Assume that the given real number is rational. This means that the number can be expressed in the form p/q where p and q belong to integers as well as are co-prime. So, 3 + 2root5 = p/q Or, 2root5 = p/q - 3 = (p -3q)/q Or,Root5 = (p-3q)/2q ....... (i)

Now, (p-3q)/2q is a rational.

So, irrational number ≠ rational number. This means root5 is rational.

But, root5 is an irrational.

How??

Assume root5 as rational. So, Root5 = a/b Where a and b are integers and co-primes. So, Squaring both sides:-5 = p²/q²So, p² =5q² ...... (ii) So, p² has 5 as a factor. So, p also has 5 as its factor for some integer c.

Now,p =5c Or, p² =25c² Putting it in (ii)

5q² =25c² Or, q² = 5c²

So, q² is a multiple of 5 So, q is also a multiple of 5.

Now, Both p and q have a common factor 5 This means they are not co-primes but it is given that they are co-primes.

Hence, it's a contradiction which has risen because of taking root5 as rational.

So, root5 is irrational.

Now,

Back to the question. From (i) :-

Root5 = (p-3q)/2q

So, This is not possible as root5 is irrational and RHS of the equation is rational.

As irrational ≠ rational.

Hence, it is a contradiction.

This has risen because of taking the given number (3 + 2root5) as rational number.

This implies that 3 + 2root5 is an irrational number.