Arrangement of the real numbers

In the set $R$ we have defined a relation of order that we denote $<$ intuitively, if $a$ and $b$ are two real numbers, we will write $a < b$ if, when drawing them on the real line, point $a$ is on the left of point $b$ . We will then say that $a$ is less than $b$ .

$a \leq b$ is usually used to indicate that number $a$ is less than or equal to $b$ . It is also said that $\leq$ is an inequality symbol and that $<$ is a strict inequality symbol.

It is said that this relation is of total order $R$ : that is, considering two different real numbers $a$ and $b$ , we always have $a < b$ or $b < a$ . Or,in other words, $a$ and $b$ are always comparable.

Example

Considering the numbers $\frac{7}{4}$ and $\frac{11}{6}$ , if we calculate its equivalent fractions with common denominator (that will be the least common multiple of both denominators), we have that: $m c m (4, 6) = m c m (2^{2}, 2 \cdot 3) = 2^{2} \cdot 3 = 12$ And therefore, we have: $\frac{7}{4} = \frac{7}{4} \cdot \frac{3}{3} = \frac{21}{12}$ $\frac{11}{6} = \frac{11}{6} \cdot \frac{2}{2} = \frac{22}{12}$

therefore, being $21 < 22$ , we have

$\frac{21}{12} < \frac{22}{12} \Rightarrow \frac{7}{4} < \frac{11}{6}$

Properties of the arrangement

The operations with real numbers and the arrangement of these are related by the following properties:

Monotonicity of the sum: an inequality won't change if adding the same value to both members, in other words, if $a < b$ then for any real number $c$ , it is satisfied that: $a + c < b + c$ Also it is satisfied if the inequality is not strict: $a \leq b \Rightarrow a + c \leq b + c .$
Monotonicity of the product by a positive number: an inequality does not change if we multiply both members by the same positive number, that is, if $a < b$ and $c$ is a positive real number $(c > 0)$ , it is satisfied: $a \cdot c < b \cdot c$ Also it is satisfied if the inequality is not strict: $a \leq b$ and $c \geq 0 \Rightarrow a \cdot c \leq b \cdot c .$
Antimonotonicity of the product by negative numbers: all inequalities change if we multiply both members by the same negative number, that is, if $a < b$ and $c$ is a negative real number $(c < 0)$ , it is satisfied that: $a \cdot c > b \cdot c$ Also it is satsifed if the inequality is not strict: $a \leq b$ and $c \leq 0 \Rightarrow a \cdot c \geq b \cdot c .$

Example

In the inequality $- 3 < 5$ if we add up $- 6$ in both members we obtain:

$- 3 + (- 6) = - 9$ and $5 + (- 6) = - 1$ , and it is verified that

$- 9 < - 1.$

If we multiply the inequality by $3$ , we have:

$- 3 \cdot 3 = - 9$ and $5 \cdot 3 = 15$ , and it is verified that

$- 9 < 15$

Finally if we multiply the inequality by $- \frac{1}{2}$ , we have:

$- 3 \cdot (- \frac{1}{2}) = \frac{3}{2}$ and $5 \cdot (- \frac{1}{2}) = - \frac{5}{2}$ , and it is verified that

$\frac{3}{2} > - \frac{5}{2}$