Math 2534 Solutions Equivalence Relations and the Pigeon Hole Principle
1) For each of the following relations defined on the set A = { 1,2,3,4,5}, determine if R
is reflexive, symmetric and/or transitive. Draw the directed graphs for each relation. If R
is an equivalence relation then represent A as a partition.
R1  {(1,1), (2, 2), (2,3)(3, 2), (3,3), (3, 4), (4,3), (5,5), (4, 4)} R,S,T A={1}  {2,3,4}  {5}
R 2  {(1, 2), (1, 4), (1,5)(2, 4), (2,5), (3, 4), (3, 5), (4,5), (4, 4)} T
R3  {(1,3), (1,5), (2, 4)(3,1), (3,5), (4, 2), (5,1), (5,3)} S
R 4  {(1,1), (2, 2), (1,3), (1,5), (3,1), (3,3), (2, 4), (5,3), (3,5), (4, 2)(4, 4), (5,1)} R, S
2) Prove the following:
A) Theorem: Define a relation R by mRn if and only if 5 m  n . R is an
equivalence relation on the positive integers.
Proof: Reflexive: To show that aRa, we must show that 5 (a  a) . We must
show that there exist an integer q such that 5q = a – a. In order for 5q = 0, q = 0
and this will satify the definition of divisible and aRa. And R is reflexive.
Symmetric: Given that aRb, we need to show that bRa. By definition of
divisible we have that 5k = a – b. By multiplying both sides by (-1) we will have
5(-k) = b-a, Let p = -k where p is an integer, then by definition of divisible we
have that bRa and R is symmetric.
Transitive: Given that aRb and bRc, we need to show that aRc. By
definition of divisible we have that 5q = a – b and 5p = b – c.
Now consider 5q + 5p = a – b + b – c = a – c. let k = q + p to get 5k = a – c.
By definition of divisible we have that 5 divides a – c and aRc. Therefore R is
transitive.
.
B) Theorem: Define a relation R by (a, b)R(c,d) if and only if ad = bc, where
a,b,c,and d are integers. R is an equivalence relation
This problem was done in class: Outline is below:
Reflexive: Need to show that (a, b)R(a, b) (Does ab = ba ?)
Symmetric: If (a, b)R(c, d) then we need to show that (c, d)R(a, b)
We are given that ad = bc and we need to show that cb = da
Transitive: If (a, b)R(c, d) and (c, d)R(e, f) then we need to show that
(a, b)R(e, f). We are given that ad = bc and cf = de. We now must show
that af = be and therefore (a, b)R(e, f).
C) Theorem: a  b mod3 is an equivalence relation on the integers.
You need to remember that an equivalent representation of a congruence relation
is that 3 (a  b) and the proof proceeds as in part A) above.
D) The Cardinality of sets forms an equivalence relation on all sets.
This one was also done in class. Remember you have to use the formal definition
of what it means for two sets to have the same cardinality. You will see some form
of this on the test.
Solution: In order to verify that Cardinality is an equivalence relation R on sets
we define R as follows: ARB iff there exist a bijection from A to B.
For R to be an equivalence relation we must verify that R is reflexive, symmetric
and transitive.
To verify that R is reflexive we must show that for each set A, ARA. This is
clearly true since the identity mapping f(ai) = ai for any i.
To verify that R is symmetric we must show that IF ARB, THEN BRA. We
assume that ARB and that there is a bijection from A to B, f : A  B. Since f is
a bijection we know that the function f-1 is guaranteed to exist and is a bijection.
Therefore f 1 : B  A and BRA.
To verify that R is transitive we must show that IF ARB AND BRC, THEN ARC.
We will assume that ARB and BRC which means that there is a bijection f from A
to B and a bijection g from B to C. We will now create a bijection from A to C.
Consider the bijections f and g then the composite ( g f ) : A  C is also a
bijection since the composition of two bijections is always a bijection.
3) If an equivalence relation is defined by the following set partition on A, then express
R as a set of ordered pairs. A  {1, 4} {3,5} {2}
R ={(1,1),(4,4),(1,4),(4,1),(3,3),(5,5)(3,5),(5,3),(2,2)}
4) If there are 500 students are in the same class, how many are guaranteed to have the
same birthday? Explain your reasoning.
 500 
Pigeon Hole Principle: 
 1  1.3698  1  1  1  2
 365 
5) You are opening a small restaurant. You have 10 tables. How many chairs do you
have to have to guarantee there will be at least 6 chairs at one of the tables.
Explain your reasoning.
x
Pigeon Hole Principle:    1  6 so x must be 51.
10 