Prove that for all real numbers x and y there is a real number $z$ such that $x + z = y - z$ This one must be pretty basic, but...


  Prove that for all real numbers $x$ and $y$ there is a real number $z$
  such that $x + z = y - z$


I am quite confused about how you need to prove this.

My attempt was:
$$\tag1 x + z = y - z$$
$$\tag2 2z = y -x $$
$$\tag3 z = \frac{y-x}{2}$$
Denominator doesn't equal to zero, hence $z$ is defined for all $x$ and $y$.

Is there anything else I need to show?
If this question is from an analysis book, you might have to explain it in addition to solving it. Strictly speaking, it is a mathematical statement. The first part of the statement, "for all $x$ and $y$," implies that values of $x$ and $y$ have already been stipulated. You have no control over them. The second part of the statement, "there is a real number $z$," implies that you only have to find one real number $z$ that satisfies $x + z = y - z$. Solving this equation for $z$ shows that you can always find this value of $z$.
If this question is from an analysis book, you might have to explain it in addition to solving it. Strictly speaking, it is a mathematical statement. The first part of the statement, "for all $x$ and $y$," implies that values of $x$ and $y$ have already been stipulated. You have no control over them. The second part of the statement, "there is a real number $z$," implies that you only have to find one real number $z$ that satisfies $x + z = y - z$. Solving this equation for $z$ shows that you can always find this value of $z$.