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