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