Definition(s) of rational numbers

Question

The definitions of rational numbers are somewhat confusing for me. The definition of rational numbers on wikipedia and most other sites is:

In mathematics, a rational number is any number that can be expressed as the quotient or fraction $\frac pq$ of two integers, with the denominator $q$ not equal to zero.

All integers can be expressed as the quotient $\frac pq$. So if they are not written in the fraction form, will they still be called rational numbers?

Another definition which I've found is:

A rational number is a quotient $\frac mn$ where $m$ and $n$ are integers and $n \neq 0$.

These definitions might be the same. However, I'm not able to figure out how! Are rational numbers those numbers which can be expressed as the quotient $\frac mn$ of two integers or those which are directly the quotient of two integers? For example, if $n$ is an integer, so is $n$ a rational number? Or is $\frac n1$ a rational number?

Answer 1

The integers are indeed rational, because each integer can be expressed as such a quotient as you describe.

The second definition seems different, at first glance, but it isn't. If a number $x$ can be expressed as such a quotient $\frac{m}{n}$, then it is such a quotient, in the sense that $\frac{m}{n}=x.$ On the other hand, such a quotient can immediately be expressed as such a quotient.

Answer 2

I hope this helps clarifing things a little bit. $\mathbb{Z}$ is built from $\mathbb{N}\times \mathbb{N}$ introducing an equivalence relation on it and $\mathbb{Q}$ is gained from $\mathbb{Z}\times (\mathbb{Z}\setminus\{0\})$ considering its quotient by another appropriete equivalence relation (details can be found everywhere, for example in Wikipedia). So, strictly speaking, it is not right to say that $\mathbb{Z}\subseteq \mathbb{Q}$ because it is not true that each element of the LHS is also an element of RHS. But we have a canonycal function $i\colon\mathbb{Z}\longrightarrow \mathbb{Q}$, mapping $n\in\mathbb{Z}$ to $\frac{n}{1}$, which is injective (therefore a bijection onto its image) and also a ring (monotone) homomorphism. Hence we identify, reasoning up to isomorphism, $\mathbb{Z}$ with $i(\mathbb{Z})$ when we want to think to $\mathbb{Z}$ as contained ("embedded") in $\mathbb{Q}$.

Answer 3

I think the problem is that if you simply state those definitions exactly as they are you'll fall in the problem of not having defined the notion of "a quotient of integers". So the good and cool definition of rationals that solve all of these problems is to introduce one equivalence relation in a certain set. I don't know if you are used to equivalence relations (or even relations at all), so I'll talk about that first.

If you have two sets $A$ and $B$ you can create a relation $R$ between them which is a subset of the cartesian product $R\subset A\times B$. Think for a minute, elements of $A\times B$ are pairs $(a,b)$ with $a\in A$ and $b\in B$, so if $(a,b)\in R$ we are telling that $a$ is in some way related to $b$ and in this case we write $aRb$. Now, an equivalence relation can be introduced between a set and itself to mimic equality, it is usually denoted $\sim$ and satisfies those properties:

1. $a\sim a$ (Reflexivity)
2. $a \sim b \Longrightarrow b \sim a$ (Symmetry)
3. $a \sim b \wedge b \sim c \Longrightarrow a\sim c$ (Transitivity)

In the third one the $\wedge$ symbol means AND. Look now that equality always obeys those three properties. So when we have a set and we want to construct a notion of the objects being equivalent without being equal we use an equivalence relation. Now, given a set $A$, an equivalence relation $\sim$ in $A$ and some element $a \in A$ the set of all other elements of $A$ equivalent to $a$ by $\sim$ is called equivalence class and denoted $\left[a\right]$. The set of all equivalence classes is called the quotient set and denoted $A/\sim$ and although the elements are sets of elements of $A$ we can usually think of $A/\sim$ as just the elements of $A$ with $\sim$ imposed on them.

Now returning to your problem! Given the set $\mathbb{Z}\times (\mathbb{Z}\setminus\{0\})$, in other words, ordered pairs of integers without any pair with $0$ in the second element we introduce the following equivalence relation $\sim$ on the set:

$$(a,b)\sim(a',b') \Longleftrightarrow ab'=a'b$$

Now stop for a while and look what we did! We are almost defining the quocient of integers. If we introduce the notation:

$$\frac{a}{b}=\left\{(a',b') \in \mathbb{Z}\times (\mathbb{Z}\setminus\{0\}) : (a',b')\sim (a,b)\right\}$$

We have exactly what the quotient is: the equivalence class of all those elements $(a,b)$ with the relation imposed. Proving this is really an equivalence relation is a good exercise. Now, look what I've said before: formally we define the quotient using equivalence classes but in practice we simply think of it as the element $(a,b) \in \mathbb{Z}\times(\mathbb{Z}\setminus\{0\})$ with the relation $\sim$ imposed.

Now, we define the set of rationals by:

$$\mathbb{Q}=(\mathbb{Z}\times(\mathbb{Z}\setminus\{0\}))/\sim$$

In other words, the set of rationals is the set of all quotients of integers, recalling that we defined the quotient as that equivalence class. With this your first definition is obviously equivalent (indeed we just made it formal using this thought) and the second is the exact same thing.

I hope this helps you somehow! Good luck!

Answer 4

rational numbers is what you get when $\mathbb{Z}$ is closed to division. Division is undefined when the divisor is $0$.

What this means is that if $p$ and $q$ are in the set, so is $p/q$, when allowed by division. The set equates to the ratio of elements of $\mathbb{Z}$.