Multiplying and adding fractions

Question

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.

Answer 1

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.

Answer 2

I'll give an abstract look at why these identities hold in arbitrary fields.

In some sense, this is the definition of addition and multiplication of fractions. Specifically, we can define division to be the multiplication of the numerator with the inverse of the denominator.

For example, we can write $\frac{a}{b}=ab^{-1}$, identifying division as taking the multiplicative inverse. Then $$ \frac{a}{b}\cdot \frac{c}{d} =(a\cdot b^{-1})\cdot (c\cdot d^{-1})$$
Now, if we want multiplication to be associative and commutative, then we would find that $$(a\cdot b^{-1})\cdot (c\cdot d^{-1})=(a\cdot c)\cdot (d^{-1}\cdot b^{-1})$$ It is a general fact that $(xy)^{-1}=y^{-1}x^{-1}$, which can be verified directly by multiplying $(xy)$ by both $y^{-1}x^{-1}$ and $(xy)^{-1}$. Then we find that $$ \frac{a}{b}\cdot \frac{c}{d}=(a\cdot c)\cdot (d^{-1}\cdot b^{-1})=(a\cdot c)\cdot (b\cdot d)^{-1}=\frac{a\cdot c}{b\cdot d}$$

Similar justifications can be given for the remaining identities. For example, $\frac{c}{c}=1$ can be verified by $\frac{c}{c}=c\cdot c^{-1}=1$. Again, this is rather definitional.

There's another way in which we can view these identities for fractions as well. There is also another approach, mirroring the construction of the integers. If we're given an integral domain (i.e. a commutative ring in which $ab=0$ implies that one of $a$ and $b$ are equal to $0$) $(R,+,\cdot,0,1)$, where $+$ is some notion of "addition", $\cdot$ some notion of "multiplication", $0$ the identity for addition, and $1$ the identity for multiplication, then we can form a field $\operatorname{Quot}(R)$ called the quotient field or fraction field of $R$.

Specifically, we define the underlying set of $\operatorname{Quot}(R)$ by the quotient $[R\times (R\setminus\{0\})]/\sim$, where $\sim$ is the equivalence relation defined by $(a,b)\sim(c,d)$ if and only if $a\cdot d=b\cdot c$. The idea is that the ordered pairs $(a,b)\in R\times (R\setminus \{0\})$ represent the fractions of elements in $R$, but we also want to identify "equivalent" fractions, and thus we introduce the equivalence relation.

We'll represent the equivalence class of an element $(a,b)$ in $\operatorname{Quot}(R)$ by $\frac{a}{b}$.

Then the definition of addition and multiplication are exactly the commonly given identities for the addition and multiplication of fractions: $$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd} \quad \text{and} \quad \frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d}$$

Answer 3

Since I do not have the required reputation to comment, I will leave my comments, which were originally intended as merely a comment, as an answer:

Contrary to Peter Baum, I believe that the question is in fact trivial, if not for the ambiguous framing of "a mathematical approach as a math student proves it." What is not "trivial," are the varying ways in which such a statement was interpreted by answerers on this question.

Or perhaps, the fact that mathematicians would prefer to resort to field theory in order to answer a question about elementary arithmetic—I believe this is what prompted Baum to write a "non-trivial" answer. Frankly, I do not know if the intent of the proofs utilizing field theory were to impress, or if the motivation was, on some level, laziness.

I would also like to point out the issues with the further ambiguity introduced by the questioner's opening statement of "I understand it," which may have further contributed to the confusion. Since the statement is highly unclear with respects to what the questioner actually "understands," and by extension, what the questioner wants.

For the sake of brevity, and as an outsider who is not a mathematician, there is no reason for me to attempt to elucidate the detailed process and conventions of a mathematical proof, or discuss any technical distinctions, which Baum has evidently already done.

I will simply present another example of an algebraic manipulation which demonstrates the Fraction Product Rule. It similarly uses merely rudimentary laws and theorems, introduces no additional variables, and does not unnecessarily resort to a different area of mathematics. The following took me a couple of minutes to find with a pen and paper, and is in my opinion, elegant:

$$ \frac{a}{b} \cdot \frac{c}{d} = \frac{\frac{c}{d}\cdot a}{\frac{d}{d}\cdot b} = \frac{\frac{ac}{d}}{\frac{bd}{d}} = \frac{ac}{bd}$$

Answer 4

I realize I am almost 8 years late, but I figured I give my two cents. Let $a$, $b$, $c$, and $d$ be elements of an arbitrary field and lets make sure that $b$ and $d$ are non-zero. $$$$Let $y=\frac{ac}{bd}$. Then $y=(ac)\cdot \left(\frac{1}{bd}\right)$ by convention. Hence $(bd)y=ac$ via multiplying both sides by $bd$. Thus we have that $b(dy)=(ac)$ by the associative property of multiplication inherent in a field. Oh, but then $dy=(ac)\left(\frac{1}{b}\right)$ as $b\neq 0$ and again we have $y=(ac)\left(\frac{1}{b}\right)\left(\frac{1}{d}\right).$ Now by a couple applications of the associative and commutative property of multiplication inherent in any field, we have that $$y=\left(a\left(\frac{1}{b}\right)\right)\left(c\left(\frac{1}{d}\right)\right).$$ Finally, by convention... $$y=\left(\frac{a}{b}\right)\cdot\left(\frac{c}{d}\right).\blacksquare$$