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.

• We define on $\Bbb R^2$ the operations $$(a,b)+(c,d)=(ad+bc,bd)$$ and $$(a,b)\times(c,d)=(ac,bd)$$ and we verify that $(\Bbb R^2,+,\times)$ is a field. – user63181 Oct 25 '14 at 22:25
• As Sami hints, we define fractions in terms of equivalence classes of ordered pairs. You may find this lecture on properties of $\mathbb{Q}$ helpful; I certainly did at one point: youtube.com/watch?v=gTgkrVATzmk – Mathemanic Oct 25 '14 at 22:29
• @SamiBenRomdhane Just what I wanted. – Dante Oct 25 '14 at 22:30
• See, not a bad question at all! – David K Oct 26 '14 at 0:00

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: $$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd} \quad \text{and} \quad \frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d}$$

• More details of the pair-based construction of the field of fractions of a domain are in prior answers, e.g. here, which expands on the details of a conceptually motivated construction from Vinberg's A course of algebra. – Bill Dubuque May 15 at 18:23
• I fear readers are being misled into thinking this response provides a valid proof of the fraction product rule. The cause for the confusion is that in elementary arithmetic, the vinculum is used to indicate division (an operator) as well as a fraction (a number). This answer proves an expression that looks exactly like the fraction product rule but is actually a proof about division. It therefore does not answer the question, which is about fractions. – Peter Baum Aug 3 at 15:26
• Also, the second half of the response incorrectly suggests that the fraction product rule is something to define rather than prove. See my post below that starts with “SUMMARY” for details. – Peter Baum Aug 3 at 15:26

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.

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 {https://www.researchgate.net/publication/342927518_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.

• I am amused by your devotion to this answer! – Baba Yaga Jul 27 at 11:58
• @Arjun I can understand why people wonder why I spent so much effort answering this question. The reason is that this question provides a great example of a problem we face in mathematics education. The fact that the question was posted over five years ago, but only recently answered with a valid proof, suggests to me that a better understanding of elementary arithmetic is needed. I think that a better understanding has implications for the way elementary arithmetic is taught. I find it a hard sell, though. Most people think that elementary arithmetic is ... well...elementary. – Peter Baum Aug 2 at 18:56
• A salute to you from my side and even I think arithmetic is elementary, it was a joke ;-) – Baba Yaga Aug 2 at 19:21
• Let's get this answer a few upvotes – Baba Yaga Aug 2 at 19:22

# Summary

• Division and fractions are fundamentally different. Division is an operation, while a fraction is a number. However, they are frequently represented by the same symbol, and the rules governing each are very similar.

• Within field theory, we cannot use the field axioms to properly define a fraction. Therefore, if we remain strictly within field theory, there can be no valid proof for the fraction product rule.

• Within field theory, we can prove an identity about division that symbolically looks exactly like the fraction product rule.

• We can form a useful bridge between field theory and elementary arithmetic by proving fraction laws that mimic elements of field theory. In particular, we find a fraction that can serve as a multiplicative inverse and prove a fraction theorem that mimics the field definition of division. Other elements, such as commutative and associate laws, are easily linked because their functions are the same within both mathematical systems.

• We use our bridge between field theory and elementary arithmetic to create a valid proof of the fraction product rule. We do this by using the bridge to convert each expression in the field theorem about division into an elementary arithmetic expression about fractions. The result is then a proof, within elementary arithmetic, of the fraction product rule.

• We can create proofs that appear short by leaving out steps and by using theorems that hide complexity. Theorems that can be used more than once in a proof are especially useful in creating short proofs.

• One simple, reasonable measure of the complexity of a proof is a count of the steps in a proof that is created strictly from axioms and uses no theorems at all.

• Although we can create equivalence classes that have characteristics similar to fractions, they do not provide a good basis for defining fractions.

• Calling the result of a valid theorem a "definition" makes the mathematics simple at the cost of obscuring the nature of the theorem as well as mathematics itself.

# Preliminary Considerations

In the exposition that follows, I will use the term "elementary arithmetic" to mean the mathematical system we first learn that involves integers and fractions, as well as the operations of addition and multiplication. All the definitions, assumptions, and theorems of elementary arithmetic that we need are presented in another section. I will keep things as simple as possible by ignoring negative numbers; they do not play a significant role in this exposition.

Let's begin with a clear definition of a field. Although there are equivalent definitions, let's use the classic definition as described at https://en.wikipedia.org/wiki/Field_(mathematics).

Formally, a field is a set F together with two binary operations on F called addition and multiplication. A binary operation on F is a mapping F $${\times}$$ F $${\rightarrow}$$ F, that is, a correspondence that associates with each ordered pair of elements of F a uniquely determined element of F. The result of the addition of a and b is called the sum of a and b, and is denoted a + b. Similarly, the result of the multiplication of a and b is called the product of a and b, and is denoted ab or a $${\cdot}$$} b. These operations are required to satisfy the following properties, referred to as field axioms. In these axioms, a, b, and c are arbitrary elements of the field F.

• Associativity of addition and multiplication: a + (b + c) = (a + b) + c, and a · (b · c) = (a · b) · c.

• Commutativity of addition and multiplication: a + b = b + a, and a · b = b · a.

• Additive and multiplicative identity: there exist two different elements 0 and 1 in F such that a + 0 = a and a · 1 = a.

• Additive inverses: for every a in F, there exists an element in F, denoted $${-}$$a, called the additive inverse of a, such that a + ($${-}$$a) = 0.

• Multiplicative inverses: for every a $${\neq}$$ 0 in F, there exists an element in F, denoted by a$$^{-1}$$ or 1/a, called the multiplicative inverse of a, such that a · a$$^{-1}$$ = 1.

• Distributivity of multiplication over addition: a · (b + c) = (a · b) + (a · c).

Now a field is an abstraction, a generalization, and a model for other mathematical structures that had been previously invented, such as those involving rational numbers, real numbers, and complex numbers. There are even simple finite fields, for example, modular arithmetic containing only the elements 0 and 1. It is easy to show that elementary arithmetic is an example of a field simply by going through each of the elements in the definition of a field listed above and verifying that it is valid elementary arithmetic. Once we have verified this, if we prove something for a field, we know that it also represents a result in elementary arithmetic.

Even though a field is an abstraction of elementary arithmetic, we need to carefully examine the relationship between them. There are elements of elementary arithmetic that we don’t find in a field, and that can cause problems. One example is the concept of number. In the field axioms, we have members of set F and, although in the elementary arithmetic field that set will be numbers, field theory cannot represent the definition of number found in elementary arithmetic. Another example is fractions, which are a subset of the numbers in elementary arithmetic. Field theory knows nothing about fractions. We cannot even create fractions within field theory if we wish to remain confined within the defining axioms of field theory. We can, however, add definitions within the concept of a field, as long as they only use the field axioms. If the definition uses something other than the field axioms, the system is then a field extension.

Even though elementary arithmetic is a field, this doesn’t mean that we can use field theory to prove everything that we can prove in elementary arithmetic. For example, both elementary arithmetic and fields have a commutative law of addition and multiplication. These are laws we can prove as theorems within elementary arithmetic but they are beyond what field theory can prove and are therefore simply stated as field axioms.

# A Problematic Attempt to Use Field Theory to Prove the Fraction Product Rule

We next look in detail at the first answer posted in response to the Multiplying and adding fractions Stack Exchange question. As a response to a request for a proof of the fraction product rule, one might incorrectly assume that the answer provides this proof, especially since the identity at the end of the proof looks exactly like the fraction product rule. As we will see, no valid proof is actually provided.

We proceed, analyzing each step.

“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.”

Comment:

• We need to be clear about “these identities.” The question explicitly states that the identities we are to prove are about fractions. The question is not about division identities, even though the same symbol is sometimes used to indicate either division or a fraction. The rules for both are similar, and people often ignore this distinction.

• We can, of course, define anything we want and do so in any way we want to. However, the identities are not “the definition of addition and multiplication of fractions” in any proper sense because we can prove these identities within elementary arithmetic. By convention, we don’t present results as definitions when we can prove those results. In a sense, definitions do make for an extremely short “proof,” but I submit that it is unlikely to provide a satisfactory response to the original question about the fraction product rule.

“Specifically, we can define division to be the multiplication of the numerator with the inverse of the denominator.”

Comments: There are a number of issues here. (Note that we are assuming b$${\neq}$$0 below.)

• The question we are being asked is about fractions, not division. If the intent of the response is to suggest the proof gives us a result regarding fractions, then that is incorrect. If the intent is to show us a result regarding division, then the question has not been answered.

• If we want to define division within a field to be $$a\cdot b^{-1}$$, that is perfectly legitimate. However, division and fractions have their own definitions within elementary arithmetic. Notice also that within elementary arithmetic, trying to divide 1 by 2 is not possible as a division operation, and is instead represented by a fraction. (We can, however, convert the fraction to other forms, such as the fraction represented by 0.5).

“For example, we can write $$\frac{\boldsymbol{a}}{\boldsymbol{b}}=\boldsymbol{a}\cdot \boldsymbol{b}^{-\mathbf{1}}$$ , identifying division as taking the multiplicative inverse. Then $$\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\left(\boldsymbol{a}\cdot \boldsymbol{b}^{-\mathbf{1}}\right)\cdot (\boldsymbol{c}\cdot \boldsymbol{d}^{-\mathbf{1}})$$

• One might easily be misled here because the given definition of division uses the same symbol $$\frac{a}{b}$$ that occurs in the original question about fractions. It’s important to remember, at this point, that in this instance the symbol $$\frac{a}{b}$$ only indicates division within a field.

• We are asked to “identify” division as “taking the multiplicative inverse.” Perhaps the intent might have been more clearly expressed as “identifying division as the product of the dividend times the multiplicative inverse of the divisor.”

We might try to make an alternative interpretation of the phrase, such as meaning that we should replace $$b^{-1}$$ with $$\frac{1}{b}$$ to create $$\frac{\boldsymbol{a}}{\boldsymbol{b}}=\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}$$ . That does not work as a definition of division, however, because it would be defining division using division. In other words, the definition would be circular.

• What happens if we try to interpret the expression $$\frac{\boldsymbol{a}}{\boldsymbol{b}}=\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}$$ as the definition of a fraction within field theory, rather than division? An attempt to use this equation to define a fraction would again be improper, because $$\frac{1}{b}$$ is represented in elementary arithmetic as a fraction, and we therefore would be defining a fraction through the use of a fraction. In other words, the definition would be circular.

• Within elementary arithmetic we will later prove the fraction relationship $$\frac{a}{b}=a\cdot \frac{1}{b}$$ using a suitable definition for a fraction.

• The result $$\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\left(\boldsymbol{a}\cdot \boldsymbol{b}^{-\mathbf{1}}\right)\cdot (\boldsymbol{c}\cdot \boldsymbol{d}^{-\mathbf{1}})$$ is valid, but it is about division, not fractions.

• As a generalization, a field definition can simply declare that we have a set F and, for each element in the set other than 0, each has a multiplicative inverse. The defining field theory set F contains all the elements we will be dealing with. Thus, hidden in this definition of a field, we find fraction-like inverses that we don’t have to justify as being numbers. In elementary arithmetic, we do have to justify calling fractions numbers. Similarly, in field theory, unlike in elementary arithmetic, we don’t have to show that multiplicative inverses have the same properties, with respect to addition and multiplication, that the other members of F have. In this way, field theory doesn’t include critical components that are needed when we create clear, complete, and valid proofs within elementary arithmetic.

“Now, if we want multiplication to be associative and commutative, then we would find that $$\left(\boldsymbol{a}\cdot \boldsymbol{b}^{-\mathbf{1}}\right)\cdot \left(\boldsymbol{c}\cdot \boldsymbol{d}^{-\mathbf{1}}\right)=(\boldsymbol{a}\cdot \boldsymbol{c})\cdot (\boldsymbol{d}^{-\mathbf{1}}\cdot \boldsymbol{b}^{-\mathbf{1}})$$

Comment: It would be clearer if all the steps were shown. Then the need for a law that allows us to remove unnecessary parentheses would be apparent.}

“It is a general fact that $$(\boldsymbol{xy})^{-\mathbf{1}}=\boldsymbol{y}^{-\mathbf{1}}\boldsymbol{x}^{-\mathbf{1}}$$, which can be verified directly by multiplying $$(\boldsymbol{xy})$$ by both $$\boldsymbol{y}^{-\mathbf{1}}\boldsymbol{x}^{-\mathbf{1}}$$ and $$(\boldsymbol{xy})^{-\mathbf{1}}$$.“

• Doing as suggested, $$\begin{equation*} \boldsymbol{x}\cdot \boldsymbol{y}\cdot \left(\boldsymbol{y}^{-\mathbf{1}}\cdot \boldsymbol{x}^{-\mathbf{1}}\right)=\boldsymbol{x}\cdot \boldsymbol{y}\cdot \boldsymbol{y}^{-\mathbf{1}}\cdot \boldsymbol{x}^{-\mathbf{1}}=\boldsymbol{x}\cdot \mathbf{1}\cdot \boldsymbol{x}^{-\mathbf{1}}=\boldsymbol{x}\cdot \boldsymbol{x}^{-\mathbf{1}}=\mathbf{1} \end{equation*}$$ and $$\begin{equation*} \left(\boldsymbol{x}\cdot \boldsymbol{y}\right)\cdot \left(\boldsymbol{x}\cdot \boldsymbol{y}\right)^{-\mathbf{1}}=\mathbf{1} \end{equation*}$$ Then by substitution we have $$\begin{equation*} \boldsymbol{x}\cdot \boldsymbol{y}\cdot \left(\boldsymbol{y}^{-\mathbf{1}}\cdot \boldsymbol{x}^{-\mathbf{1}}\right)=\left(\boldsymbol{x}\cdot \boldsymbol{y}\right)\cdot \left(\boldsymbol{x}\cdot \boldsymbol{y}\right)^{-\mathbf{1}} \end{equation*}$$ Then multiplying both sides by $$\left(x\cdot y\right)^{-1}$$ we obtain $$\begin{equation*} \left(\boldsymbol{x}\cdot \boldsymbol{y}\right)^{-\mathbf{1}}\cdot \boldsymbol{x}\cdot \boldsymbol{y}\cdot \left(\boldsymbol{y}^{-\mathbf{1}}\cdot \boldsymbol{x}^{-\mathbf{1}}\right)=\left(\boldsymbol{x}\cdot \boldsymbol{y}\right)^{-\mathbf{1}}\cdot \left(\boldsymbol{x}\cdot \boldsymbol{y}\right)\cdot \left(\boldsymbol{x}\cdot \boldsymbol{y}\right)^{-\mathbf{1}} \end{equation*}$$ Then by the definition of a multiplicative inverse $$\begin{equation*} \mathbf{1}\cdot \left(\boldsymbol{y}^{-\mathbf{1}}\cdot \boldsymbol{x}^{-\mathbf{1}}\right)=\mathbf{1}\cdot \left(\boldsymbol{x}\cdot \boldsymbol{y}\right)^{-\mathbf{1}} \end{equation*}$$ and because 1 is the multiplicative identity $$\begin{equation*} \left(y^{-1}\cdot x^{-1}\right)=\left(x\cdot y\right)^{-1} \end{equation*}$$ So, this identity about multiplicative inverses can be verified, although it takes steps to do so that were not explicitly shown.

• Now let’s examine what this looks like within elementary arithmetic where fractions reside.

We can use Law 1 (Given a$${\neq}$$0, $$\frac{\boldsymbol{a}}{\boldsymbol{a}}=1$$) and Theorem 2 (Given b$${\neq}$$0, $$\boldsymbol{c}\cdot \frac{\boldsymbol{a}}{\boldsymbol{b}}=\frac{\boldsymbol{c}\cdot \boldsymbol{a}}{\boldsymbol{b}})$$ to prove that within elementary arithmetic the multiplicative inverse of a is the fraction} $$\frac{1}{a}$$. Details about this law and theorem are provided in another section. The proof is short: $$\begin{equation*} \boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{a}}=\frac{\boldsymbol{a}}{\boldsymbol{a}}=\mathbf{1} \end{equation*}$$

• In elementary arithmetic where we have defined a fraction, we also want to show

$$\begin{equation*} \frac{\mathbf{1}}{\boldsymbol{a}\cdot \boldsymbol{b}}=\frac{\mathbf{1}}{\boldsymbol{a}}\cdot \frac{\mathbf{1}}{\boldsymbol{b}} \end{equation*}$$
Now with our multiplicative inverse written as fractions, we proceed just as we did with our field proof: $$\begin{equation*} \boldsymbol{a}\cdot \boldsymbol{b}\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{a}}\right)=\boldsymbol{a}\cdot \boldsymbol{b}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{a}}=\boldsymbol{a}\cdot \mathbf{1}\cdot \frac{\mathbf{1}}{\boldsymbol{a}}=\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{a}}=\mathbf{1} \end{equation*}$$ and

$$\begin{equation*} \boldsymbol{a}\cdot \boldsymbol{b}\cdot \frac{\mathbf{1}}{\boldsymbol{a}\cdot \boldsymbol{b}}=\mathbf{1} \end{equation*}$$ and then by substitution using the previous two expressions equal to 1 we have

\begin{align*} \boldsymbol{a}\cdot \boldsymbol{b}\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{a}}\right)&=\boldsymbol{a}\cdot \boldsymbol{b}\cdot \frac{\mathbf{1}}{\boldsymbol{a}\cdot \boldsymbol{b}} \\ \frac{\mathbf{1}}{\boldsymbol{a}\cdot \boldsymbol{b}}\cdot \boldsymbol{a}\cdot \boldsymbol{b}\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{a}}\right)&=\frac{\mathbf{1}}{\boldsymbol{a}\cdot \boldsymbol{b}}\cdot \boldsymbol{a}\cdot \boldsymbol{b}\cdot \frac{\mathbf{1}}{\boldsymbol{a}\cdot \boldsymbol{b}} \\ \mathbf{1}\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{a}}\right)&=\mathbf{1}\cdot \frac{\mathbf{1}}{\boldsymbol{a}\cdot \boldsymbol{b}} \end{align*}

$$\begin{equation*} \frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{a}}=\frac{\mathbf{1}}{\boldsymbol{a}\cdot \boldsymbol{b}} \end{equation*}$$

• So, although we have not defined a fraction in a field, fractions are properly defined in elementary arithmetic. We therefore use the field proof about division to guide our analogous fraction theorem within elementary arithmetic.

“Then we find that $$\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\boldsymbol{d}^{-\mathbf{1}}\cdot \boldsymbol{b}^{-\mathbf{1}}\right)=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\boldsymbol{b}\cdot \boldsymbol{d}\right)^{-\mathbf{1}}=\frac{\boldsymbol{a}\cdot \boldsymbol{c}}{\boldsymbol{b}\cdot \boldsymbol{d}}$$

• Filling in some of the missing parts of the proof,

\begin{align*} \frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}&=\left(\boldsymbol{a}\cdot \boldsymbol{b}^{-\mathbf{1}}\right)\cdot \left(\boldsymbol{c}\cdot \boldsymbol{d}^{-\mathbf{1}}\right) \\ &=\boldsymbol{a}\cdot \boldsymbol{b}^{-\mathbf{1}}\cdot \boldsymbol{c}\cdot \boldsymbol{d}^{-\mathbf{1}} \\ &=\boldsymbol{a}\cdot \left(\boldsymbol{b}^{-\mathbf{1}}\cdot \boldsymbol{c}\right)\cdot \boldsymbol{d}^{-\mathbf{1}} \\ &=\boldsymbol{a}\cdot \left(\boldsymbol{c}\cdot \boldsymbol{b}^{-\mathbf{1}}\right)\cdot \boldsymbol{d}^{-\mathbf{1}} \\ &=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left({\boldsymbol{b}^{-\mathbf{1}}}\cdot \boldsymbol{d}^{-\mathbf{1}}\right) \\ &=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\boldsymbol{d}^{-\mathbf{1}}\cdot \boldsymbol{b}^{-\mathbf{1}}\right) \\ &=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\boldsymbol{b}\cdot \boldsymbol{d}\right)^{-\mathbf{1}} \\ &=\frac{\boldsymbol{a}\cdot \boldsymbol{c}}{\boldsymbol{b}\cdot \boldsymbol{d}} \end{align*}

• In terms of fractions within elementary arithmetic we have

$$\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\left(\frac{\boldsymbol{a}\cdot \mathbf{1}}{\boldsymbol{b}}\right)\cdot \left(\frac{\boldsymbol{c}\cdot \mathbf{1}}{\boldsymbol{d}}\right)$$ (multiplicative identity)

$$=\left(\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}\right)\cdot \left(\boldsymbol{c}\cdot \frac{\mathbf{1}}{\boldsymbol{d}}\right)$$ (Theorem 2, constant times a fraction)

$$=\boldsymbol{a}\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \boldsymbol{c}\right)\cdot \frac{\mathbf{1}}{\boldsymbol{d}}$$ (associative law of multiplication)

$$=\boldsymbol{a}\cdot \left(\boldsymbol{c}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}\right)\cdot \frac{\mathbf{1}}{\boldsymbol{d}}$$ (commutative law of multiplication)

$$=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{b}}\cdot \frac{\mathbf{1}}{\boldsymbol{d}}\right)$$ (associative law)

$$=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \left(\frac{\mathbf{1}}{\boldsymbol{d}}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}\right)=$$ (commutative law of multiplication)

$$=\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \frac{\mathbf{1}}{\boldsymbol{b}\cdot \boldsymbol{d}}=$$ (proved in the previous comment)

$$=\frac{\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)\cdot \mathbf{1}}{\boldsymbol{b}\cdot \boldsymbol{d}}$$ (Theorem 2, constant times a fraction)

$$=\frac{\left(\boldsymbol{a}\cdot \boldsymbol{c}\right)}{\boldsymbol{b}\cdot \boldsymbol{d}}$$ (multiplicative identity)

$$=\frac{\boldsymbol{a}\cdot \boldsymbol{c}}{\boldsymbol{b}\cdot \boldsymbol{d}}$$ (removing unnecessary parentheses)

• Combining all the steps above, we can use the field proof about division as a guide for our elementary arithmetic proof of the fraction product rule. We simply replace field theory expressions at each step with the necessary elementary arithmetic expressions. When all the steps are written out, we see that the proofs end up being quite long. Proofs can often be shortened by hiding complexity in reusable theorems.

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

• In our previous elementary arithmetic development, the result $$\frac{\boldsymbol{c}}{\boldsymbol{c}}=\mathbf{1}$$ was Law 1, rather than something to prove. The basis for the proof of this identity in field theory is a field theory definition $$\frac{\boldsymbol{a}}{\boldsymbol{b}}=\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}$$ . We find the same expression in elementary arithmetic, although it is a theorem about fractions. We prove this theorem in a later section and call it Theorem 9.

• The expression is not “rather definitional” in elementary arithmetic because} $$\frac{\boldsymbol{a}}{\boldsymbol{b}}=\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}$$ is a theorem we can prove in elementary arithmetic.

“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....” (See Hayden post above for the rest of the text).”

• It is not clear from the phrase “... another way in which we can view these identities for fractions....” whether the identities given for the field are to be viewed as fractions or simply that there is another way to view the identities, specifically as part of an attempt to define fractions. In any case, as previously explained, they cannot be fraction identities within a proper field.

• The idea of defining a fraction as an infinite set of ordered pairs is problematic for a number of reasons:

• [o] Fractions are considered numbers and numbers are closely associated with counting. Sets contain elements that are not necessarily counted. Although some consider that we can define a number as a set, I do not hold this view. I believe this notion arose because some mathematicians couldn’t think of anything else they could use to define number and so simply put into service a mathematical object they were familiar with. The structure they created did have some of the characteristics of a fraction. I believe the best way to define number is using constructs from a theory of thought.

• [o] Not only is this definition problematic because a set is involved, it is made more difficult to understand because the set is infinite, and infinity raises its own set of issues.

• [o] This definition is probably not the way we would want to introduce the relatively simple mathematical idea of a fraction.

• [o] This definition of a fraction is not consistent with the historical development of the notion, and thus does not contain the intuitive characteristics of a fraction that are needed to fully understand the concept.

• [o] We don’t have to define a fraction as an equivalence class; there are other readily available definitions that I submit are more suitable. Given those definitions, one can then prove what we need about equivalent fractions.

“Then the definition of addition and multiplication are exactly the commonly given identities for the addition and multiplication of fractions:

$$\frac{\boldsymbol{a}}{\boldsymbol{b}}+\frac{\boldsymbol{c}}{\boldsymbol{d}}=\frac{\boldsymbol{ad}+\boldsymbol{bc}}{\boldsymbol{bd}}$$ and $$\frac{\boldsymbol{a}}{\boldsymbol{b}}\cdot \frac{\boldsymbol{c}}{\boldsymbol{d}}=\frac{\boldsymbol{a}\cdot \boldsymbol{b}}{\boldsymbol{c}\cdot \boldsymbol{d}}$$

Comment: You will find in my previous post the primitive definitions and assumptions needed to prove these results, rather than present them as definitions. These equations are fine if all you want to do is follow rules so you can apply fractions to practical tasks. However, if we want to fully understand fractions and the beautiful mathematics created through the use of logical proofs, then we will need something more than definitions.

## The General use of Field Theory to Prove Results in Specific Fields

Because we have shown that much of elementary arithmetic forms a field, any field result will apply in some way to elementary arithmetic. We ran into problems, however, because not every theorem in elementary arithmetic can be proved within field theory. In our case, fractions are not part of the structure known as a field. Although a theorem in field theory looked like the fraction product rule, it is actually a theorem about division. We were, however, able to circumvent this problem by using the field theory proof about division as a guide to create a valid proof for the fraction product rule. To do this, we had to show how we could replace elements in the field proof with valid expressions involving fractions in elementary arithmetic. In particular,

• We replaced the multiplicative inverse of a that is expressed as $$a^{-1}$$ within the field proof with the elementary arithmetic fraction $$\frac{1}{a}$$ . We validated the expression in elementary arithmetic by proving that the fraction was the required multiplicative inverse for a.

• We replaced the division operation defined by $$a\cdot b^{-1}$$ within the field proof with the elementary arithmetic fraction $$\frac{a}{b}$$ . We validated the expression in elementary arithmetic by proving the fraction theorem $$\frac{\boldsymbol{a}}{\boldsymbol{b}}=\boldsymbol{a}\cdot \frac{\mathbf{1}}{\boldsymbol{b}}$$ .

The other elements in the field proof could be kept the same because symbolically they represent valid expressions and transformations in elementary arithmetic. The result of all these substitutions was our desired valid proof and it was entirely within elementary arithmetic.

It is then conceivable that we could apply this approach to other theorems in other mathematical structures that also happen to be fields. We only have to provide the proper linkage, so that at each step, we are assured that any field operations will be valid within the target mathematical structure after the substitution.

What have we gained when we do this? Not necessarily a great deal. After all, setting up the proper linkage is the key and once we know what the correspondence is, we might as well just actually perform the entire proof within elementary arithmetic without actually making substitutions. There can sometimes be a benefit though. Just the fact that a theorem is expressed within the language of a particular mathematical structure sometimes suggests a different proof, and each proof can have its own set of advantages and disadvantages.

## Proof Length Issues

We might be misled when comparing the field theorem about division with one of the previous fraction product rule proofs into thinking that it is possible to prove the fraction product rule in a very simple and short way. This is an illusion because of the following:

• Although the field theorem created expressions that looked like the fraction product rule, it was about the division operation and not fractions. Thus, it wasn’t a valid proof of the fraction product rule, and any steps that would be needed to make it valid were omitted.

• The field proof left out many steps that are required if we wish to express a complete proof where each dependency is articulated.

• If we explicitly show and justify the substitutions needed to convert the field theorem about division into a valid proof of the fraction product rule, the process adds to the complexity of the proof.

• If we use definitions to express relationships that can be validated with a proof, the result looks simple, but I do not believe this should be considered a legitimate way to shorten a proof.

• Theorems can be created that can shorten proofs if those theorems can be used more than once in a proof. Although a legitimate strategy in terms of shortening a proof, it introduces its own complexity, calling into question how we should legitimately measure the length and complexity of a proof. More information about this issue can be found in Proof 4 in my following post.

I am currently beyond the 30,000 character limit, so I will post 2 of the actual complete alternative fraction product rule proofs in my next post.

## 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.