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So for any $x \in \mathbb{R}$ I have to prove that a unique $y \in \mathbb{R}$ exists such that $x^2 y = x - y$.

What I did was this:

I showed that such a $y$ exists by performing algebra to get $y = \frac{x}{x^2+1}$. Then by noting that $x^2 + 1 \ne 0$, $y \in \mathbb{R}$ since real numbers are closed under non-zero division. Afterwards I assumed another value $z$ that has the property $x^2z = x - z$. Performing the same process I get $z = \frac{x}{x^2+1}$ which is the same value as $y$. Therefore $y=z$ and that the value of $y$ is unique.

Is this a sound proof or not? Anything wrong with it?

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    I think that is fine! – 2017-02-27
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    Do I have to add that $y \ne z$ when I assumed $z$? – 2017-02-27
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    There is no need for that statement. It is sufficient to prove that for any other $z$ satisfying $x^2z = x-z$, $y=z$. If you start with $y\neq z$, then the above proves that $y=z$, which then you would address as a contradiction, only a slight change in the method of solving. You have caught the problem by the throat, and probably strangled it completely. There's nothing more to add in the answer. It's excellent. – 2017-02-27
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    I agree with the above comments. This looks solid! – 2017-02-27
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    Okay thanks! @астонвіллаолофмэллбэрг if you would be so kind tho to add an answer so that this question wouldn't be unanswered. – 2017-02-27

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