Proving that fractions are the same as division

[logicgate]

TL;DR Summary

I'm trying to prove that fractions are the same as division using the axiom of multiplicative inverse.

So using the multiplicative inverse axiom we have :
1) x . x^-1 = 1
2) x . (1/x) = 1
I have no idea why do mathematicians define the multiplicative inverse of a number x to be the "fraction" 1/x.
But I know for sure that multiplying any number a for example by the multiplicative inverse of x is the same as "a divided by x"

 

[PeroK]

TL;DR Summary: I'm trying to prove that fractions are the same as division using the axiom of multiplicative inverse.

So using the multiplicative inverse axiom we have :
1) x . x^-1 = 1
2) x . (1/x) = 1
I have no idea why do mathematicians define the multiplicative inverse of a number x to be the "fraction" 1/x.
But I know for sure that multiplying any number a for example by the multiplicative inverse of x is the same as "a divided by x"

If we start with the integers, and assume we can add and multiple them, we can ask:

Given two integers $a$ and $b$, is there a number $c$ such that $a \times c = b$?

Sometimes there is a suitable integer and sometimes there isn't. If we want that equation to be solvable for all non-zero $a$, then we have to introduce some new numbers that aren't integers. So, by definition, we say:

$c \equiv \frac b a$ is the number such that $a \times c = b$.

And that effectively defines the rational numbers, as quotients of integers.

Then we have a special case where $b = 1$, hence $c = \frac 1 a$ and $c \times a = 1$. In this case, we call $c$ the multiplicative inverse of $a$ and we also write $c = a^{-1}$.

That leads more generally to the equation: $\frac b a = ba^{-1}$.