Question

Prove that 3√5- 2 is an irrational number.​

Answer 1

Answer:

Assume that 3√5−2 is rational

3√5−2 = a/b where a and b are co- prime integers

3√5 = a - 2b / b

√5= a- 2b / 3b

RHS is of the form p/ q therefore, LHS is also rational. But it contradicts the fact that root 5 is irrational due to our incorrect assumption that 3√5−2 is rational. Hence, 3√5−2 is irrational

Answer 2

Step-by-step explanation:

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.

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