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

[Simfish]

Homework Statement

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

Homework Equations

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.

 

[statdad]

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

<br /> \sqrt 3 + \sqrt 2 = \frac a b<br />

Square each side

<br /> 3 + 2\sqrt 6 + 2 = \frac{a^2}{b^2}<br />

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.

 

[morphism]

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

\sqrt{2}+\sqrt{3} is NOT \frac{1}{\sqrt{3}-\sqrt{2}}

which it is.

 

[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.

 

[HallsofIvy]

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.

 

[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 :)