Prove $x=\sqrt{3} - \sqrt{2}$ is irrational

Question

The question is: prove $x=\sqrt{3} - \sqrt{2}$ is not rational.

I can "prove" the above (ie. I saw the answer in my book) but can't quite understand it. $x = \sqrt{3} - \sqrt{2}$, $x+\sqrt{2}=\sqrt{3}$, $(x + \sqrt{2})^2 = 3$, $x^2+2\sqrt{2}x+2 = 3$, $\sqrt{2} = \frac{3-x^2-2}{2x}$.

This is a contradiction as $\sqrt{2}$ is not rational. Ok, I understand it's a contradiction, but it contradicts what?

I mean this proof didn't start with "let's assume [...]" so I don't know is the assumptions that is being contradicted. This is probably a very basic question, but can someone please explain where in the proof did they assume $\sqrt{3} - \sqrt{2}$ is rational?

Thanks!

Answer 1

The assumption is right at the start: assume that $x=\sqrt{3}-\sqrt{2}$ is rational. Then follow through the algebraic manipulations that you've listed until you reach $$ \sqrt{2} = \frac{3-x^2-2}{2x}$$

Since $x$ is assumed to be rational, both the numerator and denominator of this fraction must be rational, which means that $\sqrt{2}$ is rational. And that's the contradiction.

(This assumes, of course, that you've seen or worked out the standard proof that $\sqrt{2}$ is irrational)

Answer 2

Also, after writing $$2\sqrt2x=1-x^2$$ we need to say that $x\neq0$(otherwise, $0=1$, which is impossible).

Thus, indeed, $$\sqrt{2}=\frac{1-x^2}{2x},$$ which is a contradiction because we assumed before that $x\in\mathbb Q$.