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This question already has an answer here:

I have seen this image and surprised that we can prove $2 + 2 = 5$. can any one tell me what is wrong with this image.

Prove that, $2+2=5$.

We know that, $2+2=4$

$$\begin{align}\Rightarrow2+2&=4-\dfrac92+\dfrac92\\\,\\ &=\sqrt{\left(4-\dfrac92\right)^2}+\dfrac92\\ &=\sqrt{16-2\cdot4\cdot\dfrac92+\left(\dfrac92\right)^2}+\dfrac92\\ &=\sqrt{-20+\left(\dfrac92\right)^2}+\dfrac92\\ &=\sqrt{(5)^2-2\cdot5\cdot\dfrac92+\left(\dfrac92\right)^2}+\dfrac92\qquad\qquad\qquad\\ &=\sqrt{\left(5-\dfrac92\right)^2}+\dfrac92\\ &=5-\dfrac92+\dfrac92\\ &=5\\\,\\&\therefore\,2+2=5\text{ (Proved)}\\ \end{align}$$

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marked as duplicate by dfeuer, Rick Decker, Dan Rust, Stefan Hamcke, user1729 Oct 1 '13 at 19:36

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    $\begingroup$ Except for $x = 0$, you can't have $\sqrt{x^2} = x$ and $\sqrt{x^2} = -x$ at the same time. (In other news, if you have $4 - \frac{9}{2} = \sqrt{\left(4-\frac92\right)^2}$, you're working with the negative-valued branch of the square root, then $\sqrt{\left(5-\frac92\right)^2} = - \left(5-\frac92\right)$.) $\endgroup$ – Daniel Fischer Oct 1 '13 at 15:46
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    $\begingroup$ $\sqrt{a}$ is only definied if $a\geq 0$ $\endgroup$ – ulead86 Oct 1 '13 at 15:47
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    $\begingroup$ $\sqrt{\left(4-\frac{9}{2}\right)^{2}} = \frac{9}{2} - 4$ $\endgroup$ – jibounet Oct 1 '13 at 15:48
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    $\begingroup$ I see this kind of "math" as being akin to wordplay. $\endgroup$ – wjm Oct 1 '13 at 15:48
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    $\begingroup$ Indeed, a lot of “proofs” of this nature work by assuming that $\sqrt{x^2}=x$ even if $x<0$. Once you have seen half a dozen of them, it gets boring really fast. $\endgroup$ – Harald Hanche-Olsen Oct 1 '13 at 15:50
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Hint: Given any real number $x,$ we have $$\sqrt{x^2}=|x|.$$

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  • $\begingroup$ Please make this a complete answer. Means I need details $\endgroup$ – user960567 Oct 1 '13 at 15:57
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    $\begingroup$ In particular, check the step where the square root first appears. Is $4-\frac92=\left|4-\frac92\right|$? $\endgroup$ – Cameron Buie Oct 1 '13 at 15:59
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Using this, not only you can prove $4=5$ but you can also prove $a=b$ for all different $a,b$. BUT all these fallacies come from a slip in part of the proof. Here the mistake comes from accepting that $\sqrt x$ can be negative. This is just wrong.

Every time you see such a thing, look at steps of proof and definition of operators carefully.

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No,its not possible

your second step is wrong

$4-4.5$ is negative number

So $\sqrt{(4-4.5)^2}$ is not possible

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