This is irrational!

Suppose both $x + y$ and $x - y$ are rational. Then by closure $(x + y) + (x - y) = 2x$ must also be rational. But this contradicts the given condition that $x$ (and hence $2x$ ) is irrational. So at most one of $x + y$ or $x - y$ is rational. This in turn implies that if $x + y$ is rational then $x - y$ is irrational, and vice versa.

Now if both $x + y$ and $x - y$ are irrational then we will require that both $x^{2} - y^{2}$ and $x^{2} + y^{2}$ be rational for the statement in question to be considered true. But if $x^{2} - y^{2}$ is rational, then $x^{2} + y^{2} = (x^{2} - y^{2}) + 2y^{2}$ will be irrational, since $y^{2}$ is irrational, (and thus so is $2y^{2}$ ), and the sum of a rational and irrational number is always irrational. Similarly if $x^{2} + y^{2}$ is rational. So in this case at most one of the given four terms can be rational.

Now suppose $x - y$ is rational. (Since $|x| \ne |y|$ we have that $x - y$ is non-zero). Then from before we already know that $x + y$ is irrational, and since the product of a non-zero rational and an irrational is always irrational we also have that $x^{2} - y^{2} = (x - y)(x + y)$ is irrational. Next, note that $x^{2} + y^{2} = (x - y)^{2} + 2xy$ . Now with $x - y$ being rational we have that $(x - y)^{2}$ is rational as well. Then since $xy$ (and thus also $2xy$ ) is given to be irrational, $x^{2} + y^{2}$ is irrational as well, (again since the sum of a rational and irrational is always irrational). A similar conclusion can be made if we had first assumed $x + y$ to be rational. So in this case at most one of the given four terms can be rational.

As we have then covered all possible cases, we conclude that the statement in question is not true.

given $x,y,x^2,y^2,xy \notin Q$

1. suppose $(x+y)$ and $(x-y) \in Q \rightarrow (x+y)+(x-y)=2x \in Q \rightarrow x \in Q \rightarrow$ contradiction
2. suppose $(x+y)$ and $(x^2-y^2)=(x+y)(x-y) \in Q \rightarrow (x+y)(x-y)/(x+y)=(x-y)\in Q \rightarrow$ contradiction by (1)
3. suppose $(x+y)$ and $(x^2+y^2) \in Q \rightarrow (x+y)^2=(x^2+2xy+y^2) \in Q \rightarrow (x^2+2x+y^2)-(x^2+y^2)=2xy \in Q \rightarrow$ contradiction
4. suppose $(x-y)$ and $(x^2-y^2)=(x+y)(x-y) \in Q \rightarrow (x+y)(x-y)/(x-y)=(x+y)\in Q \rightarrow$ contradiction by (1)
5. suppose $(x-y)$ and $(x^2+y^2) \in Q \rightarrow (x-y)^2=(x^2-2xy+y^2) \in Q \rightarrow (x^2-2xy+y^2)-(x^2+y^2)=-2xy \in Q \rightarrow$ contradiction
6. suppose $(x^2-y^2)$ and $(x^2+y^2) \in Q \rightarrow (x^2+y^2)+(x^2-y^2)=2x^2 \in Q \rightarrow$ contradiction