Re: [理工] [離散]-證明等價關係

看板Grad-ProbAsk作者 (sodas)時間16年前 (2010/03/31 01:16), 編輯推噓6(602)
留言8則, 3人參與, 最新討論串2/3 (看更多)
※ 引述《fj90406 (阿亮)》之銘言: : Let A = Z+ * Z+ a relation R on A as follows: : (x,y)R(u,v) if and only if x*v=y*u. : (1) Prove that R is an equivalent relation. For All x,y ∈Z+, x*y=y*x --> (x,y)R(x,y) --> Reflexive For All x,y,z,w ∈Z+, if x*w=y*z, then z*y=w*x --> ∀(x,y)R(z,w) ∃(z,w)R(x,y) --> Symmetric For All x,y,z,w,a,b∈Z+, if ∃(x,y)R(z,w)^(z,w)R(a,b) --> x*w=y*z ^ z*b=w*a ,,,,, No Transitive So, this is not an equivalent relation. : (2) Computer the equivalence class of (5,4) Not exists. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.39.16.163

03/31 05:18, , 1F
transitive應該成立吧...x*w*z*b=y*z*w*a->(x,y)R(a,b)
03/31 05:18, 1F

03/31 05:21, , 2F
ER(5,4)={(x,y)|4x-5y=0}
03/31 05:21, 2F

03/31 06:44, , 3F
哦 sorry 樓上對 昨天想睡覺了@@
03/31 06:44, 3F

03/31 08:27, , 4F
問sswy7121大大 第2題那個通式 會滿足反身性嗎?
03/31 08:27, 4F

03/31 08:27, , 5F
找不到(x,y) 然後x=y的
03/31 08:27, 5F

03/31 08:33, , 6F
順便感謝sodas2002大大 想睡覺還打那多回答我 給個讚@@
03/31 08:33, 6F

03/31 19:21, , 7F
他的element的定義應該是(x,y)...你可能有點誤解囉...
03/31 19:21, 7F

03/31 19:39, , 8F
所以(5,4)R(5,4)具有反身性...
03/31 19:39, 8F
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文章代碼(AID): #1BiZ7i8O (Grad-ProbAsk)