Algebra rationality problem

Algebra Level 3

Let $a$ and $b$ be rational where a≠b. If either $\sqrt a + \sqrt b$ or $\sqrt a - \sqrt b$ is rational or both are, what can be said about the numbers $\sqrt a$ and $\sqrt b$ ?

a^{1/2}

is irrational

b^{1/2}

is rational They are both irrational They are both rational

a^{1/2}

is rational

b^{1/2}

is irrational Inconclusive

1 solution

Nick Kent
Aug 9, 2019

Since $a-b=\left( \sqrt { a } +\sqrt { b } \right) \left( \sqrt { a } -\sqrt { b } \right)$ , if either $\sqrt { a } +\sqrt { b }$ or $\sqrt { a } -\sqrt { b }$ is rational, then the other is too (because $a-b$ is rational). That means their sum and difference are also rational: $\left( \sqrt { a } +\sqrt { b } \right) +\left( \sqrt { a } -\sqrt { b } \right) =2\sqrt { a }$ , $\left( \sqrt { a } +\sqrt { b } \right) -\left( \sqrt { a } -\sqrt { b } \right) =2\sqrt { b }$ . Therefore, both $\sqrt { a }$ and $\sqrt { b }$ are rational

The answer should be "inconclusive". If $a \neq b$ then your argument works fine. But the wording of the problem allows $a=b$ , which trivially allows the square roots to be irrational.

Chris Lewis - 1 year, 10 months ago

Algebra rationality problem

1 solution

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