Question

prove that 3 + 2√5 is irrational number​

Answer 1

Answer:

here √5 is irrational so the whole equation is irrational

Answer 2

★ SOLUTION ★

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 + 2√5 = p/q

Or,

⇒ 2√5 = p/q - 3 = (p -3q)/q Or,

⇒ √5 = (p-3q)/2q ....... (i)

Assume √5 as rational. So, √5 = a/b

Where a and b are integers and co-primes.

So,

Squaring both sides:-

⇒ 5 = p²/q²

⇒ 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

⇒ p² =25c²

Putting it in eq (ii)

⇒ q² =25c²

⇒ 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 √5 as rational.

So, √5 is irrational.

Now,

From eq (i) :-

⇒ √5 = (p-3q)/2q

So, This is not possible as √5 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.

∴ 3 + 2√5 is an irrational number.