Research Space for Linear Algebra & Discrete Mathematics
跟老師上課講的例題完全一樣滴題目有說是Boolean algebra, 因此一定lub和glb以及complement這些東西一定是well-defined老師上課時只是轉成我們習慣的代數系統符號這種tuple的符號定義不同的書常常會有差異
但老師當年出的解題書的解答跟上課的不一樣！我真的可以用上課的例題的解法嗎？
沒問題的, 並且我覺得那樣解還比當年書上的寫法來得更直覺因為題目就是請你證明a的補元素的補元素為a但要注意的是, 老師上課時只證一行是因為他先說明a為bar(a)之complement然後再說明因為complement具唯一性, 所以bar(bar(a)) = a所以筆記裡寫的"唯一性"那一段一定要證出來(應該就是你上課筆記裡, 記在這個Example上面的那個定理)只要有證出來就一定沒問題
呃...是假設b,c都是a的complement那個嗎？a join b=1、a meet b =0a join c=1、a meet c =0b = b meet (a join c) =(b meet a) join (b meet c) = 0 join (b meet c) = (a meet c) join (b meet c) =(a join b) meet c = c
嗯嗯沒錯
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跟老師上課講的例題完全一樣滴
題目有說是Boolean algebra,
因此一定lub和glb以及complement這些東西一定是well-defined
老師上課時只是轉成我們習慣的代數系統符號
這種tuple的符號定義不同的書常常會有差異
但老師當年出的解題書的解答跟上課的不一樣！我真的可以用上課的例題的解法嗎？
沒問題的, 並且我覺得那樣解還比當年書上的寫法來得更直覺
因為題目就是請你證明a的補元素的補元素為a
但要注意的是, 老師上課時只證一行
是因為他先說明a為bar(a)之complement
然後再說明因為complement具唯一性, 所以bar(bar(a)) = a
所以筆記裡寫的"唯一性"那一段一定要證出來
(應該就是你上課筆記裡, 記在這個Example上面的那個定理)
只要有證出來就一定沒問題
呃...是假設b,c都是a的complement那個嗎？
a join b=1、a meet b =0
a join c=1、a meet c =0
b = b meet (a join c)
=(b meet a) join (b meet c)
= 0 join (b meet c)
= (a meet c) join (b meet c)
=(a join b) meet c
= c
嗯嗯沒錯
A tech writer who is passionate to write about streaming devices and its channel programs. Apart from my work, I am interested in music and sketching. Are you a person who needs updates about the new releases and is interested in tech blogs?
Have a look at my recent blogs:
Roku com link activate code>
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How to Activate Roku?>
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Roku activation Link >
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roku.com >
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roku.com link >
roku link >
roku setup >
www.roku.com >
my.roku.com >
www.roku.com/support >
my roku account >
張貼留言