Question

Prove that : 3+√5 is an irrational number​

Answer 1

Step-by-step explanation:

3 + √5 is an irrational number.

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

So it can be written in the form a/b

√3 + √5 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving

√3 + √5 = a/b

On squaring both sides we get,

(√3 + √5)² = (a/b)²

√3² + √5² + 2(√5)(√3) = a²/b²

3 + 5 + 2√15 = a²/b²

8 + 2√15 = a²/b²

2√15 = a²/b² – 8

√15 = (a²- 8b²)/2b

a, b are integers then (a²-8b²)/2b is a rational number.

Then √15 is also a rational number.

But this contradicts the fact that √15 is an irrational number.

Our assumption is incorrect

√3 + √5 is an irrational number.

Hence, proved.

hope it helps

Answer 2

Step-by-step explanation:

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

Now,

3+5=ba [Here a and b are co-prime numbers]

5=[(ba)−3]

5=[(ba−3b)]

Here, [(ba−3b)] is a rational number.

But we know that 5 is an irrational number.

So, [(ba−3b)] is also a irrational number.

So, our assumption is wrong.

3+5 is an irrational number.

Hence, proved.