Prove --- Modulus
2012-05-12 8:11 pm
Show that the following law of cancellation holds for congruence withrespect to a prime modulus:If ab ≡ acand a ≢ 0, then b ≡ c.
回答 (4)
2012-05-12 8:16 pm
✔ 最佳答案
ab ≡ ac (mod p)=> ab - ac = np
=> a(b - c) = np
=> (b - c) = np/a
Since a ≢ 0 and a ∤ p, a should divide n or n/a = k where k is a constant
This means that b - c = kp and thus b ≡ c (mod p)
2012-05-13 1:33 am
a ≢ 0 (mod p) 表示 xa ≡ 1 (mod p) 恆有解 x。
ab ≡ ac (mod p)
兩邊左乘上 x 即得：b ≡ c (mod p)。
2012-05-12 17:38:27 補充：
欲證 003 意見中之第一行：
(a, p) = 1 可得 xa + yp = 1 for some integers x, y.〔由 Euclid algorithm 得出〕
取 (mod p) 即得 xa ≡ 1 (mod p)。
ab ≡ ac (mod p)
兩邊左乘上 x 即得：b ≡ c (mod p)。
2012-05-12 17:38:27 補充：
欲證 003 意見中之第一行：
(a, p) = 1 可得 xa + yp = 1 for some integers x, y.〔由 Euclid algorithm 得出〕
取 (mod p) 即得 xa ≡ 1 (mod p)。
2012-05-12 8:57 pm
THX...............
2012-05-12 8:47 pm
' | ' means divisible
' ∤ ' means not divisible
(p,a) means H.C.F.
' ∤ ' means not divisible
(p,a) means H.C.F.
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