Question

Prove that 2 + 3√5 is an irrational number

Answer 1

Refer to the answer below :

Answer 2

Let us assume that 2 + 3√5 is a rational number.

=> 2 + 3√5 = a/b

Here.. a and b are co-prime numbers.

Now squaring on both sides

=> (2 + 3√5)² = (a/b)2

(a + b)² = a² + b² + 2ab

So,

=> (2)² + (3√5)² + 2(2)(3√5) = a²/b²

==> 4 + 9(5) + 4(3√5) = a²/b²

=> 4 + 45 + 12√5 = a²/b²

=> 49 + 12√5 = a²/b²

=> 12√5 = a²/b² - 49

=> 12√5 = (a² - 49b²)/b²

=> √5 = (a² - 49b²)/12b²

Here.. (a² - 49b²)/12b² is a rational number.

And √5 = (a² - 49b²)/12b². Means √5 is also a rational number.

But we know that, √5 is an irrational number.

This contradicts that our assumption is wrong.

2 + 3√5 is an irrational number.

Hence Proved