Why are rational numbers (Q) the obvious expansion of integers (Z)?

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BS"D

If we take the set of natural numbers [N] as an example, expanding it to include zero and the negatives so that we get [Z] sounds (and feels) like the next logical step, so I'm okay with that.

But expending [Z] so that we get [Q], that doesn't sit right with me, mostly because of how Q is defined. It seems arbitrary to me. Why choose a quotient of two integers? Why not multiplication? Or, for that matter, any other valid mathematical expression.

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This is all within the more general context of arbitrary rings and fields. In those, we can define other kinds of multiplication, and other kinds of cancellation. We can also build other structures on the integers: we could include an irrational number, or do many other things that would result in something different. We primarily construct the rational numbers from the integers because it is a useful thing to be able to solve an equation like $m\times n=p$.

The integers, $\mathbb{Z}$, have the mathematical structure of a ring. This means that we can add two integers together and we will still get another integer, and we can subtract in order to "undo" addition. However, we can also multiply two integers and the result will always be an integer, but it isn't always possible to "undo" multiplication: Dividing an integer by an integer won't always result in an integer! So, in order to be able to undo multiplication, we construct the rational numbers, $\mathbb{Q}$, which has the special property that it is a field and also that it is the smallest field (the 'smallest' structure) containing the integers which allows us to divide.

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If we take the set of natural numbers [N] as an example, expanding it to include zero and the negatives so that we get [Z] sounds (and feels) like the next logical step

Consider that  "$\,\mathbb{Z}\,$ is to $\,\mathbb{N}\,$ for addition"  precisely what  "$\,\mathbb{Q}\,$ is to$\,\mathbb{Z}\,$ for multiplication" . The former allows equations like $\,x+a=b\,$ to be solved for arbitrary $\,\forall a,b \in \mathbb{N}\,$. The latter allows equations like $\,x\cdot c=d\,$ to be solved for arbitrary $\,\forall c \ne 0,d \in \mathbb{Z}\,$. So it is a similar "logical step" in that way.

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Passing from $\mathbf N$ (which, incidentally, contains $0$) to $\mathbf Z$ is done, mathematically, to be able to solve linear equations of the form $$a+x=b$$ in all cases. This means one tries to have a domain on which the function $x\mapsto x+a$ is invertible, whatever $a$.

The abstract generalisation is the so-called group defined by a cancellative monoid.

Quite similarly, for multiplication on $\mathbf Z$, one extends it to be able to solve equations of the form $$ ax=b $$ in all cases for which it makes sense, i.e. when $a\ne 0$. The problem has the same solution for all (commutative) integral domains leading to their field of fractions, and is a little more complex for non-integral domains, leading to their total ring of fractions.