Whats wrong with this proof?

Question

Theorem: $x$ is a real number with $x \neq 1.$ If $\frac {x^2+1}{x-1} =x$, then $x=-1$.

If we suppose that $x=-1$. Then $\frac {x^2+1}{x-1} = \frac {(-1)^2+1}{-1-1} = \frac {2}{-2} = -1 = x$

I would say that the prove is not correct in the second step when we have $(-1)^2$, but I am not completely sure.

I appreciate your answer!!!

I can't understand what the first line is saying, and in the second line, $(-1)^2$ is not wrong at all. — JiminP, Aug 02 '13 at 12:52
You haven't proven anything, you've just shown that $x=-1$ solves $\frac{x^2+1}{x-1}=x$, which it does. — Owen Sizemore, Aug 02 '13 at 12:54
Where is the proof? What we have is first the statement of a (false) theorem then a (correct) implication vaguely related to the situation in the theorem, then some comments on the implication. No proof. — Did, Aug 02 '13 at 12:55
You proved "If $x= -1$, then $\frac{x^2 + 1}{x-1} = x$.". Now prove "If $\frac{x^2 + 1}{x-1} = x$, then $x = -1$.". — k.stm, Aug 02 '13 at 12:55

score 4 · Accepted Answer · answered Aug 02 '13 at 12:59

4

$$\frac{x^2+1}{x-1}=x\iff x^2+1=x^2-x\iff x=-1$$

And indeed, substituting:

$$\frac{(-1)^2+1}{-1-1}=\frac2{-2}=-1\;(=x)$$

So everything's fine...what (and perhaps why) do you think is wrong here?

answered Aug 02 '13 at 12:59

DonAntonio

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One problem was that the OP stated the forward implication and then proved (correctly) the backward implication, which is a common enough error that you might want to draw attention to it. – mdp Aug 02 '13 at 13:01
1

I thought that something must be wrong because the exercise asks "What`s wrong with the following proof?" – user2051347 Aug 02 '13 at 13:01
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@mgelius It might have been helpful to mention that the theorem/proof came from such an exercise in the original question. As it happens I've kind of answered it for you: the proof by itself doesn't have a mistake in it, but it's a proof of a different theorem. – mdp Aug 02 '13 at 13:04
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@mgelius The proof that DonAntonio gave proves it. Are you asking for an alternative proof? – rurouniwallace Aug 02 '13 at 13:08

score 2 · Answer 2 · answered Aug 02 '13 at 16:30

While the computations are correct, the logic is flawed. The claim that should be proved is "If $A$ then $B$". The proof runs "Suppose $B$. Then ... and hence $A$", which shows the wrong direction! (Alright, in the end $A\iff B$ is true, but that doesn't matter for the - lack of - correctness of the proof).

Whats wrong with this proof?

2 Answers2