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I am completely lost on where to go with this one. Can someone provide help with how to do this? I am trying to understand uniqueness.

Here it is:

Prove that for every real number x, if x DOES NOT equal 0 and x DOES NOT equal 1 
then there exists a unique real number such that y/x = y - x
Gerry Myerson
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user131091
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1 Answers1

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cross multiplying $$ y = x y - x^2 \Rightarrow y = \frac{x^2}{x-1}$$ You now see why $x=0$ and $x=1$ are ruled out.

You can finish the rest

user44197
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    It's important that your $\Rightarrow$s are really $\Leftrightarrow$s, including the first step where you multiplied through by $x$. ($\Rightarrow$ gives uniqueness, $\Leftarrow$ gives existence.) – Clive Newstead Feb 24 '14 at 04:41
  • I'm confused though. If we plugged in 0 wouldn't that evaluate to 0^2 / -1) which would then go to 0? thus being a real number – user131091 Feb 24 '14 at 04:43
  • In the original equation, $\frac{y}{x}$ would be undefined if $x=0$. – JRN Feb 24 '14 at 04:44
  • Well you can cross multiply only if $x \ne 0$. Or else you end up with $\frac 0 0$. Who knows what that is! – user44197 Feb 24 '14 at 04:45
  • @CliveNewstead I agree... I wanted to give a broad hint without actually giving the full solution. To go both ways you have to make sure you do not cross multiply by zero and that is why I left the final comment. – user44197 Feb 24 '14 at 04:46