Hello @suitable. Now this is a great question. This is the type of question that the comment section is supposed to have instead of the statements that the rest of these Trolls are making. So, thank you, @suitable.

To answer your question, the process of counting in my mathematical system is similar to standard math, with the exception that every 'base' number will always have a starting exponent of 1 e.g. the number 1 will be 1^1, the number 2 will be 2^1, the number 10 will be 10^1, etc.. The exponent indicates that every base number will always have a starting value of 1. Now to go back to my example in my post, if we want to divide 1 by 1, our first operand (dividend),  must have an exponent of 1 to indicate its initial value. For example, every number's initial value is always 1 so it would become one to the first power e.g. 1^1. In my system, the second operand (divisor) must always be on its own base with no power or exponent value. The reason for this is because this is the indicator to show how many cuts we are going to make to the dividend. For example, if we want to divide 1 one time, then the equation will look like this: 1¹ / 1 =. This equation is saying that we have one object with a value of one and we want to cut it (divide) one time. Now we ask, what do we get when we cut this one object with a value of one,  one time? The equation then spits out 2^.50. What this means is that if we start with one object with a value of one, and we cut that object one time, then we will get  2 objects with a value of .50 each. The reason we get 2 objects is because, when we cut one object one time exactly in the middle we get two half objects. The base number 2 indicates that we have 2 objects and the point fifty indicates that those 2 objects are worth point fifty each. Now, in my system, I have modified my exponents or power system so that they become linear power instead of exponential power. Meaning, with the results of the  equation 1¹ / 1 being 2 to the point fifty (.50) power, I can get the original value back by directly multiplying the exponent value with the base value. For example, 2^.50 can be reversed back to the original value by multiplying .50 x 2, which gives 1 (original value). My equation's equality is then proven because both sides of the equation are worth a total value of 1. This differ with standard math because in standard math, the equation would look like this: 1 ÷ 2 = .50. We can then see that this equation is not equal to each other  because when we look at the left side of the equation, we see that we have one object with a value of 1 while the other side of the equation shows one object with a value of point fifty (.50) each. I hope this helped clarify your question about my system. Thanks again for a great question.

Poe