# $2+2 = 5$? 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
Aug 1, 2013 at 19:15
• $4 - \frac{9}{2} < 0$. Aug 1, 2013 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. Aug 1, 2013 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. Aug 1, 2013 at 19:23
• This is so 1984-ish! - the book for those who don't know what I'm talking about. Aug 1, 2013 at 21:25

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 $+2$? Aug 1, 2013 at 19:34
• @Ramit No, it does not. It is actually a function. Aug 1, 2013 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. Aug 1, 2013 at 19:35
• @ColeJohnson, because, sadly, those teachers are inadequately trained. May 22, 2014 at 0:40
• @Ramit $\sqrt{4}$ is the principal square root which is a function. Aug 16, 2019 at 15:28

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.

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

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. May 1, 2014 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.
– user312097
Aug 29, 2017 at 22:22

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.

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

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
Aug 1, 2013 at 21:15

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

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:

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

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.