Operations on intervals

The following are common arithmetic operations on musical intervals. If you're unfamiliar with the operations, you're encouraged to verify the examples for interval operation rules with a calculator.

Stacking and unstacking

Stacking two intervals feels perceptually like we are adding two distances, but it corresponds to multiplying two frequency ratios. The logarithm function is the bridge between frequency space and pitch space:

${\log }_{2}a+{\log }_{2}b={\log }_2(ab)$

for any two frequency ratios a and b. The above equation tells us that the sum of the size (in octaves) of a and the size (in octaves) of b is equal to the size in octaves of the product of two frequency ratios, ab. Hence, stacking corresponds to multiplying frequency ratios in linear (frequency) space, and adding cent values in logarithmic (pitch) space.

To convert octaves to cents, we simply multiply both sides by 1200:

$1200{\log }_{2}a+1200{\log }_{2}b=1200{\log }_2(ab).$

As a consequence, since the inverse operation of multiplication is division, dividing frequency ratios corresponds to subtracting their perceptual sizes:

$1200{\log }_{2}a-1200{\log }_{2}b=1200{\log }_2(a/b).$

Examples

1. $1200{\log }_2(3/2)-1200{\log }_2(5/4)=1200{\log }_2(6/5)$
2. $1200{\log }_2(3/2)+1200{\log }_2(3/2)=1200{\log }_2(9/4)$
3. $1200{\log }_2(16/9)-1200{\log }_2(3/2)=1200{\log }_2(32/27)$

Logarithmic multiplication and division

From the log rule

$n{\log }_{2}r={\log }_2(r^n)$

we see that stacking n copies of the ratio r corresponds to taking the nth power of r:

$n(1200{\log }_{2}r)=1200{\log }_2(r^n).$

The analogous holds for dividing a ratio into n equal parts:

$\frac{1}{n}(1200{\log }_{2}r)=1200{\log }_2(r^{1/n}).$

There is another operation that can be called "logarithmic division", which is finding the perceptual ratio of two intervals. This is given by the base change operation:

$\frac{1200{\log }_{2}r}{1200{\log }_{2}s}={\log }_{s}r.$

Division by 0 is not allowed, so you can't take ${\log }_{1/1}.$

Examples

1. $2\cdot 1200{\log }_2(4/3)=1200{\log }_2((4/3{)}^2)=1200{\log }_2(16/9)$
2. $2\cdot 1200{\log }_2(9/8)=1200{\log }_2((9/8{)}^2)=1200{\log }_2(81/64)$
3. $\frac{1}{13}\cdot 1200{\log }_2(2/1)=1200{\log }_2(2^{1/13})$
4. $\frac{1200{\log }_2(81/64)}{1200{\log }_2(9/8)}={\log }_{9/8}(81/64)=2$

Reduction

Reduction is the pitch equivalent of modular arithmetic ("clock arithmetic"). Reduction is more of a fundamentally pitch-space operation, so mathematically we can just denote it using $mod$ or $\mathrm{\%}$.

The most common reduction operation is octave reduction, or the operation $x\mapsto x\mathrm{\%}1200\mathrm{c}$, so let's talk about that first. You can visualize octave reduction as wrapping pitch space into a circle of octave-equivalent "pitch classes".

The way to compute the octave reduction of an interval x is:

1. If x > 0 cents, keep (logarithmically) subtracting 1200c from the cents value of x until x is greater than or equal to 0 cents and (strictly) less than 1200 cents.
2. If x < 0 cents, keep (logarithmically) adding 1200c from the cents value of x until the value of x is greater than or equal to 0 cents and (strictly) less than 1200 cents.

For reduction by any other interval, just substitute the cents value of the interval you wish to reduce by for 1200 cents.