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Compound Inequalities



Summary Sets

Set Notation

We are already somewhat familiar with set notation: we can pick a solution set from a replacement set. The answer is a finite number of points, written in brackets; for example, {1, 2, 3}. This is the set containing only the elements 1, 2, and 3. But how do we pick a solution set from all the real numbers?

When we solve the inequality 3x < 6, we end up with the answer x < 2. We can write this answer in set notation. We write {x : x < 2} or {x| x < 2}, and we say "the set of all x such that x is less than 2." This is the set of all real numbers that satisfy the equation 3x < 6.

Here are some examples of sets in set notation:

  1. {x : x > 5}The set of all x such that x is greater than 5.
  2. {y| y≠4}The set of all y such that y is not equal to 4.
  3. {2, 3, 4}The set containing only the members 2, 3, and 4.
  4. {z : z2≥16}The set of all z such that x squared is greater than or equal to 16.
  5. òThe null set.

Venn Diagrams

A Venn diagram is a representation of two or more sets that shows the relationship between them. A Venn diagram looks like two (or more) overlapping circles. Each circle represents a set. Members that both sets have in common are placed in the overlap, and members that one set has but the other doesn't are placed in the non-overlapping part of their respective sets. Here is an example of a Venn diagram for the sets A = {1, 2, 3, 4, 5, 6} and B = {0, 2, 4, 6}:

Venn Diagram
Note that circle A contains all the members of A and circle B contains all the members of B. The overlap contains all the members they have in common.

To draw a Venn diagram for two sets, first place all the members that are in both sets in the center ("overlap") of the diagram. Then, place all the remaining members of set A in the non-overlap of circle A. Place all the remaining members of set B in the non-overlap of circle B.

Some sets will have no members in common. If this is the case, the overlap of the two circles will be empty. In some instances, all the members of one set will be contained in the other set. In this case, the non-overlap of one set will be empty.