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

Question

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} $)

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 03:45
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 08:53
@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. — jimjim, Apr 25 '12 at 09:20
@GerryMyerson : I consider Bill's answer to be sufficiently noncomputational, maybe my usage of "noncomputational" is not correct in this context. — jimjim, Apr 25 '12 at 11:31
See also https://math.stackexchange.com/questions/530778 and the generalization in the accepted answer to https://math.stackexchange.com/questions/951609 — Watson, Dec 25 '16 at 14:20
Possible duplicate of Can $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m$ are perfect squares? — Watson, Nov 25 '18 at 17:43

Bill Dubuque · Accepted Answer · 2024-08-25T21:01:56.287

25

Hint $\rm\ \ \ \color{#c00}{\underbrace{\sqrt{a}-\!\sqrt{b}}_{\rm \textstyle w}} \,=\, \dfrac{\,a\,-\,b}{\underbrace{\color{#0a0}{\sqrt{a}+\!\sqrt{b}}}_{\textstyle\color{#0a0}{\bar w\in \Bbb Q}}}\in\Bbb Q\ $ $\Rightarrow\ {\rm \dfrac{\color{#c00}w\!+\!\color{#0a0}{\bar w}}2} $ $\rm = \sqrt{a}\,\in \Bbb Q$

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

Generally if a field $\rm F $ has two $\rm F$-linear independent combinations of $\rm\ \sqrt{a},\ \sqrt{b}\ $ then we can solve for $\rm\ \sqrt{a},\ \sqrt{b}\ $ in $\rm F\,$ (e.g. in the Primitive Element Theorem).

edited Aug 25 '24 at 21:01

answered Apr 25 '12 at 03:30

Bill Dubuque

282,220

2

It's more a complete solution than a hint :) – Najib Idrissi Apr 25 '12 at 10:04
The case $a=b=0$ has to be done separately (otherwise we cannot write down the fraction here). – Martin Brandenburg Jun 05 '21 at 16:29
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@Martin Yes, of course (generally hints leave trivial degenerate cases for the reader). – Bill Dubuque Jun 05 '21 at 19:33

score 6 · Answer 2 · answered Apr 25 '12 at 03:24

6

$$\sqrt a + \sqrt b \in \mathbb{Q} \Rightarrow \sqrt a + \sqrt b = \dfrac{p}{q}$$

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

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

answered Apr 25 '12 at 03:24

The Chaz 2.0

10,606

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 03:40
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You omitted a crucial step: solving your last equation for $,\sqrt b,$ (to infer $\sqrt b\in \Bbb Q),$ works only because its coef $,2p/q\neq 0,$ (by $,\color{#c00}{p/q = \sqrt a + \sqrt b > 0}).,$ Lacking that, the above method would also wrongly deduce that $,\sqrt a - \sqrt b \not\in \Bbb Q,,$ which is false if $,a = b,$ (so $,p/q = 0).,$ The crucial role played by such $\rm\color{#c00}{positivity}$ is highlighted in the ordered field generalization linked in my answer. $\ \ $ – Bill Dubuque Aug 25 '24 at 22:14

score 1 · Answer 3 · answered Apr 25 '12 at 03:41

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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.

answered Apr 25 '12 at 03:41

Alexander L

380

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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 04:03
Me neither! Let's hope you avoid the serial downvoters :) – The Chaz 2.0 Apr 25 '12 at 04:04
@TheChaz The downvote is a bit puzzling. I'd ask for an explanation. – Bill Dubuque Apr 25 '12 at 04:24
@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 04:28

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

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