Question

prove that 3+3√5 is an irrational no​

Answer 1

$\LARGE{ \underline{\underline{ \sf{Required \: answer:}}}}$

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!!!

Answer 2

Answer:

$\huge \bf \: given$

• prove that 3+3√5

$\huge \bf \: solution$

Let assume that 3+3√5 is a rational number.

$3 + 3 \sqrt{5}$

$3 + 3 \sqrt{5} = \frac{p}{q}$

$3 \sqrt{5} = \frac{p}{q} + 3$

$3 \sqrt{5} = \frac{p \div q - 3 }{3}$

$\sqrt{5} = \frac{p-3q}{9}$

So, p-3q / 9 is rational number.

$but \: \sqrt{5} \: is \: not \: rational \: number$

$so \: 3 + 3 \sqrt{5} \: is \: irrational$