Question

prove that 3√3/5 is irrational

Answer 1

$let \: \frac{3 \sqrt{3} }{5} \: be \: a \: rational \: number. \\ then \\ \frac{5}{3} \: is \: also \: a \: rational \: number. \\ \frac{3 \sqrt{3} }{5} \times \frac{5}{3} \: \: \: (product \: of \: two \: rational \: is \: rational) \\ = \sqrt{3} \: \: \: rational \\ but \: this \: contradicts \: the \: fact \: that \: \sqrt{3} \: is \: irrational. \\ this \: contradiction \: arises \: by \: assuming \: \frac{3 \sqrt{3} }{5} \: a \: rational \: number . \\ hence \: \frac{3 \sqrt{3} }{5} \: is \: irrational.$

Answer 2

To prove:

• 3+3√5 is an irrational no

Proof:

The proof below is based on assumption and contradiction.

Let us assume on an contrary that 3 + 3√5 is rational.

Then,

⇒ 3 + 3√5 = p/q

Where p, q are integers and q is not equals to 0.

⇒ 3√5 = p/q - 3

⇒ 3√5 = p - 3q / 3

⇒ √5 = p - 3q / 9

Since p, q, 3 and 9 are rational numbers, p - 3q / 9 is also a rational number.

But this contradicts the fact that √5 is an irrational number. This contradiction has arisen due to our wrong assumption that 3 + 3√5 is a rational number.

Thus,

3 + 3√5 is a irrational number. Hence, proved!!!