Rational Roots

Age 16 to 18

Challenge Level Yellow star

David sent in this solution, using the hints we gave you.

$\sqrt{a}+\sqrt{b}$ rational

$\Rightarrow (\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{a b}$ rational

$\Rightarrow 2\sqrt{a b}$ rational

$\Rightarrow \sqrt{a b}$ rational

$\Rightarrow a+\sqrt{a b}$ rational
i.e.,

$\sqrt{a}(\sqrt{a}+\sqrt{b})$ rational

$\Rightarrow \sqrt{a}$ rational (and so

$a$ is a square)

$\Rightarrow \sqrt{b}$ is also rational and hence

$b$ is a square.

For the second part:

$\sqrt{a}+\sqrt{b}+\sqrt{c}$ rational

$\Rightarrow (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$ rational

$\Rightarrow \sqrt{a b}+\sqrt{b c}+\sqrt{c a}$ rational

$(\sqrt{a b}+\sqrt{b c}+\sqrt{c a})^2=a b+b c+c a+2\sqrt{a b c}(\sqrt{a}+\sqrt{b}+ \sqrt{c})$

$\sqrt{a b c}$ is also rational

$\Rightarrow \sqrt{a}(\sqrt{a b}+\sqrt{b c}+\sqrt{c a})-\sqrt{a b c}=a(\sqrt{b} +\sqrt{c})$ is rational

$\sqrt{b}+\sqrt{c}$ is rational and

$\sqrt{a}$ is rational, and use the above.

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