Solution verification: Proving that if $a,b \in \Bbb Q$, then $\sqrt a+\sqrt b \in \mathbb Q\Leftrightarrow \sqrt a, \sqrt b \in \mathbb Q$ In an exercise I'm asked to prove the following:

Let $a,b \in \mathbb Q$. Prove that $\sqrt a+\sqrt b \in \mathbb Q$ if and only of $\sqrt a, \sqrt b \in \mathbb Q$.

This is how I approached the problem:

Proving that $\sqrt a, \sqrt b \in \mathbb Q \Rightarrow \sqrt a+\sqrt b \in \mathbb Q$ is trivial, because the sum of two rational numbers is rational.
Now, let's assume that $\sqrt a+\sqrt b \in \mathbb Q$. So we have:
$$\sqrt a+\sqrt b =\frac{\alpha}{\beta}$$
for some $\alpha,\beta \in \Bbb Z$.
If only one of the square roots is rational and the other irrational this is false, because the sum of a rational number with an irrational number is allways irrational, so we have two possible scenarios:
$$\sqrt a , \sqrt b \in \Bbb Q \vee \sqrt a , \sqrt b \in \Bbb R \setminus \Bbb Q$$
So we have:
$$\begin{align}
 & \sqrt a+\sqrt b =\frac{\alpha}{\beta} 
\\
 & \Leftrightarrow \left(\sqrt a+\sqrt b \right)^2 = \frac{\alpha^2}{\beta^2}
\\
 & \Leftrightarrow |a| + |b| + 2\sqrt a \sqrt b = \frac{\alpha^2}{\beta^2}
\\
& \Leftrightarrow  \sqrt a \sqrt b = \underbrace{\frac{1}{2}\left(\frac{\alpha^2}{\beta^2} -|a| - |b|\right)}_{\in \Bbb Q \text{, because } a,b \in \Bbb Q}
\end{align}$$
So we have that $\sqrt a+\sqrt b \in \mathbb Q$ and $\sqrt a \sqrt b \in \mathbb Q$. Working with the second expression we have, for some $p,q \in \Bbb Z$:
$$\begin{align}
\sqrt a \sqrt b = \frac{p}{q} \Leftrightarrow \sqrt b = \frac{p}{q} \frac{1}{\sqrt a }
\end{align}$$
If we use this expression for $\sqrt b$ in the first expression we have:
$$\begin{align}
 & \sqrt a+\frac{p}{q} \frac{1}{\sqrt a} =\frac{\alpha}{\beta} 
\\
 & \Leftrightarrow \sqrt a+\frac{p}{q} \frac{\sqrt a}{|a|} =\frac{\alpha}{\beta}
\\
& \Leftrightarrow \frac{(q |a| + p) \sqrt a}{ q |a|} =\frac{\alpha}{\beta}
\\
& \Leftrightarrow \sqrt a =\underbrace{\frac{\alpha}{\beta} \frac{q |a|}{q |a| + p}}_{\in \Bbb Q}
\end{align}$$
Because $\sqrt a \in \Bbb Q$ this means that $\sqrt b \in \Bbb b$ because either they are both irrational or rational as I concluded in the beginning.

My first question is: Is my proof correct? How can I improve it. This seems like a simple problem and I think that I over-complicated the solution so, let me know of other ways of solving this.
 A: Your proof looks basically correct, except you should account for the possibility $\sqrt{a} = 0$ before you divide by it, as well as other cases where you divide by various values. I suggest you initially handle the cases of $\sqrt{a}$ and/or $\sqrt{b}$ being $0$ to make the rest of the proof somewhat simpler.
As for any other ways to improve your proof, after your line $\sqrt a+\sqrt b =\frac{\alpha}{\beta}$, I believe it'll be somewhat simpler & easier if you first move $\sqrt{a}$ or $\sqrt{b}$ to the right side before squaring. That way, you'll get an expression dealing with just a rational multiple of $\sqrt{a}$ or $\sqrt{b}$ equaling a rational value. With non-$0$ values for the square roots, the multiplier of the square root will be non-$0$ as well, so you can show $\sqrt{a}$ or $\sqrt{b}$ is a rational, and then state you can do the same thing for the other square root.
A: One way to consider it is there are three cases

*

*both $\sqrt a,\sqrt b$ are rational and there for pretty trivially $\sqrt a+\sqrt b$ is rational.


*one of $\sqrt a, \sqrt b$ is rational and the other is not then $\sqrt a +\sqrt b$ is irrational as a rational plus an irrational is irrational. (if $r$ is rational and $y$ is irrational then $r+y$ being rational would imply $(r+y) +(-r) = y$ would be rational which is a contradiction.


*That leaves proving if $\sqrt a, \sqrt b$ are both irrational then $\sqrt a + \sqrt b$ is irrational.   This is the heart of the proof.
let $\sqrt a + \sqrt b = w$ and so $\sqrt a = w -\sqrt b$ so $a = (w-\sqrt b)= w^2 -2w\sqrt b + b$ so $\sqrt b = \frac {w^2+b -a}{2w}$ (assuming $w \ne 0$).  If $w$ is rational then we get a contradiction that $\sqrt a$ is rational.
So of the three cases:

*

*$\sqrt a, \sqrt b$ both rational $\implies \sqrt a + \sqrt b$ is rational.


*$\sqrt a, \sqrt b$ are rational and irrational $\implies  \sqrt a + \sqrt b$ is irrational.


*$\sqrt a, \sqrt b$ are  irrational $\implies  \sqrt a + \sqrt b$ is irrational.
we have $\sqrt a, \sqrt b \in \mathbb Q \iff \sqrt a+\sqrt b\in \mathbb Q$>
A: The proof looks good. Another approach that I saw recently on this site is to calculate
$$ {\sqrt{a}-\sqrt{b}}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$$
so
$$\sqrt{a}= \frac{1}{2}\left( \sqrt{a}+\sqrt{b}+ \frac{a-b}{\sqrt{a}+\sqrt{b}} \right)$$
Note that this statement can be vastly generalized: if $\sum_{i=1}^k \sqrt[n_i]{a_i}$  ($a_i$ positive rationals, $n_i$ natural numbers) is rational, then all the $\sqrt[n_i]{a_i}$ are again rational.
