$2+2 = 5$? error in proof

Question

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

Where did I go wrong

Answer 1

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

Answer 2

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 $4$.

Answer 3

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

It's not true.

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

Answer 4

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.

Answer 5

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.