Chapter 3 Inequalities in one Variable

Section 3.1 Inequalities and their Solution Sets

3.1.2 Solving simple Inequalities

If the variable occurs isolated in the inequality, the solution set is an interval, see also info box 1.1.5:

Info 3.1.3

The solved inequalities have the following intervals as their solution sets:

$x < a$ has the solution set $] - \infty; a [$ , i.e. all $x$ less than $a$ .
$x \leq a$ has the solution set $] - \infty; a]$ , i.e. all $x$ less than or equal to $a$ .
$x > a$ has the solution set $] a; \infty [$ , i.e. all $x$ greater than $a$ .
$x \geq a$ has the solution set $[a; \infty [$ , i.e. all $x$ greater than or equal to $a$ .

Here,

x

is the variable and

a

is a specific value. If the variable does not occur in the inequality anymore, the solution set is either

ℝ =] - \infty; \infty [

if the inequality is satisfied, or the empty set

{}

if the inequality is not satisfied.

The symbol

\infty

means infinity. A finite interval has the form

] a; b [

which reads "all numbers between

a

and

b

". If the interval is bounded only on one side, we can write the symbol

\infty

(right-hand side) or

- \infty

(left-hand side) as the other bound.
As in the case of equations, one tries to find a solved inequality by applying transformations that do not change the solution set. The solution set can be read off from the solved inequality.

Info 3.1.4

To obtain a solved inequality from an unsolved inequality the following equivalent transformations are allowed:

adding a constant to both sides of the inequality: $a < b$ is equivalent to $a + c < b + c$ .
multiplying both sides of the inequality by a positive constant: $a < b$ is equivalent to $a \cdot c < b \cdot c$ if $c > 0$ .
multiplying both sides of the inequality by a negative constant and inverting the comparator: $a < b$ is equivalent to $a \cdot c > b \cdot c$ if $c < 0$ .

Example 3.1.5

The inequality

- \frac{3}{4} x - \frac{1}{2} < 2

is solved stepwise by the above transformations:

\begin{matrix} - \frac{3}{4} x - \frac{1}{2} < 2 ‖ + \frac{1}{2} \\ \Leftrightarrow & - \frac{3}{4} x < 2 + \frac{1}{2} ‖ \cdot (- \frac{4}{3}) \\ \Leftrightarrow & x > - \frac{4}{3} (2 + \frac{1}{2}) ‖ simplifying \\ \Leftrightarrow & x > - \frac{20}{6} = - \frac{10}{3} . \end{matrix}

So, the initial inequality has the solution set

] - \frac{10}{3}; \infty [

. Importantly, multiplying the inequality by the negative number

- \frac{4}{3}

inverts the comparator.

Exercise 3.1.6

Are the following inequalities true or false?

		$\frac{1}{2} > 1 - \frac{1}{3}$
		$a^{2} \geq 2 a b - b^{2}$ (where $a$ and $b$ are unknown numbers)
		$\frac{1}{2} < \frac{2}{3} < \frac{3}{4}$
		Let $a < b$ , then also $a^{2} < b^{2}$ .

The first inequality can be simplified to

\frac{1}{2} > \frac{2}{3}

, which, after multiplying by

6

, is equivalent to

3 > 4

. This statement is false. The second inequality can be simplified by collecting all numbers on the left-hand side:

a^{2} - 2 a b + b^{2} \geq 0

. Since

a^{2} - 2 a b + b^{2} = (a - b)^{2}

, this statement is true for all

a

and

b

. Multiplying the third chain of inequalities by the least common denominator

12

results in the chain of inequalities

6 < 8 < 9

. This statement is true. In contrast, the last statement is false, since for example, for

a = - 1

and

b = 1

, the term

a^{2} = 1

is not less than

b^{2} = 1

. Taking the square of terms is not an equivalent transformation.

Exercise 3.1.7

Find the solution sets of the following inequalities.

$2 x + 1 > 3 x - 1$ has the solution interval $L$ $=$ .
$- 3 x - \frac{1}{2} \leq x + \frac{1}{2}$ has the solution interval $L$ $=$ .
$x - \frac{1}{2} \leq x + \frac{1}{2}$ has the solution interval $L$ $=$ .

Enter the intervals in the form

[a; b]

] a; b]

, etc., for the interval boundaries also fractions and infinity or -infinity can be used. Take care whether the interval boundaries are included or excluded.

Transformation of the first inequality results in

\begin{matrix} 2 x + 1 > 3 x - 1 ‖ + 1 \\ \Leftrightarrow & 2 x + 2 > 3 x ‖ - 2 x \\ \Leftrightarrow & 2 > x \end{matrix}

and hence the solution interval is

L =] - \infty; 2 [

. Transformation of the second inequality results in

\begin{matrix} - 3 x - \frac{1}{2} \leq x + \frac{1}{2} ‖ + 3 x - \frac{1}{2} \\ \Leftrightarrow & - 1 \leq 4 x ‖ \cdot \frac{1}{4} \\ \Leftrightarrow & - \frac{1}{4} \leq x \end{matrix}

and hence

L = [- \frac{1}{4}; \infty [

. Transformation of the third inequality results in

\begin{matrix} x - \frac{1}{2} \leq x + \frac{1}{2} ‖ - x \\ \Leftrightarrow & - \frac{1}{2} \leq \frac{1}{2} . \end{matrix}

This statement does not depend on

x \in ℝ

and is always true, thus the solution set is

L = ℝ =] - \infty; \infty [

Info 3.1.8

An inequality in one variable

x

is linear if on both sides of the inequality only multiples of

x

and constants occur. Each linear inequality can be transformed into a solved inequality by one of the equivalent transformations described in the info box 3.1.3.

Onlinebrückenkurs Mathematik

1. Elementary Arithmetic

2. Equations in one Variable

3. Inequalities in one Variable

4. System of Linear Equations

5. Geometry

6. Elementary Functions

7. Differential Calculus

8. Integral Calculus

9. Objects in the Two-Dimensional Coordinate System

10. Basic Concepts of Descriptive Vector Geometry

11. Language of Descriptive Statistics