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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$?$

Edit:

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.

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  • $\begingroup$ 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

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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 $y=3$.

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

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