Understanding the Basics of Inequalities

Inequalities are foundational in algebra, analysis, and mathematical reasoning as they provide a precise method to compare real numbers, algebraic expressions, and functions. An inequality asserts that one mathematical quantity is greater than, less than, or otherwise unequal to another, using specific relational symbols.

Formal Structure and Symbols for Mathematical Inequalities

Let $a$ and $b$ be real numbers. The symbols $>$, $<$, $\geq$, $\leq$, and $\neq$ are used to compare $a$ and $b$ as follows:

Strict Inequality: $a > b$ means $a$ is strictly greater than $b$; $a < b$ means $a$ is strictly less than $b$.

Non-Strict (Weak) Inequality: $a \geq b$ means $a$ is greater than or equal to $b$; $a \leq b$ means $a$ is less than or equal to $b$.

Inequality of Distinctness: $a \neq b$ means $a$ and $b$ are not equal.

Classification: Linear and Non-Linear Inequalities

Linear Inequality in One Variable: An inequality of the form $ax + b \gt 0$, $ax + b \geq 0$, $ax + b \lt 0$, or $ax + b \leq 0$, where $a$, $b$ are real constants and $a \neq 0$, is called a linear inequality in one variable.

Linear Inequality in Two Variables: An inequality of the form $ax + by + c \gt 0$ or similar, where $a$, $b$, $c \in \mathbb{R}$ and at least one of $a$, $b$ is non-zero, is called a linear inequality in two variables.

Non-linear inequalities involve degrees greater than one, or more complex structures, such as quadratic or rational inequalities.

Addition and Subtraction Properties of Inequalities

Addition Property: If $a \gt b$ and $c$ is any real number, then $a + c \gt b + c$. This property also holds for the symbols $\geq$, $<$, and $\leq$ analogously.

Proof: Given $a > b$, by the definition of the order relation in $\mathbb{R}$, adding the same real number $c$ to both sides preserves the inequality: $a + c > b + c$.

Subtraction Property: If $a > b$ and $c$ is any real number, then $a - c \gt b - c$.

This follows as $a - c = a + (-c)$ and $b - c = b + (-c)$. Then by the addition property, $a + (-c) > b + (-c)$.

Multiplicative and Divisional Properties for Inequalities

Multiplication by Positive Numbers: If $a > b$ and $k > 0$, then $ak > bk$. The sense of the inequality is preserved.

Proof: Since $k > 0$, multiplying each side of $a > b$ by $k$ gives $a \cdot k \gt b \cdot k$.

Multiplication by Negative Numbers: If $a > b$ and $k < 0$, then $ak < bk$, so the sense of the inequality is reversed.

Proof: Let $k = -|k| < 0$. Multiplying both sides by $k$ reverses the inequality: $a \cdot k < b \cdot k$.

Division by Non-Zero Numbers: If $a > b$ and $k > 0$, then $\dfrac{a}{k} \gt \dfrac{b}{k}$. If $k < 0$, then $\dfrac{a}{k} \lt \dfrac{b}{k}$.

Transitive and Converse Properties of Inequalities

Transitive Property: For real numbers $a$, $b$, $c$, if $a \geq b$ and $b \geq c$, then $a \geq c$. If at least one inequality is strict, then $a \gt c$.

Converse (Reverse) Property: If $a \gt b$, then $b \lt a$. More generally, the direction of an inequality is reversed if the order of the terms is reversed.

Solving Linear Inequality in One Variable: Stepwise Method

Consider the inequality $2x + 5 < 4x - 15$. Each algebraic transformation must preserve the inequality.

Step 1: Subtract $2x$ from both sides.

$2x + 5 - 2x < 4x - 15 - 2x$

$5 < 2x - 15$

Step 2: Add $15$ to both sides.

$5 + 15 < 2x - 15 + 15$

$20 < 2x$

Step 3: Divide both sides by $2$ (a positive number).

$\dfrac{20}{2} < \dfrac{2x}{2}$

$10 < x$

Result: $x > 10$.

Solving Linear Inequality with Fractions: Worked Example

Given: $\dfrac{y-3}{2} \geq \dfrac{5}{4}$

Step 1: Multiply both sides by $4$. $4 > 0$ retains the direction.

$4 \cdot \dfrac{y-3}{2} \geq 4 \cdot \dfrac{5}{4}$

$2(y-3) \geq 5$

Step 2: Expand the left side: $2y - 6 \geq 5$

Step 3: Add $6$ to both sides.

$2y - 6 + 6 \geq 5 + 6$

$2y \geq 11$

Step 4: Divide both sides by $2$.

$\dfrac{2y}{2} \geq \dfrac{11}{2}$

$y \geq 5.5$

Graphical Representation of Linear Inequalities in Two Variables

Let $2y \geq 4x - 8$. First, isolate $y$:

$2y \geq 4x - 8$

Divide both sides by $2$:

$\dfrac{2y}{2} \geq \dfrac{4x-8}{2}$

$y \geq 2x - 4$

The boundary is the straight line $y = 2x - 4$. The solution set for $y \geq 2x - 4$ consists of all points in the coordinate plane that lie on or above this line. For detailed methodology on graphing inequalities, see Understanding Sets, Relations, And Functions.

Case Distinction: Inequality Involving Multiplication and Division by Negatives

When both sides of an inequality are multiplied or divided by a negative number, the direction must be reversed. Consider the inequality $x - 5 > 3$.

Step 1: Subtract $x$ from both sides: $x - x - 5 > 3 - x$ yields $-5 > 3 - x$.

Step 2: Subtract $3$ from both sides: $-5 - 3 > 3 - x - 3$ yields $-8 > -x$.

Step 3: Multiply both sides by $-1$ ($-1 < 0$) and reverse the direction:

$(-8) \cdot (-1) < (-x) \cdot (-1)$ yields $8 < x$.

Result: $x > 8$.

Logical Combination and Transitivity in Chains of Inequalities

Given a chain $a > b \geq c > d$, one can deduce $a > d$ by transitivity. However, if any link is non-strict ($\geq$), then the final relation is non-strict unless all intermediate relations are strict.

For details on induction in inequalities, refer to Understanding Mathematical Induction.

Worked Example: Deduction from a Chain of Inequalities

Given: $h < a < t = g > u \geq v \geq b$. Determine whether $t > b$ and $g > h$ are true.

$t = g \gt u \geq v \geq b$. Since $t = g$, and $g \gt u \geq v \geq b$, $t \gt b$.

Also, $h < a < t = g$. Therefore, $h < a < t$ and $t = g \implies h < g$, so $g > h$.

Result: Both $t > b$ and $g > h$ are definitely true.

Further Reading: Advanced Inequalities and Olympiad Concepts

Beyond linear inequalities, the study extends to quadratic, polynomial, rational, and classical inequalities such as the Cauchy-Schwarz, AM-GM, and triangle inequalities.