<s>[INST] Rule for multiplying rational numbers (follow-up) I was thinking about the rule where
$\space\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$ and why it worked, and I found this question: Rule for multiplying rational numbers.
The best answer was:

Did:
One knows that $a/b$ may be defined as the number $x$ such that
$bx=a$, and that $c/d$ may be defined as the number $y$ such that
$dy=c$.
If one wants the multiplication on these objects to be associative and
commutative as it is on the integers, one should ask that
$ac=(bx)(dy)=(bd)(xy)$ hence that the object $xy$ fits the definition
of $(ac)/(bd)$.

And I was thinking, if in the end we get $\frac{(xy)(bd)}{bd}=\frac{ac}{bd}$, how do we know how to simplify the two $(bd)$s here? Don't we need the property itself $\space\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$ to know how to do that?
$$\frac{(xy)(bd)}{bd}=\frac{(xy)\cdot (bd)}{1\cdot (bd)}=\frac{xy}{1}\cdot\frac{bd}{bd}=\frac{xy}{1}\cdot 1=\frac{xy}{1}=xy$$
So do we just assume the property is true inductively (not mathematically inductively), or am I missing something?
PS: As a beginner, I may say some ridiculous things, excuse me for that.
 [/INST] A completely formal proof depends on the particular axioms and definitions you work with, but as far as the intuition of it goes, consider the following points.


*

*The fraction $\frac{1}{b}$ denotes the multiplicative inverse of $b \ne 0\,$, which is to say that $x=\frac{1}{b}$ is the unique solution to $b \cdot x = 1\,$. The fraction $\frac{a}{b}$ can be thought of as shorthand for $a \cdot \frac{1}{b}\,$.

*Multiplying together $b \cdot \frac{1}{b}=1$ and $d \cdot \frac{1}{d}=1$ gives $b \cdot \frac{1}{b} \cdot d \cdot \frac{1}{d}=1\,$. Since multiplication is commutative, this can be rewritten as $(b \cdot d) \cdot (\frac{1}{b} \cdot \frac{1}{d})=1\,$. But the latter means that $\frac{1}{b} \cdot \frac{1}{d}$ is the multiplicative inverse of $b\cdot d\,$, which proves that $\frac{1}{b} \cdot \frac{1}{d} = \frac{1}{b \,\cdot\, d}\,$.
Therefore $\;\frac{a}{b}\cdot \frac{c}{d} \stackrel{\;(1)\;}{=} a \cdot \frac{1}{b} \cdot c \cdot \frac{1}{d} = (a \cdot c) \cdot (\frac{1}{b} \cdot \frac{1}{d}) \stackrel{\;(2)\;}{=} (a \cdot c) \cdot \frac{1}{b \,\cdot\, d} \stackrel{\;(1)\;}{=} \frac{a \,\cdot c}{b \,\cdot\, d}\,$.
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