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I've been reading Velleman's How to Prove it and I'm having a good time with the book up to now. However, I've been really stuck in the exercice 10 (Ch 3, Sec 3.2):

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

I've tried to prove the theorem by conditional, contraposition and contradiction but nothing seem's to result out of it. In any case (particularly by contradiction) it seems difficult to deal with negative statements ($x\neq0$,$y\neq3$) as my sole givens to complete the proof. By the way, I'm new in the forum.

Bruno Bentzen
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6 Answers6

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Do some algebraic manipulation; $$y(x^2+2)=3x^2+2y \iff 2y+x^2y-3x^2-2y=0 \iff x^2y-3x^2=0 \iff x^2(y-3)=0 \iff x=0$$(Discarded because of the hypothesis) or $$y=3$$

The denominator cannot be zero because $x \in \mathbb{R}$

UserX
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Just with these equivalences

$$y=\frac{3x^2+2y}{x^2+2}\iff y(x^2+2)=3x^2+2y\iff x^2(y-3)=0$$ we have

  • if $x=0$ then the equality holds for all $y$ and
  • if $x\ne0$ then $y=3$.
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Just simplify the equation:

If we have

$$ y = \frac{3x^2 + 2y}{x^2 + 2}, $$ then by multiplying through by $x^2 + 2$ this simplifies to $$ yx^2 + 2y = 3x^2 + 2y. $$ Can you finish it off from here?

CameronJWhitehead
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As $x^2+2\ge 2>0$, multiplicatin with $x^2+2$ is an equivalence transform, so $$(x^2+2)y=3x^2+2y. $$ Subtract $3x^2+2y$ from both sides: $$ (x^2+2)y-3x^2-2y=0,$$ simplify: $$ x^2(y-3)=0.$$ What does it mean if a product is zero?

Hagen von Eitzen
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You are contradicting the wrong condition.

Instead of assuming $y\ne3$ you have to assume $y\ne\frac{3x^2+2y}{x^2+2}$.

If $y\ne\frac{3x^2+2y}{x^2+2}$ then $y=3$:

$3\ne\frac{3x^2+6}{x^2+2}\iff3(x^2+2)\ne3x^2+6\iff0\ne0$

which is a contradiction, thus our assumption is wrong.

Narmer
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    Narmer, I'm afraid I didn't understand your point. By proceeding the proof by contradiction in this case, we negate the conclusion of the theorem, which is of the form $ If$ $P$ $then$ $Q$. Therefore, we have its corresponding negation, $P$ $and$ $not$ $Q$, i.e. $y=\frac{3x^2+2y}{x^2+2}$ and not $y=3$. – Bruno Bentzen Sep 08 '14 at 14:16
  • Yep, you're right. I got twisted in the proof and assumed that the assumption is wrong and not the conclusion, my bad! Too much work i guess... – Narmer Sep 08 '14 at 14:20
  • Not your fault! I've been strugling with this proof for 2 days and I was almost crazy too...! I'd never wonder the strategy wasn't about its logical structure but just simplify the equation! – Bruno Bentzen Sep 08 '14 at 14:25
  • My first approach was to extract Y from $\frac{3x^2+2y}{x^2+2}$ getting $y=3$, so the assumption by contradiction of $y\ne3$ was good. But then the "other Y" (left hand of the equation) showed up messing everything and resulting in my answer – Narmer Sep 08 '14 at 14:29
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Another variant: $$\begin{align}y&=\frac {3x^2+2y}{x^2+2}\\ y\left(1-\frac 2{x^2+2}\right)&=\frac {3x^2}{x^2+2}\\ y\left(\frac {x^2}{x^2+2}\right)&=3\left(\frac {x^2}{x^2+2}\right)\\ \because x\neq0 \therefore y&=3 \qquad \blacksquare \end{align}$$

Hypergeometricx
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