# 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*}

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It'd be nice if the equations were aligned by = sign (as someone thought it to be worth the effort to rewrite it to tex - it was an image before) – MightyPork May 9 '14 at 19:16
I can't roll it back but you could just edit back to your way. – Zook May 9 '14 at 19:23
@Zook I'm fine with this, but it'd be easier to read with $=$'s aligned. I'm fairly new to this formatting, so I dunno how to do it. – MightyPork May 9 '14 at 19:24
heh I figured it out :D – MightyPork May 9 '14 at 19:26
Also note that if $\pi=3$, then you've multiplied both sides by $0$ in step three. Seems odd to do in the middle of a proof. – Keen May 9 '14 at 19:57

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 $x = (\pi + 3) / 2$.

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Makes sense, but why can you just omit the abs value in the last step? Not knowing what $\pi$ or $x$ equals to (hypothetically). – MightyPork May 9 '14 at 18:54
@MightyPork Indeed, I dropped the absolute value bars because I do know what $\pi$ is. It's pretty easy to show that $\pi > 3$ (inscribe a hexagon inside a circle, and look at the "circumferences" to do this). Then once you know $\pi > 3$, it's immediate that $\pi > x$, since it's the average of $\pi$ with a strictly smaller number. – user61527 May 9 '14 at 18:56
Ok, thanks for clearing it out - I'll accept after it lets me to – MightyPork May 9 '14 at 18:57
In fact, we only need to use $\pi \neq 3,$ a weaker condition than $\pi > 3.$ Getting the squares on one side gives $(3 - x)^2 - (\pi - x)^2 = 0.$ Now factor the left hand side (difference of squares) to get $(3 - x + \pi - x)(3 - x - \pi + x) = 0,$ or $(3 + \pi - 2x)(3 - \pi) = 0.$ At this point we can make use of the fact that $3 - \pi \neq 0$ to get $3 + \pi - 2x = 0,$ which is clearly equivalent to what we started with. – Dave L. Renfro May 9 '14 at 19:23
@DaveL.Renfro: We don't even need $\pi\neq 3$. All we need is that, as $(\pi + 3)/2$ is the average of the values, $x$ lies between $\pi$ and $3$ (and "possibly" equals both). This tells us that $$\operatorname{sgn}(x-\pi) = - \operatorname{sgn}(x-3)$$ so that, since $|x-\pi| = |x-3|$, we deduce $$x-\pi = -(x-3)$$ – Blue May 9 '14 at 19:38

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.

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Ok.. this is stupid, but.. can we just pick one of those and claim that $\pi = 3$? I'm not really sure what to make out of this now. Obviously $\pi \neq 3$, but if we didn't know that... (btw, how comes your score is so low?) – MightyPork May 9 '14 at 19:14
Well, I assumed that "pi" is known, so pi = 3 cannot be the correct solution. But T.Bongers pointed out correctly that pi > x > 3, which means that after removing the absolute value bars, you automatically get the -+ combination I spoke of in my answer. – SDV May 9 '14 at 19:24
Additionally, the solution presented in the puzzle assumes two unknowns: $x$ and $\pi$. Without a second equation, you can't solve for one of the unknowns. In other words, the two combinations that give a definite solution would have to be wrong. – Sam Dickson May 9 '14 at 20:11
@MightyPork (2 comments up) if you have 200 or more reputation on one SE site, then you will get an automatic 100 reputation bonus on each new SE site you sign up on after that. – David Z May 10 '14 at 7:59
@SDV BTW, $\pm a=\pm b$ means that $+a=+b\vee -a=-b$, which is wrong in current context. Right way to write the equality would be $\pm a=b$. – Ruslan May 10 '14 at 13:25

Up to step 4 seems perfectly valid.

But I don't see where exactly between steps 4 and 5 that we've established that 2πx is interchangeable with 9 on the left side.

Though we seem to then depend on that assumption on the right side as well to substitute in -2πx for the -9 we got fairly, and all the following steps all seem to continue on that interesting assumption.

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The values aren't interchangeable. Two simple steps have merely been combined, namely: Add $9$ to both sides; subtract $2\pi x$ from both sides. – Blue May 12 '14 at 19:53