一言筆發
:^(
卡利古拉
又黎料啦
等學野
劉明
當integer係group,division咪唔會close,會走左去rational number
你依句其實唔太make sense
所謂嘅除法係整個inverse俾個乘法，用你個例子，如果淨係睇integer，你唔會搵到一個integer令到2x嗰個integer=1
但唔代表咁樣就定義唔到除法人類就係construct rational number嚟解決依個問題 :^(
remark: 用啲數學terms,construct rational number嘅方法就係localization of Z/construct field of fraction of Z,再fancy啲可以睇做consider grothendieck group of (Z*,multiplication)
:^(
Michaelzaki
Contrust field之後就睇唔明
利伸 睇緊elementary real analysis 所以明construct field是乜
有得乘未必有得除#hehe :^(
又黎料啦
等學野
劉明
當integer係group,division咪唔會close,會走左去rational number
你依句其實唔太make sense
所謂嘅除法係整個inverse俾個乘法，用你個例子，如果淨係睇integer，你唔會搵到一個integer令到2x嗰個integer=1
但唔代表咁樣就定義唔到除法人類就係construct rational number嚟解決依個問題 :^(
remark: 用啲數學terms,construct rational number嘅方法就係localization of Z/construct field of fraction of Z,再fancy啲可以睇做consider grothendieck group of (Z*,multiplication)
Contrust field之後就睇唔明
:^(
利伸 睇緊elementary real analysis 所以明construct field是乜
我沒有放棄
有得乘未必有得除#hehe :^(
又黎料啦
等學野
劉明
當integer係group,division咪唔會close,會走左去rational number
你依句其實唔太make sense
所謂嘅除法係整個inverse俾個乘法，用你個例子，如果淨係睇integer，你唔會搵到一個integer令到2x嗰個integer=1
但唔代表咁樣就定義唔到除法人類就係construct rational number嚟解決依個問題 :^(
remark: 用啲數學terms,construct rational number嘅方法就係localization of Z/construct field of fraction of Z,再fancy啲可以睇做consider grothendieck group of (Z*,multiplication)
Contrust field之後就睇唔明
利伸 睇緊elementary real analysis 所以明construct field是乜
:^(
:^(
:^(
雙失廢柴
我又拋下磚先
其實根據定義 0^a for all real a 係咪都係undefined?
因為x^y:=exp(y ln(x))
就算for a>=0, 因為ln(0)=negative infinity
所以0^a都係undefined?
不過係咪數佬為左方便起見 先define 0^0=1, 0^a=0 for a>0?
其實根據定義 0^a for all real a 係咪都係undefined?
因為x^y:=exp(y ln(x))
就算for a>=0, 因為ln(0)=negative infinity
所以0^a都係undefined?
不過係咪數佬為左方便起見 先define 0^0=1, 0^a=0 for a>0?
雙失廢柴
咁即係要argue continuity?
我又拋下磚先
其實根據定義 0^a for all real a 係咪都係undefined?
因為x^y:=exp(y ln(x))
就算for a>=0, 因為ln(0)=negative infinity
所以0^a都係undefined?
不過係咪數佬為左方便起見 先define 0^0=1, 0^a=0 for a>0?
但係lim->0 e^(xlnx)=1
咁即係要argue continuity?
:^(
雙失廢柴
但係可以就咁塞個not real number落去?
Take limit嘅話又好似答唔到個問題咁
我又拋下磚先
其實根據定義 0^a for all real a 係咪都係undefined?
因為x^y:=exp(y ln(x))
就算for a>=0, 因為ln(0)=negative infinity
所以0^a都係undefined?
不過係咪數佬為左方便起見 先define 0^0=1, 0^a=0 for a>0?
但係lim->0 e^(xlnx)=1
咁即係要argue continuity?
你可以理解成exp(-infinity)=0
實際上一般好少會理0^a as a function in a
就算睇x^y,一般都係fix x或者fix y再討論
但係可以就咁塞個not real number落去?
Take limit嘅話又好似答唔到個問題咁
Michaelzaki
又黎料啦
等學野
劉明
當integer係group,division咪唔會close,會走左去rational number
你依句其實唔太make sense
所謂嘅除法係整個inverse俾個乘法，用你個例子，如果淨係睇integer，你唔會搵到一個integer令到2x嗰個integer=1
但唔代表咁樣就定義唔到除法人類就係construct rational number嚟解決依個問題 :^(
remark: 用啲數學terms,construct rational number嘅方法就係localization of Z/construct field of fraction of Z,再fancy啲可以睇做consider grothendieck group of (Z*,multiplication)
Contrust field之後就睇唔明
利伸 睇緊elementary real analysis 所以明construct field是乜
呢啲abstract algebra嚟 :^(
:^(
Michaelzaki
問下先 我見本real analysis書有construct fields of real number from rational numbers
咁construct fields of rational numbers from integers唔係都應該係同一個idea
咁construct fields of rational numbers from integers唔係都應該係同一個idea
:^(
Michaelzaki
I mean 方法可能唔同 但basic concept都係一樣？
問下先 我見本real analysis書有construct fields of real number from rational numbers
咁construct fields of rational numbers from integers唔係都應該係同一個idea? :^(
I mean 方法可能唔同 但basic concept都係一樣？
Adam_Smith
劉明
當integer係group,division咪唔會close,會走左去rational number
你依句其實唔太make sense
所謂嘅除法係整個inverse俾個乘法，用你個例子，如果淨係睇integer，你唔會搵到一個integer令到2x嗰個integer=1
但唔代表咁樣就定義唔到除法人類就係construct rational number嚟解決依個問題 :^(
remark: 用啲數學terms,construct rational number嘅方法就係localization of Z/construct field of fraction of Z,再fancy啲可以睇做consider grothendieck group of (Z*,multiplication)
Contrust field之後就睇唔明
利伸 睇緊elementary real analysis 所以明construct field是乜
呢啲abstract algebra嚟 :^(
:^(
Michaelzaki
問下先 我見本real analysis書有construct fields of real number from rational numbers
咁construct fields of rational numbers from integers唔係都應該係同一個idea? :^(
Dedekind cut?
Real number係靠整個partition符合well ordering principle,archimedean property.
btw CS construct integer個方法都幾得意,{},{{}},{{{}}}...
:^(