How to prove this theorem? If I use contrapositive, the theorem becomes "If at least one of the numbers $x+\sqrt{2}$ and $x^2 - 2$ is rational, $x$ is rational". I have no idea how to prove the two numbers are rational.
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1$x^2-2=(x+\sqrt2)(x-\sqrt2)$. – David Mitra Oct 02 '16 at 10:26
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If both $x+\sqrt{2}$ and $x^2-2$ are rational, then the quotient $$\frac{x^2-2}{x+\sqrt{2}}=x-\sqrt{2}$$ is rational, and hence $$x=\frac{1}{2}((x+\sqrt{2})+(x-\sqrt{2}))$$ is rational. Contradiction.
Damian Reding
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Well, $x+\sqrt{2}$ and $x-\sqrt{2}$ differ by $2\sqrt{2}$ so at most one of them can be rational. If only one of them is rational, their product $x^2-2$ is irrational. If $x^2-2$ is rational, both $x-\sqrt{2}$ and $x+\sqrt{2}$ must be irrational since they cannot both be rational.
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"If only one of them is rational, their product $x^2-2$ is irrational." Not true: try $x=\sqrt2$. – Did Oct 02 '16 at 15:47