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

\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

• 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

In the first line you have $4-4.5=\sqrt{(4-4.5)^2}$, which isn't true, because $-0.5\neq 0.5$.

• 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

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.

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 $4$.

• 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 '14 at 12:49
• 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. – A---B Aug 29 '17 at 22:22

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

$\sqrt{\left(4 - \frac 9 2 \right)^2} = 4 - \frac 9 2 = -0.5.$

It's not true.

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

• 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.

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

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