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I came across this funny proof-


$$=4+\frac 92-\frac 92$$

$$=\sqrt{(4-\frac 92)^2}+\frac 92$$

$$=\sqrt{16+\frac{81}{4}-36}+\frac 92$$

$$=\sqrt{25+\frac {81}{4}-45}+\frac 92$$

$$=\sqrt{(5-\frac 92)^2}+\frac 92$$

$$=5-\frac 92+\frac 92$$


I suspect,that the error is in the second line where operation on negative is done before positive violating $BODMAS$ rule.But,$\sqrt{(4-\frac 92)^2}=4-\frac 92$ and the similar solving continues.So, where is the real error?

Thanks for any help!!


marked as duplicate by Daniel Fischer Nov 24 '15 at 8:56

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  • 26
    $\begingroup$ $\sqrt{(4-\frac 92)^2}=-(4-\frac 92)$. $\endgroup$ – mathlove Nov 23 '15 at 18:11
  • 4
    $\begingroup$ when you take the square root of $\sqrt{(5-\frac{9}{2})^2}$ take $-5+\frac{9}{2}$ instead of $5-\frac{9}{2}$ $\endgroup$ – Gune Nov 23 '15 at 18:16
  • 3
    $\begingroup$ @RPH: While that would prevent this particular "proof" from demonstrating a fallacy, it involves the use on two occasions of a non-standard evaluation of the square root to represent the negative value whose square is the argument. $\endgroup$ – Brian Tung Nov 23 '15 at 22:40

The error is going from the second to third line. $$\sqrt{x^2}=|x|$$ so $$\sqrt{\left(4-\frac{9}{2}\right)^2}=\left|4-\frac{9}{2}\right|\neq 4-\frac{9}{2}$$

  • $\begingroup$ It's not really the second line...its the third last line I suppose when I am removing the square root... $\endgroup$ – tatan Nov 23 '15 at 18:15
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    $\begingroup$ The error is the claim that the second and third line are equal. The third-to-last line and second-to-last line are indeed equal, because $\sqrt{(5-4.5)^2}=|5-4.5|=5-4.5$. $\endgroup$ – vadim123 Nov 23 '15 at 18:16
  • $\begingroup$ while this post is generally right I think my post gets better into the point $\endgroup$ – Dac0 Nov 23 '15 at 22:57
  • 1
    $\begingroup$ Specifically: $(4 - \frac 9 2) < 0$ so $4 - \frac 9 2 = -|4 - \frac 9 2| = - \sqrt{(4 - \frac 9 2)^2}$. So the third line is simply not true. Everything past that is valid. If you had continued with the negative square root. You would have concluded $-\sqrt{ (5 - \frac 9 2)^2} + \frac 9 2 = -(5 - \frac 9 2) + \frac 9 2 = -5 + \frac 9 2 + \frac 9 2 = 5 + 9 = 4$. $\endgroup$ – fleablood Nov 24 '15 at 22:44

The error comes in when $4 - \frac92$ is squared, which causes $-\frac12$ to become $+\frac14$, and the square root taken from this is erroneously taken as $+\frac12$.


The equality $$\sqrt{\left(4-\frac92\right)^2} = \sqrt{\left(5-\frac92\right)^2}$$ is equivalent to $$\sqrt{\left(-\frac12\right)^2} = \sqrt{\left(\frac12\right)^2}$$ which of course is true, but that does not imply $$-\frac12 = \frac12$$

Adding the first square and square root is invalid, this equality:

$$4+\frac 92-\frac 92$$ $$=\sqrt{\left(4-\frac 92\right)^2}+\frac 92$$

does not hold.


The effective problem seems to be in switching between solutions, but I'd like to point out the relevance of the numbers involved in the game (using other numbers it wouldn't have the same effect): $$ (4- \frac{9}{2})^2=(16+\frac{81}{4}-36) $$ $$^{(*)}(16+9-9+\frac{81}{4}-36)= (25+\frac{81}{4}-45) $$ $$(25+\frac{81}{4}-45)=(5-\frac{9}{2})^2 $$

The game is based on the fact that the numbers involved are $4$, $5$ and $\frac{9}{2}$ indeed is $$4-\frac{9}{2}=-0.5$$ $$-4+\frac{9}{2}=0.5$$ while $$5-\frac{9}{2}=0.5$$ $$-5+\frac{9}{2}=-0.5$$ So what's happening is that adding and subtracting 9 in (*) is sufficient to confuse and switch between the two solution 0.5 and -0.5 when ricomposing the binomial square

  • 3
    $\begingroup$ The error is elsewhere, as described in other answers. It is in fact true that (4 - 9/2)^2 = (5 - 9/2)^2 $\endgroup$ – Lily Chung Nov 23 '15 at 23:55
  • $\begingroup$ Yes indeed it's just what I pointed out... $\endgroup$ – Dac0 Nov 24 '15 at 7:52
  • $\begingroup$ Although adding and subtracting $9$ does, as you say, conceal the subterfuge, note that we could immediately evaluate both $\left(4-\frac{9}{2}\right)^2$ and $\left(5-\frac{9}{2}\right)^2$ as $\frac{1}{4}$, and then there is nothing in the line you've marked $(*)$ that would be wrong, and yet the overall demonstration would still be a fallacy. To the extent that a fallacy is explained by a logical error, it must be the implication that $\sqrt{x^2} = x$ necessarily. What you've pointed out is relevant to how the game is played, but it is not where the foul is committed, as it were. $\endgroup$ – Brian Tung Nov 24 '15 at 22:40
  • $\begingroup$ Yes you're right I was pointing out the subterfuge that allows the camuflage of the fallacy. The same fallacy with other numbers it wouldn't have the same effect. I think that's a psychological point that needs to be pointed out. The numbers involved are not casual $\endgroup$ – Dac0 Nov 24 '15 at 22:55

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