In mathematics, particularly in set theory and algebra, a relation is a way to describe a connection or association between elements of two sets. Relations form the foundation for many advanced topics, including functions, equivalence relations, and orderings.
Given two sets \( A \) and \( B \), a relation \( R \) from \( A \) to \( B \) is a subset of the Cartesian product \( A \times B \). Formally,
\[R \subseteq A \times B\]where
\[A \times B = { (a, b) : a \in A, b \in B }\]Each element of \( R \) is an ordered pair \( (a, b) \) indicating that \( a \) is related to \( b \) under the relation \( R \).
- The domain of \( R \) is the set of all first elements of the ordered pairs in \( R \):
\[\text{Domain}(R) = { a \in A : \exists b \in B, (a, b) \in R }\]- The range of \( R \) is the set of all second elements of the ordered pairs in \( R \):
\[\text{Range}(R) = { b \in B : \exists a \in A, (a, b) \in R }\]Relations can be represented in multiple ways:
Example: Let \( A = {1, 2} \), \( B = {3, 4} \), and \( R = {(1, 3), (2, 4)} \).Matrix: 3 41 [1 0]2 [0 1]When \( A = B \), relations on \( A \) can have special properties:
Relations satisfying all three properties are called equivalence relations. These partition the set into equivalence classes.
Relations are subsets of Cartesian products, so set operations apply:
Also, the power set \( \mathcal{P}(S) \) of a set \( S \) is the set of all subsets of \( S \). If \( S \) has \( n \) elements, then
\[|\mathcal{P}(S)| = 2^n\]Some key properties and theorems useful in solving problems:
The Cartesian product \( A \times B \) is the set of all ordered pairs \( (a, b) \) with \( a \in A \) and \( b \in B \). Its cardinality is
\[n(A \times B) = n(A) \times n(B)\]Note that \( A \times B \neq B \times A \) in general, as ordered pairs are ordered.
The symmetric difference \( A \oplus B \) is defined as
\[A \oplus B = (A - B) \cup (B - A)\]It contains elements in either \( A \) or \( B \), but not in both.
------Problem: Let \( A = {1, 2, 3} \), \( B = {4, 5} \), and relation \( R = {(1, 4), (2, 5), (3, 4)} \). Find the domain and range of \( R \).
Solution:
Domain is the set of all first elements of ordered pairs:
\[ \text{Domain}(R) = {1, 2, 3} \]Range is the set of all second elements:
\[ \text{Range}(R) = {4, 5} \]Thus, domain = \( {1, 2, 3} \), range = \( {4, 5} \).
Difficulty: Easy
Problem: Find the number of subsets of \( S = {a, b, c, d} \).
Solution:
Since \( n(S) = 4 \), the number of subsets is
\[ |\mathcal{P}(S)| = 2^4 = 16 \]Difficulty: Easy
Problem: If \( n(A) = 37 \), \( n(B) = 25 \), and \( n(A \cup B) = 50 \), find \( n(A \cap B) \).
Solution: Using the formula
\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]Substitute values:
\[ 50 = 37 + 25 - n(A \cap B) \] \[ n(A \cap B) = 37 + 25 - 50 = 12 \]Difficulty: Medium
Problem: If \( n(A) = 4 \) and \( n(B) = 3 \), what is the maximum value of \( n(A \cap B) \)?
Solution:
The intersection cannot have more elements than the smaller set:
\[ \max n(A \cap B) = \min(n(A), n(B)) = 3 \]Difficulty: Easy
Problem: Let \( U = {1, 2, 3, 4, 5, 6, 7, 8, 9} \), \( A = {1, 2, 3, 4} \), and \( B = {2, 4, 6, 8} \). Find \( (A \cup B)' \) (complement relative to \( U \)).
Solution:
First find \( A \cup B \):
\[ A \cup B = {1, 2, 3, 4, 6, 8} \]Complement relative to \( U \) is elements in \( U \) not in \( A \cup B \):
\[ (A \cup B)' = U - (A \cup B) = {5, 7, 9} \]Difficulty: Medium
Problem: If \( A = {1, 2, 5, 6} \) and \( B = {1, 2, 3} \), find \( (A \times B) \cap (B \times A) \).
Solution:
Pairs in \( A \times B \) are all \( (a,b) \) with \( a \in A \), \( b \in B \).
Pairs in \( B \times A \) are all \( (b,a) \) with \( b \in B \), \( a \in A \).
The intersection contains pairs \( (x,y) \) such that \( (x,y) \in A \times B \) and \( (x,y) \in B \times A \).
But \( (x,y) \in B \times A \) means \( x \in B \) and \( y \in A \).
So for \( (x,y) \) to be in both, \( x \in A \cap B \) and \( y \in A \cap B \).
Find \( A \cap B = {1, 2} \).
Thus, intersection is
\[ (A \times B) \cap (B \times A) = {(1,1), (1,2), (2,1), (2,2)} \]Difficulty: Medium
Progress tracking is paywalled — subscribe to mark subtopics as understood and save your streak.
Go to practice →