<s>[INST] Multiplying and adding fractions Why multiplying fractions is equal to multiply the tops, multiply the bottoms?
$$\frac{a}{b}\times \frac{c}{d}=\frac{a\times c}{b \times d},$$
And why
$$\frac{a}{b}\times \frac{c}{c}=\frac{a}{b},$$
Also why
$$\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}.$$
I understand it, but I want a mathematical approach as a math student proves it. Also I want to know the mathematics topic of this question (number theory, logic, etc). A full answer is not necessary. Just a reference.
 [/INST] 
Complete Alternative Fraction Product Rule Proofs
In a previous post I showed how to create a legitimate elementary arithmetic proof of the fraction product rule using a field theorem about division. That proof was spread out amongst discussions of a number of issues, and some of the steps needed were not explicitly shown. I therefore have written out the complete proof below.
I will not duplicate the fundamentals required for this proof because they can be found in the my first post, which also contains another proof of the fraction product rule.
Theorems to Make Shorter Proofs
Theorem 2 (constant times a fraction)
Given b${\neq}$0,
\begin{equation*}
\boldsymbol{c}\cdot \frac{\boldsymbol{a}}{\boldsymbol{b}}=\frac{\boldsymbol{c}\cdot \boldsymbol{a}}{\boldsymbol{b}}.
\end{equation*}
Proof: (See my first post)
Notice that here we are assuming that the variables in the above theorem are non-negative integers. We will, of course, want these variables to eventually be other numbers, such as fractions. In fact, the theorem we are trying to prove, that
\begin{equation*}
\frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d},
\end{equation*}
can be viewed as part of that process. In particular, we are replacing the variable c in Theorem 2 (constant times a fraction) with a fraction.
Because we assume the multiplicative commutative law, in our proof we also consider the following as justified by Theorem 2 (constant times a fraction):
Given b${\neq}$0,
\begin{equation*}
\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \mathbf{c}=\frac{\boldsymbol{a}\cdot \boldsymbol{c}}{\boldsymbol{b}}.
\end{equation*}
Theorem 9 (fraction as product and inverse)
For b${\neq}$0,
\begin{equation*}
\frac{\boldsymbol{a}}{\boldsymbol{b}}=\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}.
\end{equation*}
Proof:
$\frac{\boldsymbol{a}\cdot \boldsymbol{c}}{\boldsymbol{b}}=a\cdot \frac{\boldsymbol{c}}{\boldsymbol{b}}$    (Theorem 2, Constant Times a Fraction)
$\frac{\boldsymbol{a}\mathbf{\cdot}\mathbf{1}}{\boldsymbol{b}}=a\cdot \frac{\mathbf{1}}{\boldsymbol{b}}$   (substituting c=1)
$\frac{\boldsymbol{a}}{\boldsymbol{b}}=a\cdot \frac{\mathbf{1}}{\boldsymbol{b}}$   (multiplicative identity)
Theorem 10: (multiplicative inverse as a fraction)
If a${\neq}$0, then the multiplicative inverse of a is the fraction $\frac{\mathbf{1}}{\boldsymbol{a}}$ .
Proof:
$a\cdot \frac{1}{a}=\frac{a\cdot 1}{a}$  (Theorem 2, Constant Times a Fraction)
$=\frac{a}{a}$   (multiplicative identity)
$=1$   (Law 1)
Theorem 11: (product of inverses)
Given a${\neq}$0 and b${\neq}$0,
\begin{equation*}
\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{a}}=\frac{\mathbf{1}}{\boldsymbol{a}\cdot \boldsymbol{b}}
\end{equation*}
Comment: note that this theorem and its proof is very similar to the fraction product law itself and the proof presented in my previous post. In this proof, factor $a\cdot b$ is first introduced followed by factor $\frac{\mathbf{1}}{\boldsymbol{a}\cdot \boldsymbol{b}}$ . In the proof of the fraction product rule in my first post, $\frac{a\cdot b}{a\cdot b}$ was introduced as the multiplicative identity and then separated during the rest of the proof.
Proof:
$a\cdot b\cdot \left(\frac{1}{b}\cdot \frac{1}{a}\right)=a\cdot b\cdot \frac{1}{b}\cdot \frac{1}{a}$   (removing unnecessary parenthesis)
$=a\cdot \left(b\cdot \frac{1}{b}\right)\cdot \frac{1}{a}$   (associative law)
$=a\cdot 1\cdot \frac{1}{a}$   (Theorem 10, multiplicative inverse as a fraction)
$=a\cdot \frac{1}{a}$   (multiplicative identity)
$=1$   (Theorem 10, multiplicative inverse as a fraction)
$=a\cdot b\cdot \frac{1}{a\cdot b}$   (Theorem 10, multiplicative inverse as a fraction)
Since $a\cdot b\cdot \left(\frac{1}{b}\cdot \frac{1}{a}\right)=a\cdot b\cdot \frac{1}{a\cdot b}$ , we next multiply both sides of this equality by $\frac{1}{a\cdot b}$ . This is possible because we are given that a${\neq}$0 and b${\neq}$0, and by the lemma we know that $a\cdot b\neq 0$. The result is
$\frac{1}{a\cdot b}\cdot a\cdot b\cdot \left(\frac{1}{b}\cdot \frac{1}{a}\right)=\frac{1}{a\cdot b}\cdot a\cdot b\cdot \frac{1}{a\cdot b}$
$1\cdot \left(\frac{1}{b}\cdot \frac{1}{a}\right)=1\cdot \frac{1}{a\cdot b}$   (Theorem 10, multiplicative inverse as a fraction)
$\left(\frac{1}{b}\cdot \frac{1}{a}\right)=\frac{1}{a\cdot b}$   (multiplicative identity)
$\frac{1}{b}\cdot \frac{1}{a}=\frac{1}{a\cdot b}$   (removing unnecessary parentheses)

Proof 3 of the Fraction Product Rule
With Theorems 9, 10, and 11, we can prove the fraction product rule as shown next.
$\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\left(\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}\right)\cdot \left(\boldsymbol{c}\cdot \frac{\mathbf{1}}{\boldsymbol{d}}\right)$   (Theorem 9, fraction as product and inverse)
$=\boldsymbol{a}\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \boldsymbol{c}\right)\cdot \frac{\mathbf{1}}{\boldsymbol{d}}$   (multiplicative associative law)
$=\boldsymbol{a}\cdot \left(\boldsymbol{c}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}\right)\cdot \frac{\mathbf{1}}{\boldsymbol{d}}$   (multiplicative commutative law)
$=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{d}}\right)$   (multiplicative associative law)
$=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}\cdot \boldsymbol{d}}\right)$   (Theorem 11, product of inverses)
$=\frac{\left(a\cdot c\right)\cdot 1}{b\cdot d}$   (Theorem 2, constant times a fraction)
$=\frac{\left(a\cdot c\right)}{b\cdot d}$   (multiplicative identity)
$=\frac{a\cdot c}{b\cdot d}$   (removing unnecessary parentheses)
Q.E.D.

Proof 4 of the Fraction Product rule
Methodology
Using the step-by-step form used in Proof 3 of the Fraction Product Rule makes it especially easy to make shorter proofs by collecting the transformations effected by contiguous proof lines and turning them into a theorem. Using Proof 3 as an example, we collect the first 5 lines to create Theorem 12.
Theorem 12 (product of fractions as a product of inverses)
Given b${\neq}$0 and d${\neq}$0
\begin{equation*}
\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}\cdot \boldsymbol{d}}\right)
\end{equation*}
Proof:
$\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\left(\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}\right)\cdot \left(\boldsymbol{c}\cdot \frac{\mathbf{1}}{\boldsymbol{d}}\right)$   (Theorem 9, fraction as product and inverse)
$=\boldsymbol{a}\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \boldsymbol{c}\right)\cdot \frac{\mathbf{1}}{\boldsymbol{d}}$   (multiplicative associative law)
$=\boldsymbol{a}\cdot \left(\boldsymbol{c}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}\right)\cdot \frac{\mathbf{1}}{\boldsymbol{d}}$   (multiplicative commutative law)
$=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{d}}\right)$   (multiplicative associative law)
$=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}\cdot \boldsymbol{d}}\right)$   (Theorem 11, product of inverses)
Q.E.D.
Next, we consolidate the remaining lines into a theorem. We need to show
\begin{equation*}
\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}\cdot \boldsymbol{d}}\right)=\frac{\boldsymbol{a}\cdot \boldsymbol{c}}{\boldsymbol{b}\cdot \boldsymbol{d}}
\end{equation*}
but this is just an application of
Theorem 9 (fraction as product and inverse):
For b${\neq}$0,
\begin{equation*}
\frac{\boldsymbol{a}}{\boldsymbol{b}}=\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}.
\end{equation*}
Thus, our Proof 4 of the Fraction Product Rule requires only two steps.

Proof 4
Given b${\neq}$0 and d${\neq}$0,
$\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}\cdot \boldsymbol{d}}\right)$   (Theorem 12, product of fractions as a product of inverses)
$=\frac{\boldsymbol{a}\cdot \boldsymbol{c}}{\boldsymbol{b}\cdot \boldsymbol{d}}$   (Theorem 9, fraction as product and inverse
Q.E.D.
Results like this suggest to me that, if we are allowed to use theorems in a proof, then the complexity should not be measured by the number of steps. This is because we can easily hide complexity within a theorem. If we must prove a theorem strictly from axioms, then the proofs can be fairly long. However, perhaps this is the best way to provide a simple measure of the complexity of a proof.
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