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

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)

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