Question

Prove that 3+√5 is irrational​

Answer 1

Hey mate♡♡

Step-by-step explanation:

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

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.

3 + √5 is a irrational number.

Hence, proved.

♡♡Hope this will helpful for you.♡♡

Answer 2

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

Now,

3 + √5 = (p ÷ q )

[Here p and q are co-prime numbers]

√5 = [(p÷ q) - 3]

√5 = [(p - 3q) ÷ q]

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

But we know that √5 is a irrational number.

So, {(p - 3q) ÷ q} is also a irrational number.

Our assumption is wrong

3 + √5 is a irrational number.

Hence, proved.