Question

prove that 3√5/2 is an irrational number????​

Answer 1

Answer:

Answer

Let us assume to the contrary that (√3+√5)² is a rational number,then there exists a and b co-prime integers such that,

(√3+√5)²=a/b

3+5+2√15=a/b

8+2√15=a/b

2√15=(a/b)-8

2√15=(a-8b)/b

√15=(a-8b)/2b

(a-8b)/2b is a rational number.

Then √15 is also a rational number

But as we know √15 is an irrational number.

This is a contradiction.

This contradiction has arisen as our assumption is wrong.

Hence (√3+√5)² is an irrational number.

Answer 2

Given: 3 + 2√5

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