Riddle with Pi = 3

This is a riddle someone posted on Google+, so please forgive it's triviality - I'm asking here because I just can't figure out what exactly is wrong, and it really bugs me ;)

I think something is not right with the square root at the end, but I'm not sure.

Here is the post

and here the riddle - "Find the mistake":

\begin{align*} x&=(\pi+3)/2\newline 2x&=\pi+3\newline 2x(\pi-3)&=(\pi+3)(\pi-3)\newline 2\pi x-6x&=\pi^2-9\newline 9-6x&=\pi^2-2\pi x\newline 9-6x+x^2&=\pi^2-2\pi x+x^2\newline (3-x)^2&=(\pi-x)^2\newline 3-x&=\pi-x\newline \pi&=3\newline \end{align*}

Solution 1:

You're quite right about where you think the error is - in fact, every other step is valid. Since $\pi > x > 3$, $3 - x$ is negative; hence, the step which involves a square root should actually read

\begin{align*} (3 - x)^2 = (\pi - x)^2 &\implies \sqrt{(3 - x)^2} = \sqrt{(\pi - x)^2} \\ &\implies |3 - x| = |\pi - x| \\ &\implies x - 3 = \pi - x \end{align*}

Upon rearranging this for $x$, we simply get back the original definition that $x = (\pi + 3) / 2$.

Solution 2:

Your instincts were right: if you take the square root on both sides on the 7th line, you will have to keep in mind that each side can have a positive or a negative algebraic sign afterwards. As such, the 8th line should look like this:

$$\pm(3 - x) = \pm(\pi - x)$$

This results in four combinations, two of which give the wrong result of "$\pi = 3$" ($++$ and $--$), while the other two ($+-$ and $-+$) will result in the 1st line, which is correct.