I Proving that fractions are the same as division

logicgate
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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"
 
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logicgate said:
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}##.
 
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This may be of interest:

https://en.m.wikipedia.org/wiki/Field_of_fractions

Every integral domain, i.e., a ring R in which a.b=0 implies a=0 or b=0, may be embedded in a field, called the Field of Fractions, through a specific process , described in the link.In your case, the Integral Domain of the Integers, can be embeded in the field of Rationals. In this case, if r is an element of said integral domain, the element (1/r) is used to denote it's multiplicative inverse in the associated field of fractions. Notice other examples like that of the Integral Domain ##\{ a+ib: a,b \in \mathbb Z \}##, has ##\{c+id: c,d \in \mathbb Q \}## as its field of fractions.
Maybe @fresh_42 can elaborate.
 
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