The given function is $F(x,y) = x/y$ for all $x$ and $y$ in the set $A$.
where $A = \{a+b \sqrt[]{2} : a,b \in \mathbb{Q}\}$ and $a$ and $b$ cant be both $0.$
So what I need is to prove $F(x,y) \in A$ for every $x,y \in A$
Here is what I tried:
Let $x = \frac{p}{q} +\frac{m}{n} \sqrt[]2$ and $y =\frac{u}{v} +\frac{k}{j} \sqrt[]2$
Then $$F(x,y) = \frac{\frac{p}{q} +\frac{m}{n} \sqrt[]2}{\frac{u}{v} +\frac{k}{j} \sqrt[]2}$$ $$F(x,y) = \frac{\frac{pn+mq \sqrt[]{2}}{qn}}{\frac{uk+kv \sqrt[]{2}}{vj}}$$ $$F(x,y) = \frac{pn+mq \sqrt[]{2}}{qn}.\frac{vj}{uk+kv \sqrt[]{2}}$$ $$F(x,y) = \frac{vjpn+vjmq \sqrt[]{2}}{qnuj+qnkv \sqrt[]{2}}$$
I got stuck here, It dont seems that the result of this function will always produce number in the format of the elements from $A$, how can I show this if it is the case?
For $y=0=0+0\sqrt{2}\in A$, the function $F$ is not well defined, but I assume you just missed that.
In general, if you have a fraction $\frac{a+b\sqrt{2}}{c+d\sqrt{2}}$ you can multiply above and below y $c-d\sqrt{2}$, which gives you
$$\frac{(ac-2bd)+(-ad+bc)\sqrt{2}}{c^2+2d^2}$$
which is clearly in $A$.