Proof that a set is a subring of the real numbers ring 
Is the set $\mathit{A}=\{x-y\sqrt[3]{2}\mid x, y\in \mathbb{Q}\}$ a subring of the ring $\mathbb{R}$?
Hi SE community, I have tried to proof that $0\in\mathit{A}$ so, $0 = x-y\sqrt[3]{2}\;$. Obviously I got a contradiction since $\sqrt[3]{2}\;$ is irrational and $x$ and $m$ rationals. I concluded that $A$ is not a subring since $0$ is not an element of it.
I am sure that $A$ is not a subring of $\mathbb{R}$ because I have checked that it is not closed under multiplication. But then I found that since  $(3)\;a+(-a) =0 \Rightarrow(x_1+x_1)+(y_1+y_1)\sqrt[3]{2}\;$, and $0 \in \mathbb{Q}, \; 0 \in A$. Then $A$ has to be a subring of $\mathbb{R}$. This last statement demolished my first proof.
I think that there is problem with statement $(3)$, I still thinking $A$ is not a subring. Please let me know why is wrong the last statement.
 A: Not a subring as you describe it, as you do not include the square of $\sqrt[3] 2.$ The word for what is missing is closure: your set is not closed under multiplication.
If you include all
$$ x + y \sqrt[3] 2 + z \sqrt[3] 4 $$
with rational $x,y,z$ you get a ring, with norm
$$ x^3 + 2y^2 + 4 z^3 -6xyz  $$
As any nonzero element has a multiplicative inverse, it is a field.
Note:
$$ \left( x + y \sqrt[3] 2 + z \sqrt[3] 4 \right) \left( \; (x^2 - 2yz) + (2z^2 - xy) \sqrt[3] 2 + (y^2-zx) \sqrt[3] 4 \right) =  x^3 + 2y^2 + 4 z^3 -6xyz$$
so
$$ \left( x + y \sqrt[3] 2 + z \sqrt[3] 4 \right) \frac{ \left( \; (x^2 - 2yz) + (2z^2 - xy) \sqrt[3] 2 + (y^2-zx) \sqrt[3] 4 \right)}{   x^3 + 2y^2 + 4 z^3 -6xyz} = 1$$
A: First note that $0 \in \mathbb{Q}$, so taking $x=y=0$ we have that $0-0\cdot \sqrt[3]{2} = 0 \in A$. I think your confusion comes from the fact that $\sqrt[3]{2}$ is irrational, as you noted, so any nonzero rational multiple of it is also irrational and any sum of it with a rational will be irrational as well. The thing to note is that this is only true for nonzero rational multiples, so we have no contradiction with writing a rational number $0$ as the difference between a rational and a $y\sqrt[3]{2}$ because in this case $y=0$.
