Inclusion of the integers in the rational

We're going to see the operations of the rational

\mathbb{Z} includes in \mathbb{Q} associating to each a\in\mathbb{Z} the equivalence class of (a, 1)

Being the sum and the product in \mathbb{Q} extensions of \mathbb{Z}

In addition, since (-a, -b)\sim(a, b), and as by definition b\not= 0, as long as we take (a, b) in \mathbb{Q}, we can assume that b > 0 (i.e. denominators of fractions can be imposed as positive)

Operation sum

It is defined by the rules of fractions, when it makes sense:

\frac{a}{b}+\frac{c}{d}=\frac{a\cdot d + b \cdot c}{b \cdot d}

and to achieve this the formal definition must be:

(a, b) + (c, d) = (a \cdot d + b\cdot c, b\cdot d)

Operation product

It is defined by the rules of fractions, when it makes sense:

\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b \cdot d}

and to achieve this the formal definition must be:

(a, b) \cdot (c, d) = (a \cdot c, b\cdot d)

Relationship of equivalence

For the definitions in \mathbb{Z}\times(\mathbb{Z}\backslash\left\{0\right\}) given above are valid in \mathbb{Q}=\mathbb{Z}\times(\mathbb{Z}\backslash \left\{0\right\})/\sim, we have to prove that they are compatible with the equivalence ratio, that is, if we have (a_1,b_1)\sim(a_2, b_2) and (c_1,d_1)\sim(c_2, d_2) it is fulfilled that:

\begin{cases} (a_1,b_1)+(c_1,d_1)\sim(a_2,b_2)+(c_2,d_2) \\ (a_1,a_1)\cdot(c_1,d_1)\sim(a_2,b_2)\cdot(c_2,d_2) \end{cases}

Order relationship

There is also to define the order relationship in \mathbb{Q}. To do this we have to express the still non existent inequality \frac{a}{b}\leq \frac{c}{d} in terms of inequalities with integers, but as we can assume that b > 0 and d > 0, is equivalent to a \cdot d \leq c \cdot b (this inequality is obtained from which we wanted to obtain by multiplying b \cdot d and because we have assumed that denominators are positive, it makes inequality not change direction). So we define:

(a, b) \leq (c, d) \Longleftrightarrow a \cdot d \leq c\cdot b \quad (b, d > 0)

It can be shown that this relationship is of order, which is compatible with the equivalence ratio (we have used it when defining \leq for positive denominators only) and extends that of \mathbb{Z}. As in \mathbb{Z}, the order relationship in \mathbb{Q} is total, and we also have positive and negative numbers

Note Once we have already made a good definition of rational numbers, we can detach ourselves from auxiliary notation (a, b) to use from now on, the usual \frac{a}{b} or a / b, where a is called numerator and b denominator (as mentioned above, it can be assumed that b is always positive). Also, instead of using \frac{a}{1}, we can just use a. And instead of using the equivalence ratio as a (a, b)\sim (c, d) we will use \frac{a}{b}=\frac{c}{d}

Note In each class of equivalent fractions of \mathbb{Q}, there is a fraction that is usually taken as a representative. It is the irreducible so-called, in which the maximum common numerator divider and denominator is worth 1. Although it is elementary, we should have developed something more arithmetic in \mathbb{N}_0 to be able to insert it here

Operation division

The properties of the sum and the product in \mathbb{Z} (associativity, commutivity, neutral elements, distributivity, etc.) extend to \mathbb{Q}, in addition, order is stable with operations in the same sense as in \mathbb{Z}. But we have a novelty regarding \mathbb{Z}, we can divide

For any \frac{a}{b}\in \mathbb{Q} not null, we can find a reverse with respect to the product. Since:

\frac{a}{b}\cdot \frac{b}{a}=\frac{a\cdot b}{b \cdot a}=\frac{1}{1}=1

The notation for the inverse of \frac{a}{b} it is (\frac{a}{b})^{-1}=\frac{b}{a}

Thanks to the reverse we can divide by a rational not null. Divide \frac{a}{b} between \frac{c}{d} is to find a rational number such that \frac{a}{b} multiplied by that rational \frac{c}{d}; this operation is immediate, just multiply \frac{a}{b} by the inverse of \frac{c}{d}:

\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot(\frac{c}{d})^{-1}=\frac{a}{b}\cdot\frac{d}{c}=\frac{a\cdot d}{b\cdot c}

The existence of inverse in the product (and therefore division) makes the algebraic structure of \mathbb{Z} (which is a ring) fall short, and that \mathbb{Q} with the sum and the product becomes a body. In addition, the order structure is stable with trades, i.e. dice x, y, t \in \mathbb{Q} it is fulfilled that:

\begin{cases}x\leq y \Rightarrow x+t \leq y + t \\ x\leq y, t \geq 0 \Rightarrow t\cdot x \leq t\cdot y \end{cases}