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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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Being, or not being, in the rationals is a computational concept, so the search for a non-computational method is doomed. –  Gerry Myerson Apr 25 '12 at 3:45
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If the term "logical argument" is meant as "a formal proof in the language of fields" then I apologize for the retag. If it meant something else, then I am not sorry - it has nothing to do with [logic]. –  Asaf Karagila Apr 25 '12 at 8:53
    
@Asaf : You are completly correct, what I meant was something similar to simple logical reasoning like "sum of rational and irrational is irrational" type of statement. Thank you for your courtesy. –  Arjang Apr 25 '12 at 9:20
    
@GerryMyerson : I consider Bill's answer to be sufficiently noncomputational, maybe my usage of "noncomputational" is not correct in this context. –  Arjang Apr 25 '12 at 11:31
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3 Answers 3

up vote 18 down vote accepted

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 $\rm = \sqrt{a}\in \mathbb Q$

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It's more a complete solution than a hint :) –  Najib Idrissi Apr 25 '12 at 10:04
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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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Sorry, I didn't read the part about not being computational. Also, this just boils down to Bill Dubuque's proof. –  Alexander L Apr 25 '12 at 4:03
    
Me neither! Let's hope you avoid the serial downvoters :) –  The Chaz 2.0 Apr 25 '12 at 4:04
    
@TheChaz The downvote is a bit puzzling. I'd ask for an explanation. –  Bill Dubuque Apr 25 '12 at 4:24
    
@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... –  The Chaz 2.0 Apr 25 '12 at 4:28
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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...) –  The Chaz 2.0 Apr 25 '12 at 3:40
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