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!
Best Answer
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)