Is there a proof that no rational number splits the octave equally?

Question

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.

Answer 1

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.

Answer 2

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.....)