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} $)

Answer 1

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.

Answer 2

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

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

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

Answer 3

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.

Answer 4

Suppose $\sqrt{a},\sqrt{b}\in\mathbb{R}-\mathbb{Q}$ for $a,b\in\mathbb{N}$. Now suppose $\sqrt{a}+\sqrt{b}=p/q$ for some $p,q\in\mathbb{Z}$ with $\gcd(p,q)=1$ and $q>0$. Then $$q(\sqrt{a}+\sqrt{b})=p\implies q^2(a+b+2\sqrt{ab})=p^2$$ $$\implies 4q^4ab=(p^2-q^2(a+b))^2$$

Hence we have $q|p^4$. But $\gcd(p,q)=1\implies\gcd(p^4,q)=1$. Now since $q$ is a divisor of $p^4$, $q\leq 1$, therefore $q=1$. Now we have $$\sqrt{a}+\sqrt{b}=p\implies a=(p-\sqrt{b})^2$$ $$2p\sqrt{b}=p^2+b-a$$ Now this implies that $\sqrt{b}$ is rational if $p\neq0$, hence $p=0$ and $0=\sqrt{a}+\sqrt{b}>0$, thus this is a contradiction. $$\square$$