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.
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.