Transcript Lesson49
Math 409/409G History of Mathematics
The Theory of Congruences
What are congruences?
If you are familiar with modular or clock arithmetic, then you already know what congruences are.
Definition:
Let
n
be a fixed positive integer. Two integers
a
be
congruent modulo n
and if
n b
are said to divides
a
–
b
.
Notation: a ≡
b
(mod
n
) or simply
a
≡
n b.
Examples:
24 ≡ 7 3 ≡ 7 31 ≡ 7 25 7 3 since 7│(24 – 3) 24 since 7│(3 – 24) 11 since 7│(-31 – 11) 12 since 7 (25 – 12)
Theorem:
a ≡ n b
if and only if
a
and
b
have the same remainder when divided by
n
.
Proof:
First assume that
a
and
b
have the same remainder when divided by
n
. Then there are integers
q a
and
b
and
Qn
+
r Q
such that . Thus
a ≡ n b
since
a
–
qn b
+
r
(
q – Q
)
n
.
Now assume that
a ≡ n b
. Then a – b for some integer
s
. And by the division
ns
algorithm there exist integers
q
,
Q
,
r
, and
R
0 such that
r
,
R
<
n. a
So
qn
+
r
and
b
Qn
+
R
with
ns
( )
r R
)
ns R
( )
R
) But –
n
that
r
– <
R r
-
R
<
n
. So
n
│(
r
0. Thus
a
and
b
– remainder when divided by
n
.
R
) implies have the same
Since adding a multiple of
n
to a number does not change the remainder when that number is divided by
n
, we have: Corollary: a ≡
n a
+
kn
for any integer
n
.
Example:
31 ≡ 7 31 + 5·7 = 4. So the remainder when -31 is divided by 7 is 4. The same result could, of course, been achieved by doing a long division problem.
7 5 31 35 4
Properties of Congruences
For integers
a
,
b
,
c
,
d
and positive integers
n
,
k
:
1. a 2. a 3. a 4. a 5. a 6. a
≡
n
≡
n
≡
n
≡
n a.
b b b
≡ ≡
n n b b
b
and ≡
n b a.
≡
n c
and
c
≡
n d
a + c
≡
n a
≡
n a + c ac
≡
n
≡
n bd.
b + d
and
b + c c.
and
ac
≡
n bc.
a k
≡
n b k
.
a
≡
n b
Proof that and
c
≡
n d
ac
≡
n bd
hyp.
ns
and
ac
b c
n sc
and
ac
(
bd
nbt
)
nsc
bt
) )
ac
n bd nt
nbt
Proof that
a
≡
n b
ac
≡
n bc
This follows from the previous property. That is,
a
≡
n b
and
c
≡
n c
ac
≡
n bc.
Or one could give a direct proof.
a
n b
ns
nsc
)
ac
n bc
Proof that
a
≡
n b
a k
≡
n b k
This statement is proved by induction. Clearly the statement holds for
k
1. Assume that it is true for some fixed
k
. Then
a k
1
a k
n
n
a b
b k
1
b k b k
by induct. h yp.
by prop . 4.
So by induction, the statement is true for all positive integers
k
.
Is 2
20
– 1 divisible by 41?
Answering this question is equivalent to finding the remainder when 2 20 – 1 is divided by 41. Using congruences makes this easy. Find the remainder when 2 20 is divided by 41 by first finding the remainder when smaller powers of 2 are divided by 41. Then you can use the property
a
≡
n b
a k
≡
n b k
to find the remainder for larger powers of 2.
2 3 8 2 4 16 2 5 32 2 6 64
You may be tempted to use 2 6 since this is the smallest power 2 3 8 2 4 16 of 2 that is larger than 41. But 2 6 64 ≡ 41 23 and 23 is a rather 2 5 2 6 large number to raise to power when using the property
a
≡
n b
a k
≡
n b
find the remainder for larger powers of 2.
k
32 64 to So use 2 5 since 2 5 32 41 – 9 ≡ 41 9 and it’s easier to find powers of -9 than powers of 23.
Is 2 20 – 1 divisible by 41?
2 5 41 9 2 10 2 20 41 81 41 41 2 40 41 1 1 So 2 20 – 1 ≡ 41 1 – 1 remainder when 2 20 0. Since the – 1 is divided by 41 is 0, the answer is yes, 2 20 – 1 is divisible by 41.
What’s the remainder when N
1! +2! + 3! + ··· + 100!
is divided by 4?
When
n
So
n
! ≡ 4 4, 4 appears in the product of
n
!. 0 when
n
4. This gives 1! +2! + 3! + ··· + 100! ≡ 4 1! +2! + 3! ≡ 4 3.
So the remainder when N is divided by 4 is 3.