# $2+2 = 5$? error in proof

I found this proof somewhere:

\begin{align} 2+2 &= 4 - \frac92 +\frac92\\ &= \sqrt{(4-\frac92)^2} +\frac92\\ &= \sqrt{16 -2\times4\times\frac92 +(\frac92)^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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You're misusing the square-root function. –  Pedro Tamaroff Aug 1 '13 at 19:15
$4 - \frac{9}{2} < 0$. –  Daniel Fischer Aug 1 '13 at 19:16
$4 - 9/2 +9/2 \neq \sqrt(4-9/2)^2 +9/2$. Typically the misatke in these fake proofs is either division by zero or using $\sqrt{x^2}=x$ for negative/complex numbers. –  N. S. Aug 1 '13 at 19:17
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. –  Thomas Andrews Aug 1 '13 at 19:23
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)? –  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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I think $\sqrt{4}$ equals $\pm2$ and not just $+2$? –  Ramit Aug 1 '13 at 19:34
@Ramit No, it does not. It is actually a function. –  Tobias Kildetoft Aug 1 '13 at 19:35
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. –  Tomas Aug 1 '13 at 19:35
Ok, I see your point. Thanks. –  Ramit Aug 1 '13 at 19:42
@ColeJohnson, because, sadly, those teachers are inadequately trained. –  vadim123 May 22 at 0:40
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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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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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While this does work, $\sqrt{a^2}=|a|$, the square root is usually taken to be a function, the principal square root. –  JMCF125 May 1 at 12:49

$\sqrt{(4-\frac92)^2} =4-\frac92=-0.5$

Its not True

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

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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 $a$. –  Pedro Tamaroff Aug 1 '13 at 21:15
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.