are rational. So,

⇒ x = 5√2 - 2√3

⇒ x² = (5√2 - 2√3)²

⇒ x² = (5√2)² + (2√3)² - (2 × 5√2 × 2√3)

⇒ x² = 50 + 12 - 20√6

⇒ x² = 62 - 20√6

⇒ 20√6 = 62 - x²

⇒ √6 = (62 - x²)/20

Here, as we found earlier that both sides of x = 5√2 - 2√3 are rational, then so are √6 = (62 - x²)/20. But this contradicts our earlier assumption that 5√2 - 2√3 is irrational, because in √6 = (62 - x²)/20, the RHS seems rational while the LHS is irrational.

Taking x = p/q,

⇒ √6 = (62 - x²)/20

⇒ √6 = (62 - (p/q)²)/20

⇒ √6 = (62 - p²/q²)/20

⇒ √6 = ((62q² - p²)/q²)/20

⇒ √6 = (62q² - p²)/q² × (1/20)

⇒ √6 = (62q² - p²)/20q²

Thus we reached our contradiction and proved that 5√2 - 2√3 is irrational.