Question

Prove that `(sqrt(2) + sqrt(3))^2` is an irrational number, given that `sqrt(6)` is an irrational number.

Answer 1

Given: `sqrt(6)` is an irrational number.

To Prove: `(sqrt(2) + sqrt(3))^2` is an irrational number.

Proof [Step-wise]:

1. Compute the square:

`(sqrt(2) + sqrt(3))^2 = 2 + 3 + 2 xx sqrt(6)` 

= `5 + 2 xx sqrt(6)`

2. Suppose, for contradiction, that `(sqrt(2) + sqrt(3))^2` is rational.

Then `5 + 2 xx sqrt(6)` is rational; so there exist integers a, b (b ≠ 0) with `5 + 2 xx sqrt(6) = a/b`.

3. Rearranging gives `2 xx sqrt(6) = a/b - 5`, so `sqrt(6) = (a - 5b)/(2b)`.

The right-hand side is a rational number (ratio of integers), so this implies `sqrt(6)` is rational.

4. This contradicts the given fact that `sqrt(6)` is irrational. 

Therefore the assumption in step 2 is false.