Help with proof for rational number problem

[iamsmooth]

Homework Statement

For all a in the set of real numbers, if a is rational, a + \sqrt{2} is irrational.
You may use that \sqrt{2} is irrational and the sum and difference of rational numbers is rational.

Homework Equations

The Attempt at a Solution

My proof seems way too simple, I don't trust it. Can anyone see anything wrong with this?

Proof:

Let a be an arbitrary rational number. Therefore a = m/n for some integers m and n (n is not 0).

Suppose a + \sqrt{2} is rational. Therefore \frac{m}{n} + \sqrt{2} = \frac{q}{r} where q and r are integers (r is not 0).

Therefore,
\sqrt{2} = \frac{q}{r} - \frac{m}{n}

BUT this is a contradiction because we know that \sqrt{2} is an irrational number. But since \frac{q}{r} is rational and \frac{m}{n} is rational, we know \frac{q}{r} - \frac{m}{n} is rational because the difference of two rationals is rational.

Therefore \frac{m}{n} + \sqrt{2} is irrational.

QED

 

[wisvuze]

yep, that's right. generally, a rational plus an irrational is an irrational. Proven in the exact same way

 

[iamsmooth]

Thank you very much :D