DISCRETE MATHEMATICS
W W L CHEN
c
±
W W L Chen, 1982, 2008.
This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990.
It is available free to all individuals, on the understanding that it is not to be used for ﬁnancial gain,
and may be downloaded and/or photocopied, with or without permission from the author.
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Chapter 4
DIVISION AND FACTORIZATION
4.1. Division
Definition.
Suppose that
a,b
∈
Z
and
a
6
= 0. Then we say that
a
divides
b
, denoted by
a

b
, if there
exists
c
∈
Z
such that
b
=
ac
. In this case, we also say that
a
is a divisor of
b
, or
b
is a multiple of
a
.
Example 4.1.1.
For every
a
∈
Z
\ {
0
}
,
a

a
and
a
 
a
.
Example 4.1.2.
For every
a
∈
Z
, 1

a
and

1

a
.
Example 4.1.3.
If
a

b
and
b

c
, then
a

c
. To see this, note that if
a

b
and
b

c
, then there exist
m,n
∈
Z
such that
b
=
am
and
c
=
bn
, so that
c
=
amn
. Clearly
mn
∈
Z
.
Example 4.1.4.
If
a

b
and
a

c
, then for every
x,y
∈
Z
,
a

(
bx
+
cy
). To see this, note that if
a

b
and
a

c
, then there exist
m,n
∈
Z
such that
b
=
am
and
c
=
an
, so that
bx
+
cy
=
amx
+
any
=
a
(
mx
+
ny
).
Clearly
mx
+
ny
∈
Z
.
PROPOSITION 4A.
Suppose that
a
∈
N
and
b
∈
Z
. Then there exist unique
q,r
∈
Z
such that
b
=
aq
+
r
and
0
≤
r < a
.
Proof.
We shall ﬁrst of all show the existence of such numbers
q,r
∈
Z
. Consider the set
S
=
{
b

as
≥
0 :
s
∈
Z
}
.
Then it is easy to see that
S
is a nonempty subset of
N
∪ {
0
}
. It follows from the Wellordering
principle that
S
has a smallest element. Let
r
be the smallest element of
S
, and let
q
∈
Z
such that
b

aq
=
r
. Clearly
r
≥
0, so it remains to show that
r < a
. Suppose on the contrary that
r
≥
a
. Then
b

a
(
q
+ 1) = (
b

aq
)

a
=
r

a
≥
0, so that
b

a
(
q
+ 1)
∈
S
. Clearly
b

a
(
q
+ 1)
< r
, contradicting
that
r
is the smallest element of
S
.
Chapter 4 : Division and Factorization
page 1 of 6
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View Full DocumentDiscrete Mathematics
c
±
W W L Chen, 1982, 2008
Next we show that such numbers
q,r
∈
Z
are unique. Suppose that
b
=
aq
1
+
r
1
=
aq
2
+
r
2
with
0
≤
r
1
< a
and 0
≤
r
2
< a
. Then
a

q
1

q
2

=

r
2

r
1

< a
. Since

q
1

q
2
 ∈
N
∪ {
0
}
, we must have

q
1

q
2

= 0, so that
q
1
=
q
2
and so
r
1
=
r
2
also.
±
Definition.
Suppose that
a
∈
N
and
a >
1. Then we say that
a
is prime if it has exactly two positive
divisors, namely 1 and
a
. We also say that
a
is composite if it is not prime.
Remark.
Note that 1 is neither prime nor composite. There is a good reason for not including 1 as a
prime. See the remark following Proposition 4D.
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 Spring '08
 RAMASWAMY
 Math, Prime number, Proposition, Greatest common divisor, Divisor, Integer factorization, smallest element

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