This page is part of a series of lessons on sets.

In these lessons, we will learn the complement of a set and the relative complement of a set.

Related Topics: More Lessons on Sets

**complement** of set A, denoted by A’ , is the set of all elements in the universal set that are not in A.
** n(A) + n(A’) = n(U) **

**Example: **

**Solution: **

a) First, list out the members of U.

U = {–4, –3, –2, –1, 0, 1, 2, 3, 4, 5, 6, 7}

P ’ = {–3, –1, 1, 3, 7} ← in U but not in P

b) Draw a Venn diagram to display the sets U , P and P ’

c) Find n(Q)

n( U ) = 12, n(Q ’ ) = 5

Use the formula:

n(Q) + n(Q ’ ) = n( U )

n(Q) = n( U ) – n(Q ’ ) = 12 – 5 = 7

**Complement and Relative Complement**

The complement of a set is the collection of all elements which are not members of that set. Although this operation appears to be straightforward, the way we define "all elements" can significantly change the results.**Learn what a complement of a set is**
**Universal set and absolute complement**
**Relative complement or difference between sets**

In these lessons, we will learn the complement of a set and the relative complement of a set.

Related Topics: More Lessons on Sets

The following diagram shows the complement of a set. Scroll down the page for more examples and solutions on the complement of a set.

TheThe number of elements of A and the number of elements of A ’ make up the total number of elements in U .

Let U = {x : x is an integer, –4 ≤ x ≤ 7}, P = {–4, –2, 0, 2, 4, 5, 6} and

Q ’ = {–3, –2, –1, 2, 3}.

a) List the elements of set P ’

b) Draw a Venn diagram to display the sets U , P and P ’

c) Find n(Q)

a) First, list out the members of U.

U = {–4, –3, –2, –1, 0, 1, 2, 3, 4, 5, 6, 7}

P ’ = {–3, –1, 1, 3, 7} ← in U but not in P

b) Draw a Venn diagram to display the sets U , P and P ’

c) Find n(Q)

n( U ) = 12, n(Q ’ ) = 5

Use the formula:

n(Q) + n(Q ’ ) = n( U )

n(Q) = n( U ) – n(Q ’ ) = 12 – 5 = 7

The complement of a set is the collection of all elements which are not members of that set. Although this operation appears to be straightforward, the way we define "all elements" can significantly change the results.

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