Just like the notes repeat after you go past B, in our system our numbers restart after 11. Due to the cyclic nature of what we're working with there are properties that we should know about our system. As a reference, we'll have the circle of notes:
The first thing we'll discuss has to do with addition in our system, in some situations everything is normal, so if we were to consider 2 + 3, then the answer to that would be 5. But things get a little more complicated once we start above 11 or below 0.
The way we deal with the more complicated situtations should become clear momentarily. If we were asked to compute 10 + 5, then normally we would answer 15, but in our system we mainly use the numbers 0 to 11, so in this situation we have to "wrap around", so it would be 10, 10 + 1 = 11, 10 + 2 = 0, 10 + 3 = 1, 10 + 4 = 2, 10 + 5 = 3, so 10 + 5 = 3 in our system.
Subtraction also has the same properties, if we did 10 - 5, then we can answer 5, like normal, but if we considered 1 - 3, then we would have 1, 1 - 1 = 0, 1 - 2 = 11, 1 - 3 = 10 so as we can see when we consider 1 - 3 we remember that our numbers "wrap" and then we can say 1 - 3 = 10.
Since we've considered subtracting numbers and subtracting a number is the same as adding the negative version, we'll also analyze negative numbers.
If we had the number -3, then we can consider that as the number which is 3 below 0. Since in our system we mainly focus on the numbers from 0 to 11, and stepping three below 0 gives us 9, so in this sense -3 and 9 are equivalent.
This means that given any number from 0 to 11, there is a negative number from -11 to 0 which is "equivalent" to it. Following a similar process to the one we just did with -3, we can figure out all the equivalences.
Putting into a circle, it would be:
The reason why equivalent elements are interesting is that we can solve questions we may encounter more efficiently, to see why this is true, we can consider 10 + 10, one method would be to add the two numbers and get 20, then realize that 20 is 8 steps above 12, and so 10 + 10 = 8. This computation required a lot of movement though, so alternatively we could have turned one of the 10's into its equivalent number and done the calculation.
So instead of 10 + 10 we would have 10 + (-2) = 10 - 2 = 8, which is a much faster way of getting to the answer.
Finally we'll define what an interval is in our system, an interval is the number of steps required to get from number to another. So the interval from 10 to 3 in our system is the number added to 10 which gets us to 3. Equivalently we're just asking what x is in the question 10 + x = 3, if you remember from earlier the answer to that question is 5.
Let's try another, if someone told us to find the interval between 5 and 9, then we're trying to figure out what x is in the equation 5 + x = 9, and so if we isolate for x, then we get x = 9 - 5 = 4, so the interval from 5 to 9 is 4. When we want to talk about intervals without having write a lot of english language we can write:
I(5, 9) = 4
Let's figure out I(9, 5) now, so we're trying to figure out what we have to add to 9 to get to 5, one might come up with the answer of 8, since first you add 3 to get to 0, and then add 5 to get to 5. In this case it's easier to see the answer could be -4 since there is less distance moved when you go in this direction. So we could also say that I(9, 5) = -4. This example shows us that I(9,5) can be written in two ways but they both mean the same thing.
Now in the most general example we could have two numbers X and Y, then the interval between them is Y - X which is right because X + (Y - X) = Y. So we have:
I(X, Y) = Y - X
We now know that given any number from 0 to 11, we have a number from -11 to 0 which is equivalent to it, we in a table we can see all the equivalences, we can also choose the one which uses the smallest numbers
The small representation is useful since it can represent a small distance offset, so if we wanted to know I(10, 0) can consider 0 - (-2) instead of 0 - 10 which is easier to figure out since the distance offset is smaller.
In this way we can make some notation which gives us the small representation of a number so that when we write s(10) it really means -2 and when we write s(-8) it really means 4.
This gives us an efficient way to compute intervals, finally we can say that:
I(X, Y) = Y - s(X)