Question

Prove the 3+ root5 is an irrational number​

Answer 1

Let us assume that 3 + √5 is a rational

number (in contradiction).

Now,

3 + √5 = (a ÷ b)

[Here a and b are co-prime numbers]

√5 = [(a ÷ b) - 3]

√5 = [(a - 3b) ÷ b]

Here, {(a - 3b) ÷ b} is a rational number.

But we know that √5 is a irrational

number.

So, {(a - 3b) ÷ b} is also a irrational

number.

So, our assumption is wrong.

Ths, 3 + √5 is a irrational number.

Answer 2

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