I am reading the second article Rational Numbers of the book "A Treatise on Advanced Calculus" by Philip Franklin.
I have mainly 3 questions regarding this article. I am writing all these $3$ question in one question because they are related to each other.

Question $1$ The author defined the equality of two positive rational numbers by the rule:
$\dfrac{a}{b}=\dfrac{a'}{b'}\ \ \ \mbox{if}\ \ \ \ \ \ ab'=a'b. \tag{2}$
Here $a,a',b,b'$ are Natural numbers.
The problem is that this definition can be proved by definition of the product of two positive rational numbers. In the article the definition of product of two positive rational numbers is give after the definition$2$.
In the definition$5$ the author defines the product of two positive rational numbers as: $$\dfrac{a}{b} \cdot \dfrac{a'}{b'}=\dfrac{aa'}{bb'} \tag{5} $$
We can use this definition to prove definition$2$. To do so let's first consider three ratioanal numbers $\dfrac{a}{b}$ , $\dfrac{b'}{a'}$ and $\dfrac{a'}{b'}$ s.t. $a,a',b,b'$ are Natural numbers.
$\dfrac{a}{b}\times \dfrac{b'}{a'}=\dfrac{ab'}{ba'}$ by definition$5$
If $ba'=ab'$ Then $\dfrac{ab'}{ba'}=1$ hence $\dfrac{a}{b}\times \dfrac{b'}{a'}=1$
Multiplying both sides with $\dfrac{a'}{b'}$ we get :
$\dfrac{a}{b}\times \dfrac{b'}{a'}\times \dfrac{a'}{b'}=1\times \dfrac{a'}{b'}$
$\implies \dfrac{a}{b}=\dfrac{a'}{b'}$
So definition$5$ implies that if $ab'=a'b$ then $\dfrac{a}{b}=\dfrac{a'}{b'}$ which is nothing but the definition$2$.

  • "The definition$2$ can be proved from the definition$5$ so why the author quote Equation$2$ as a definition?"
  • "What if we reject definition$2$?"

Question $2$
Author has quoted an important result in equation$3$ as:
"We identify certain rational numbers with integers by regarding $$a=\dfrac{a}{1}\tag{3}"$$
How this relation is identified? Uptill equation$3$ the author had not defined the multiplication of two positive rational numbers. Does equation $1$ and $2$ imply equation$3$. Or equation$3$ is in itself a definition(this doesn't seem to be the case)?
The best I could understand this is as:
$\dfrac{a}{1}=x$(say) is a rational number which satisfies the relation $a=x\times 1$. We have not defined the rule for multiplication of $x$ with $1$ when $x$ is a rational number. But we see that $x=a$ is also a solution of relation $a=x\times 1$ because $a=a\times 1$.
So if $\dfrac{a}{1}$ is just the solution of $a=x\times 1$ then $a=\dfrac{a}{1}$ is an identity not an equality because for every positive number $a$ , $a$ is equal to $\dfrac{a}{1}$, that is we should have $a \equiv \dfrac{a}{1}$ not $a=\dfrac{a}{1}$.

Question $3$
Why we defined different operations on rationals number this way, that is the rules described in the article from equation $(4)$ to equation$(13)$.
e.g. The multiplication of two positive rational numbers is as: $\dfrac{a}{b} \cdot \dfrac{a'}{b'}=\dfrac{aa'}{bb'} $. An applied Mathematician may answer this by saying:
" Let $\dfrac{a}{b}$ and $\dfrac{a'}{b'}$ be the magnitude of the length and breadth of a rectangle then $\dfrac{aa'}{bb'}$ represents its area that's why we defined multiplication of two positive rational numbers this way."
On the other hand a pure Mathematician may answer this as:
"Let $x=\dfrac{a}{b}$ and $y=\dfrac{a'}{b'}$ be the two positive rational numbers s.t. $a,a',b,b'$ are natural numbers. By defining the multiplication of two positive rational numbers this way(as defined in equation$5$) we recognize that the commutative law and the associative law will hold good. That's why we defined it to be this way.

  • In reality who, how, when and why defined these definitions/rules(equation$1$ to equation$13$)?
  • Are the reason behind these definition to be of this kind pure mathematical or applied mathematical or both?

Please give me a link or tell me further reference on the historical perspective on these definitions. I want to study the history of these things in detail.

  • 1
    $\begingroup$ It's impossible to prove definition (2) from the definition of a product. In fact, it's impossible to prove definition (2) at all. Why? Because it's a definition. It isn't even really "true", it's just declared to be one of the rules. You can't prove a definition any more than you can prove that bishops can only move diagonally in chess. $\endgroup$
    – Jack M
    Commented Mar 9, 2014 at 14:25
  • $\begingroup$ @JackM I think definition-2 is a corollary of definition-5, so we do not need to define this as a rule. $\endgroup$
    – user103816
    Commented Mar 16, 2014 at 6:36
  • $\begingroup$ Let's start from scratch. This is the book. We define positive rational numbers as pairs of positive integers, with addition defined by (4) and multiplication by (5). Now suppose you don't use the definition (2) of equality. What do you do next? $\endgroup$ Commented Mar 30, 2014 at 13:16
  • $\begingroup$ @ShreevatsaR, I will prove that the two rational numbers $\dfrac{a}{b}$ and $\dfrac{a'}{b'}$ both represents the solution of $a=xb$ and $a'=xb'$, that is $\dfrac{a}{b}=\dfrac{a'}{b'}=x$ if $ab'=a'b$ . So they are equal. So we do not need definition-2 and we can continue to study rational numbers. $\endgroup$
    – user103816
    Commented Mar 30, 2014 at 13:31
  • 2
    $\begingroup$ @Anupam: So please read my answer, and tell me if you still have the same question. By the way, the confusion due to notation, in your proof, has crept in when you write "if $ab' = ba'$, then $\dfrac{ab'}{ba'}=1$". Now, it is true with the definition of integer division (where a solution exists), but this is different from saying that the rational number given by the pair $(ab', ba')$ is the same as the rational number $1$, that is, $(1, 1)$. In general you cannot say that the rational number given by some $(m, m)$ is the same as $1$, without using definition $(2)$. $\endgroup$ Commented Mar 30, 2014 at 15:05

3 Answers 3


Here's another attempt at doing the whole thing, without the possibly confusing notation.

Let's define a kind of object that we'll call a "prn". A "prn" is defined as an ordered pair $(a, b)$ (with some additional structure that we'll define below), and to make it clear that it's a "prn" and not just an ordered pair, we'll denote $(a \star b)$. We can define addition and multiplication of "prn"s:

  • Addition $\oplus$ is defined as: $$(a \star b) \oplus (a' \star b') = ((ab' + a'b) \star bb') \tag 4$$
  • Multiplication $\otimes$ is defined as: $$(a \star b) \times (a' \star b') = (aa' \star bb'). \tag 5$$

We also define an embedding of the positive integers into our structure of "prn"s:

  • The positive integer $a$ corresponds to a particular "prn" $$(a \star 1). \tag 3$$

Now so far, we have not said what it means for two "prn"s to be equal. We can believe that $(a \star b) = (a' \star b')$ when $a = a'$ and $b = b'$, but are there other cases when $(a \star b) = (a' \star b')$?

For example, the integer $1$ corresponds to $(1 \star 1)$, and $(a \star b) \otimes (b \star a) = (ab \star ab)$. Are these two the same?

Nothing we have done so far indicates that they are!

Now, if we want $(a \star b)$ to correspond to the notion of "a solution to $a = xb$", then we are forced to add a definition remedying the situation: we'll have to define

$$(a \star b) = (a' \star b') \quad \text{ if } ab' = a'b \tag 2$$

Suppose we don't adopt this definition $(2)$, then is there a sense in which $(a \star b)$ is a solution to $a = xb$? Well, firstly $a = xb$ doesn't make sense when $x$ is a "prn", since we have defined $xy$ only when $x$ and $y$ are both integers, and we have defined $x \otimes y$ only when $x$ and $y$ are both "prn"s. But using $(3)$ to identify the positive integer $a$ with the "prn" $(a \star 1)$, and to identify the positive integer $b$ with the "prn" $(b \star 1)$, we can wonder whether $(a \star 1) = (a \star b) \otimes (b \star 1)$, i.e., whether it's a solution when we use the "multiplication" for the "prn" type instead of the multiplication for the positive integer type (where there is no solution). All we can say is that $$(a \star b) \otimes (b \star 1) = (ab \star b) \overset{?}{=} (a \star 1).$$

So it turns out that our $(a \star b)$ is not a solution to $(a \star 1) = x \otimes (b \star 1)$ either, unless we adopt $(2)$ which makes $(ab \star b) = (a \star 1)$, and finally (but only if we do so!) can we say that $x = (a \star b)$ is a solution to $a = x \cdot b$, in the sense that in the "prn" space, $x$ is a solution to $(a \star 1) = x \otimes (b \star 1)$, and by $(3)$, we have decided to let $(a \star 1)$ stand for $a$, and $(b \star 1)$ stand for $b$.

Of course, after we see all this work out so nicely, we can use the notation $\frac{a}{b}$ instead of $(a \star b)$, and (because it worked out) we can also say that it's actually the same as division. We call the structure "prn" as "positive rational numbers" (hey they are actually numbers now, not pairs of numbers with some weird addition/multiplication/equality rules), and drop the special symbols for $\oplus$ and $\otimes$ because it's understood what we mean.

Edit: In one sense, your argument holds to the following extent: instead of definition $(2)$, if we adopt the weaker equivalence that $$(a \star a) = (1, 1) \quad \text{ for any $a$} \tag {2'}$$ then your proof is that whenever $ab' = a'b$, then $$\begin{align} (a \star b) &= (a \star b) \otimes (1 \star 1) \\ &= (a \star b) \otimes (b'a' \star a'b') \\ &= (ab'a' \star ba'b') \\ &= (ab' \star ba') \otimes (a' \star b') \\ &= (1, 1) \otimes (a' \star b') \\ &= (a' \star b') \end{align}$$

So the weaker $(2')$ gives $(2)$ too! But it's just that you were implicitly assuming $(2')$ and then with the clever proof above deriving $(2)$, but in fact we need to assume either one or the other.

Brief answers to your other questions:

Question 2:: Yes, $(3)$, that $a = (a \star 1)$, is a definition, not something we can derive. It's a way of going between integers and "prns" (positive natural numbers): when $a = xb$ has no solution in the integers, we want to be able to say that the "corresponding" $(a \star 1) = x \otimes (b \star 1)$ has a solution in the "prn"s.

Question 3: Both your motivations are valid; another is that if we want to say $a/b$ (our $(a \star b)$ above) is a solution to $a = xb$, then we are forced to define it that way. Indeed, the fact that all approaches lead to the same structure/properties is a sign to us that we're probably doing things right.

Of course, historically, people already had an intuitive sense of fractions, and this formalization came later, to capture this intuitive sense that applies to all those motivations, and surely others as well.

  • $\begingroup$ I think I was presuming that (h,h) is the solution of $h=xh$ but I cannot do so because I will have to prove that $h=(h,h)\times h$ is true but this cannot be further simplified so cannot be proved. Thankyou very much for your constructive comments and this answer. You were right from the beginning. . Thankyou very much. I cannot accept your answer because I am not satisfied for Q3 part of my question. $\endgroup$
    – user103816
    Commented Mar 30, 2014 at 15:40
  • $\begingroup$ @Anupam: Yes, by the way that assumption is enough actually, in place of $(2)$. That's what your proof shows. I've updated the answer. $\endgroup$ Commented Mar 30, 2014 at 16:36
  • 1
    $\begingroup$ Very nice, gets the point across better than my own answer. $\endgroup$ Commented Mar 30, 2014 at 22:35
  • 1
    $\begingroup$ @Anupam: Yes you are correct. Another way of looking at it: the structure we have established for the rational numbers, showing that $(a\star b)$ (in the notation of the answer) is a solution (in the rational-number space) to $a=xb$, justifies calling it $\frac{a}{b}$, which, when $b$ divides $a$, is the same notation we use for the integer $x$ such that $a=xb$. As the integers embed cleanly into the rational numbers, we can stop distinguishing between them, and consider integers a subset of the rationals. (This is similar to how we define complex numbers $(a,b)$ in terms of real numbers.) $\endgroup$ Commented Mar 31, 2014 at 18:21
  • 1
    $\begingroup$ @Anupam: I don't think there exists a detailed history of rational numbers, because the intuitive concept of "fractions" is ancient, possibly prehistoric. So it's not any particular mathematician who one day came up with this formalization out of thin air, it's just a matter of refining the concept that already exists and is well-known at some level of formality. But you may like to follow the references from en.wikipedia.org/wiki/Fraction_(mathematics)#History (current version). $\endgroup$ Commented Mar 31, 2014 at 18:27

Question 1

You need to be slightly more careful about what you can and can't do. Def. 1 doesn't allow you to conclude that $\frac{a}{a}=1$, only that $\frac{a}{a}=\frac{1}{1}$, and so your proof doesn't work. You won't be able to derive the definition of a fraction from the definition of multiplication. (Indeed, if you have no rule telling you when things are equal, how can you hope to conclude anything at all?)

Question 2

This is in fact a definition. Until now, you only know how to talk about things that look like $\frac{a}{1}$. There's no mention of how to interpret this as an integer. This definition creates a link between fractions and integers, and allows us to think of $\frac{a}{1}$ and $a$ as the same thing.

Question 3

So, I think you will understand rational numbers best if you study the concept of an equivalence class. Given a set $A$ of things, you can define an equivalence relation (lets call it ~) on these things by giving an explicit rule that tells you when $x$ and $y \in A$ are equivalent (we write $x \sim y$). This equivalence relation splits $A$ up into equivalence classes - we say that the equivalence relation partitions the set $A$. This means that everything in $A$ is in some equivalence class, all the elements in a given equivalence class are equivalent to each other, and no two elements from different equivalence classes are equivalent.

In the case of the rational numbers, we consider the set of all formal fractions: $A=\{\frac{a}{b} $such that $a$ and $b$ are integers$\}$. We then define the equivalence relation ~ on $A$ by $\frac{a}{b} \sim \frac{a^\prime}{b^\prime}$ if $ab^\prime=a^\prime b$. From now on, we will only be considering the equivalences classes, which correspond to fractions as we usually understand them. (We think of $\frac{1}{2}$ and $\frac{2}{4}$ as the same thing because they actually represent the same equivalence class.) When we "simplify" a fraction, what we are really doing is choosing another representative in its equivalence class. By the properties of natural numbers under division, we can always choose a representative with coprime numerator and denominator.

Multiplication is then defined between equivalence classes of fractions.

The point/motivation of doing this is that it extends the integers from a ring to a field. Explaining why having a field is desirable is beyond the scope of this answer, but simply put - it's a structure with interesting/useful algebraic properties.

Summary: We consider the set of fractions, and obtain rational numbers as equivalence classes of fractions under the equivalence relation defined in Q1. We also observe that we can identify rational numbers with 1 in the denominator with integers, using the definition in Q2. In this way, we have "extended" the integers to a larger set of objects that contains it, the rational numbers.

  • $\begingroup$ Why I can not conclude from definition-1, $\dfrac{a}{a}=1$. We know that $a=a\times 1$ then by definition 1 $\dfrac{a}{a}=1$. Also you do not explain "who and when defined these definitions/rules(equation1 to equation13) this way?" $\endgroup$
    – user103816
    Commented Mar 16, 2014 at 6:26
  • $\begingroup$ You only know how to equate fractions with other fractions. 1 is not a fraction. $a=a\times 1$, but this does not imply $\frac{a}{a}=1$ until you introduce the definition in question 2. My answer to question 3 motivates these rules. Trying to identify a single person who "first came up with the idea" is not a fruitful question: firstly, we don't know enough about history to figure this out, and secondly, mathematical ideas take a long time to mature - there is no single "first person". $\endgroup$ Commented Mar 16, 2014 at 8:16
  • $\begingroup$ Why equation-1 does not imply $\dfrac{a}{a}=1$? . Defination-1 is : $\dfrac{a}{b}=x$ if $bx=a$, right? Now this imply $\dfrac{a}{a}=1$ because $a\times 1=a$. What's wrong there? $\endgroup$
    – user103816
    Commented Mar 16, 2014 at 8:32
  • $\begingroup$ Equation 1 gives a rule for equating fractions, and fractions only, i.e. two numbers, one on top, one on bottom. 1 is not of this form, and so you can't apply the rule to it! $\endgroup$ Commented Mar 16, 2014 at 8:52
  • $\begingroup$ Equation-1 is nothing but the definition of division. It is not only valid for equating fractions, it is valid for for every $a,b,x \in \mathbb{N}$ So the two numbers ,one on top, one on bottom can be same. $\endgroup$
    – user103816
    Commented Mar 16, 2014 at 8:58

The main motivation of defining the rationals this way is of course the attempt to make sense of division, since none of the integers except $1$ have a multiplicative inverse. So we start with definition 2, but the symbols used are misleading, since they suggest that division is already taking place, which is not the case. Remember, until now we only know the integers, so we don't know how to make sense of an expression like $\frac{1}{a}$. Now, formally, a rational number $p$ is an ordered pair $(a,b)$ where $a$ and $b$ are integers. We then define an equivalence relation by saying that $(a,b)\sim (a',b')$ iff $ab'=a'b$. We can use this to embed the integers in the rationals by identifying an integer $a$ with $(a,1)$. This makes sense, since $(a,1)=(b,1)$ iff $a=a1=b1=b$. Note however, that $a\neq(a,1)$, since the right hand side is an ordered pair and the left hand side is not.

So, until now we only have a set partitioned into equivalence classes. We also need to define addition and multiplication. But this is easy, we just define as we always intuitively did. Now, what you are trying to do, is define multiplication without defining the equivalence classes. But that doesn't work. Of course, we can define the multiplication. But we cannot derive the equivalence classes from the multiplication. Your proof doesn't work, since we cannot conclude $\frac{a}{a}=1$. We need the equivalence relation to tell us that $(a,a)= (1,1)$. We could try and take the naive stance, but this takes us nowhere, since we cannot make sense of the left hand side. As I said before, if we are only working with the integers, we don't know what $\frac{1}{a}$ is, since this is not an integer. So we really need definition 2.

About your last question, I don't think an applied and pure mathematician would take these different definitions. As I see it, the other "definition" is not a definition at all. It is a motivation for defining the rationals this way, because if we do this, the rationals behave exactly as we expect them to. And that is exactly the reason for using this construction. It is short, elegant, makes everything rigorous and the objects we defined behave exactly as we expected according to our intuition.

  • $\begingroup$ Are "equivalence" and "equality" one and the same thing? And why we cannot conclude $\frac{a}{a}=1$ $\endgroup$
    – user103816
    Commented Mar 30, 2014 at 9:55
  • $\begingroup$ @Anupam: What does $\frac{a}{a}$ mean? Have you defined division of natural numbers? And have you defined that a rational number $\frac{p}{q}$ is the same as the result of dividing the number $p$ by the number $q$? $\endgroup$ Commented Mar 30, 2014 at 10:23
  • $\begingroup$ @ShreevatsaR, Re: definition 1, $\frac{a}{b}=x$ if $bx=a.$ (Using def-1) $\frac{a}{a}=1$ because $a\cdot 1=a.$ See $\frac{a}{a}$ is nothing but the solution of the equation $a=xa.$ The author at the very beginning of his article assumes the division of positive integers as the starting point of the article. $\endgroup$
    – user103816
    Commented Mar 30, 2014 at 10:33
  • $\begingroup$ @Anupam This doesn't work, since the equation doesn't always have a solution in the integers, therefore doesn't always exist if you haven't yet constructed the rationals. $\endgroup$ Commented Mar 30, 2014 at 10:43
  • $\begingroup$ Also, regarding equivalence, it is not the same as equality. Sometimes we want to say that certain elements of a set are essentially the same, even though they are not equal. So we define an equivalence relation to formalise this. However, because the definition of an equivalence relation is quite formal, we tend to "forget" the equivalence relation after we defined it and say that two elements of the same class are equal, even though we should strictly say the classes they represent are equal. $\endgroup$ Commented Mar 30, 2014 at 10:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .