Not a duplicate of

Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$

Prove that for any real numbers $x$ and $y$ if $x \neq 0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$.

This is exercise $3.2.10$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

Suppose that $x$ and $y$ are real numbers. Prove that if $x\neq0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$.

Here is my proof:

Proof. We will prove the contrapositive. Suppose $y=\frac{3x^2+2y}{x^2+2}$ and $y\neq3$. Suppose $x=0$. Then substituting $x=0$ into $y=\frac{3x^2+2y}{x^2+2}$ we obtain $y-y=0$ which means that $y$ can be any number and in particular $y=3$ which contradicts the assumption that $y\neq 3$. Thus $x\neq 0$. Therefore if $x\neq0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$. $Q.E.D.$

Is my proof valid$?$


I was reviewing the material today and I noticed a fatal error in the above proof. I am not allowed to assume $y\neq3$ and conclude $y=3$. So the above proof is certainly not valid.

Proof. Suppose $x\neq0$. Suppose $y=\frac{3x^2+2y}{x^2+2}$. Simplifying $y=\frac{3x^2+2y}{x^2+2}$ we obtain $(y-3)x^2=0$. Since $x\neq 0$ and $(y-3)x^2=0$, then $y-3=0$ which is equivalent to $y=3$. Thus if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$. Therefore if $x\neq0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$. $Q.E.D.$

I think this one should be valid.

Thanks for your attention.

  • $\begingroup$ Welcome to Mathematics Stack Exchange. The contrapositive should be if $y\ne3$ then $x=0$ $\endgroup$ Jun 21, 2020 at 13:16
  • $\begingroup$ Thank you. I think that the conclusion of the theorem is of the form $P\rightarrow(Q\rightarrow R)$. So shouldn't the contrapositive be $\lnot(Q\rightarrow R)\rightarrow \lnot P?$ $\endgroup$ Jun 21, 2020 at 13:20

1 Answer 1


You could have said this:

if $y=\dfrac{3x^2+2y}{x^2+2}$, then $y(x^2+2)=3x^2+2y$, so $(y-3)x^2=0$, so $x=0$ or $y=3$.

  • $\begingroup$ Is your answer an alternative or my answer is problematic? $\endgroup$ Jun 21, 2020 at 13:13
  • 1
    $\begingroup$ I gave an alternative $\endgroup$ Jun 21, 2020 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.