Help with proof for rational number problem

[iamsmooth]

Homework Statement

For all a in the set of real numbers, if a is rational, [tex]a + \sqrt{2}[/tex] is irrational.
You may use that [itex]\sqrt{2}[/itex] 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 [itex]a + \sqrt{2}[/itex] is rational. Therefore [tex]\frac{m}{n} + \sqrt{2} = \frac{q}{r}[/tex] where q and r are integers (r is not 0).

Therefore,
[tex] \sqrt{2} = \frac{q}{r} - \frac{m}{n}[/tex]

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

Therefore [tex]\frac{m}{n} + \sqrt{2}[/tex] 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