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In music circles, when the topic of tuning comes up, it is said that there is no rational number that splits the octave (the interval between a musical pitch and another with double its frequency, for example 440 Hz and 880 Hz) equally, but I have never seen a proof for this, so I wonder if it has been proven, or just universally accepted. Is it maybe that it's too simple to write down and it's expected that one can just do it themselves? I have only finished high school maths (SQA Advanced Higher mathematics) so I haven't got any experience constructing proofs, otherwise I would try it myself.

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  • $\begingroup$ Are you familiar with any mathematical expressions that are connected to 'octave'? If so, can you also write those in the question? $\endgroup$
    – Ken Hung
    May 11 at 6:49
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    $\begingroup$ Why it's impossible to tune a piano $\endgroup$ May 11 at 6:50
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    $\begingroup$ The ratio of the half octave interval would need to be the square root of $2$. This is typically the first proof of irrationality that a maths student meets. $\endgroup$
    – badjohn
    May 11 at 6:51
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    $\begingroup$ To split the octave "equally" into $n$ parts, you need a rational number equal to $\root{n}\of2$, which is impossible. $\endgroup$ May 11 at 6:59
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    $\begingroup$ "How do you know that is has to be 2–√n" That's pretty much the definition of what you are talking about, isn't it. You want a note that vibrates and $x$ times 440, and then if you vibrate again at $x\times x$ and then vibrate again at $x\times x\times x$ and you do this $n$ times it vibrates and 880. If each note vibrates at $x$ times the previous note then the $n$th not vibrates and $x^n$ times the first note. For that to be a full octave we need $x^n = 2$. That's the parameter of the question. $\endgroup$
    – fleablood
    May 11 at 7:09
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To prove a statement you must state it.

So....

Definition: A musical interval is two notes where the higher note vibrates at a set frequency rate faster than the lower one.

An octave is an interval where the frequency rate is $2$.

Observation: If you compound an interval upon itself so that you have a base lower note, and a second note that vibrates at the set frequency more than a note the vaibrates at a set frequency above the low note, this new interval has a frequency that is the original frequency squared. That is to say if you have three notes: a base note $m$ a second note $n$ so that $m$ and $n$ is an interval of a frequency of $x$, and third note $u$ so that $n$ and $u$ is an interval of a frequency of $x$, then $m$ and $u$ is an interval of frequency $x^2$.

This should be clear as the vibration of $n$ is $x$ times the vibration of $m$, and the vibration of $u$ is $x$ times the vibration of $n$ then the vibration of $u$ is $x$ times $x$ times the vibration of $m$. That is $x^2$ times the vibration of $m$.

If you repeat the procedure a positive integer $k$ times, the resulting interval will be an interval with a frequency of the original frequency to the $k$th power.

Definition: An interval with a set frequency rate between $1$ and $2$ is said to split an octave if compounding the interval some positive integers of times creates an interval that is a positive integer of octaves.

Claim: There is no interval with rational frequency that splits the octave.

Pf: Suppose we have an interval with a frequency of $x; 1< x < 2$, that when compounded $k$ times because an interval of $m$ octaves.

That means $x^k = 2^m$. If $x$ is rational then there exist integers $a, b$ so that $x =\frac ab$ and $a$ and $b$ have now factors in common. So $(\frac ab)^k = 2^m$.

So $\frac {a^k}{b^k} = 2^m$

So $a^k = 2^m b^k$. This means $2$ is a factor of $a^k$ and therefore, because $2$ is prime, $2$ must be a factor of $a$. Any other prime factor, $p$, of $a$ must be a prime factor of $2^mb^k$ but it can't be $2$ so it must be a prime factor of $b^k$ so it must be a factor of $b$. But that's impossible as we assumed $a$ and $b$ had no factors in common.

So $a$ has no other prime factors but $2$.

Meanwhile any prime factor of $b$ must be factor of $a^k$ and thus of $a$. But we assumed $a$ and $b$ had not factors in common.

So $b$ has no prime factors at all. ANd $a$ has only $2$ as a prime factor.

So $b = 1$ and $a = 2^u$ for some positive integer $u$. So that means $x = \frac ab = \frac {2^u}1 = 2^u$. But then $x = 2^u \ge 2$. So the interval of $x$ is not less than an octave. In fact, $x$ must be $u$ octaves.

So no interval "splits an octave".

=========== old answer below =====

The major fifth vibrates and $\frac 32$. And a major fourth ant $\frac 43$.

So a major fourth over a major fifth vibrates at $\frac 32\cdot \frac 43=2$ and forms an octave.

A major fifth over a major fifth vibrates at $\frac 32\cdot \frac 32 =\frac 94= 2\frac 14$ and is more than an octave.

The circle of fifths actually fails. C to G to D to A to E to B to F sharp to C is 7 major fifths and it's supposed to be 4 octaves. But it is actually $(\frac 32)^7 = \frac {2187}{128}=17\frac {11}{128}$ whereas 4 octaves are $2^4 = 16$.

Now to make a chord that vibrates at $x$ and have a chord above that vibrate and $x \times x$ and have that be a perfect octave is to have $x^2 =2$. There is no rational frequency that does that. (Although that is the frequency the news was broadcast on Futurama.....)

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If $x$ split up the octave in equal parts then from 440Hz to the middle would be a step of $x$ and from the middle to 880Hz would again be a step of $x$. So the doubling would be two consecutive steps of $x$, so $x \times x = x^2$, and we'd have $x^2 = 2$.

No rational number can satisfy this equation: suppose $\frac{p}{q}$ did, where $p,q$ are integers without a common divisor (so the fraction is in its lowest terms). Then $(\frac{p}{q})^2 = \frac{p^2}{q^2}=2$ and so $2q^2 =p^2$. So $p^2$ is even (being $2q^2$) so $p$ is even and then $p^2$ is also disivible by $4$ so $q^2$ is even (divide $p^2=2q^2$ by two on both sides, and at least one factor $2$ remains in $\frac{p^2}{2}$) and so $q$ is even, but then we have a contradiction with $p$ and $q$ having no divisor in common... So $\frac{p}{q}$ cannot exist and $x$ with $x^2=2$ is irrational.

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