Download Lecture 6: Sept. 23

SUBTRACTION AND SETS
If B is a subset of A, then A = B U (A-B)
and n(A) = n(B) + n(A-B).
Thus n(A-B)=n(A) – n(B),
justifying the double use of “-“
and the informal words “take away”.
In general, we have n(A-B) = n(A)-n(AB).
We can use this for counting problems such
as: “If there are eight math teachers in the
school, and four also teach history, how many
teachers teach math and not history?”
We can use this to prove the inclusionexclusion rule, a more general rule that holds
for all sets A,B.
AUB is the disjoint
union of
A-B,
AB, and B-A.
n(AUB) = (n(A) –n(AB)) + n(AB) + (n(B)-n(AB))
= n(A) + n(B) – n(AB).
We can use this for counting problems
such as: “If Sandra has three coats that she
can wear in the summer and seven that she
can wear in the winter, including two allseason coats, how many coats does she own?”
RULES FOR SUBTRACTION
* a-0 = a.
*Subtraction is NOT commutative.
Indeed, within the whole numbers, only one of
a-b, b-a
exist (unless a=b).
*Subtraction is NOT associative; we have
to leave the parentheses on!
(3-2)-1 = 0
while 3-(2-1) = 2.
However, if a-(b-c) exists in the whole
numbers, so does (a+c)-b; and they are equal.
(6) MULTIPLICATION.
The Cartesian product of two sets A,B
consists of all ordered pairs (a,b), aA, bB.
If n(A) = a, and n(B) = b, then we define
axb (also written ab) to be n(AxB).
We can use this for counting problems
that involve counting pairs of objects, one
from each of two sets. For instance:
“If there are three first courses and
five desserts on the menu, how many
different meals can I choose?”
PROBLEM! Does n(AxB) depend on our
choice of A and B?
Suppose n(A) =n(A’) and n(B)=n(B’). Then
there are bijections f:AA’ and g:BB’.
There is a bijection fg:(AxB)  (A’ x B’)
(a,b)
(fa,gb)
EXAMPLE:
{Adam,Bill,Carl} {Apple, Berry,Chocolate}
x
x
{Smith,Thorpe}
{Squares, Tart}
Adam Smith Adam Thorpe
Bill Smith Bill Thorpe
Carl Smith Carl Thorpe
Apple Squares Apple Tart
Berry Squares Berry Tart
Choc. Squares Choc. Tart
One possible bijective relation = “has the same initials”
Conclusion: multiplication of sets gives us a
well-defined operation on set sizes.
Properties of Multiplication.
mx0 = 0
(Zero Property)
If ab = 0 then a=0 or b=0
(Zero Factor Property)
mx1 = m
ab = a+a+...+a
b times
ab = ba
(ab)c = a(bc)
(Identity Property)
(Repeated Addition Property)
(Commutative Property)
(Associative Property)
(a+b)c = ac + bc (Distributive Property)
If ab = ac, then b=c
(Cancellative Property)
m x 0 = 0 (Zero Property)
Proof: mx0 = n(M x Z) where n(M)=m and
n(Z)=0. The only set with n(Z) = 0 is the
empty set .
Mx is the set of all pairs (m,z) with
mM, z. But there are no z in , so
there are no pairs (m,z) like this.
So Mx = , and n(Mx) = 0.
If ab = 0 then a=0 or b=0
(Zero Factor Property)
Proof (by indirect method!)
Suppose a>0 and b>0 . Then there are
sets A and B with a=n(A), b=n(B), and neither
of these are the empty set.
So there are elements iA, jB.
So (i,j) AxB and AxB is not empty.
m x 1 = m (Identity Property)
Proof: Let M = {0,1,...m-1} be a set with m
elements, and S = {} be a set with 1
element. Then
MxS = {(0,), (1,),...(m-1,)}
and there is a bijection from M to this set,
i
(i,)
ab = a+a+...+a
b times
(Repeated Addition
Property)
ab = n(AxB), where A is any set with
n(A)=a and we can take B = {1,2,...,b}.
We can consider AxB as the set
(Apple,1),(Avocado,1),...,(Artichoke,1)
(Apple,2),(Avocado,2)...,(Artichoke,2)
...
...
...
(Apple,b),(Avocado,b),...,(Artichoke,b)
which is the union of b sets
(Apple,1),(Avocado,1),...,(Artichoke,1)
(Apple,2),(Avocado,2),...,(Artichoke,2)
each of size a. They’re disjoint because
the second terms are all different. So
ab = a+a+...+a, adding b terms.
ac + bc = (a+b)c (Distributive Property)
-this follows from the Repeated Addition
Property.
ac+ bc = (c+c+...+c)+(c+c+...+c)
a times
b times
= c+c+
...
a+b times
+c
If ab = ac and a0, then b=c
(Cancellative Property)
Either bc or bc.
Suppose bc.
Then b =c+d
(addition/comparison)
ab = ac + ad
(distributivity)
ac = ac + ad
(given: ab=ac)
ad = 0
(addition/comparison)
a=0 or d=0 (Zero Factor Property)
But a0, so d=0
(given)
So b = c+0 = c.
ab = ba (Commutative Property)
Let n(A) = a and n(B) = b.
ab = n(AxB)
ba = n(BxA)
AxB is the set of all pairs (i,j), iA, jB
BxA is the set of all pairs (i,j), iA, jB
and there is a bijection between these sets,
the switch map (i,j)
(j,i).
Thus ab = n(AxB) = n(BxA) = ba.
(ab)c = a(bc) (Associative Property)
(ab)c = n((AxB)xC); a(bc) = n(Ax(BxC)).
(AxB)xC is the set of all pairs ((i,j),k)
where (i,j) is in AxB and k is in C.
Ax(BxC) is the set of all pairs (i,(j,k))
where i is in A and (j,k) is in BxC.
These are NOT the same sets, but there
is a bijection ((i,j),k)
(i,(j,k)) that
just rearranges the parentheses.
So (ab)c = n((AxB)xC) = n(Ax(BxC)) = a(bc).
(7) DIVISION:
Division is another partial operation, like
subtraction. When ab = c, we define c/a = b
provided that a0.
In these cases, division is well-defined:
if a0, and ab=c, ad=c, then c=b.
(This does not work when a=0; division by 0 is
NOT well-defined.)
Division, like subtraction, is not
commutative; in fact, unless a=b, at most one
of a/b, b/a exists over the whole numbers.
Division, like subtraction, is not
associative; but IF a/(b/c) exists, it is
equal to ac/b and this exists.
If a/c + b/c exists, it equals (a+b)/c and
this exists.
DIVISION WITH REMAINDER
For any two natural numbers a,b there always
exist two natural numbers q (quotient) and
r (remainder) such that
(i) 0  r < b
(ii) a = qb + r
and this pair (q,r) is unique with these
properties.
(8) EXPONENTS
We write ab for axax...xa
b times
This can be shown to be n(Ab) where Ab
is the set of b-tuples from A, with repetition
allowed.
Similarly, if B is a b-element set, we can
write AB for the set of all functions B-->A.
b-tuple from A: chooses an element of A for
each of b positions
bijection
function B-->A: chooses an element of A for
each of the b elements of B.
EG: {Ann,Bill,Carol}4 =
{(Ann,Ann,Ann,Ann), (Ann,Ann,Ann,Bill),...
...(Carol,Carol,Carol,Carol)}
and this set has 81 = 34 elements.
This technique is used in counting
questions such as: “In how many ways can we
distribute four distinct prizes among Ann,Bill,
and Carol, if any person may receive multiple
prizes or no prize?”
Rules for Exponents:
a1 = a
ab x ac
(ab)c
(a 1-tuple from A is essentially
just an element of A.)
(Try some examples! For instance,
what is a2 x a3 ? Use repeated
multiplication.)
(Try some examples! For
instance, what is (a2 )3 ? )
What should a0 be? Why?