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3. APPLICATIONS OF NUMBER THEORY
163
3. Applications of Number Theory
3.1. Representation of Integers.
Theorem
3.1.1
.
Given an integer
b >
1
, every positive integer
n
can be expresses
uniquely as
n
=
a
k
b
k
+
a
k

1
b
k

1
+
···
+
a
1
b
+
a
0
,
where
k
≥
0
,
0
≤
a
0
,a
1
,a
2
,...,a
k
< b
, and are all integers.
Definition
3.1.1
.
Base
b
expansion of
n
is
(
a
k
a
k

1
···
a
1
a
0
)
b
if the
a
i
are as
described in Theorem 3.1.1.
Example
3.1.1
.
Here are examples of common expansions other than the more
familiar decimal expansion.
•
Binary expansion
is the base 2 expansion.
•
Octal expansion
is the base 8 expansion.
•
Hexadecimal expansion
is base 16 expansion. The symbols A through F
are used to represent 10 through 15 in the expansion.
Discussion
Theorem 3.1.1 asserts that each positive integer
n
can be expressed uniquely as
a linear combination of powers of a ﬁxed integer
b >
1. The coeﬃcients in the
linear combination must be less than
b
and must be greater than or equal to zero.
These coeﬃcients are, by deﬁnition, the digits of the
base
b
expansion of
n
,
n
=
(
a
k
a
k

1
...a
1
a
0
)
b
.
3.2. Constructing Base
b
Expansion of
n
.
Use the division algorithm to get
the base
b
expansion of
n
:
1.
n
=
bq
1
+
a
0
, 0
≤
a
0
< b
and
q
1
< n
.
2.
q
1
=
bq
2
+
a
1
, 0
≤
a
1
< b
and
q
2
< q
1
.
3.
q
2
=
bq
3
+
a
2
, 0
≤
a
2
< b
and
q
3
< q
2
.
4. etc. until
q
i
= 0.
Then
n
= (
a
k
a
k

1
...a
1
a
0
)
b
.
Example
3.2.1
.
Find the binary expansion of 42.
Solution: We can use the division algorithm to get the
a
i
’s.
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164
42 = 2(21) + 0
21 = 2(10) + 1
10 = 2(5) + 0
5 = 2(2) + 1
2 = 2(1) + 0
1 = 2(0) + 1
This gives us
42 = (1)(2
5
) + (0)(2
4
) + (1)(2
3
) + (0)(2
2
) + (1)(2
1
) + 0
.
Thus the
binary expansion of 42 is
(101010)
2
.
Example
3.2.2
.
Find the hexadecimal expansion of 42.
Solution: This time we use 16 for
b
.
42 = 16(2) + 10
2 = 16(0) + 2
So the hexadecimal expansion of 42 is
(2
A
)
16
(recall we use
A
= 10
in hexadecimal
notation).
Example
3.2.3
.
Find the decimal notation of the octal representation
(1024)
8
.
(1024)
8
= 1(8
3
) + 0(8
2
) + 2(8
1
) + 4 = 532
3.3. Cancellation in Congruences.
Theorem
3.3.1
.
Suppose
GCD(
c,m
) = 1
and
ac
≡
bc
(
mod
m
)
. Then
a
≡
b
(
mod
m
)
.
Proof.
Suppose GCD(
c,m
) = 1 and
ac
≡
bc
(mod
m
). (We may assume
m >
1 so that
c
6
= 0.) Then
ac

bc
=
c
(
a

b
) =
km
for some integer
k
. This implies
c

km.
Since GCD(
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 Fall '08
 PenelopeKirby

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