Question

Prove that `sqrt(3)` is an irrational number.
(Hint: Follow the method that we have used to prove `sqrt(2)` ∉ Q)

Answer 1

Suppose that `sqrt(3)` is rational P

Then `sqrt(3) = "P"/"q"`  ......(where p and q are integers which are co-prime)

⇒ p = `sqrt(3)` q

⇒ p² = 3q²   ......(1)

`"p"^2/3` = q²

⇒ 3 is a factor of p

So let p = 3c

Substitutiing p= 3c in (1) w get

(3c)² = 3q²

⇒ 9c² = 3q²

⇒ c² = `(3"q"^2)/9`

= `"q"^2/3`

⇒ 3 is a factor of q also

So 3 is a factor ofp and q which is a contradiction.

⇒ `sqrt(3)` is not a rational number

⇒ `sqrt(3)` is an irrational number