Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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} $)

share|cite|improve this question
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
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
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 $\rm = \sqrt{a}\in \mathbb Q$

share|cite|improve this answer
It's more a complete solution than a hint :) – Najib Idrissi Apr 25 '12 at 10:04

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

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

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

share|cite|improve this answer
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.