# Does this proof for irrationality of sqrt(2)+sqrt(3) work?

1. Oct 1, 2008

### Simfish

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

Prove that $$\sqrt{2}+\sqrt{3}$$ is irrational.

2. Relevant equations

3. The attempt at a solution

So we know that $$(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2}) = 1$$. But a rational number must be of the form a/b, and if (a/b)c = 1, the only number c that works (for rational numbers) is c = b/a in reduced form due to unique inverses for rational numbers. But here we have a value of c that is NOT of the form c = b/a. And so once we prove that $$\sqrt{2}+\sqrt{3}$$ is NOT $$\frac{1}{\sqrt{3}-\sqrt{2}}$$, we can only conclude that $$(\sqrt{2}+\sqrt{3})$$ is irrational.

2. Oct 1, 2008

No, this won't do it, because by claiming that $$c = \sqrt 3 - \sqrt 2$$ is not a rational number you are making an assumption of precisely the type of thing you are trying to prove.

As an alternate start, try this. Assume, to the contrary, that $$\sqrt 2 + \sqrt 3$$ is rational. Then there are integers $$, a, b$$ such that

$$\sqrt 3 + \sqrt 2 = \frac a b$$

Square each side

$$3 + 2\sqrt 6 + 2 = \frac{a^2}{b^2}$$

Rewrite this to isolate $$\sqrt 6$$ - you will see that you assumption says something about the type of number $$\sqrt 6$$ is. Go from there.

3. Oct 1, 2008

### morphism

Actually I don't even understand what the OP is asking. It seems as though he/she is trying to prove this:
which it is.

4. Oct 1, 2008

### Simfish

Okay thanks to you both! Sorry I was wrong - yeah - my argument was circular (and was contingent on a wrong assumption too - one that I forgot to check). So i'll go along with your suggestion.

5. Oct 1, 2008

### HallsofIvy

Staff Emeritus
Do you already know that $1/\sqrt{2}$ and $1/\sqrt{3}$ are irrational- or, equivalently, that $\sqrt{2}$ and $\sqrt{3}$ are irrational are you allowed to use those facts? If so, that simplifies the problem enormously.

6. Oct 1, 2008

### Simfish

Yeah, I already know from the book (apostol math analysis) that $$\sqrt{2}$$ is irrational. But I'm not sure whether or not I know that $$\sqrt{6}$$ is irrational or not - but this can be easily proven using the same techniques as those involved in $$\sqrt{2}$$. I got it now - thanks :)