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I stumbled across the following computation proving $2+2=5$

calculation proving 2+2=5

Clearly it doesn't, but where is the mistake? I expect that it's a simple one, but I'm even simpler and don't really understand the application of the binomial form to this...

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    $\begingroup$ There are many such posts with "fake proofs" that are fun to review, for example; math.stackexchange.com/questions/438/… $\endgroup$
    – Moo
    Jan 2 '17 at 0:54
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    $\begingroup$ $(-1)^2 = 1^2 \implies -1 = 1$! $\endgroup$
    – user14972
    Jan 2 '17 at 1:49
  • $\begingroup$ Don't think that the edit is appropriate. Someone is suggesting that I replace the actual question with the answer from a previous question, which would kinda make my not knowing the answer initially to my question moot. Perhaps it sits better on philosophy.SE :-) $\endgroup$
    – Paul Uszak
    Jan 2 '17 at 2:21
  • $\begingroup$ Why do you believe $\sqrt{a^2} = a$? It doesn't. Example: $\sqrt{(4-9/2)^2} = 1/2 != 4 - 9/2$. It doesn't. Why did you think it does? $\endgroup$
    – fleablood
    Jan 2 '17 at 3:47
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    $\begingroup$ One way to find where the error is, would be to see on which line does it stop saying true things and on what line does it start to say false things. Lines 1-4 are all true. Lines 5-7 are all false. So 4 must somehow not imply 5. 4 says $ (-1/2)^2 = (1/2)^2$ and line 5 says $-1/2 = 1/2$. So can you see 4 does not imply 5? $\endgroup$
    – fleablood
    Jan 2 '17 at 7:25
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The error is in the step where the derivation goes from $$\left(4-\frac{9}{2}\right)^2 = \left(5-\frac{9}{2}\right)^2$$ to $$\left(4-\frac{9}{2}\right) = \left(5-\frac{9}{2}\right)$$

In general, if $a^2=b^2$ it is not necessarily true that $a=b$; all you can conclude is that either $a=b$ or $a=-b$. In this case, the latter is true, because $\left(4-\frac{9}{2}\right) = -\frac{1}{2}$ and $\left(5-\frac{9}{2}\right) = \frac{1}{2}$. Once you have written down the (false) equation $-\frac{1}{2} = \frac{1}{2}$ it is easy to derive any false conclusion you want.

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  • $\begingroup$ I am curious why this was downvoted. Does the downvoter have any feedback or comment on how it could be improved? $\endgroup$
    – mweiss
    Jan 2 '17 at 2:38
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Because $x^2=y^2\Leftrightarrow x=y$ is not true.

Also $\sqrt{x^2}=x$ is wrong: actually, $\sqrt{x^2}=|x|$.

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    $\begingroup$ In particular, $4-9/2<0$ and is not the square root of the LHS. $\endgroup$
    – mrob
    Jan 2 '17 at 0:41
  • $\begingroup$ @mrob $\sqrt{\left(4-\frac{9}{2}\right)^2}=|4-\frac{9}{2}|=0.5$ $\endgroup$ Jan 2 '17 at 0:43
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$9^2 = (-9)^2$
But $9 \ne -9$

If we had taken as $\left|4 - \dfrac{9}{2} \right|=\left|5 - \dfrac{9}{2} \right|$
$0.5=0.5$

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