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

Question

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?

It's just fine. What made you think that you needed to do more? — José Carlos Santos, Jul 30 '19 at 18:33
@JoséCarlosSantos The textbook I'm currently studying normally requires me to prove a little bit more complicated conjectures. So I'd thought this one was not an exception. — Stokolos Ilya, Jul 30 '19 at 18:36
You found the answer. Now you need to show it satisfies the equation. So basically you do your steps backward. And, for rigorous proof, you actually need only this second half. — GEdgar, Jul 30 '19 at 21:31
You proved that the only possible answer is $z=(y-x)/2$, but you did not prove that $(y-x)/2$ is indeed a number that satisfies the required condition. For example, can you solve the equation $\frac{1}{x}=\frac{1}{x^2+x}$? — Luiz Cordeiro, Jul 30 '19 at 21:57

score -1 · Answer 1 · answered Jul 30 '19 at 21:28

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 $z$.

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

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