Equality of positive rational numbers.

Solution 1:

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.

Solution 2:

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.