page 1 of Section 6.1
CHAPTER 6 INDUCTION
SECTION 6.1 PROOF BY (WEAK) INDUCTION
principle of mathematical induction
Start with a conjecture about an arbitrary positive integer n like one of these:
The sum of the first n odd integers is n2.
If a planar connected graph has n edges then V - E + F = 2.
Mathematical induction is a method for proving such conjectures true for all n. (Of
course if the conjecture isn't true then the attempt to prove it by induction will
fail.) The catch is that induction doesn't tell you how to make the conjecture in
the first place.
Given a statement involving an arbitrary integer n, the principle of induction
says that if you can carry out the following two steps then the statement holds
for all n ≥ 1.
(I) Prove that the statement holds for n = 1
(II) (the inductive step) Prove that if the statement holds for one particular
fixed (but arbitrary) value of n, say n=k (this is called the induction hypothesis),
then it also holds for the value n=k+1.
If you want to prove that a statement is true for say n ≥ 6 (instead of n ≥ 1)
then, at step I, prove that it holds for n = 6. And continue with the same step II
The induction principle is often viewed as the falling domino principle. If you
can topple the first domino (Part I) and if every falling domino knocks over the
next one (Part II) then all the dominos fall. Neither I nor II alone is sufficient.
Part I without II guarantees only that the first domino falls. Part II without Part
I is like being all dressed up with no place to go (no trigger).
example 1
I'll prove that the sum of the first n odd integers is n2, i.e., that
(1)
1 + 3 + 5 + ... + (2n-1) = n2
for n = 1, 2, 3, 4, ...
Part I Show that it's true for n = 1
Of course it is.
If n = 1 then the LHS of (1) is 1 and the RHS of (1) is 12 which is also 1. QED
Part II
Assume that it's true for n=k; i.e., assume that for a fixed k,
1 + 3 + 5 + ... + 2k-1 = k2
(the induction hypothesis)
Show that it's true for n=k+1; i.e., show that
(2)
1 + 3 + 5 + ... + 2k-1 + 2k+1 = (k+1)2
Here's the proof:
1 + 3 + 5 + ... +
2k+1 =
1 + 3 + 5 + ... + 2k-1
+
2k+1
this is k 2
by the induction hypothesis
warning The reason here is not "by induction" and it is not
"by part I".
The reason is "by the induction hypothesis", i.e., by the
assumption in part II that the statement is true for n=k.
Don't forget to state the induction hypothesis at the beginning of part II.
= k2 + 2k+1
= (k+1)2
page 2 of Section 6.1
This proves part II
And now that I and II are proved, the statement in (1) is true for all n, by
induction.
warning In part II, you are not assuming that n = k. You are assuming that the
statement (which in this case is "the sum of the first n odd integers is n2) is true for
n = k.
warning about style
In example 1, to prove (2) in part II, it is neither good style nor good logic to
write like this:
1 + 3 + 5 + ... +
1 + 3 + 5 + ... + 2n-1 +
2n+1 = (n+1)2
2n+1
= (n+1)2
Don't write like this. You will lose
points on an exam if you do.
n2 by induction hyp
n2 + 2n + 1 = (n+1)2
(n+1)2 = (n+1)2
TRUE
First of all, it's silly to keep repeating the (n+1)2 on the right side when it
would be more efficient to write like this.
1 + 3 + 5 + ... + 2n-1
+
2n+1
n2 by induction hyp
Write like this instead
= n2 + 2n+1
= (n+1)2
And secondly, any ''proof'' in mathematics that begins with what you want to prove and
ends with TRUE is at best badly written and at worst incorrect and drives me crazy.
If a statement leads to something true that does mean that the statement is true.
With this ''method'' I can prove that 3 = 4:
3 = 4 (start with what you want to prove)
4 = 3 (just reverse the preceding line)
7 = 7
(add the two preceding lines)
TRUE
So conclude
that 3 = 4
(??????)
What you should do to prove an identity is begin with the lefthand side and work on
it until you get the righthand side (as in the second box above), or visa versa. Or
work on each side separately until they turn into the same thing. But don't start
with what you are trying to prove and end up with something like 7 = 7, 1 = 1,
(n+1)2 = (n+1)2 etc. because that kind of mathematical writing is not good style.
page 3 of Section 6.1
example 2
The generalized De Morgan law says that for n = 2, 3, ...,
(3)
A1 ∪ A2 ∪ ... ∪ An
Here's a proof by induction.
= A1
∩
A2
∩ ... ∩
(There are also non-inductive proofs.)
Show that (3) holds for n = 2, i.e., show that
Part I
I think it's obviously true by looking at a
Venn diagram.
Both
A1 ∪ A2 and A1
be the shaded area in Fig 1
An
∩
A2
A1 ∪ A2
A1 ∪ A2 ∪ ... ∪ Ak = A1 ∩ A2
Now show that (3) holds for k + 1 sets.
∩ ... ∩
A1
∩
A2 .
A1
turn out to
A2
Suppose (3) holds for k sets, i.e., suppose that
Part II
=
FIG 1
Ak (the induction hypothesis)
A1 ∪ A2 ∪ ... ∪ Ak ∪ Ak+1
=
(A1 ∪ A2 ∪ ... ∪ Ak) ∪ Ak+1
by grouping
=
A1 ∪ A2 ∪ ... ∪ Ak
by part I (the k=2 part)
=
( A1
=
∩
A2
A1 ∩ A2
So Part II is proved.
∩
∩ ... ∩
∩ ... ∩
Ak+1
Ak
Ak
)
∩
∩
Ak+1
Ak+1
by the induction hypothesis
by ungrouping
example 3
Here's the proof by induction that 5n - 4n - 1 is divisible by 16 for n = 1,2,3,...
Part I
If n = 1 then 5n - 4n - 1 is 0 and 0 is divisible by 16
Part II Assume it's true for n=k; i.e., assume that 5k - 4k - 1 is divisible by 16
(the induction hypothesis).
warning
The induction hypothesis is not "assume true for all k".
It's "assume true for the one particular value n=k".
Try to show that it's true for k+1, i.e., show that 5k+1 - 4(k+1)- 1 is divisible
by 16.
By clever algebra you can write 5k+1 - 4(k+1) - 1 in terms of 5k - 4k - 1 and make
use of the induction hypothesis:
page 4 of Section 6.1
5k+1 - 4(k+1) - 1
= (by algebra)
5(5k - 4k - 1)
+
divisible by 16 by the
induction hypothesis
16k
clearly divisible
by 16
So 5k+1 - 4(k+1) - 1 is the sum of two things each of which is divisible by 16 so
it's divisible by 16 also. End of part II.
warning
1. In a proof by induction, Part II is not to "prove it for n=k+1". In fact if
you could prove it for k+1 directly then you wouldn't need an induction proof to
begin with. Part II says to prove it for n=k+1 assuming that it's true for n=k, which is
very different from just proving it for n=k+1.
Don't forget to have an induction hypothesis in Part II.
2. The induction hypothesis is not "assume true for all k" (as a matter of fact
that's what you're trying to prove overall). The induction hypothesis is to assume
it's true for a particular (but arbitrary) k.
3. The induction hypothesis is not "assume n = k".
The induction hypothesis is "assume it is true for n = k" where you fill in the
appropriate "it".
PROBLEMS FOR SECTION 6.1
Do everything by induction although other methods of proof may also work and might
even be better.
And in each induction proof, state exactly what it is that you're proving in each
part.
1
1. Show that the sum of the first n integers is 2 n(n+1); i.e., show that
n(n+1)
1 + 2 + 3 + ... + n =
for n ≥ 1.
2
n(n+1)(2n+1)
2. Show that the sum of the first n squares is
, i.e., show that
6
12 + 22 + ... + n2
=
n(n+1)(2n+1)
6
3. Show that 13 + 23 + 33 + ... + n3 =
[
n(n+1)
2
for n ≥ 1.
]
2
for n ≥ 1.
4. Show that 11n - 4n is divisible by 7 for n = 1,2,3,...
2n
5. Show that 2
- 1 is divisible by 3 for all positive integers n.
6. A post office sells only 5Æ and 9Æ stamps.
Show that any postage of 35Æ or more can be paid using only these stamps.
Suggestion In Part II consider two cases for the induction hypothesis:
case 1 Postage of nÆ can be paid using only 5Æ stamps.
case 2 Postage for nÆ requires at least one 9Æ stamp.
7. Show that 2n > n for n ≥ 1.
8. (a) Use induction to show that n2 + 5n + 1 is odd for n = 1, 2, 3, ...
(b) Since part (a) showed that n2 + 5n + 1 is always odd, you couldn't possibly show
by induction (or any other method) that it is even for all n. But show that part II
of the induction argument still works and it's only part I that fails.
page 5 of Section 6.1
n
9. Show that (cos œ + i sin œ) = cos nœ + i sin nœ for all n ≥ 1
where i is the imaginary number satisfying i2 = -1.
For Reference: sin(x + y) = sin x cos y + cos x sin y
cos(x + y) = cos x cos y - sin x sin y
n
n
n
n
10. Show that (0) + (1) + ... + (n) = 2 for n = 1,2,3,...
Use Pascal's identity from Section 1.9.
page 1 of Section 6.2
SECTION 6.2 STRONG INDUCTION
principle of strong induction
Given a statement involving an arbitrary integer n, the principle of strong
induction says that if you can carry out the following two steps then the
statement holds for all n ≥ 1.
(I) Prove that the statement holds for n = 1.
(II) Prove that if the statement holds for n = 1, 2,..., (i.e., not just for n = k
but for all integers from 1 up through k) (this is called the strong induction
hypothesis), then it also holds for n=k+1.
If you want to prove that a statement is true for say n ≥ 6 (instead of n ≥ 1)
then, at step I, prove that it holds for n = 6. And at step II show that if it
holds for n = 6, ..., k then it also holds for n = k+1.
When would you use strong induction rather than ordinary induction?
If you are assuming "it" is true for n = k and that isn't enough to show that it is
true for n = k+1, maybe going back and assuming "it" is true for n = 1, 2, ..., k
will help. You are entitled to change your induction hypothesis in midstream.
example 1
I want to show that in a planar connected graph, V - E + F = 2.
I'm going to try it by induction on E.
I'll start out using plain induction.
Part I Show that V - E + F = 2 in any planar connected graph where E = 1 (i.e.,
one edge)
Figs 1 and 2 show the only two planar connected graphs with E=1.
•
•
•
FIG 1
FIG 2
In Fig 1, E = 1, V = 2, F = 1 (the exterior face) so V - E + F = 2 - 1 + 1 = 2
In Fig 2, E = 1, V = 1, F = 2 so V - E + F = 1 - 1 + 2 = 2
Part II Assume that V - E + F = 2 in any planar connected graph with k edges (this
is the induction hypothesis).
I want to show that in any planar connected graph with k + 1 edges, V - E + F = 2.
Start with a planar connected graph with k+1 edges.
Delete an edge from this graph so that the new graph has k edges.
case 1 The new graph is still connected.
Imagine that we had deleted say edge BD or edge BC or
edge GF in Fig 3.
B•
A•
•
D
E
•
•
C
FIG 3
•
F
G
•
•H
page 2 of Section 6.2
Let Enew, Vnew, Fnew be the number of edges, vertices, faces in the new graph.
The new graph has k edges, is still connected and of course is still planar. So by
the induction hypothesis
(1)
Vnew - Enew + Fnew = 2
But
(2a)
(2b)
(2c)
Vnew = original V
Enew = original E - 1
Fnew = original F - 1
If an edge like BD in Fig 3 had been deleted, two interior faces are
thrown together so there is one less face. If an edge like BC is deleted
an interior face and the exterior face are thrown together. In each
instance, after deleting the edge there is one less face than before.
Substitute (2a,b,c) into (1):
(3)
V - (E-1) + (F-1) = 2
V - E + 1 + F - 1 = 2
V - E + F = 2
QED
case 2 The new graph is not connected.
Look at Fig 3 and imagine deleting say edge CE (Fig 4)
B•
A•
•
D
E
•
•
C
Component 1
•
F
G
•
•H
Component 2
Fig 4
Then we have two planar connected components.
Let E1, V1, F1 be the number of edges, vertices, faces in Component 1.
Let E2, V2, F2 be the number of edges, vertices, faces in Component 2.
But E1 is not necessarily k so I can't use the induction hypothesis on Component 1.
But I do know that E1 is between 1 and k. So I will change my induction hypothesis:
Assume that V - E + F = 2 in any planar connected graph where the number of edges is
1, 2, ..., k (strong induction hypothesis).
Now by the strong induction hypothesis,
(4)
V1 - E1 + F1 = 2
And similarly
(5)
V2 - E2 + F2 = 2
page 3 of Section 6.2
Add (4) and (5):
(6)
V1 + V2 - (E1 + E2) + F1 + F2 = 4
But
(7a)
(7b)
(7c)
V1 + V2 = the original V
E1 + E2 = original E - 1 (since we deleted an edge)
F1 + F2 = original F + 1
Look at Fig 4 to see that F1 includes an exterior face and F2 also
includes an exterior face so F1 + F2 includes an exterior face twice
whereas the F from Fig 3 includes the exterior face only once.
Substitute (7a,b,c) into (6):
V - (E - 1) + F + 1 = 4
V - E + 1 + F + 1 = 4
V - E + F = 2
QED
example 1 continued
I can make plain induction work by revising part II a little. (I didn't have to use
strong induction.)
Part II Assume that V - E + F = 2 in any planar connected graph with k edges (this
is the induction hypothesis).
I want to show that in any planar connected graph with k + 1 edges, V - E + F = 2.
Start with a planar connected graph with k+1 edges.
case 1 It has a cycle (as in Fig 3 where there are several cycles).
Delete an edge from one of the cycles.
The new graph is still connected: If you had planned to travel from here to there
on a path that included the deleted edge you can still travel from here to there by
going around the "other" way around the cycle.
Imagine that in Fig 3 you had planned to go from C to G using path
CEG, but edge EG was deleted from the cycle EGFE.. You could still get
from C to G on path CEFG
So the new graph is connected, still planar, and has k edges. And we can use the
induction hypothesis as in case 1 above (just repeat lines (1)-(3) )to conclude
that in the original graph (with k+1) edges, V - E + F = 2
case 2 It does not have a cycle.
Then the graph is connected, has no cycles, so it is a tree.
We know that in a tree V = E + 1 and F = 1 so
V - E + F = E+1 - E + 1 = 2 QED
(Didn't need any induction hypothesis in this case)
page 4 of Section 6.2
PROBLEMS FOR SECTION 6.2
1.
A sequence of numbers is defined as follows:
F1 = 1
F2 = 1
For n ≥ 3,
Show that
(•)
Fn = Fn-1 + Fn-2
 1 + √ 5  n-1


 2

Fn ≤
for n = 1,2,3,...
using the following version of induction (sort of inbetween weak and strong):
Part I Show that (•) is true for n = 1 and n = 2
Part II Show that if (•) is true for n=k-1 and n=k then it's true for n=k+1.
Use the algebraic identity
 1 + √ 5  2
1 + √ 5


=
+ 1
 2

2
2. The primes are the integers larger than 1 which can't be divided by anything
except themselves and 1. For example, 2,5,7,11,13 are the first five primes.
The problem is to show that every integer ≥ 2 is either prime or can be factored
into a product of primes (e.g., 12 = 3…2…2)
(a) Try to prove it by ordinary induction but get stuck.
(b) Try it using strong induction.
3. The problem is to use induction on E to show that in a tree, V = E + 1.
(a) Do Part I.
(b) For Part II you want to show that if V = E + 1 in any tree where E = k (the
induction hypothesis) then V = E + 1 in any tree where E = k+1.
To do this, start with a tree with k+1 edges
•
•
•
•
•
•
•
•
•
•
•
and delete an "outer" edge along with its "outer" vertex too, e.g., like this:
•
•
•
•
• DEL
ET
•E
•
•
DELETE
•
•
•
•
•
•
•
•
TE
LE
DE
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Now keep going.
(c) Suppose that in Part II you start with a tree with k+1 edges and delete an edge
that is not an "outer" edge (without removing any vertices) like this:
page 5 of Section 6.2
•
•
•
•
•
•
•
delete
•
•
•
•
You are not left with one tree with k edges to which the induction hypothesis
applies as in part (b). In the diagram above you would be left with two trees each
of which has less than k edges.
So ordinary induction would not work.
Do it using strong induction.
page 6 of Section 6.2
The new graph is still connected: If you had planned to travel from here to
there on a path that included the deleted edge (e.g., here to A to B to C to there
in Fig 5), you can still travel from here to there by going around the "other" way
around the cycle (cf. here AGFEDC there)
F•
G•
•
Here
•E
• D
•
A
•
B
delete
FIG 5
•
C
• There
page 1 of Section 7.1
CHAPTER 7 RELATIONS
SECTION 7.1 INTRODUCTION
I'll show you some examples of relations before getting to the formal definition.
example 1
Look at all the courses offered at the university. Let R denote the relation "is a
prerequisite for". Then
Physics 1 R Physics 2
Physics 2 R Physics 3
English 1 Rı Physics 1
(Rı means is not a prerequisite for)
etc.
example 2
Start with the set of all people. Let R denote the relation "is a sibling of".
Then
Princess Anne R Prince Charles
Ronald Reagan Rı Queen Elizabeth
Prince William R Prince Harry etc.
example 3
Start with the set of non—negative integers. Let R mean "is 1 larger than". Then
4 R 3,
3 Rı 4,
4 Rı 2,
7 R 6
etc.
definition of a relation
Start with a set A, called the universe.
A relation on A is a collection of ordered pairs (x,y) where x and y are in A.
If (x,y) is in the collection then we write x R y.
If (x,y) is not in the collection then we write x Rı y.
For example, the relation "is 1 more than" defined on the universe of non—negative
integers contains the pairs (1,0), (2,1), (3,2) etc.
If A = {a,b,c,d}, some relations on A are
R1 = { (a,a),(b,b), (c,c), (d,d) }
R2 = { (a,d) }
R3 = { (a,b), (b,c), (c,d) }
For instance
a R1 a,
b R1 b,
a R2 d,
a Rı2 a,
a R3 b,
a Rı3 c
etc.
In general, if A is a set then the set of all ordered pairs (x,y) where x and y are
in A is called A ≈ A so a relation is often referred to as a subset of A ≈ A.
reflexive and antireflexive relations
Let R be a relation on a universe A.
R is called reflexive if x R x for all x in the universe
R is called antireflexive if x notR x for all x in the universe.
In other words, to decide if a relation is reflexive, look at the universe. For
every x in the universe you must have x R x. For antireflexive you must never have
x R x.
For example, if the universe is the set of all people in the world then "is the
same height as" is reflexive since each person is the same height as herself; "is
taller than" is antireflexive" since no person is taller than herself; "thinks
highly of" is neither reflexive nor antireflexive since some people think highly of
themselves and some don't.
page 2 of Section 7.1
symmetric and antisymmetric relations
A relation R is called symmetric if the following holds for all x and y in the
universe:
If x R y then y R x.
In other words, for every pair in R, the "reverse" pair is also in R.
I'd like to define an antisymmetric relation as one where pairings never reverse.
But with that definition, an antisymmetric relation could never allow x R x since
that kind of pairing always reverses. Rather than force an antisymmetric relation to
exclude x R x, I'll apply "never reverse" only to pairings x R y where x ≠ y. In
other words:
A relation R is called antisymmetric if the following holds:
if x ≠ y and x R y then y notR x.
To decide about these properties, look at all the pairs (x,y) in the relation
where x ≠ y. If (y,x) is always in there too then the relation is symmetric; if it's
never in there then the relation is antisymmetric.
Here's another way to look at it.
Suppose x and y are two different members of the universe.
A symmetric relation includes either both (x,y) and (y,x) or includes neither.
An antisymmetric relation includes either (x,y) or (y,x) or neither but not both.
The relation "is married to" is symmetric since whenever x is married to y then y
is also married to x.
The relation "is a subset of" is antisymmetric because for x ≠ y you can't have
both x ⊆ y and y ⊆ x.
The relation "is taller than" is antisymmetric because you can't have x taller
than y and y taller than x
The relation "admires" is not symmetric (it's possible for x to admire y while y
does not admire x) and it's not antisymmetric (it's possible for x to admire y and y
to admire x)
transitive relations
A relation R is called transitive if the following holds:
if x R y and y R z then x R z.
For example, the relation "is taller than" is transitive: if x is taller than y and
y is taller than z then x must be taller than z.
The relation "admires" is not transitive: if x admires y and y admires z then it
is not necessarily true that x admires z.
true by default
If you're checking for antisymmetry you have to check that if x R y where x ≠ y
then y notR x. If there never is an x R y to begin with then you have no checking
to do and the relation is called antisymmetric by default.
And if you're checking for transitivity you have to look at all hookups where
x R y and y R z (here, x,y,z don't have to be different) and check that you also have
x R z. If there are no hookups at all then there is nothing to check and the
relation is called transitive by default.
example 4
Let
A = {5,6,7,8,9}
(the universe)
Define a relation R on A as follows: R = { (5,5), (6,6) }
In other words, 5 R 5, 6 R 6 and no other pairs of A are related.
R is not reflexive because you don't have 7 R 7 (or 8 R 8 or 9 R 9).
R is not antireflexive because you do have 5 R 5 (and 6 R 6).
R is symmetric because every pair reverses.
page 3 of Section 7.1
R is antisymmetric by default There's no danger of having both x R y and y R x
where x ≠ y because there aren't any of those x R y 's to begin with.
R is transitive because whenever x R y and y R z (which only happens when
5 R 5 and 5 R 5 and when 6 R 6 and 6 R 6) then you also have x R z.
warning
Symmetry does not mean that you always have x R y and y R x. It means that if you
have x R y then you must also have y R x, i.e., you can't have one pairing without
the other. But a symmetric relation can have neither x R y nor y R x.
example 5
Let the universe be {5,6,7,8} and let R = { (5,6) }, i.e., 5 R 6 and that's all.
R
R
R
R
R
is
is
is
is
is
not reflexive since you don't have 5 R 5 for instance.
antireflexive since you never have x R x
not symmetric since you have 5 R 6 but not 6 R 5
antisymmetric because you don't have the reverse of 5 R 6
transitive (by default) since you never have x R y and y R z.
the digraph of a relation
A relation on a set may be pictured as a directed graph.
The vertices represent the elements in the set and an edge is drawn from x to y if
x R y
If A = {a,b,c,d,e } and a R b, b R c, c R d then Fig 1 shows the corresponding
digraph.
Fig 2 shows the relation "is smaller than" on the set {Dog, Horse, Elephant}
•e
b•
a•
•H
•d
•c
FIG 1
•E
D•
FIG 2
PROBLEMS FOR SECTION 7.1
1. Decide if the relation is reflexive, antireflexive, symmetric, antisymm,
transitive.
For the first two, the universe is the set of all people.
For the next three, the universe is the set of integers.
For the last three, the universe is the set of all lines in the plane.
(a)
(b)
(c)
(d)
(e)
(f)
x R y
x R y
x R y
x R y
x R y
x R y
(i.e.
(g) x R y
(h) x R y
if x is a descendent of y
if x has the same color eyes as y
if xy is even
if x + y < 10
if x < y
if line x has no points in common with line y
x is parallel to but not coincident with y)
if x is parallel to or coincident with y
if x is perpendicular to y
2. Define a relation R as follows on the universe {rock, scissors, paper}.
rock R scissors
scissors R paper
paper R rock
Is it reflexive? antireflexive? symmetric? antisymmetric? transitive?
page 4 of Section 7.1
3. Let A = {1,2,3,4}.
Are the following relations on A reflexive? antireflexive? symm? antisymm? trans?
(a) R = { (1,1), (2,1) }
(b) R = { (1,1), (3,3), (4,4) }
(c) R = { (1,2), (2,3), (1,3), (4,2) }
4. Let A = {a,b}.
First decide how many relations can be defined on A and then write them all out.
For each one decide if it is reflexive, antireflexive, symmetric, antisymm, trans
5. Let A
(a) How
(b) How
(c) How
(d) How
(e) How
= {x1, x2, x3, x4, x5} be the universe. Consider relations on A.
many relations are there.
many reflexive relations are there .
may relations are there which are reflexive and contain (x2, x4).
many symmetric relations are there.
many relations are symmetric and reflexive.
6. What's special about the digraph of a relation if the relation is
(a) symmetric (b) reflexive (c) transitive
7.
If
If
(a)
(b)
(c)
(d)
you
you
Can
Can
Can
Can
answer YES then find a specific example.
answer NO then explain why not.
a relation be both symmetric and antisymmetric.
a relation be both reflexive and antireflexive.
a relation be neither symmetric nor antisymmetric
a relation be neither reflexive nor antireflexive.
page 1 of Section 7.2
SECTION 7.2 EQUIVALENCE RELATIONS
definition and notation
A relation R on a set A is an equivalence relation if R is reflexive, symmetric and
transitive. The generic symbol for an equiv relation is ‡
Here are some examples:
is the same age as (for people)
begins with the same letter as (for words)
has the same square as (for integers)
is the same color as (for marbles)
equivalence classes and partitions
Let ‡ be an equivalence relation on a set A. If x is in A the equivalence class of
x is the set of all elements in A equivalent to x. The equivalence class of x
contains (at least) x itself since x ‡ x by reflexivity.
Look at the equivalence relation "begins with the same letter as". The
equivalence class of the word quiet consists of all words beginning with q. The
equiv class of the word down consists of all words beginning with d. All in all
there are 26 equivalence classes (Fig 1); they are non-overlapping and exhaustive
(i.e., every word is in exactly one of them). Fig 1 is called a partition of the set
of words.
In general a partition of a set A is a collection of non—empty disjoint subsets
(called cells or blocks of the partition) whose union is A. Fig 2 shows a partition
of {a,b,c,d,e,f,g,h,i,j,k,m} with 7 cells.
calor
apple
ie
c
lump
ark
...
...
baby
...
bump
...
FIG 1
b
•
•k
d•
•i
zoo
zebr
a
...
a
•
•e
•g
•f
•h
•j
•
c
•
m
FIG 2
Equivalence relations and partitions are interchangeable ideas as follows:
(1) Every equivalence relation on a set A induces a partition of A
(2) Every partition of a set A induces an equivalence relation on A
In particular if you begin with an equivalence relation then the equivalence
classes are a partition of A. For instance, the equivalence relation "begins with
the same letter as" produced the partition in Fig 1.
And if you begin with a partition of A then "is in the same cell as" is an
equivalence relation. The partition in Fig 2 produces the equivalence relation where
b ‡ k, d ‡ h, e ‡ g ‡ f, c ‡  (and everything is ‡ to itself).
example 1
Look at the equivalence relation "has the same square as" for integers.
equivalence classes are {0}, {-1,1}, {-2,2}, {-3,3} ,..
The
page 2 of Section 7.2
example 2
Look at the equivalence relation "is the same age as"
equivalence classes are
for children. The
under a year
1 year olds
2 year olds
»
17 year olds
congruence module m (a famous equivalence relation)
Let A be the set of non—negative integers. Let
i.e.,
x R y if x-y is divisible by 3;
x R y if x and y have the same remainder when divided by 3
The relation is called congruence modulo 3.
For example 14 R 8 (and 8 R 14) because 14 - 8 = 6 which is divisible by 3 or
equivalently because 14 and 8 have the same remainder, namely 2, when divided by 3.
Usually we write 14 ≡ 8 mod 3.
Congruence mod 3 is an equivalence relation. There are 3 equivalence classes since
every integer is congruent mod 3 to exactly one of 0,1,2. The equivalence classes
are
{0,3,6,9,12,...}
{1,4,7,10,13,...}
{2,5,8,11,14,...}.
Similarly x ≡ y mod 6 if x and y have the same remainder when the y are divided
by 6. There are 6 equivalence classes since every integer is congruent mod 6 to
exactly one of 0,1,2,3,4,5. The equivalence classes are
{0,6,12,18,...}
{1,7,13,19,...}
{2,8,14,20,...}
{3,9,15,21,...}
(4,10,16,22,...}
{5,11,17,23,...}
In general
x ≡ y mod m if x and y have the same remainder when they are divided by m
or equivalently
x ≡ y mod m if x - y is divisible by m.
There are m equivalence classes since every integer is congruent mod m to exactly
one of 0,1,2,...,m-1.
page 3 of Section 7.2
PROBLEMS FOR SECTION 7.2
1. Is it an equivalence relation? If so describe the equivalence classes.
If not, why not?
(a) same name as
(b) same major as
(c) has a class in common with
(d) has the same two initials as
(e) lives in the same state as
(f) lives within 30 miles of
(g) has the same social security number as
(h) equals (for integers)
(i) has the same first four letters as (for words with at least 4 letters)
(j) has the same y—coordinate as (for points in a 2-dim coord system)
(k) has the same marital status as
() is married to
2. Let the universe be the set of nonnegative integers.
Is R an equivalence relation?
If so, find the equivalence classes.
(a) n R k if sin nπ = sin kπ
(b) n R k if cos nπ = cos kπ
3. Find all the equivalence relations on {a, b, c} by drawing the corresponding
partitions.
4. According to (1), equivalence classes are supposed be be a partition of the
universe. So they shouldn't overlap. How can you be so sure they don't.
5. A relation R is called circular if it has the following property
If a R b and b R c then c R a
(a) Show that if R is reflexive and circular then R is an equivalence relation.
(b) Show that if R is an equivalence relation then R is circular.