Question

prove that 2√3+3√5 is irrational number​

Answer 1

Let us assume that $2 \sqrt{3} + 3 \sqrt{5}$ is a rational numbers.

So,

They can be expressed in the form of p/q where p and q are integers and q≠0.

Also,

Both p and q are co-primes.

Now,

$2 \sqrt{3} + 3 \sqrt{5} = \frac{p}{q} \\ \\ \\ {(2 \sqrt{3} + 3 \sqrt{5})}^{2} = \frac{{p}^{2}}{{q}^{2}} \\ \\ \\12 + 45 + 12 \sqrt{15} = \frac{{p}^{2}}{{q}^{2}} \\ \\ \\ 12 \sqrt{15} = \frac{{p}^{2}}{{q}^{2}} - 57 \\ \\ \\ \sqrt{15} = \frac{{p}^{2} - 57 {q}^{2}}{12{q}^{2}}$

We know that

$\sqrt{15} is \: an \: irrational \: number$

But,

The RHS is a rational numbers as it is in the form of p/q and q≠0

Thus ,

It is a contradiction as both LHS and RHS can never be equal.

This has arose because we took

$2 \sqrt{3} + 3 \sqrt{5}$ as a rational number.

Thus,

$2 \sqrt{3} + 3 \sqrt{5}$ is an irrational number.

Hence, proved!

Answer 2

hope this helps you
please mark as brainily