数论——整除

发布于:2023-01-10 ⋅ 阅读:(345) ⋅ 点赞:(0)

设 $ a $ 是非零整数, b b b 是整数并且有 a × q = b a \times q = b a×q=b 。记作 a ∣ b a | b ab

  1. a ∣ b a | b ab b ∣ c b | c bc ,则 a ∣ c a | c ac

  2. a ∣ b a | b ab a ∣ c a | c ac ,则 a ∣ ( b × x + c × y ) a | ( b \times x + c \times y ) a(b×x+c×y)

证明:

a × q = b  ,  a × p = c a \times q = b \text{ , } a \times p = c a×q=b , a×p=c

b × x = a × q × x  ,  c × y = a × p × y b \times x = a \times q \times x \text{ , } c \times y = a \times p \times y b×x=a×q×x , c×y=a×p×y

所以 b × x + c × y = a × ( q × x + p × y ) b \times x + c \times y = a \times (q \times x + p \times y) b×x+c×y=a×(q×x+p×y)

a ∣ ( b × x + c × y ) a | ( b \times x + c \times y ) a(b×x+c×y)

  1. a ∣ b a|b ab m ≠ 0 m \ne 0 m=0 ,则 ( a × m ) ∣ ( b × m ) (a \times m)|(b \times m) (a×m)(b×m)

  2. a ∣ n , b ∣ n a|n , b|n an,bn , 并满足 a × x + b × y = 1 a \times x + b \times y = 1 a×x+b×y=1 ,则 ( a × b ) ∣ n (a \times b ) | n (a×b)n

证明:

a × q = n  ,  b × p = n  ,  a × x × n + b × y × n = n a \times q = n \text{ , } b \times p = n \text{ , } a \times x \times n + b \times y \times n = n a×q=n , b×p=n , a×x×n+b×y×n=n

所以 a × x × b × p + b × y × a × q = n a \times x \times b \times p + b \times y \times a \times q = n a×x×b×p+b×y×a×q=n

所以 a × b × ( x × p + y × q ) = n a \times b \times (x \times p + y \times q) = n a×b×(x×p+y×q)=n

所以 ( a × b ) ∣ n (a \times b ) | n (a×b)n

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