1.整除性:
m
\
n
⇔
m
>
0
,
并
且
m
可
以
整
除
n
m \backslash n \Leftrightarrow m > 0,并且m可以整除n
m \ n ⇔ m > 0 , 并 且 m 可 以 整 除 n
欧几里得算法求解最大公约数
gcd
(
0
,
n
)
=
n
\operatorname{gcd}(0, n)=n
g c d ( 0 , n ) = n
gcd
(
m
,
n
)
=
gcd
(
n
 
m
o
d
 
m
,
m
)
,
m
>
0
\operatorname{gcd}(m, n)=\operatorname{gcd}(n \bmod m, m), \quad m>0
g c d ( m , n ) = g c d ( n m o d m , m ) , m > 0
2.素数: eg:证明有无限多个素数 假设仅有有限个素数,比如k个:2,3,5,…,
P
k
P_k
P k .那么考虑
M
=
2
×
3
×
5
×
⋯
×
P
k
+
1
M=2 \times 3 \times 5 \times \cdots \times P_{k}+1
M = 2 × 3 × 5 × ⋯ × P k + 1 我们发现M不能被这已有的k个素数中的任何一个整除,那么肯定有另一个素数可以整除M,或许M本身就是一个素数。因此与假设相矛盾!所以存在无穷多个素数。
3.阶乘的因子: 斯特林公式:
n
!
∼
2
π
n
(
n
e
)
n
n ! \sim \sqrt{2 \pi n}\left(\frac{n}{\mathrm{e}}\right)^{n}
n ! ∼ 2 π n
( e n ) n 确定能整除n!的p的最大幂: 因此,
10
!
10!
1 0 ! 能被
2
8
2^{8}
2 8 整除
4.互素:
m
⊥
n
m \perp n
m ⊥ n : m与n互素
m
⊥
n
⇔
m
,
n
是
整
数
,
且
g
c
d
(
m
,
n
)
=
1
m \perp n \Leftrightarrow m, n是整数,且gcd(m,n)=1
m ⊥ n ⇔ m , n 是 整 数 , 且 g c d ( m , n ) = 1
m
/
gcd
(
m
,
n
)
⊥
n
/
gcd
(
m
,
n
)
m / \operatorname{gcd}(m, n) \perp n / \operatorname{gcd}(m, n)
m / g c d ( m , n ) ⊥ n / g c d ( m , n )
k
⊥
m
且
k
⊥
n
⇔
k
⊥
m
n
k \perp m \mathrm且 k \perp n \Leftrightarrow k \perp m n
k ⊥ m 且 k ⊥ n ⇔ k ⊥ m n
从两个分数
(
0
1
,
1
0
)
\left(\frac{0}{1}, \frac{1}{0}\right)
( 1 0 , 0 1 ) 开始,依次在两个相邻的分数
m
n
\frac{m}{n}
n m 和
m
′
n
′
\frac{m^{\prime}}{n^{\prime}}
n ′ m ′ 之间插入
m
+
m
′
n
+
n
′
\frac{m+m^{\prime}}{n+n^{\prime}}
n + n ′ m + m ′ ,
m
+
m
′
n
+
n
′
\frac{m+m^{\prime}}{n+n^{\prime}}
n + n ′ m + m ′ 称为中位分数。 如果
m
/
n
m / n
m / n 和
m
′
/
n
′
m^{\prime} / n^{\prime}
m ′ / n ′ 是相邻分数,那么有
m
′
n
−
m
n
′
=
1
m^{\prime} n-m n^{\prime}=1
m ′ n − m n ′ = 1 用如下形式来表示,第一列表示
m
/
n
m / n
m / n ,可见分子在下,分母在上,第二列也一样:
M
(
S
)
=
(
n
n
′
m
m
′
)
M(S)=\left( \begin{array}{ll}{n} & {n^{\prime}} \\ {m} & {m^{\prime}}\end{array}\right)
M ( S ) = ( n m n ′ m ′ )
M
(
S
L
)
=
(
n
n
+
n
′
m
m
+
m
′
)
=
(
n
n
′
m
m
′
)
(
1
1
0
1
)
=
M
(
S
)
(
1
1
0
1
)
M(S L)=\left( \begin{array}{cc}{n} & {n+n^{\prime}} \\ {m} & {m+m^{\prime}}\end{array}\right)=\left( \begin{array}{cc}{n} & {n^{\prime}} \\ {m} & {m^{\prime}}\end{array}\right) \left( \begin{array}{ll}{1} & {1} \\ {0} & {1}\end{array}\right)=M(S) \left( \begin{array}{ll}{1} & {1} \\ {0} & {1}\end{array}\right)
M ( S L ) = ( n m n + n ′ m + m ′ ) = ( n m n ′ m ′ ) ( 1 0 1 1 ) = M ( S ) ( 1 0 1 1 )
M
(
S
R
)
=
(
n
+
n
′
n
′
m
+
m
′
m
′
)
=
M
(
S
)
(
1
0
1
1
)
M(S R)=\left( \begin{array}{cc}{n+n^{\prime}} & {n^{\prime}} \\ {m+m^{\prime}} & {m^{\prime}}\end{array}\right)=M(S) \left( \begin{array}{ll}{1} & {0} \\ {1} & {1}\end{array}\right)
M ( S R ) = ( n + n ′ m + m ′ n ′ m ′ ) = M ( S ) ( 1 1 0 1 ) 可将L和R定义为:
L
=
(
1
1
0
1
)
,
R
=
(
1
0
1
1
)
L=\left( \begin{array}{ll}{1} & {1} \\ {0} & {1}\end{array}\right), \quad R=\left( \begin{array}{ll}{1} & {0} \\ {1} & {1}\end{array}\right)
L = ( 1 0 1 1 ) , R = ( 1 1 0 1 )
S
=
(
n
n
′
m
m
′
)
;
R
S
=
(
n
n
′
m
+
n
m
′
+
n
′
)
S=\left( \begin{array}{cc}{n} & {n^{\prime}} \\ {m} & {m^{\prime}}\end{array}\right) ; \quad R S=\left( \begin{array}{cc}{n} & {n^{\prime}} \\ {m+n} & {m^{\prime}+n^{\prime}}\end{array}\right)
S = ( n m n ′ m ′ ) ; R S = ( n m + n n ′ m ′ + n ′ )
⇒
\Rightarrow
⇒
f
(
S
)
=
(
m
+
m
′
)
/
(
n
+
n
′
)
f(S)=\left(m+m^{\prime}\right) /\left(n+n^{\prime}\right)
f ( S ) = ( m + m ′ ) / ( n + n ′ ) ,
f
(
R
S
)
=
(
(
m
+
n
)
+
(
m
′
+
n
′
)
)
/
(
n
+
n
′
)
f(R S)=\left((m+n)+\left(m^{\prime}+n^{\prime}\right)\right) /\left(n+n^{\prime}\right)
f ( R S ) = ( ( m + n ) + ( m ′ + n ′ ) ) / ( n + n ′ )
⇒
\Rightarrow
⇒
f
(
R
S
)
=
f
(
S
)
+
1
f(R S)=f(S)+1
f ( R S ) = f ( S ) + 1
⇒
\Rightarrow
⇒ 往右走意味着将分母+分子作为分子 同理,
L
S
=
(
m
+
n
m
′
+
n
′
m
m
′
)
\quad L S=\left( \begin{array}{cc}{m+n} & {m^{\prime}+n^{\prime}} \\ {m} & {m^{\prime}}\end{array}\right)
L S = ( m + n m m ′ + n ′ m ′ )
f
(
L
S
)
=
(
m
+
m
′
)
/
(
(
m
+
n
)
+
(
m
′
+
n
′
)
)
f(LS)= \left(m+m^{\prime}\right)/\left((m+n)+\left(m^{\prime}+n^{\prime}\right)\right)
f ( L S ) = ( m + m ′ ) / ( ( m + n ) + ( m ′ + n ′ ) )
⇒
\Rightarrow
⇒ 往左走意味着将分子+分母作为分母 5.mod:同余关系
a
≡
b
(
 
m
o
d
 
m
)
⇔
a
 
m
o
d
 
m
=
b
 
m
o
d
 
m
a \equiv b \quad(\bmod m) \quad \Leftrightarrow \quad a \bmod m=b \bmod m
a ≡ b ( m o d m ) ⇔ a m o d m = b m o d m
⇔
m
∣
(
a
−
b
)
\Leftrightarrow m|(a-b)
⇔ m ∣ ( a − b )
⇔
\Leftrightarrow
⇔ a-b is the multiple of m
a
≡
b
且
c
≡
d
⇒
a
+
c
≡
b
+
d
(
 
m
o
d
 
m
)
a \equiv b \quad \mathrm且\quad c \equiv d \quad \Rightarrow \quad a+c \equiv b+d \quad(\bmod m)
a ≡ b 且 c ≡ d ⇒ a + c ≡ b + d ( m o d m )
a
≡
b
且
c
≡
d
⇒
a
−
c
≡
b
−
d
(
 
m
o
d
 
m
)
a \equiv b \quad \mathrm且\quad c \equiv d \quad \Rightarrow \quad a-c \equiv b-d \quad(\bmod m)
a ≡ b 且 c ≡ d ⇒ a − c ≡ b − d ( m o d m ) if
a
d
≡
b
d
(
 
m
o
d
 
m
)
a d \equiv b d(\bmod m)
a d ≡ b d ( m o d m ) ,we can’t always conclude that
a
≡
b
a \equiv b
a ≡ b 比如:
3
×
2
≡
5
×
2
(
 
m
o
d
 
4
)
3 \times 2 \equiv 5 \times 2(\bmod 4)
3 × 2 ≡ 5 × 2 ( m o d 4 ) 但是
3
̸
≡
5
3 \not\equiv5
3 ̸ ≡ 5 但是如果d和m互素,那么有
a
d
≡
b
d
⇔
a
≡
b
(
 
m
o
d
 
m
)
,
a
,
b
,
d
,
m
,
d
⊥
m
a d \equiv b d \quad \Leftrightarrow \quad a \equiv b \quad(\bmod m),a,b,d,m,d \perp m
a d ≡ b d ⇔ a ≡ b ( m o d m ) , a , b , d , m , d ⊥ m
⇒
m
∣
d
(
a
−
b
)
\Rightarrow m|d(a-b)
⇒ m ∣ d ( a − b )
⇒
m
∣
(
a
−
b
)
\Rightarrow m|(a-b)
⇒ m ∣ ( a − b ) 将除法应用到同余式的另一种方法,对模作除法:
a
d
≡
b
d
(
 
m
o
d
 
m
d
)
⇔
a
≡
b
(
 
m
o
d
 
m
)
,
d
≠
0
a d \equiv b d \quad(\bmod m d) \quad \Leftrightarrow \quad a \equiv b \quad(\bmod m), d \neq 0
a d ≡ b d ( m o d m d ) ⇔ a ≡ b ( m o d m ) , d ̸ = 0