1.整除性:
m \ n ⇔ m > 0 , 並 且 m 可 以 整 除 n m \backslash n \Leftrightarrow m > 0,並且m可以整除n m \ n ⇔ m > 0 , 並 且 m 可 以 整 除 n
歐幾里得算法求解最大公約數
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.素數:
eg:證明有無限多個素數
假設僅有有限個素數,比如k個:2,3,5,…,P k P_k P k .那麼考慮 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
我們發現M不能被這已有的k個素數中的任何一個整除,那麼肯定有另一個素數可以整除M,或許M本身就是一個素數。因此與假設相矛盾!所以存在無窮多個素數。
3.階乘的因子:
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
確定能整除n!的p的最大冪:
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與n互素
m ⊥ n ⇔ m , n 是 整 數 , 且 g c d ( m , n ) = 1 m \perp n \Leftrightarrow m, n是整數,且gcd(m,n)=1 m ⊥ n ⇔ m , n 是 整 數 , 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
從兩個分數( 0 1 , 1 0 ) \left(\frac{0}{1}, \frac{1}{0}\right) ( 1 0 , 0 1 ) 開始,依次在兩個相鄰的分數m n \frac{m}{n} n m 和m ′ n ′ \frac{m^{\prime}}{n^{\prime}} n ′ m ′ 之間插入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 ′ 稱爲中位分數。
如果m / n m / n m / n 和m ′ / n ′ m^{\prime} / n^{\prime} m ′ / n ′ 是相鄰分數,那麼有m ′ n − m n ′ = 1 m^{\prime} n-m n^{\prime}=1 m ′ n − m n ′ = 1
用如下形式來表示,第一列表示m / n m / n m / n ,可見分子在下,分母在上,第二列也一樣:
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 )
可將L和R定義爲:
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 ⇒ 往右走意味着將分母+分子作爲分子
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 ⇒ 往左走意味着將分子+分母作爲分母
5.mod:同餘關係
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,那麼有
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 )
將除法應用到同餘式的另一種方法,對模作除法:
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