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Below is a theorem I was asked to prove from Velleman's book How to Prove It, along with my proof. My question is regarding the last step of the uniqueness portion of the proof where I am required to divide by $(y + 1)$. Since $y$ can be any number, how do I handle the case where $y = -1$? Is this proof still valid in its current form? If not, how would I fix it?

Theorem. For every real number $y$, there is a unique real number $x$ such that $xy + x - 4 = 4y$.

Proof. Let $y$ be an arbitrary real number, and let $x = 4$. Then
$$xy + x - 4 = 4y + 4 - 4 = 4y.$$

To see that $x$ is unique, let $z$ be an arbitrary real number and suppose $zy + z - 4 = 4y$. Then $\ zy + z = 4y + 4$, so $z(y + 1) = 4y + 4$. If $y + 1 = 0$, then $z$ is undefined, but otherwise $z = \ \frac{4y + 4}{y + 1} = 4 = x$.

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As you saw, the case $y=-1$ is special. If $y=-1$ then every $x$ works (just plug into the equation, the left side and right side are both equal to $-4$ for any $x$), so we do not have uniqueness. The "theorem" is false.

André Nicolas
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  • Ah, right! I should have gone back to the original equation and plugged in $y = -1$. Then the "theorem" would have obviously been false. Thank you for your help. –  Jul 15 '16 at 12:11
  • @jmh: You are welcome. When you reached the stage $z(y+1)=4y+4$, then things naturally break into two cases. Note the minor typo "$zy+z=4y-4$" in the post. – André Nicolas Jul 15 '16 at 12:53
  • Fixed the typo. Thanks again. –  Jul 15 '16 at 14:00