Chapter 1.2 Intervals and Inequalities

The goal of this section is to visualize real numbers so that we can solve more complicated problems. We first begin with visualizing real numbers using a numberline.

Unions and Intersections of Intervals

View the set \(\mathbb R\), the set of Real Numbers, as a line in the following way:

  • mark the real number \(0\) as a point on the line,
  • mark the number \(1\) as a point to the right of \(0\).

View the natural numbers, \(\mathbb N\), the integers, \(\mathbb Z\), and the rational numbers, \(\mathbb Q\), as subsets of \(\mathbb R\).

It can be visualized liked this:

Visualizing sets allows us to use spatial reasoning to solve computational problems—to “compute without computing”.

To do this, we will describes certain subsets of real numbers that we will use not just in this class, but in other mathematics classes. The first are called unbounded intervals.

Unbounded Intervals

For any real number \(a\), there are four unbounded intervals with endpoint \(a\):

  • \((-\infty, a)= \{x \in \mathbb R\colon x<a\}\) the set of all real numbers less than \(a\).

  • \((-\infty, a]= \{x \in \mathbb R\colon x\leq a\}\) the set of all real numbers less than or equal to \(a\).

  • \((a, \infty)= \{x \in \mathbb R\colon x>a\}\) the set of all real numbers greater than \(a\).

  • \([a, \infty)= \{x \in \mathbb R\colon x\ge a\}\) the set of all real numbers greater than or equal to \(a\).

When we write unbounded intervals, we are describing sets of all real numbers that satisfy a specific inequality. Here are some concrete examples.

Example 1

The set of all real numbers greater than \(5\) is the unbounded interval \((5,\infty)\).

The unbounded interval \((-\infty,-3]\) is the set of all real numbers less than or equal to \(-3\).

The set of all real numbers \(x\) that satisfy the inequality \(x\leq 4\) is the unbounded interval \((-\infty,4]\)

To describe the set of all real numbers that are greater than \(-3\) and less than \(5\), we represent this by intersecting the unbounded intervals \((-3,\infty)\) and \((-\infty,5)\) to the form the set. In this particular example, this is an example of an open interval.

Open and Closed Interval

Suppose that \(a<b\).

There are four intervals with endpoints \(a\) and \(b\). Two of these are the open interval \((a,b)\) and the closed interval \([a,b]\), where \[(a, b) = (a,\infty) \cap (-\infty, b)\] and \[[a,b] = [a,\infty) \cap (-\infty, b].\]

Here are two examples.

Example 2

The set of all real numbers greater than \(-3\) and less than \(5\) is the open interval \((-3,5)\).

The closed interval \([-2,5]\) is the set of all real numbers greater than or equal to \(-2\) and less than or equal to \(5\).

The set of all real numbers \(x\) that satisfy the inequality \(-10\leq x\leq 1\) is the closed interval \([-10,1].\)

To describe all real numbers \(x\) that satisfy inequalities of the form \(a\leq x<b\) or \(a<x\leq b\), we can take two unbounded intervals and intersect them to produce what is called a half-open interval.

Half-Open Intervals

The other two are the half-open intervals, given by:

  • the half open interval with left endpoint \([a, b)= \{x \in \mathbb R\colon a\leq x<b\}\)

  • the half open interval with right endpoint \((a, b]= \{x \in \mathbb R\colon a<x\leq b\}\).

Let’s practice visualizing intervals.

Example 3

Sketch on the number line the intervals

  1. \((-2, 3)\)
  2. \([-2, 3)\)
  3. \((-2, 3]\)
  4. \([-2, 3]\)

Part a

Part b

Part c

Part d

What does the union, intersection and set difference of two intervals look like? Use the next example to see this.

Example 4

Describe the sets

  1. \((4, 12)\cup (-3, 7)\)
  2. \((4, 12)\cap (-3, 7)\)
  3. \((4,12)\cap (-3,7)\).
  1. Graph each interval and take the union.

Notice that the number \(4\) and \(7\) are included in the final answer. It is because \(4\) is included in \((-3,7)\) while \(7\) was included in \((4,12)\). For unions, the number must be in at least one of the intervals. The final answer is \((-3,12)\).

  1. Graph each interval and take the intersection.

Notice that the number \(4\) is not included in the final answer. It is because \(4\) is not in \((4,12)\). For intersections, the number must be in both of the intervals. The final answer is \((4,7)\).

  1. Graph the interval \((4,12)\) and \((-3,7)\) right underneath it. Remove all real numbers in \((4,12)\) that are also in \((-3,7)\). The final answer is \([7,12)\). The number \(7\) is included in the final answer because the number \(7\) is in the interval \((4,12)\) but not in the interval \((-3,7)\).

What if we need to calculate a mix of unions and intersections of intervals? The principle of decomposition can help us make the problem manageable, which we demonstrate below.

Example 5

Describe the set \[\big([-3, 1)\cup (5, 8]\big) \cap \big((-1, 6]\cup (7, 12]\big).\]

Use decomposition to break up into manageable parts. Take \(A=[-3,1)\cup(5,8]\) and \(B=(-1,6]\cup(7,12]\). To solve \(\big([-3, 1)\cup (5, 8]\big) \cap \big((-1, 6]\cup (7, 12]\big)\) is the same as solving \(A\cap B\). Figure out what \(A\) and \(B\) are.

The solution \(A\) is \([-3,1)\cup(5,8]\).

The solution \(B\) is \((-1,6]\cup (7,12]\).

The final answer is thus \((-1,1)\cup(5,6]\cup(7,8].\)

Remember that intervals like we’ve seen above are really just inequalities. Identify the inequalities with the appropriate intervals.

Example 6

When possible, match the inequalities with the sets sketched below:

  1. \(x\ge 2\)

  2. \(x< 6\)

  3. \(x\ge 2\) and \(x<6\)

  4. \(x> 2\) and \(x\leq6\)

  5. \(x\leq 2\) or \(x< 6\)

  6. \(x\geq 2\) or \(x\ge 6\)

  7. \(x< 2\) and \(x\leq 6\)

  8. \(x< 2\) and \(x\ge 6\)

For parts c through h, draw it on a numberline and simplify.

  1. Matches with \(B\)
  2. Matches with \(D\).
  3. Matches with \(A\).
  4. Matches with \(F\).
  5. Matches with \(D\).
  6. Matches with \(B\).
  7. Has no matches
  8. Has no matches.

To summarize, “and” functions like an intersection and “or” functions like a union.

Multiple Linear Inequalities

Sketching intervals on the real line is a strategy for solving multiple linear inequalities. Each inequality determines an interval. Take unions and intersections of the intervals as required to find the solutions to the original problem.

First, let’s understand why \(x< y\) implies that \(-x>-y\).

Example 7

Why does multiplying both sides of an inequality by a negative number flip the inequality?

Start with the inequality \(x<y\): \[x<y.\] Subtract both sides by \(x\): \[\begin{align*} x&<y\\ x-x&<y-x\\ 0&<y-x. \end{align*}\] Subtract both sides by \(y\): \[\begin{align*} 0&<y-x\\ 0-y&<y-y-x\\ -y&<0-x\\ -y&<-x. \end{align*}\] Hence \(x<y\) and \(-y<-x\) and \(-x>-y\) are equivalent.

We solve the following double-sided inequality.

Example 8

Find all \(x\) satisfying \[x -2 \ge 3x - 6.\]

Solve the inequality by doing some rewriting \[\begin{align*} x-2&\geq 3x-6\\ x-2+6&\geq 3x-6+6 &&\text{ add }6\\ x+4&\geq 3x \\ x-x+4&\geq 3x-x &&\text{ subtract }x\\ 4&\geq 2x \\ 4\cdot \frac{1}{2}&\geq 2x\cdot\frac{1}{2} &&\text{ multiply by } 1/2 \text{ or divide by }2\\ 2\geq x \end{align*}\]

The solution is all \(x\) less than or equal to \(2\). In interval notation it is \((-\infty,2]\).

We solve the following “and” compound inequality.

Example 9

Find all \(x\) satisfying \[x -3 \ge 2x - 6\] and \[2x > x +1.\]

To figure out what \(x\) satisfy the requirements, use decomposition. Take \(A\) to be the solutions to \(x-3\geq 2x-6\) and \(B\) to be the solutions to \(2x>x+1\). To solve \[x -3 \ge 2x - 6\] and \[2x > x +1,\] identify what \(A\) and \(B\) are and then intersect \(A\) and \(B\). The solution set \(A\) is: \[\begin{align*} x-3&\geq 2x-6\\ x-3+6&\geq 2x-6+6 &&\text{ add }6\\ x+3&\geq 2x \\ x-x+3&\geq 2x-x &&\text{ subtract }x\\ 3&\geq x \\ \end{align*}\] The solution set \(B\) is: \[\begin{align*} 2x&\geq x+1\\ 2x-x&\geq x-x+1&&\text{ subtract }x\\ x&\geq 1 \\ \end{align*}\]

We need all solutions to \(3\geq x\) and \(x>1\).

The solution to the original problem is \((1,3].\)

We solve the following “or” compound inequality

Example 10

Find all \(x\) satisfying \[2x -3 \le x - 5\] or \[ 3x -4 > x +2.\]

To figure out what \(x\) satisfy the requirements, use decomposition. Take \(A\) to be the solutions to \(2x-3\leq x-5\) and \(B\) to be the solutions to \(3x-4>x+2\). To solve \[2x -3 \le x - 5\] or \[ 3x -4 > x +2,\] identify what \(A\) and \(B\) are and then union \(A\) and \(B\). The solution set \(A\) is: \[\begin{align*} 2x-3&\leq x-5\\ 2x-x-3&\leq x-x-5 &&\text{ subtract }x\\ x-3&\leq -5 \\ x-3+3&\leq -5+3 &&\text{ add }3\\ x&\leq -2 \\ \end{align*}\] The solution set \(B\) is: \[\begin{align*} 3x-4&> x+2\\ 3x-x-4&> x-x+2&&\text{ subtract }x\\ 2x-4&> 2 \\ 2x-4+4&> 2+4 &&\text{ add }3\\ 2x&>6\\ 2x\cdot \frac{1}{2}&>6\cdot\frac{1}{2} &&\text{ divide by }2\\ x&>3 \end{align*}\]

We need all solutions to \(x\leq -2\) or \(x>3\).

The solution to the original problem is \((-\infty,-2]\cup(3,\infty).\)

We solve the following problem using the principle of decomposition.

Example 11

For any real number \(x\) take Condition 1 to be the condition: \[2x +3 \ge x - 2\] and \[5x +1 < 3x +9.\] For any real number \(x\) take Condition 2 to be the condition: \[x \geq 3\] or \[ 4x -1 < 3x -2.\] Find all real \(x\) that satisfies Condition 1 and Condition 2.

Use decomposition again.

Take \(C_1\) to be all \(x\) that satisfy \(2x+3\geq x-2\) and \(5x+1<3x+9\).

Take \(C_2\) to be all \(x\) that satisfy \(x\geq 3\) or \(4x-1<3x-2\).

Simplify \(C_1\) and \(C_2\) as much as possible by rewriting the inequalities.

The solution set to \(C_1\) is \(-5\leq x<4\) or \([-5,4]\). The solution set to \(C_2\) is \(x<-1\) or \(x\geq 3\)

Now determine \(C_1\cap C_2\).

The solution to the original problem is \([-5,-1)\cup[3,4).\)

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