Question

Use method of contradiction to show that √3 and √5 are irrational numbers.

Answer 1

Let us suppose that √3 and √5 are rational numbers.

∴ √3 = `a/b` and √5 = `x/y`     (Where a, b ∈ 7 and b, y ≠ 0 x, y)

Squaring both sides,

3 = `a^2/b^2`, 5 = `x^2/y^2`

3b² = a², 5y² = x²

⇒ a² and x² are odd as 3b² and 5y² are odd.

⇒ a and x are odd               ...(1)

Let a = 3c, x = 5z

a² = 9c², x² = 25z²

3b² = 9c², 5y² = 25z²        ...(From equation)

⇒ b² = 3c², y² = 5z²

⇒ b² and y² are odd as 3c² and 5z² are odd.

⇒ b and y are odd                 ...(2)

From equation (1) and (2) we get a, b, x, y are odd integers.

i.e., a, b, and x, y have common factors 3 and 5 this contradicts our assumption that `a/b` and `x/y` are rational i.e, a, b and x, y do not have any common factors other than.

⇒ `a/b` and `x/y` is not rational.

⇒ √3 and √5 and are irrational.