Subsets, Closure, Proof $X$ is a subset of $\mathbb{R}$. It contains $\sqrt{3}$ and $\sqrt{2}$ and is closed under addition, subtraction and multiplication.


*

*Prove that $X$ contains $\sqrt{8}$.

*Prove that $X$ contains $1$.

*Prove that $X$ contains $\frac{1}{\sqrt{2}+1}$.

*Is it true that $X$ is necessarily a field?


So far I have only been successful with 1 and 2. I've just used the elements that are already in the group and performed operations on the them to try and get $\sqrt{8}$ and $1$. 
 A: You had already solved parts 1 and 2.
For 3, note that $$\frac1{\sqrt 2 +1}=\frac{\sqrt 2 -1}{(\sqrt 2 +1)(\sqrt 2 -1)}=\sqrt 2 -1$$
and so is $\in X$ as difference of $\sqrt 2$ and $1$ (which you know is $\in X$ by part 2).
Nevertheless, we can only say that $X$ is a ring (with $1$ per part 2).
(Edited after hint from Carsten S)
We easily see that the following sets have the described property for such a set $X$:
$$ X_1=\{\,a+b\sqrt 2+c\sqrt 3+d\sqrt 6\mid a,b,c,d\in\Bbb Z\,\}$$
and
$$ X_2=\{\,a+b\sqrt 2+c\sqrt 3+d\sqrt 6\mid a,b,c,d\in\Bbb Q\,\}.$$
While $X_2$ is a field, $X_1$ is not. To see this, note that even in $X_2$, i.e., with rational coefficients, the coefficients $a,b,c,d$ for $\alpha\in X_2$ are uniquely determined (we express this fact as "$1,\sqrt 2,\sqrt 3,\sqrt 6$ are linearly independent over $\Bbb Q$"). It follows that the multiplicative invers of $\sqrt 2$, which is $0+\frac12\sqrt 2+0\sqrt 3+0\sqrt 6\in X_2$, is not in $X_1$. So $X_1$ is not a field.
A: A hint for (3) has already been given in the comments: Make the denominator rational. (If that is how this is usually expressed in English.)
For (4) you could show that the set of all $a+b\sqrt2+c\sqrt3+d\sqrt6$ with $a, b, c, d$ integers has the given property and that it is not a field.
