Question

2. Prove that 3 + 2
$\sqrt{5 \\ \\ }$
5 is irrational​

Answer 1

Answer:

Step-by-step explanation:

Let us assume that 3 + 2√5 is a rational number.

So, it can be written in the form a/b

3 + 2√5 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving 3 + 2√5 = a/b we get,

=>2√5 = a/b – 3

=>2√5 = (a-3b)/b

=>√5 = (a-3b)/2b

This shows (a-3b)/2b is a rational number. But we know that √5 is an irrational number.

So, it contradicts our assumption. Our assumption of 3 + 2√5 is a rational number is incorrect.

3 + 2√5 is an irrational number

Hence proved

Answer 2

Given :

To prove: 3 + 2√5 is an irrational number.

Proof :

Let us assume that 3 + 2√5 is a rational number.

So, it can be written in the form a/b 3 + 2√5 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving 3 + 2√5 = a/b we get,

=>2√5 = a/b – 3

=>2√5 = (a-3b)/b

=>√5 = (a-3b)/2b

This shows (a-3b)/2b is a rational number. But we know that √5 is an irrational number.

So, it contradicts our assumption. Our assumption of 3 + 2√5 is a rational number is incorrect.

3 + 2√5 is an irrational number

Hence proved