Can you give a good rational approximation of

$$\sqrt[3]2 \times \sqrt[4]2$$

without doing any significant maths?


I've left out one appropriate tag because it would be too much of a hint.

Attribution: Mine.

  • 2
    $\begingroup$ Nice Puzzle! Would it be better if 'significant' were 'decimal'? $\endgroup$ – user121330 Mar 26 at 17:06

The expression can be written as

$\sqrt[3]{2}\times \sqrt[4]{2}=2^{1/3}\times 2^{1/4}=2^{7/12}=(\sqrt[12]{2})^7$.

The key insight is that $\sqrt[12]{2}$ is

exactly the ratio of frequencies between two consecutive notes on a piano! That's because a piano is tuned per 12 equal temperament: the whole octave (which should double the frequency all in all) is split into 12 equal parts, which means the ratio between two consecutive notes needs to be the twelfth root of 2.

This means the seventh power of this corresponds to

moving seven keys to the right. But that's a perfect fifth; and the ratio for a perfect fifth (at least in string instruments that can be perfectly tuned to impose this ratio) is in fact $3:2$.


$3/2$ should be a fairly close rational approximation.

  • 2
    $\begingroup$ Gah, just barely beat me to it! Good answer, +1. (You may want to mention equal temperament, though?) $\endgroup$ – Deusovi Mar 26 at 8:05
  • $\begingroup$ @Deusovi done. My music theory knowledge is limited (to that one minutephysics video on piano tunings :P), so let me know if I've said something too wrong. $\endgroup$ – Ankoganit Mar 26 at 8:11
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    $\begingroup$ Unless your piano tuner has a taste for the exotic I'd have to argue that a perfect fifth doesn't actually exist on a modern piano. But petty quibbles aside, well done! $\endgroup$ – Albert.Lang Mar 26 at 8:55
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    $\begingroup$ And to complete this answer, the actual value is approximately 1.4983, so it is pretty close indeed. $\endgroup$ – justhalf Mar 27 at 6:18
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    $\begingroup$ FWIW, here's a related answer I wrote a couple of months ago: math.stackexchange.com/a/3983124/207316 Also see en.wikipedia.org/wiki/Pythagorean_comma $\endgroup$ – PM 2Ring Mar 27 at 11:38

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