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

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.

• Welcome to Mathematics Stack Exchange. The contrapositive should be if $y\ne3$ then $x=0$ Jun 21, 2020 at 13:16
• 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 $\lnot(Q\rightarrow R)\rightarrow \lnot P?$ Jun 21, 2020 at 13:20
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 $$y=3$$.