Question

prove that 3+√5 is an irrational




please give me answer​

Answer 1

Let us assume to the contrary that 3+√5 is rational.

That is we can find coprime a& b( b≠0)

Such that 3+√5 =a/b.

3+√5 = a/b =>√5 = a/b - 3

=>√5=a-3 b/b

Since a & b are integers, a-3b/b is rational.

So√5 is rational. But this contradicts the fact that √5 is irrational. So our assumption is wrong.

Therefore 3+√5 is irrational.

Answer 2

Answer:

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+root5 is an irrational number.

Hence, proved.