Ex :
Find
=
23
24 =
23
84
1cm (84 , 24)
.
.
3
.
.
7
all
3
↓
lam(84
,
24)
=
2 371
.
showing
primer
factorization
=
168
in
any
raised
prince
to its
largest exponent
gcd and lcm
Theorem: Let 𝑎, 𝑏 ∈ ℤ, with one of them not
zero. Then,
𝑎𝑏 = gcd 𝑎, 𝑏 ∙ lcm 𝑎, 𝑏
Er :
Determine Com (18
gcd(18 135)
,
=
,
97
138)
lam (18
,
135)
18
=
-
136
gad (18 138)
,
=
278
To
15
&
The Euclidean algorithm
Theorem: Let 𝑎, 𝑏 ∈ ℤ+ . If using the division
algorithm,
𝑎 = 𝑞1 𝑏 + 𝑟1 , with 0 ≤ 𝑟1 < 𝑏
𝑏 = 𝑞2 𝑟1 + 𝑟2 , with 0 ≤ 𝑟2 < 𝑟1
𝑟1 = 𝑞3 𝑟2 + 𝑟3 , with 0 ≤ 𝑟3 < 𝑟2
⋮
𝑟𝑛−3 = 𝑞𝑛−1 𝑟𝑛−2 + 𝑟𝑛−1 , with 0 ≤ 𝑟𝑛−1 < 𝑟𝑛−2
𝑟𝑛−2 = 𝑞𝑛 𝑟𝑛−1 + 0
then gcd 𝑎, 𝑏 = 𝑟𝑛−1 .
Ex :
Determine
a
gad (460
+divided
2202
465
E
Sedivior
=
465
=
345
b
2206
,
4
.
+
r
,
-rem
345
rz
V
1
.
120
+
My
345
=
120
-
2
+
10s
r4
120
105
=
=
105
18
.
1
+
18 +
.
7
+
0
gcd(465 2205)
,
g(
-stop
=
10
.
Bezout’s theorem
Theorem: Let 𝑎, 𝑏 ∈ ℤ+ , and let 𝑑 = gcd 𝑎, 𝑏 .
Then there exist integers s and t, such that
𝑑 = 𝑎𝑠 + 𝑏𝑡
That is, d is a linear combination of a and b.
Ex
Express god (465 , 2205)
:
By Bezout's Theorem ,
Site
2205
=
465
②
465
=
345
③
345
⑪
120
105
18
=
100
=
2206)
,
=
465s
+
2205t
for
=
120 3
348-
,
2
-
120
+
+
10s
.
1
+
18 +
.
7
+
0
10
=
16
1 +
.
g(
-stop
,
120
gad (465
465 and 2205
346
+
1
.
120
=
=
4
.
gcd(465 2205)
①
=
linear combination of
a
.
①
105
15
as
15
120 - 105 .
=
f
③
345
=
16 =
120
468
=
120
2
.
100 =
+
105
340
=
120 2
-
.
↓
②
(348
-
345
1
:
-
120
·
2) 1
348
-
1
.
120
+
2
.
-
.
-
1207120
+
120
=
468
=
345 -
-
↳
(468
15 =
=
0
348
-
465
:
3
-
·
1) 3
.
348
.
-
348
.
1
465
=
.
3
-
345
-
3
-
348-1
=
4
2208
=
465
.
4
+
3457348
=
2206
-
465
-
4
↳
16
=
=
468
3
·
465
-
-
(2208
-
465
19-2208 4
.
-
4) 4
.
=
465 3
.
-
2208
.
4 +
465 13
-
=
:
18
=
465-19
J
+
2200(
-
4)
Greatest common divisor
Theorem: Let 𝑎, 𝑏 ∈ ℤ+ , such that 1 = gcd 𝑎, 𝑏 ,
and let 𝑐 ∈ ℤ. If 𝑎|𝑏𝑐, then 𝑎|𝑐.
Corollary: If p is a prime and 𝑝|𝑎1 𝑎2 ⋯ 𝑎𝑛 ,
where 𝑎𝑖 is an integer, for 1 ≤ 𝑖 ≤ 𝑛, then 𝑝|𝑎𝑗
for some 1 ≤ 𝑗 ≤ 𝑛.
Pf of Thm
.
:
From Becort's Theorem
1 =
as
+
bt
,
,
gad (a b)
Since
,
=
as
+
As ala
+b =
and
c
)
c(as + +b) =
=
albe
,
then
1
,
then
for ste
E
E
1
=
M
c
=
a(c)
+
(b)t
alacob //
-
Greatest common divisor
6
=
0 (modb)
5 <
:
2
=
2
=
(nod
6)
(nob)
=
.
0
x
0
Theorem: Let 𝑎, 𝑏, 𝑐 ∈ ℤ, and let 𝑚 ∈ ℤ+ . If
𝑎𝑐 ≡ 𝑏𝑐 mod 𝑚 and gcd 𝑐, 𝑚 = 1, then
𝑎 ≡ 𝑏 mod 𝑚 .
gcd(3 6)
,
61z
-
=
3 f)
0X
Pf :
As
=
ac
By previous
mla-b
bc (mod m)
theorem
.
Hence
,
,
,
since
a
=
then
m/ac-bc
m)c(a- b)
b (modu)
/I
and
=
c
(a
-
b) .
god (cm)
=
1
,
then
Multiplicative inverses mod m
2x
zx
x
3
=
+.
=
=
2
3
Z
3
-
t
2
.?
modm
=
inverse of
Def: Let 𝑎, 𝑏 ∈ ℤ and let 𝑚 ∈ ℤ+ . We say that a
and b are multiplicative inverses mod m if
2
𝑎𝑏 ≡ 1 mod 𝑚
or equivalently,
𝑎𝑏 mod 𝑚 = 1 mod 𝑚
Ex :
3
and I are
multiplicative
3
.
2
=
6
. 2
3
inverses
mods =
=
61
(nodb)
mod
=
o
,
1
because
1
E
:
I
does not
have
a
multiplicative
inverse
mod 4
Multiplicative inverses mod m
Theorem: Let 𝑎 ∈ ℤ and let 𝑚 ∈ ℤ+ . If
gcd 𝑎, 𝑚 = 1 and 𝑚 > 1, then a has a
multiplicative inverse mod m. Furthermore, this
inverse is unique mod m. That is, if 𝑏1 and 𝑏2
are two multiplicative inverses mod m of a, then
𝑏1 ≡ 𝑏2 mod 𝑚 .
Ex
:
Find
the
multiplicative
①
1201
=
100
②
1001
=
200
200
=
1
gad (1001
By
②
①
.
1
+
8
+
1
200
+
0
-
-
1201)
,
=
200
1201
.
ga
+
1 must in of
=
1
,
+
5
.
mod
1001
200
Bezout's Theorem
1001
of
inverse
1
1001
=
1001
=
12017
+
5
200
-
.
mod
1001
exists
1201
for test
,
·
f
1201
=
1001
2005
1 +
.
200
1201
=
b
1006-(1201-600 1)-6
d
1001
.
6
-
1201
.
8
=
6001-12066
=
.
=
1001 .
-
1001
.
6 +
1201(
5
mult
>
-
-
.
100ls
+
=
5)
inv
-
of 1001
mod
Y
1 =
1001
=
6
is
a
multiplicative
-
6
1001
+
.
6
inverse
120(-r)
(mod
of
1201)
1001
mod
1201
1201