The goal of this lesson is to familiarize the reader with the properties of operations of rational numbers, and before that, how we construct and define this specified set.
Rational numbers
On the set of natural numbers we could not define the operation $’-‘$ for all two natural numbers. Also, in the set $\mathbb{Z}$ we can not find the number $a \in \mathbb{Z}$ such that $y = a \cdot x, \forall x, y \in \mathbb{Z}$. Right here we find the motivation for expanding the set of integers $\mathbb{Z}$.
We consider the cartesian product $\mathbb{Z} \times \mathbb{Z}^{+}$ with a relation $\sim$ defined on it as follows. For two pairs $(a, b), (c, d) \in \mathbb{Z} \times \mathbb{Z}^{+}$ we say that they are in the relation $ \sim$ if the following is valid:
$$(a, b) \sim (c, d) \Longleftrightarrow ad = bc.$$
$\sim$ defines an equivalence relation on $\mathbb{Z} \times \mathbb{Z}^{+}$.
A quotient set $\mathbb{Q} = \mathbb{Z} \times \mathbb{Z}^{+}/_{\sim}$ is called a set of rational numbers and its elements rational numbers, that is
$$\mathbb{Q} =\mathbb{Z} \times \mathbb{Z}^{+}/_{\sim} = \{\frac{x}{y} : x, y \in \mathbb{Z} \times \mathbb{Z}^{+} \}.$$
We define $[(x, y)] := \frac{x}{y}, (y \neq 0)$.
Addition of rational numbers
Definition 1. The addition operation $+$ on $\mathbb{Q}$ is defined as follows:
$$+: \left (\mathbb{Z} \times \mathbb{Z}^{+} \right) \times \left (\mathbb{Z} \times \mathbb{Z}^{+} \right) \to \mathbb{Z} \times \mathbb{Z}^{+},$$
$$[(x, y)] + [(u, w)] = [(x \cdot w + y \cdot u, y \cdot w)].$$
In fraction notation:
$$\frac{x}{y} + \frac{u}{w} = \frac{x \cdot w + y \cdot u}{y \cdot w}.$$
The operation of addition on $\mathbb{Q}$ is well defined and closed:
$$\forall x, y \in \mathbb{Q}: x + y \in \mathbb{Q}.$$
Theorem 1.(Commutativity) For any $x, y \in \mathbb{Q}$ the following is valid:
$$x + y = y + x.$$
Proof.
Let $[(a,b)] = x$ and $[(c,d)]= y \in \mathbb{Q}$. Then we have
$$x + y = [(a,b)] + [(c,d)] $$
$$ = [(a \cdot d + b \cdot c, b \cdot d)]$$
$$y+x = [(c,d)] + [(a,b)] $$
$$ = [(c \cdot b +d \cdot a, d \cdot b)] $$
$$= [(a \cdot d + b \cdot c, b \cdot d)]. $$
We can notice that the left and right side of equality are equal, therefore, the commutativity of addition of rational numbers is proven.
Theorem 2. (Associativity) For any $x, y, z \in \mathbb{Q}$ the following is valid:
$$(x + y) + z = x + (y + z).$$
Proof.
Let $x = [(a, b)], y = [(c, d)], z= [(e, f)] \in \mathbb{Q}$. Then we have:
$$(x + y) + z = ([(a, b)] + [(c, d)]) + [(e, f)] $$
$$= ([(a \cdot d + b \cdot c, b \cdot d)]) + [(e, f)] $$
$$ = \left [ \left ( \left ( a \cdot d + b \cdot c\right) \cdot f + \left (b \cdot d \right ) \cdot e, \left (b \cdot d \right) \cdot f\right ) \right] $$
$$ =\left[ \left( \left( \left( a \cdot d \right) \cdot f + \left ( b \cdot c\right) \cdot f \right) + \left(b \cdot d \right) \cdot d, \left (b \cdot d \right) \cdot f\right ) \right]$$
$$x+ (y +z) = [(a, b)] + \left ([(c, d)] + [(e, f)] \right) $$
$$ = [(a, b)] + ([(c \cdot f + e \cdot d, d \cdot f)]) $$
$$=[(a \cdot (d \cdot f) + b \cdot (c \cdot f + e \cdot d), b \cdot (d \cdot f))] $$
$$=\left[ \left( a \cdot (d \cdot f) + (b \cdot (c \cdot f) + b \cdot (e \cdot d), b \cdot (d \cdot f) \right) \right] $$
$$ =\left[ \left( (a \cdot d) \cdot f + ((b \cdot c) \cdot f + (b \cdot d) \cdot e, (b \cdot d) \cdot f \right) \right]$$
$$= \left[ \left( \left( \left( a \cdot d \right) \cdot f + \left ( b \cdot c\right) \cdot f \right) + \left(b \cdot d \right) \cdot d, \left (b \cdot d \right) \cdot f\right ) \right]. $$
We obtain that $(x + y) + z = x + (y + z), \forall x, y, z \in \mathbb{Q}$, therefore, the associativity of addition of rational numbers is valid.
Theorem 3. There exists $e \in \mathbb{Q}$ such that $\forall x \in \mathbb{Q}$ the following is valid:
$$e + x = x + e = x.$$
Proof.
Before the proof, we need to highlight two lemma without proving.
Lemma 4. For any $x \in \mathbb{Z}$ is valid $x \cdot 1 = x$.
Lemma 5. For any $x \in \mathbb{Z}$ is valid $x \cdot 0 = 0$.
Let $e=[(0,1)]$ and $x = [(a, b)]$. Then we have:
$$e + x = [(0,1)] + [(a, b)] $$
$$ = [(0 \cdot b + 1 \cdot a, 1 \cdot b)] $$
$$= [(0 \cdot b + a \cdot 1, b \cdot 1)] $$
$$= [(0 \cdot b + a, b)] $$
$$= [(a, b)] $$
$$= x$$
$$a + e = [(a, b)] + [(0,1)]$$
$$= [(a \cdot 1 + b \cdot 0, b \cdot 1)] $$
$$= [(a + b \cdot 0, b)] $$
$$ = [(a, b)] $$
$$= x. $$
The both side of equality are equal to $x$, it follows that the statement of the theorem is true.
The element $e = [(0,1)] = \frac{0}{1} = 0$ is called the identity element for addition in the set $\mathbb{Q}$.
Theorem 4. For any $x \in \mathbb{Q}$ there exists $-x \in \mathbb{Q}$ such that
$$x+ (-x) = (-x) + x = e.$$
Proof.
Let $x= [(a, b)], -x= [(-a, b)], e=[(0,1)] \in \mathbb{Q}$. Then we have:
$$x + (-x) = [(a, b)] + [(-a, b)] $$
$$ = [(a \cdot b + b \cdot (-a), b\cdot b)] $$
$$= [(0, b \cdot b)]$$
$$(-x) + x = [(-a, b)] + [(a, b)] $$
$$ = [((-a) \cdot b + a \cdot b, b \cdot b)] $$
$$= [(0, b \cdot b)]$$
What is remains is to prove that $[(0, b \cdot b)] = [(0,1)]=e$.
According to the definition of the set of rational numbers, the relation $\sim$ is defined as $(a, b) \sim (c, d) \Leftrightarrow a \cdot d = b \cdot c$. We conclude that pairs $(0,1)$ and $(0, b \cdot b)$ are in the relation because $0 \cdot (b \cdot b) =1 \cdot 0$, that is $0= 0$, is valid. Therefore, $[(0, b \cdot b)] = [(0,1)]$.
This means, $x=[(a, b)]$ has an addition inverse in $\mathbb{Q}$: $-x= [(-a, b)]$.
We have proven that on the set of rational numbers are valid properties of associativity and commutativity of addition, there exists the identity element for addition and an addition inverse, therefore, the ordered pair $(\mathbb{Q}, +)$ has a structure of the Abelian group.
Properties of multiplication in $\mathbb{Q}$
Definition 2.
The multiplication operation $\cdot$ on $\mathbb{Q}$ is defined as follows:
$$+: \left (\mathbb{Z} \times \mathbb{Z}^{+} \right) \times \left (\mathbb{Z} \times \mathbb{Z}^{+} \right) \to \mathbb{Z} \times \mathbb{Z}^{+},$$
$$[(a, b)] \cdot [(c, d)] = [(a \cdot c, b \cdot d)], \quad [(a, b)], [(c, d)] \in \mathbb{Z} \times \mathbb{Z}^{+}.$$
The operation of multiplication on $\mathbb{Q}$ is well defined and closed:
$$\forall x, y \in \mathbb{Q}: x \cdot y \in \mathbb{Q}.$$
Theorem 5. (Commutativity of multiplication) For any $x, y \in \mathbb{Q}$ is valid $x \cdot y = y \cdot x$.
Proof.
Let $x = [(a, b)], y=[(c, d)] \in \mathbb{Q}$. Then we have
$$x \cdot y = [(a, b)] \cdot [(c, d)] $$
$$= [(a \cdot c, b \cdot d)] \in \mathbb{Q}$$
$$y \cdot a = [(c, d)] \cdot [(a, b)] $$
$$ = [(c \cdot a, d \cdot b)] $$
$$= [(a \cdot c, b \cdot d)].$$
Since the left and the right side of equality are equal, it follows the statement of the theorem.
We will mention the following properties of multiplication in $\mathbb{Q}$ without proof.
Theorem 6. (Associativity of multiplication) For any $x, y, z \in \mathbb{Q}$ the following is valid:
$$(x \cdot y) \cdot z = x \cdot (y \cdot z).$$
Theorem 7. There exists $e \in \mathbb{Q}$ such that for every $x \in \mathbb{Q}$ the following is valid:
$$x \cdot e = e \cdot x = x.$$
The element $e \in \mathbb{Q}$ is called the identity element for multiplication in $\mathbb{Q}$, whereby $e=[(1,1)]$. If we were writing in the form of a fraction, then $e=\frac{1}{1} = 1$ is a neutral element.
Theorem 8. For any $x \in \mathbb{Q} /\{0\}$ there exists $x^{-1} \in \mathbb{Q}$ such that the following is valid:
$$x \cdot x^{-1} = x^{-1} \cdot x = e.$$
The element $x^{-1} \in \mathbb{Q}$ is called a multiplicative inverse for a number $x= [(a, b)] \in \mathbb{Q} /\{0\}$, whereby:
$(1.)$ If $a> 0$ then $x^{-1} = [(b, a)]$,
$(2.)$ if $a < 0$ then $x^{-1} = [(-b, -a)]$.
Theorem 9. There is no a multiplicative inverse for a $0 \in \mathbb{Q}$.
Since $0 \in \mathbb{Q}$ has no a multiplicative inverse, the ordered pair $(\mathbb{Q}, \cdot)$ is a commutative monoid, however, $(\mathbb{Q} /\{0\}, \cdot)$ is an Abelian (commutative) group.
Theorem 10. (Distributivity) For all $x, y, z \in \mathbb{Q}$ the following is valid:
$$(x + y) \cdot z = x \cdot z + y \cdot z.$$
Ordering on $\mathbb{Q}$
Let $[(a, b)], [(c, d)] \in \mathbb{Q}$. Assuming $b, d > 0 \in \mathbb{Z}$, we say that $[(a, b)] < [(c, d)]$ if and only if $ad < bc$. With $<$ we denoted our ordering relation in $\mathbb{Q}$, which is different from the usual ordering in $\mathbb{Z}$.
Theorem 11. If $(a, b) \sim (c, d)$ and $(a’, b’) \sim (c’, d’)$, then
$$[(a, b)] < [(c, d)] \Longleftrightarrow [(a’, b’)] < [(c’, d’)],$$
that is, ordering on $\mathbb{Q}$ is well defined.
Theorem 12. The relation $<$ is an ordering of the rational numbers.
Theorem 13. The rational numbers form an ordered field.
Density property of rational numbers
If $x $and $y$ are rational numbers such that $x<y$, then there exists a rational number $ a$ such that $x < a <y$.
The simplest, we can take the arithmetic mean of the numbers $x$ and $y$. The specified property is another property that distinguishes a set of integers from the set of rational numbers.
