I Proving that fractions are the same as division

AI Thread Summary
The discussion explores the relationship between fractions and division through the multiplicative inverse axiom, stating that multiplying by the inverse of a number is equivalent to division. It questions why mathematicians define the multiplicative inverse of a number x as the fraction 1/x. The conversation highlights that introducing rational numbers allows for solving equations like a × c = b for integers a and b. The definition of rational numbers as quotients of integers is established, leading to the conclusion that c = b/a can be expressed as c = a^-1 when b equals 1. The thread also mentions the concept of embedding integral domains into fields of fractions, emphasizing the broader mathematical context.
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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