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Inspired by this, I was wondering if there is a simple logical argument to

Show that $ a,b, \sqrt{a}+ \sqrt{b} \in\mathbb Q \implies \sqrt{a},\sqrt{b} \in\mathbb Q $

Note that the original link is using a computational method, where as I am looking for a simple logical argument.

I tried (unjutifiably) to argue that if some of two square roots is rational then each one is rational, this is a different than the (incorrect) argument that if sum of two algebraic numbers is rational then each one is rational ( counter example $a=1-\sqrt{2},b= 1+\sqrt{2} $)

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    Possible duplicate of [Can $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m$ are perfect squares?](https://math.stackexchange.com/questions/457382/can-sqrtn-sqrtm-be-rational-if-neither-n-m-are-perfect-squares)2018-11-25

3 Answers 3

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Hint $\rm\ \sqrt{a}-\sqrt{b}\: = \dfrac{a-b}{\sqrt{a}+\sqrt{b}}\ $ so $\rm\ \sqrt{a}+\sqrt{b}\in\mathbb Q\:\Rightarrow\:\sqrt{a}-\sqrt{b}\in\mathbb Q\:\Rightarrow\:$ sum/2 $\rm = \sqrt{a}\in \mathbb Q$

Remark $ $ This generalizes to a positive sum of any number of square roots over an ordered field.

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    It's more a complete solution than a hint :)2012-04-25
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$\sqrt a + \sqrt b \in \mathbb{Q} \Rightarrow \sqrt a + \sqrt b = \dfrac{p}{q}$

$\sqrt a = \dfrac{p}{q} - \sqrt b$ $a = \dfrac{p^2}{q^2} - 2 \cdot \dfrac{p}{q} \sqrt b + b$

So if $a,b$ are rational, this forces their square roots to be also.

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    Note: This is arguably "computational". I just now checked out the link in the OP (a rare occurrence...)2012-04-25
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Note that $a+b=(\sqrt{a}+\sqrt{b})^2-2\sqrt{ab}$ and since both a+b and $\sqrt{a}+\sqrt{b}$ are rational, we may claim that $\sqrt{ab}$ is also rational. These remind us of the quadratic: $(x-\sqrt{a})(x-\sqrt{b})=x^2-2x(\sqrt{a}+\sqrt{b})+\sqrt{ab}$ Solving for x gives us $x=\frac{\sqrt{a}+\sqrt{b}+-\sqrt{a+b-2\sqrt{ab}}}{2}$ which implies that if $\sqrt{a}-\sqrt{b}$ is rational, then so are $\sqrt{a}$ and $\sqrt{b}$. Since $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$ and both $a-b$ and $\sqrt{a}+\sqrt{b}$ are rational, we may conclude that both $\sqrt{a}$ and $\sqrt{b}$ are rational.

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    @Bill: I know you would! Honestly, I'll probably just chalk it up to the same serial downvoter(s) that have been targeting me of late. It's not a big deal, and I'm kinda surprised that it doesn't happen *more often*, considering my abrasive tendencies...2012-04-25