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.

The question requests the following:
Part I - a proof for
\begin{equation*}
\frac{a}{b}\times \frac{c}{d}=\frac{a\times c}{b\times d}
\end{equation*}
Part II - a proof for
\begin{equation*}
\frac{a}{b}\times \frac{c}{c}=\frac{a}{b}
\end{equation*}
Part III - a proof for
\begin{equation*}
\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}
\end{equation*}
Part IV - What is "the mathematics topic of this question (number theory, logic, etc)."
Part V - a reference if the answer is not a full answer.
ANSWERS IN REVERSE ORDER:
Part V - Below you will find full answers to the questions but I have included a reference at the end of this post for related information about this topic.
Part IV - The mathematics topic is elementary arithmetic.  Discussions and proofs within this topic sometimes utilize concepts and results from other areas, such as the foundations of mathematics, algebra, and logic.
In order to prove the identities of Parts I, II, and III, we need a definition, some postulates, and some simple results we can call "laws."
Fundamentals Required for the Proofs
We need to start with a valid definition of a fraction if we are to provide a complete, valid mathematical proof. The following definition will suffice for our purpose here:
Definition: a fraction is represented symbolically as $\frac{a}{b}$, where a is a non-negative integer called the numerator and b is a positive integer called the denominator. A fraction is the value of a within the context that this value is the count of units such that b units are required for the sum to be 1.
Note how closely this definition follows the usual physical model of a fraction in terms of slices of pie. Also notice that this definition provides justification for saying that a fraction is a number, since the numerator is a number and all the fraction does is provide context.
The Fundamentals we will need follow:

*

*Redundant or unnecessary parentheses can be removed, e.g., ((a)) is the same as (a) and in some cases (a) the same as a.


*The multiplicative identity law $1 \cdot a = a$.  Since we are assuming that multiplication is commutative, this also gives $a = a \cdot 1$.


*The multiplicative associate law extended to include fractions, i.e., $(a\cdot b)\cdot c = a\cdot(b\cdot c)$.


*The multiplicative commutative law extended to include fractions, i.e., $a\cdot b=b\cdot a$, where we allow the variables to be fractions.


*Law 1 (multiplicative identity as a fraction)
Given $a\neq0$,
\begin{equation*}
\frac{\boldsymbol{a}}{\boldsymbol{a}}=\mathbf{1}.
\end{equation*}
Proof: This law follows directly from our definition of a fraction.
Reminder: this is a law about fractions, not a law about division.


*Law 2 ($0$ as a fraction)
Given $a\neq0$,
\begin{equation*}
\frac{0}{a}=0.
\end{equation*}
Proof: Using our definition of a fraction, this law is saying that if we have no small units, the size of which is such that a of them are required for the sum to be 1, then we have no larger units either.  If our laws and theorems are to apply only to positive integers, then we don't need this law.  This is a law about fractions, not a law about division.


*Law 3 (fraction addition with common denominator)
Given $c\neq0$,
\begin{equation*}
\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}.
\end{equation*}
Proof: This law closely follows the definition of a fraction. Here the context provided by the denominators for each of the fractions is the same. The two terms on the left side of the equation each express the individual units to be counted and does so as individual fractions. Using the plus sign extended to fractions just means that we are to add the units represented by the fraction within the context of the unit size expressed by the denominator. (If we like, we can make this extension of addition to fractions an explicit law.) The right side expresses the same thing but uses a single fraction to express the sum. The right side is also within the same unit size context, so the same units are being counted on both sides of the equation.


*We also need an extension of our simple definition of multiplication so that the right factor can be a fraction:
Definition: Multiplication Extended to Fractions
Given $b\neq0$ and $X$ as a non-negative integer,
\begin{equation*}
X\cdot\left(\frac{a}{b}\right)=\boldsymbol{X}\cdot\frac{a}{b}=\underset{X \text{ copies}}{\underbrace{\frac{a}{b}+\frac{a}{b}+\ldots +\frac{a}{b}}} ,
\end{equation*}
and to be clear, for $X<3$,
\begin{align*}
2\cdot\left(\frac{a}{b}\right)&=2\cdot\frac{a}{b}=\frac{a}{b}+\frac{a}{b} \\
1\cdot\frac{a}{b}&=\frac{a}{b} \\
0\cdot\frac{a}{b}&=0. 
\end{align*}


*When working with the product of two fractions, we will need the following:
Lemma: $a\neq0$ and $b\neq0$ implies $a\cdot b\neq0$.
Proof: By the definition of multiplication, $a \cdot b$ is the sum of $1$ or more copies of $b$.  Variable $b$, in turn, represents a sum of $b$ copies of the number $1$.  In total, there are $a\cdot b$ ones added together, which is just a positive number and therefore not zero.
That is all we really need, although we can make our proof shorter with the following theorem.
Theorem 2 (constant times a fraction)
Given $b\neq0$,
\begin{equation*}
c\cdot\frac{a}{b}=\frac{c\cdot a}{b}.
\end{equation*}
Proof:
Case: $c=0$.

By our definition of multiplication extended to fractions,
\begin{equation*}
0\cdot\frac{a}{b}=0.
\end{equation*}
By a property of zero,  $0\cdot a=0$, and using these equal expressions in the numerator of fractions with the same denominator, we have the identity
\begin{equation*}
\frac{0\cdot a}{b}=\frac{0}{b}.
\end{equation*}
By law 2 ($0$ as a fraction),
\begin{equation*}
=0.
\end{equation*}
Therefore, since both expressions are equal to zero,
\begin{equation*}
0\cdot\frac{a}{b}=\frac{0\cdot a}{b}.
\end{equation*}

Case: $c=1$.

By our definition of multiplication extended to fractions,
\begin{equation*}
1\cdot\frac{a}{b}=\frac{a}{b}.
\end{equation*}
By a property of the multiplicative identity,
\begin{equation*}
\frac{1\cdot a}{b}=\frac{a}{b}.
\end{equation*}
Then, since both expressions are equal to $\frac{a}{b}$,
\begin{equation*}
1\cdot\frac{a}{b}=\frac{1\cdot a}{b}.
\end{equation*}

Case $c=2$:

By our definition of multiplication extended to fractions,
\begin{equation*}
2\cdot\frac{a}{b}=\frac{a}{b}+\frac{a}{b}.
\end{equation*}
By Law $3$ (addition of fractions with a common denominator)
\begin{equation*}
=\frac{a+a}{b}.
\end{equation*}
By the definition of multiplication
\begin{equation*}
=\frac{2\cdot a}{b}.
\end{equation*}

Case $c>2$:

By our definition of multiplication extended to fractions, we have
\begin{equation*}
c\cdot\frac{a}{b}=\underset{c\text{ copies}}{\underbrace{\frac{a}{b}+\frac{a}{b}+\ldots +\frac{a}{b}.}}
\end{equation*}
Using Law 3 (fraction addition with common denominator) applied to our first 2 addends, we have
\begin{equation*}
c\cdot\frac{a}{b}=\frac{a+a}{b}+\underset{c-2\text{ copies}}{\underbrace{\frac{a}{b}+\frac{a}{b}+\ldots +\frac{a}{b}}} ,
\end{equation*}
$\text{and we}$ continue this application of Law 3 until we have
\begin{equation*}
c\cdot\frac{a}{b}=\frac{\overset{c}{\overbrace{a+a+\ldots +a}}}{b},
\end{equation*}
and then using the definition of multiplication,
\begin{equation*}
=\frac{c\cdot a}{b},
\end{equation*}
so that

\begin{equation*}
c\cdot\frac{a}{b}=\frac{c\cdot a}{b}.
\end{equation*} .
Q.E.D.

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 $b\neq0$,
\begin{equation*}
\frac{a}{b}\cdot c=\frac{a\cdot c}{b}.
\end{equation*}
Proof of Part III
\begin{equation*}
\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}
\end{equation*}
Proof: this is Law 3
Proof of Part II
\begin{equation*}
\frac{a}{b}\times \frac{c}{c}=\frac{a}{b}
\end{equation*}
Proof:
$\frac{a}{b}\cdot \frac{c}{c}=\frac{a}{b}\cdot 1$   by Law 1 (multiplicative identity as a fraction)
$=\frac{a\cdot 1}{b}$   by Theorem 2 (constant times a fraction)
$=\frac{a}{b}$   by the multiplicative identity law
Proof of Part I
We wish to prove the following:
Let $a, b, c$, and $d$ be non-negative integers, with $b\neq0$ and $d\neq0$.
Then
\begin{equation*}
\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}.
\end{equation*}
$\frac{a}{b}\cdot \frac{c}{d}$
$=\frac{1\cdot a}{b}\cdot\frac{c}{d}$ multiplicative identity law
$=\left(\frac{1\cdot a}{b}\right)\cdot \frac{c}{d}$ multiplicative associate law
$=\left(1\cdot \frac{a}{b}\right)\cdot \frac{c}{d}$ Theorem 2 (constant times a fraction)
Noting by our Lemma that $b\cdot d{\neq}0$,
= $\left(\frac{b\cdot d}{b\cdot d}\cdot \frac{a}{b}\right)\cdot \frac{c}{d}$    multiplicative identity as a fraction
Here we have just introduced the heart of the matter in the form of the factor $\frac{b\cdot d}{b\cdot d}$, which allows us essentially to first scale by $b \cdot d$ and then reverse scale by the same factor later in the process. Continuing with these formal steps:
$=\left(\left(\frac{b\cdot d}{b\cdot d}\right)\cdot \frac{a}{b}\right)\cdot \frac{c}{d}$ multiplicative associate law
$=\left(\left(\frac{1\cdot b\cdot d}{b\cdot d}\right)\cdot \frac{a}{b}\right)\cdot \frac{c}{d}$ multiplicative identity law
$=\left(\left(\frac{1\cdot \left(b\cdot d\right)}{b\cdot d}\right)\cdot \frac{a}{b}\right)\cdot \frac{c}{d}$ multiplicative associate law
$=\left(\left(\frac{1}{b\cdot d}\cdot \left(b\cdot d\right)\right) \cdot \frac{a}{b}\right)\cdot \frac{c}{d}$ Theorem 2 (constant times a fraction)
$=\left(\frac{1}{b\cdot d}\cdot \left(\left(b\cdot d\right)\cdot \frac{a}{b}\right)\right)\cdot \frac{c}{d}$ multiplicative associate law
$=\left(\frac{1}{bd}\cdot \left(\frac{\left(b\mathrm{*}d\right)\mathrm{*}a}{b}\right)\right)$ * $\frac{c}{d}$ Theorem 2 (constant times a fraction)
$=\left(\frac{1}{b*d}\mathrm{*}\left(\frac{b\mathrm{*}\left(\mathrm{d*}a\right)}{b}\right)\right)$ * $\frac{c}{d}$ multiplicative associate law
$=\left(\frac{1}{b*d}\mathrm{*}\left(\frac{b}{b}\mathrm{*}\left(\mathrm{d*}a\right)\right) \right)$ * $\frac{c}{d}$ Theorem 2 (constant times a fraction)
$=\left(\frac{1}{b*d}\mathrm{*}\left(1\mathrm{*}\left(\mathrm{d*}a\right)\right) \right)$ * $\frac{c}{d}$ multiplicative identity as a fraction
$=\left(\frac{1}{b*d}\mathrm{*}\left( \left(\mathrm{d*}a\right)\right) \right)$ * $\frac{c}{d}$ multiplicative identity law
$=\left(\frac{1}{b*d}\mathrm{*}\left(\mathrm{d*}a \right) \right)$ * $\frac{c}{d}$ removing redundant parentheses
$=\left(\frac{1}{b\mathrm{*}d}\mathrm{*}\left( \left(\mathrm{d*}a\right)\mathrm{*}\frac{c}{d}\right)\right)$ multiplicative associate law
$=\left(\frac{1}{b*d}\mathrm{*}\left(\frac{\left(\mathrm{d*}a\right)\mathrm{*}c}{d}\right)\right)$ Theorem 2 (constant times a fraction)
$=\left(\frac{1}{b*d}\mathrm{*}\left(\frac{\mathrm{d*}\left(a\mathrm{*}c\right)}{d}\right)\right)$ multiplicative associate law
$=\left(\frac{1}{b*d}\mathrm{*}\left(\frac{\mathrm{d}}{d}\mathrm{*}\left(a\mathrm{*}c\right)\right) \right)$ Theorem 2 (constant times a fraction)
$=\left(\frac{1}{b*d}\mathrm{*}\left(1\mathrm{*}\left(a\mathrm{*}c\right)\right) \right)$ multiplicative identity as a fraction
$=\left(\frac{1}{b*d}\mathrm{*}\left(\left(a\mathrm{*}c\right)\right) \right)$ multiplicative identity law
$=\left(\frac{1}{b*d}\mathrm{*}\left(a\mathrm{*}c\right) \right)$ removing redundant parentheses
$=\left(\frac{1\mathrm{*}\left(a\mathrm{*}c\right)}{b*d}\right)$ Theorem 2 (constant times a fraction)
$=\left(\frac{\left(a\mathrm{*}c\right)}{b*d}\right)$ multiplicative identity law
$=\frac{ a*c}{b*d}$    removing unnecessary parentheses
Therefore,
\begin{equation*}
\frac{a}{b}\text{ * }\frac{c}{d}=\frac{a\mathrm{*}c}{b\mathrm{*}d}
\end{equation*}
Q.E.D.
There appears to be issues with all of the other "proofs" currently on this page as well as on the "duplicate" at Understanding the multiplication of fractions. The fraction product rule question is not trivial, but a 38-page document on ResearchGate can be found that describes in detail problems with the current responses here and on the "duplicate" page. That document also provides 2 complete and valid proofs of the fraction product rule as well as a description that gives an intuitive understanding of the rule. The document can be found at On the Fraction Product Rule
.
It should be noted that I do not consider the proofs above to be the most elegant proofs that are possible.  One way we can create shorter proofs is by hiding the complexity of the proofs within other theorems.  Also be aware that by developing a very different definition for a fraction, it is possible to create much clearer and shorter proofs.  In that case, some of the complexity is "hidden" in a clear and concise definition of number.
