Arrangement of the real numbers

Arrangement of the real numbers

In the set $\mathbb{R}$ 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 $\mathbb{R}$: 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.

Considering the numbers $\dfrac{7}{4}$ and $\dfrac{11}{6}$, if we calculate its equivalent fractions with common denominator (that will be the least common multiple of both denominators), we have that: $$mcm(4,6)=mcm(2^2, 2\cdot3)=2^2\cdot3=12$$ And therefore, we have: $$\dfrac{7}{4}=\dfrac{7}{4}\cdot\dfrac{3}{3}=\dfrac{21}{12}$$ $$\dfrac{11}{6}=\dfrac{11}{6}\cdot\dfrac{2}{2}=\dfrac{22}{12}$$

therefore, being $21 < 22$, we have

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

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 $c\leq 0 \Rightarrow a\cdot c \geq b\cdot c.$

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 $5\cdot3=15$, and it is verified that

$$-9 < 15$$

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

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

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