+
×
N
n
∈ N ↔ n := 1 + 1 +· · · + 1 . n
N = 1, 2, 3,
{
Z Z :=
−N ∪ {0}N
−N := {−n : n ∈ N } . {·· · , −2, −1, 0, 1, 2, ···}. Z = {·· Q :=
a : a, b b
Q
∈ Z, b = 0
.
√ 2 R Q
�������� �������� ��������
������ ����� ��� �� �� ���� ����
Z � �������� �������� ����� � � � � ���� ��� �� �� �� � � � � N � �������� �������� ����
��� �� �� � � � �
R\Q �������� �������� ���������� ������ ��
···} .
a q
r=1 r = 3
−
b > 0
r a = qb + r, 0
a = 9 a= 9
b
b = 4 b = 4
≤ r < b.
9 = (2)(4) + 1 9 = ( 3)(4) + 3
−
q r
q =
−
q = 2 3
−
r
b N S S := a
{ − nb | n ∈ Z} = {a, a ± b, a ± 2b, ···} .
n :=
t
0 N S
∩
r q
r
−| a|
∈Z
t := a
∈ N ∩ S
∈ N ∩ S
a = qb + r
− (−|a|)(b) = a + |a|b >
N r = a r r
≥0 ≥ b
r1
− qb ≥ 0
∈ S ∩ N
r1 = a
− (q + 1)b = r − b < r S ∩ N.
r 0 q
≤r
q 1
a = qb + r = q 1 b + r1 , 0 r q = q 1
− r = (q − q )b 1
≤ r, r < b. 1
q = q 1 , r1 r = q q 1 b
1
|q − q | ≥ 1
| − | | − | ≥b
1
r
b
r1
r1
q = q 1
r = r 1.
b a r = q + , b b q =
q =
a b
r b
−
0
r b
< 1.
a b
a b
−
a = 27 b = 12 q = −27 = 2 14 = 12 r = a qb = 27 ( 3)(12) = 27 + 36 = 9
− −
≤
− −−
−
−3
r
b
b=0
b > 0 b < 0 a q
r
a = qb + r, 0
b>0
|b| = b b < 0
q
≤ r < |b|.
|b|
r
b
a = q b + r, 0
||
q =
− q
≤ r < |b|. |b| = −b
a = q b + r =
||
b < 0
−q (−b) + r = qb + r, 0 ≤ r < |b|. |b| b
=
b a r = q + b b q =
−1
q =
a b
−
r b
≥ rb > −1,
,0
a b
a b
1 q = −7 = 0 a = 2 a = 61 a = 59
b = 7 < 0 a = 1 r = a qb = 1 0 = 1 1 = (0)( 7) + 1 = 27 = 1 q = −2 r = 2 (1)( 7 ) = 5 −7 q = 61 = 8 67 = 8 r = 61 ( 8)( 7) = 5 −7 q = −59 = 8 17 = 9 r = 59 (9)( 7) = 4 −7
− − −
−
−
− − − −− − − − −
a
−
−
−
a(a2 + 2)/3
b = 3 q r a = 3q + r r = 0, 1 2 r=0 a = 3q a(a2 + 2)/3 3q (9q 2 + 2)/3 = q (9q 2 + 2) r = 1, a = 3q + 1 a(a2 + 2)/3 (3q + 1)(9q 2 + 6q + 1 + 2)/3 = (3q + 1)3(3q 2 + 2q + 1)/3 = (3q + 1)(3q 2 + 2q + 1) r = 2 a = 3q + 2 a(a2 + 2)/3 (3q + 2)(9q 2 + 12q + 4 + 2)/3 = (3q + 2)3(3q 2 + 4q + 2)/3 = (3q + 2)(3q 2 + 4q + 2) a =
−1, 0, 1, 2, 3
a
b = 4
a = 4q + r r = 0, 1, 2, 3 r = 0 n = 4(4q 2 ) 0 16q 2 + 8q + 1 = 4(4q 2 + 2q ) + 1 n = 16q 2 + 16q + 4 = 4(4q 2 + 4q + 4) n = 16q 2 + 24q + 9 = 4(4q 2 + 6q + 2) + 1
q r n := a 2 . r = 1 r = 2
n = r=3
11, 111, 1111, 11111,
···
11
=
8+3
111
=
108 + 3
1111
=
1008 + 3
11111
=
10008 + 3
··· 1111 4
··· 111 = 1000 ··· 108 + 3
4k + 3
r = 0 a
b
1000
··· 108
b a=0 a b
b 12 = 4 3 10 = 3c 3 10
·
c 4 12
|
ab
a
|
a
b
c b a
−a
a
| | |
a0
aa
|
|
10 = 3c
b
a
∈ Z.
a=0
|
a,b,c a1
b
ab
b
a 0, 1 a, a a
∈Z
a =
|
ab
cd
| a|b ab
| b|c
±1 ac|bd a|c
ba
|
a =
±b |a| < |b|
|
ab
b=0
| a|b
a|c a|(bx + cy) a = ± 1 → a|1 a|1 k ∈ Z 1 = ka ±1
b = ac a
c
a1
k = 1, a = 1 1
| ↔ a = ±
x
k=
−1, a = −1
y
a1
| → a =
ab
cd
|
k1 , k2
∈ Z
|
b = k 1 a
d = k 2 c
bd = (k1 k2 )ac, ac bd
|
ab
bc k1, k2 Z c = k 2 b = k 2(k1 a) = (k1 k2 a) a = k1 b b = k2 a k1 k2 = 1 k1 = k2 = 1
|
(k1k2)(ab) a = b
±
|
∈
b = ac b=0 c 1 b = 0 b = k 1 a
|a||c|
c
||≥
∈ Z.
b = k1 a
c = k2 b ab =
k1 = k2 =
−1
|b| = |ac| = |c| = 0 |b| = |a||c| ≥ |a| x, y ∈ Z
c = k 2 a
bx + cy = k 1 ax + k2 ay = (k1 x + k2 y)a a (bx + cy )
|
a
a bk , k = 1,
|
··· , n
a (b1 x1 + b2 x2 +
|
x1 , x 2 ,
a
·· · + b x ) n n
·· · , x
n
b a
b
d da
db
|
|
ca
|
cb
|
c
≤d d
d
d
a
b
d = gcd(a, b). 12
1, 2, 3, 4, 6, 12
30
1, 2, 3, 5, 6, 10, 15, 30. gcd(12, 30) = 6
1, 2, 3, 6
a
gcd(a, b) = gcd(a, b) = gcd( a, b) = gcd( a, b).
−
a y
x
−
− −
b
gcd(a, b) = ax + by.
gcd( 12, 30) = 6 = ( 12)2 + 30 1
− gcd(−8, −36)
− · 4 = (−8)4 + (−36)(−1).
=
d = gcd(a, b) a a
b
b
x
y
S
a S = au + bv au + bv
{
|
≥ 1, u , v ∈ Z }
b
b
a=0 a = au +b 0 S u = 1 a S d = gcd(a, b) d S
−
a
||
S x, y
∈
ax + by da
|
r
u = 1
· ∈
a
d
∈
d Z
a = qd + r, 0 r > 0
≤r
0 < r = a
− qd = a − q (ax + by) = a(1 − qx) + b(−qy) ∈ S.
∈ S
d r = 0
da
|
db
b
d
d
|
cd
c
|
a
≤
ca d
a
b c c ax + b
cb
|
|
|
d = gcd(a, b)
b T = ax + by x, y
| ∈ Z}
{
d = gcd(a, b) da T x0 , y0
|
db
d (ax + by )
|
x, y Z d = ax0 + by 0
|
d
∈ Z
∈
d nd = n(ax0 + by0 ) = a(nx0 ) + b(ay0 ) d a
T. b
gcd(a, b) = 1 (3, 5), (5, 9)
−
q
r = 0
r
S
d = r
( 27, 35)
− −
∈ T.
a
b
x, y
ax + by = 1 a
b
gcd(a, b) = 1
x, y 1 = ax +by. gcd(a, b) = d = 1 da d = 1
db
|
d = gcd(a, b)
gcd
|
ax +by = 1 d (ax + by = 1) d1
|
|
a d
, db = 1. x d
a d
y
ax + by = d x+ db y = 1
a d
b d
a b
a
8 12
b gcd(8, 12) = 4
2 3
→ gcd(2, 3) = 1
gcd(a, b) = 1 ac
| a|bc
bc
ab c
|
|
ac
|
r gcd(a, b) = 1
s c = ar = bs 1 = ax + by
x, y c = c 1 = c(ax + by) = acx + bcy = a(bs)x + b(ar)y = ab(sx + ry),
·
ab c
|
c = c 1 = c(ax + by) = acx + bcy.
·
a ac
|
a bc
|
a (acx + bcy)
|
ac
|
a
q r r = 1
a a+2 a+4
a = 3q + r r = 0, 1, 2
a = 3q + 1
r=0
a a = 3q
3 a3
|
↔
a + 2 = 3q + 1 + 2 = 3(q + 1), r = 2
3 (a + 2)
|
a = 3q + 2
↔
a + 4 = 3q + 2 + 4 = 3(q + 2), 3 (a + 4)
|
a
q
∈ Z
a = 2q + r a(a + 1) 2) = (2q + 1)(q + 1)2
2 a(a + 1)
|
b=2 r = 0 a = 2q k = 1 a(a + 1) = (2q + 1)(2q +
r = 0, 1
a 2a 2 a(a + 1)
|
a+1
2 (a + 1)
|
|
a a(a + 1)(a + 2)
|
n a
gcd(a, n + a) n
|
d = gcd(a, a + n)
da d (a + n) d d ((a)( 1) + (a + n)(1)) = n
|
|
n=1
| −
(a, a + 1) = 1
a = qb + r
gcd(a, b) = gcd(b, r) b
qb + r = a b
r = a r
r
− qb
a, b
a b, r
a a
b
≥ b > 0
a q 1 a = q 1b + r1 , 0
r1 = 0
r1
≤ r < b. 1
gcd(a, b) = b r2
q 2
r1 = 0
b = q 2 r1 + r2 , 0 r2 = 0
b
≤r
2
r2 = 0,
≤r
3
r1
< r1 .
gcd(a, b) = r1 q 3 r3 r1 = q 3 r2 + r3 , 0
b
< r2 .
r1
r2
a = q 1 b + r1 , 0 < r1 < b b = q 2 r1 + r2, 0 < r2 < r1 r1 = q 3 r2 + r3, 0 < r3 < r2 rn−2 = q nrn−1 + rn , 0 < rn < rn−1 rn−1 = q n+1rn + 0.
gcd(a, b) = gcd(b, r1 ) = gcd(r1 , r2 ) =
a = 1492
· ·· = gcd(r
n−1
, rn ) = gcd(rn , 0) = r n.
b = 1066 1492 = 1 1066 + 426 1066 = 426 = 214 = 212 =
· 2 · 426 + 214 1 · 214 + 212 1 · 212 + 2 106 · 2 + 0.
d = gcd(1492, 1066) = 2
a
b a
m am
|
bm
|
c > 0
ac
|
bc
|
m
≤c
b
m b
a m m
a
(a, b)
m =
−12
(12, 30) = 60
d = gcd(a, b) r s
60, 120, 180, . . .
30
a
m =
(a, b) = gcdaba,b . ( )
b a = dr
ab d
= as → m = a(ds) d
m
b
a
b = ds
m =
(dr)b = rb, d
b
m c
a
b
c = au
c = bv x
v
u d = ax + by
y
ab d
m =
c cd c(ax + by) c c = = = x + y = vx + uy m ab ab b a mc
m
|
m =
ab d
↔
≤ c (a, b) =
m ab
gcd(a,b)
a gcd(a, b) = 1.
∈ Z, a
.
b
b
(a, b) = ab
12378 = 4 3054 + 162
·
3054 = 18 162 + 138
·
162 = 1 138 + 24
· 5 · 24 + 18 1 · 18 + 6 3·6+0
138 = 24 = 18 =
gcd(3054, 12378) = 6 (3054, 12378) =
d
a, b
c
3054 12378 = 6300402. 6
·
d = gcd(a,b,c) c d
c
d a, d b
gcd(39, 42, 54) = 3
gcd(a1 , a2 ,
dc
| |
≤
a1 ,
|
gcd(49, 210, 350) = 7
··· a
k
··· , a ) = gcd(gcd(a , a ) , a , ··· , a ) . 1
k
2
3
k
gcd(a1 , a2 , a3 ) = gcd (a1 , gcd(a2 , a3)) .
d1
d = gcd(a1 , a2 , a3 ) d (a2 x + a 3 y)
|
d a1 , d a2
|
|
x
d a3 y
|
d gcd(a2 , a3 ) := d1 = gcd(a2 , a3 )
|
d a1 d d1
|
c a1 a1 , a2
|
d c d1
|
a3 a1
a1
|
d1 c
c d1 c a2 c a3 d = gcd(a1 , a2 , a3 ) c d d1 d = gcd(a1 , d1 ) = gcd (a1, gcd(a2 , a3 )) .
|
d = gcd(a1, a2 ,
|
|
≤
c d
· ·· , a ) k
d2 = gcd(a1 , a2 ) d3 = gcd(d1, a3 ) d4 = gcd(d2, a4 ) dk = gcd(dk−2 , ak ) d = d k .
12
d = gcd(36, 24, 54, 27) d2 = gcd(36, 24) = d3 = gcd(12, 54) = 6 d = d 4 = gcd(6, 27) = 3
(a,b,c) =
abc . gcd(a,b,c)
ax + by = c
a, b
c (x, y) 3x + 6y = 18
3 4 + 6 1 = 18
· · 3 · (−6) + 6 · 6 3 · 10 + 6 · (−2)
= 18 = 18
(4, 1), ( 6, 6), (10, 2)
−
−
2x + 10y = 17
(x, y) 2x + 10y = 17
a, b
c
a
b
d = gcd(a, b) ax + by = c dc
|
b x = x 0 + n, y = y 0 d
− da n, n ∈ Z
(x0 , y0 ) T = ax + by x, y d = gcd(a, b) K = d T dc c = kd K x, y Z ax + by = c (x0, y0 )
{
Z kd k
}
{ | ∈ Z} c
∈ T
K = T
∈
|
| ∈
∈
ax0 + by0 = c. x = x 0 + db n, y = y 0
b a x0 + n + b y0 d
−
a d
−
n, n
∈Z
a ab n = ax 0 + by0 + n d d
(x, y) (x, y)
n
− abd n = ax + by 0
0
= c,
∈Z
ax + by = c = ax 0 + by0 a(x a (x d a
0
=
0
=
−b(y − y ) − db (y − y ). 0
0
b
b d
= |(x0 − x ) (x − x ) b d
−x ) −x )
b = 0
a d
(x x0 ). x x0 = db n
0
− −
0
a
gcd( ad , db ) = 1 x = x0 + db n b (y y0) = ad db n d
→ − −
· →
| b d
a d
(x n Z y = y 0 ad n
∈ −
−x ) 0
b
x = x 0 + bn, y = y 0
− an, n ∈ Z
(x0 , y0 )
d = gcd(a, b) d c dc
|
c = kd v
w
av + bw = d
akv + bkw = kd a(kv) + b(kw) = c.
k
x0 = kv
y0 = kw
172x + 20y = 1000.
gcd(172, 20) 172 = 8 20 + 12 20 = 12 = 8 =
· 1 · 12 + 8 1·8+4 2·4+0
gcd(172, 20) = 4
4 1000 1000 = 250 4
|
4 = 12 = = = =
−8 12 − (20 − 1 · 12) 2 · 12 − 20 2(172 − 8 · 20) − 20 2 · 172 + (−17) · 20. 250
500 172 + ( 4250) 20 = 1000.
·
x0 = 500
−
y0 =
−4250
·
·
20 t = 500 + 5t 4 172 4250 t = 4250 4
x = 500 + y =
−
−
−
− 43t
t
500 + 5t > 0
−4250 − 43t t > t<
−
4250 43
=
−98 t = −99
36 43
−
500 5
> 0 = 100 t
−
x = 500 + 5( 99) = 5 y =
− −4250 − 43(−99) = 7.
x
y
700x + 1300y = 100000 x
≥0
& y
≥ 0
y > x 7x + 13y = 1000 gcd(7, 13) = 1
1 = 7 2 + 13 ( 1).
·
·−
1000
x0 = 2000
1000 = 7 2000 + 13 ( 1000)
·
y0 =
·−
−1000.
x = 2000 + 13n y = x 2000 + 13n y
−1000 − 7n, n ∈ Z.
≥0
≥ 0 → n ≥ − 2000 ≈ −153.84 → n = {−153, −152, −151, ···}. 13 n
≥ 0
−1000 − 7n ≥ 0 → n ≤ − 1000 ≈ −142.85 → n = {··· , −145, −144, −143}. 7 y>x
−1000 − 7n > 2000 + 13n → n < −150 → n = {··· , −153, −152, −151}. n n =
{−153, −152, −151} n
n
x + y
− − −
n 153 152 151
37
x y x + y 11 71 82 24 64 88 37 57 94
57
x = i + jt y = k + mt i,j,k,m
∈ Z. 6x + 5y = 171, x, y > 0 gcd(6, 5) = 1 x 171
x =
− 5y = 28 + 3 − 5y .
6
6
p
p =
3−5y 6
∈Z
x = 28+ p
y
p y=
3
− 6 p = − p + 3 − p . 5
5
q
q =
3−q 5
∈Z
y = p + q
−
p=3
− 5q
x = 28 + (3 y = x>0 6q
− 3 > 0 →
− 5q ) = 31 − 5q −(3 − 5q ) + q = 6q − 3.
31 5q > 0 q > 21
−
→ q <
31 5
= 6 15 q
∈ {1, 2, 3, 4, 5, 6}
y>0
