It's great that you have discovered the secret of subtraction which is equivalent to negation and addition. Similarly, you could make the same case division which is equivalent to reciprocal and multiplication. Why would you need division?

However, I make two arguments:

**Argument 1:**

How would you solve the following problem without the concept of subtraction?

Anand had 10 apples, he gave 3 apples to Mike. How many apples does Anand have now?

This is not a simple negation and addition problem. -3 apples does not have a real life connotation, unless we come up with innovative ways of rewording the original problem.

For example, Anand had -3 apples, does not make much sense by itself, does it?

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**Argument 2:**

Think of subtraction and division as training wheels. They are absolutely necessary when you start learning to ride a bike. But completely unnecessary when you have mastered that skill. Subtraction and division are the training wheels.

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Now to commutation. I am curious to know why you think commutation is not needed. Is it because you think commutative property is obvious? There are many operations that do not commute.

Biggest examples of operations that do not commute: Subtraction and Division However, writing their equivalent representations absolutely commutes, because multiplication commutes.

To come up with more examples where commutative property does not hold, we have to formalize the definition.

Let's say we have a function with two arguments: f(x, y). Then f(x, y) is commutative if f(x, y) = f(y, x), i.e. when it is irrelevant which argument is first and which is second. If f(x, y) = x + y, then it absolutely commutes.

Here are some examples of functions. f(x, y) = x + y where x and y are strings and "+" operation is string concatenation. Then f("COM", "MUTE") = "COMMUTE" is absolutely not equal to f("MUTE", "COM") = "MUTECOM".

Know that an operator satisfies the commutative property can be used to simplify how things are done. Easiest example, since addition is commutative and associative, you know that you can add in any order. Therefore, you get the famous sum of first n-integers formula exploiting this property \sum_{k=1}^n k = \frac{n(n+1)}{2} .