推 ddczx:1.T 2.F 令gcd(m,n)=d, d|m,d|n 故d|n-m 09/18 00:47
→ movo11:第一題true怎麼推導? 還有第二題是true喔 09/18 10:23
推 ddczx:看漏gcd(a,b)=1了... 1. 證 p->q = 證 ~q -> ~p 09/18 11:32
→ ddczx:設n非1or質數,則n=xy for some x,y=/=1,此時令a=x,b=y得矛盾 09/18 11:33
推 ddczx:2.gcd(a,b)=1,故a,b最小公倍數是ab,又a|c,b|c ,c是a,b之 09/18 11:38
→ ddczx:公倍數為最小公倍數ab之倍數,故ab|c 09/18 11:39
推 a00000816:令gcd(m,n)=k,k|m&k|n,令m=p*k,n=q*k(p,q∈Z),欲證k|n-m 09/18 22:43
→ a00000816:->k|(q*k-p*k)->k(q-p),故k|n-m 09/18 22:44
推 a00000816:阿沒看到樓上已經解了XD 09/18 22:58
推 a1098137129:可是第一題假如說代數字進去 a=6 b=12 n=6 09/27 01:58
→ a1098137129:這樣答案是F吧? 還是我題目有看漏的? 09/27 01:58