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$$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\ &= \sqrt{\left(4-\frac92\right)^2} +\frac92\\ &= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\ &= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\ &= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\ &= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\ &= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\ &= \sqrt {\left(5-\frac92\right)^2} +\frac92\\ &= 5 + \frac92 - \frac92 \\ &= 5\end{align}$$

Where did I go wrong

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    $\begingroup$ You're misusing the square-root function. $\endgroup$ – Pedro Tamaroff Aug 1 '13 at 19:15
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    $\begingroup$ $4 - \frac{9}{2} < 0$. $\endgroup$ – Daniel Fischer Aug 1 '13 at 19:16
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    $\begingroup$ $4 - 9/2 +9/2 \neq \sqrt(4-9/2)^2 +9/2$. Typically the misatke in these fake proofs is either division by zero or using $\sqrt{x^2}=x$ for negative/complex numbers. $\endgroup$ – N. S. Aug 1 '13 at 19:17
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    $\begingroup$ The reason this "works" is that we don't intuitively see that $4<9/2$. You could do the same with $9/2$ replaced by $5$, and you'd immediately see the problem. $\endgroup$ – Thomas Andrews Aug 1 '13 at 19:23
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    $\begingroup$ How did $\sqrt{16 -2\times4\times\frac92 +(\frac92)^2}$ turn into $\left(\sqrt{16 -36 + (\frac92)^2}\right)^2$ (with that superscript $2$ OUTSIDE the parentheses)? $\endgroup$ – Michael Hardy Aug 1 '13 at 23:18
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In the first line you have $4-4.5=\sqrt{(4-4.5)^2}$, which isn't true, because $-0.5\neq 0.5$.

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  • $\begingroup$ I think $\sqrt{4}$ equals $\pm2$ and not just $+2$? $\endgroup$ – Ramit Aug 1 '13 at 19:34
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    $\begingroup$ @Ramit No, it does not. It is actually a function. $\endgroup$ – Tobias Kildetoft Aug 1 '13 at 19:35
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    $\begingroup$ No, $\sqrt{4}$ is positive, thats the definition of the square root, opposed to "$2$ and $-2$ are square roots of four", which is true. $\endgroup$ – Tomas Aug 1 '13 at 19:35
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    $\begingroup$ @ColeJohnson, because, sadly, those teachers are inadequately trained. $\endgroup$ – vadim123 May 22 '14 at 0:40
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    $\begingroup$ @Ramit $\sqrt{4}$ is the principal square root which is a function. $\endgroup$ – zixuan Aug 16 '19 at 15:28
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Here's what your "proof" would look like correcting all the errors. As you can see, it's not nearly as impressive as a proof that 2+2=5.

$$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\ &= -\sqrt{(4-\frac92)^2} +\frac92\\ &= -\sqrt{16 -2\times4\times\frac92 +(\frac92)^2} + \frac92\\ &= \left(-\sqrt{16 -36 + (\frac92)^2}\right) +\frac92\\ &= \left(-\sqrt {-20 +(\frac92)^2}\right) + \frac92\\ &= -\sqrt{25-45 +(\frac92)^2} +\frac92\\ &= -\sqrt {5^2 -2\times5\times\frac92 + (\frac92) ^2} + \frac92\\ &= -\sqrt {(5-\frac92)^2} +\frac92\\ &= -5 + \frac92 + \frac92 \\ &= -5+9\end{align}$$

For reference, the most serious mistake was in the 2nd line. In general, it's not true that $\sqrt{x^2} = x$, but rather $\sqrt{x^2} = |x|$. For $x=4-\frac92<0$, you need to keep track of the extra minus sign coming from the absolute value. Other than that, there were some obvious typos that I've corrected.

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Basically, your proof is saying $$ -0.5=\sqrt{(-0.5)^2}=\sqrt{0.5^2}=0.5 $$ Now find the error.

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Apart from the other answers, even at the last, $\sqrt{(5-\frac92)^2}=\pm(5-\frac92)$. with + it is wrong.

With $-(5-\frac92)$, that is $-5+\frac92$, adding the other $\frac92$ from the original equation, we do get $4$.

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    $\begingroup$ While this does work, $\sqrt{a^2}=|a|$, the square root is usually taken to be a function, the principal square root. $\endgroup$ – JMCF125 May 1 '14 at 12:49
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    $\begingroup$ This is incorrect. $\sqrt{x} \ne \pm x$. This works because that -ve sign is missing from the start which you are adding in the end with this fallacy. $\endgroup$ – A---B Aug 29 '17 at 22:22
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$\sqrt{\left(4 - \frac 9 2 \right)^2} = 4 - \frac 9 2 = -0.5.$

It's not true.

If $a \geq 0$, then $\sqrt{a} \geq 0$.

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  • $\begingroup$ You're repeating what has already been said. I wouldn't write $a\geq 0\implies \sqrt a\geq 0$. The "real" square-root is defined only for positive entries, so one should just say $\sqrt a \geq 0$ for any $a$. $\endgroup$ – Pedro Tamaroff Aug 1 '13 at 21:15
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How did $\sqrt{16 -2\times4\times\frac92 +(\frac92)^2}$ turn into $\left(\sqrt{16 -36 + (\frac92)^2}\right)^2$?

Then later, you seem to assume that since $\left(4-\frac92\right)^2$ is the same as $\left(5-\frac92\right)^2$, it follows that $4-\frac92=5-\frac92$. Like saying that since $3^2=(-3)^2$, it follows that $3=-3$. A well known mistake.

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$$\sqrt{a^2}=|a|\not=a$$ Watch the signs.

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Clearly, $\sqrt{(4-\frac{9}{2})^2}$ is what? Well, $4-\frac{9}{2}$ is $-\frac{1}{2}$. But we all know $(-\frac{1}{2})^2$ is $\frac{1}{4}$ since $(-x)^2=x^2$. That means that $\sqrt{(4-\frac{9}{2})^2}$ is $\frac{1}{2}$. So the assumption that $\sqrt{(4-\frac{9}{2})^2} = 4 - \frac{9}{2}$ leads to $-\frac{1}{2}=\frac{1}{2}$ which is clearly wrong.

So the square root of a positive number squared is a positive number and the square root of a negative positive number squared is also a positive number. So if we assume $\sqrt{(-x)^2} = -x$ to $x = -x$ which is false. Note: x is positive, not negative. This is the correct statement:

$\sqrt{(-x)^2} = \sqrt{x^2} = x \not= -x$

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In the very first line, you assumed that $a=|a|=\sqrt{a^2}.$ But this is not true in general. In particular, it is false for negative $a.$ And in fact, $$4-\frac92<0,$$ so in this case we do not have that $$4-\frac92=\sqrt{\left(4-\frac92\right)^2}=\left|4-\frac92\right|$$ as assumed.

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