I need to prove this equivalence
Let $a,b \in \mathbb{Q^{+}}$ then $\sqrt{a} + \sqrt{b} \in \mathbb{Q} $ if and only if $\sqrt{a} \in \mathbb{Q}$ and $\sqrt{b} \in \mathbb{Q}$
I have already proved the reciprocal (the obvious part) but I am having difficulties with the other part
Please any hint will be appreciated.
Thanks in advance.
Let $x = \sqrt{a} + \sqrt{b}$ and suppose $x \in \Bbb Q$. If $a = b$, then $x = 2\sqrt{a}$, so $\sqrt{b} = \sqrt{a} = \frac{x}{2} \in \Bbb Q$. Now suppose $a \neq b$. Then $\frac{1}{x} = \frac{\sqrt{a} - \sqrt{b}}{a - b} \in \Bbb Q$. So $y:= \sqrt{a} - \sqrt{b} = \frac{a - b}{x}\in \Bbb Q$. Now prove the result by considering $x + y$ and $x - y$.
If $\sqrt{a}+\sqrt{b}\in\Bbb Q$ then $\Bbb Q\ni \frac{a-b}{\sqrt{a}+\sqrt{b}}=\sqrt{a}-\sqrt{b}$, so $(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})$ and $(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})$ are rational.
If $\sqrt{a}+\frac{c}{d}=\sqrt{b}$ with $\frac{c}{d}\neq 0$ then after squaring $$ a+2\sqrt{a}\frac{c}{d}+\frac{c^2}{d^2}=b, $$ and hence $\sqrt{a}\in \mathbb{Q}$. The same argument works for $\sqrt{b}$.
Look at powers of $\sqrt{a}+\sqrt{b}$ and then solve a system of linear equations.
Suppose that $\sqrt{a}+\sqrt{b}\in\mathbb{Q}$. Let $x=\sqrt{a}+\sqrt{b}$. Then,
$$x^3=a\sqrt{a}+3a\sqrt{b}+3b\sqrt{a}+b\sqrt{b}.$$
In other words,
$$x^3=(a+3b)\sqrt{a}+(3a+b)\sqrt{b}.$$
Note that $a+3b$ and $3a+b$ are never zero. Also, they are equal only when $a=b$. If $a=b$, then $2\sqrt{a}\in\mathbb{Q}$, so $\sqrt{a}\in\mathbb{Q}$. If $a\not=b$, then you can write $\sqrt{a}$ in terms of $x$ and $x^3$ (with some coefficients).
Key Idea $ $ If field F has $2\,$ F-linear independent combinations of $\rm\, \sqrt{a},\ \sqrt{b}\, $ then we can solve for $\rm\, \sqrt{a},\ \sqrt{b}\, $ in F. For example, the Primitive Element Theorem works that way, obtaining two such independent combinations by Pigeonholing the infinite set $\rm\ F(\sqrt{a} + r\ \sqrt{b}),\ r \in F,\ |F| = \infty,\,$ into the finitely many fields between F and $\rm\ F(\sqrt{a}, \sqrt{b}),\,$ e.g. see this proof.
In this case it is simplest to notice $\rm\ F = \mathbb Q(\sqrt{a} + \sqrt{b})\ $ contains the independent $\rm\ \sqrt{a} - \sqrt{b}\ $ since
$$\rm \sqrt{a}\ -\ \sqrt{b}\ =\ \dfrac{\ a\,-\,b}{\sqrt{a}+\sqrt{b}}\ \in\ F = \mathbb Q(\sqrt{a}+\sqrt{b}) $$
To be explicit, notice that $\rm\, u = \sqrt{a}+\sqrt{b},\ v = \sqrt{a}-\sqrt{b}\in F\, $ so solving the linear system for the roots yields $\rm\, \sqrt{a}\ =\ (u+v)/2,\ \ \sqrt{b}\ =\ (u-v)/2,\, $ both of which are clearly $\rm\,\in F,\,$ since $\rm\,u,v\in F\,$ and $\rm\,2\ne 0\,$ in $\rm\:F,\:$ so $\rm\,1/2\:\in F.\,$ This works over any field where $\rm\,2\ne 0\,\,$ i.e. where the determinant (here $\,2)\,$ of the linear system is invertible, i.e. where the linear combinations $\rm\:u,v\:$ of the square-roots are linearly independent over the base field.
