# Help with proof for rational number problem

1. Apr 25, 2010

### iamsmooth

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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

2. Apr 25, 2010

### wisvuze

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

3. Apr 25, 2010

### iamsmooth

Thank you very much :D