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$\phi(\pi(\phi^\pi)) = 1$

I saw it on an expired flier for a lecture at the university. I don't know what $\phi$ is, so I tried asking Wolfram Alpha to solve $x \pi x^\pi = 1$ and it gave me a bunch of results with $i$, and I don't know what that is either.

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    $\begingroup$ Could you give us the title of the lecture? The $ \ \pi \ $ in that equation is likely not the number, but the name of a function or mapping... $\endgroup$ Commented Jul 9, 2014 at 21:24
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    $\begingroup$ Something about teaching computers to resolve ambiguities. The flier was expired, so we had to throw it away. $\endgroup$
    – Mr. Brooks
    Commented Jul 9, 2014 at 21:25
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    $\begingroup$ @Mr.Brooks That then makes perfect sense as the computer needs to distinguish between the number $\pi$ and the function $\pi$, and the same goes for $\phi$. $\endgroup$
    – Lord Soth
    Commented Jul 9, 2014 at 21:26
  • $\begingroup$ @Mr.Brooks So I guess the equation does not mean anything by itself but it is an example of an ambiguous equation. $\endgroup$
    – Lord Soth
    Commented Jul 9, 2014 at 21:26
  • $\begingroup$ @Mr.Brooks Not to mention whether these are multiplications (your interpretation) or functions, etc. $\endgroup$
    – Lord Soth
    Commented Jul 9, 2014 at 21:28

1 Answer 1

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It's a joke based on the use of the $\phi$ function (Euler's totient function), the $\pi$ function (the prime counting function), the constant $\phi$ (the golden ratio), and the constant $\pi$. Note $\phi^\pi\approx 4.5$, so there are two primes less than $\phi^\pi$ (they are $2$ and $3$), so $\pi(\phi^\pi)=2$. There is only one positive integer less than or equal to $2$ which is also relatively prime to $2$ (this number is $1$), so $\phi(2)=1$. Hence we have

$$\phi(\pi(\phi^\pi))=\phi(2)=1$$

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  • $\begingroup$ Thanks! I thought someone else would be better able to interpret it than I would; I was thinking some sort of algebraic mapping was being referred to (rather than something from number theory)... $\endgroup$ Commented Jul 9, 2014 at 21:40
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    $\begingroup$ Brilliant! I only have one thing to add, since the asker mentioned Wolfram|Alpha: I have a feeling that one day soon, Wolfram Mathematica will be able to instinctively resolve the symbolic ambiguities in that equation and understand it internally as "EulerPhi[PrimePi[GoldenRatio^Pi]]". $\endgroup$ Commented Jul 9, 2014 at 22:50

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