8

I've read somewhere that no integer power of $1.5$ can ever equal any integer power of $2$ (besides the zeroth power, of course). It makes sense, but how is this proven?

(The application here is music. Octaves have a frequency ratio of $2:1$. Pure $5$ths have a $3:2$ frequency ratio. This issue—octaves never lining up with fifths—leads to either the compromise of temperaments or awful sounding intervals that prevent modulation.)

Bill Dubuque
  • 272,048
trw
  • 1,063
  • 2
    You probably mean integer powers? – Martin R Feb 02 '19 at 14:19
  • Yes, I do indeed. – trw Feb 02 '19 at 14:20
  • This isn't a full answer, but you might also be interested in Størmer's theorem, which finds the "closest" you can get two independent musical intervals by raising them to positive integer powers. ("closest" with regards to linear frequency, rather than the more commonly used logarithmic frequency. You can get arbitrarily close by using log frequency since log2(1.5) is irrational) – Ryan Feb 02 '19 at 16:02

5 Answers5

20

If $\left(\frac32\right)^n=2^m$, then $3^n=2^{m+n}$. For all but trivial cases, the right hand side is even, whereas the left is always odd.

Hagen von Eitzen
  • 374,180
5

We have $$1.5^n=\left(\frac{3}{2}\right)^n=\frac{3^n}{2^n}$$ For this to equal a power of 2 we need $$\frac{3^n}{2^n}=2^k$$ $$3^n=2^k \cdot 2^n = 2^{(n+k)}$$ And as no power of 3 is equal to a power of 2 ($2^a$ and $3^b$ are relatively prime), we cannot have $1.5^a=2^b$.

Mutantoe
  • 708
Peter Foreman
  • 19,947
3

Apart from $1.5^0$, no integer power of $1.5$ is an integer. We have $1.5=\frac32$, and the fundamental theorem of arithmetic shows that $\frac{3^n}{2^n}$ can never be simplified.

Arthur
  • 199,419
3

You know that $1.5=\frac{3}{2}$.

To have a power of 1.5 equal to a power of 2 means that $$\left(\dfrac{3}{2}\right)^m=2^n$$ for some integers $m,n\in\mathbb Z$.

This means $2^{m+n}=3^m$. Since $2$ and $3$ are prime numbers, this can't hold unless $m=n=0$ (see the fundamental theorem of number theory).

Scientifica
  • 8,781
2

More generally: if an integer power $\,n>0\,$ of a fraction $\,q\,$ is an integer $\,a,\,$ then $\,q\,$ is an integer, by the Rational Root Test, i.e. $\,q\,$ is a root of the $\rm\color{#c00}{monic}$ $\,f(x) \:\!=\:\! \color{#c00}{1}\:\!x^n - a\,$ so writing $\,q = c/d\,$ in least terms, RRT $\Rightarrow d\mid \color{#c00}1,\,$ so $\,d=1,\,$ so $\,q=c/d\in\Bbb Z$.

Generally RRT shows that a fractional root of any integer coefficient polynomial $\,f(x)\neq 0\,$ with $\color{#c00}{\rm lead\ coeff= 1}\,$ must be an integer. Above is the special case of $f$ being binomial (two terms).

Remark $ $ The arguments in the other answers are essentially special cases of the (short and simple) proof of RRT. The above is but a glimpse of the key role that RRT plays in factorization and number theory - which is clarified when one studies more general number rings (e.g. domains with unique prime factorization must be integrally closed, i.e. they must satisfy said $\color{#c00}{\rm monic}$ case of RRT).

Bill Dubuque
  • 272,048