# Interval Notation and Inequalities

## Introduction

Inequalities allow us to talk about all of the numbers between two real numbers a and b. They are frequently used when talking about safe temperature ranges for food. They can also be used when creating a modeling function since they tell you for which input values the function makes sense.

## Notation

The collection of real numbers strictly between a and b is denoted (a, b), and is called an open interval. Another way to describe this collection of numbers is that it is all x such that ${\displaystyle a The collection of numbers between a and b, inclusive, is denoted [a, b] and is called a closed interval. Another way to describe this collection of numbers is that it is all x such that ${\displaystyle a\leq x\leq b}$ If we only want one of the endpoints we call the interval a half open interval. The notation corresponds as follows: [a, b) for ${\displaystyle a\leq x and (a, b] for ${\displaystyle a

If we want to include ${\displaystyle \infty {\text{ or }}-\infty }$ we use parenthesis for any side of the inequality with an infinity. For example, if we wanted to denote ${\displaystyle a\leq \infty }$ we would use ${\displaystyle [a,\infty )}$

## Properties

Before we start working on examples we need to discuss the properties. For the following properties a, b, and c are real number, unless otherwise stated.

1) ${\displaystyle a^{2}\geq 0}$ for any real number a.

2) If ${\displaystyle a

3) If ${\displaystyle a and if ${\displaystyle c>0{\text{, then }}ac

4) If ${\displaystyle a and if ${\displaystyle c<0{\text{, then }}ac>bc}$

5) If ${\displaystyle a>0{\text{, then }}{\frac {1}{a}}>0}$

6) If ${\displaystyle a<0{\text{, then }}{\frac {1}{a}}<0}$

Just like equations, there are some operations that leave the inequality direction unchanged.

1) Simplifying either side of the inequality

2) Adding or subtracting the same expression from both sides of the inequality

3) Multiplying both sides of the inequality by the same positive expression

## Examples

This section comprises some examples of problems involving inequalities that may arise

Solve the following inequalities 1) ${\displaystyle 3x+5\leq 6x+14}$

2) ${\displaystyle -3<2x+5\leq 10}$

3) ${\displaystyle {\frac {1}{3x-5}}>0}$

Solutions:

1)

${\displaystyle {\begin{array}{rcl}3x+5&\leq &6x+14\\-9&\leq &3x\\-3&\leq &x\end{array}}}$

When solving inequalities, the final answer is sometimes required to be in interval notation. For this problem that is ${\displaystyle [-3,\infty )}$

2) Here we can solve each inequality individually, and x has to satisfy both inequalities. Thus, we have to solve ${\displaystyle -3<2x+5{\text{ and }}2x+5\leq 10}$

For the first, we get -8 < 2x and -4 < x. For the last one we have ${\displaystyle 2x\leq 5{\text{ and }}x\leq {\frac {5}{2}}}$

So ${\displaystyle -4 In interval notation, ${\displaystyle (-4,{\frac {5}{2}}]}$

3) For this problem we need the numerator and denominator to both be positive or both be negative. So we want to solve when 3x - 5 > 0. Notice that we do not include 3x - 5 = 0 since we cannot divide by 0. Solving this inequality we find ${\displaystyle x>{\frac {5}{3}}}$. In interval notation we have ${\displaystyle ({\frac {5}{3}},\infty )}$

## Absolute Value and Inequalities

Inequalities involving absolute values are presented separately since the inequalities ${\displaystyle \left|u\right| are equivalent to ${\displaystyle -a, while ${\displaystyle \,\left|u\right|>a}$ is equivalent to u < -a or u > a. In all of those inequalities ${\displaystyle <{\text{ or }}>}$ can be replaced with ${\displaystyle \leq {\text{ or }}\geq }$, respectively. Once we replace the inequality involving absolute values with an equivalent inequality we can solve it the same way we solved the other inequalities before.

Example: Solve the inequality ${\displaystyle \left|4x+6\right|\leq 5}$

We use the first equivalence to create two connected inequalities: ${\displaystyle -5\leq 4x+6\leq 5}$. We now separate the two inequalities and solve them individually reaching our final answer. Splitting the inequalities we get ${\displaystyle -5\leq 4x+6{\text{ and }}4x+6\leq 5}$. Solving the first inequality we find ${\displaystyle {\frac {-11}{4}}\leq x}$. The second inequality yields the fact that ${\displaystyle x\leq {\frac {-1}{4}}}$. Putting these together we can conclude, ${\displaystyle {\frac {-11}{6}}\leq x\leq {\frac {-1}{4}}}$ In interval notation the solution is ${\displaystyle \left[{\frac {-11}{4}},{\frac {-1}{4}}\right]}$