(=>) x = k1m1+a1 且 x = k2m2+a2, for some k1,k2 => a1-a2 = k2m2-k1m1 => 存在 t 使得 a1-a2 = t*gcd(m1,m2) => gcd(m1,m2) | a1-a2
(<=) a1-a2 = k*gcd(m1,m2), for some k 因為 gcd(m1,m2)=sm1+tm2, for some s,t => a1-a2 = k(sm1+tm2) => a1-ksm1 = a2+ktm2 所以取 x = a1-ksm1 = a2+ktm2, x = a1 mod m1 且 x = a2 mod m2
1 則留言:
(=>)
x = k1m1+a1 且 x = k2m2+a2, for some k1,k2
=> a1-a2 = k2m2-k1m1
=> 存在 t 使得 a1-a2 = t*gcd(m1,m2)
=> gcd(m1,m2) | a1-a2
(<=)
a1-a2 = k*gcd(m1,m2), for some k
因為 gcd(m1,m2)=sm1+tm2, for some s,t
=> a1-a2 = k(sm1+tm2)
=> a1-ksm1 = a2+ktm2
所以取 x = a1-ksm1 = a2+ktm2,
x = a1 mod m1 且 x = a2 mod m2
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