我正式宣佈 0.999999........=1 不服來辯

Andrew_D 2017-3-7 23:03:37

我反對

0.9999999...x10 咁少左一個小數位囉點減到齊頭
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你咁即係話0.999 x10 =9.999 邊忽合理
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...係無限個小數位

無限同無限減一個冇分別？
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首先你要搞清楚咩叫無限

所以我講黎係想提你你o既講法係有assumption

有討論空間唔好以為一定係真理
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無限唔係一個數值
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你改變唔到無限㗎喎
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你肯定你對無限的定義絕對正確？

淨係嘈無限的定義已經可以嘈幾個世紀

我甚至可以質疑人作為有限的生物係唔會理解到咩叫無限

巴打涉獵少少哲學會明我講咩

你睇完幾次數學危機，明白左呢堆真理先講啦
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讀數D人真係好固執

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-🐣- 2017-3-7 23:04:05 156789=4564032456
不服來辯

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米奇老尾 2017-3-7 23:07:32

No doubt 嘅野d人都可以討論咁耐
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睇倒呢句笑左

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一舊膠買起你 2017-3-7 23:07:44

0.999999999999=/=1
但係0.999999999999.........=1
話1-0.99999...=0.000....1個啲不如去讀多次小學
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佢所謂既0.000....1其實係0.1,0.01,0.001...既limit

宵闇の舞 2017-3-7 23:07:45

lim 1 - x = 1
x→0

上面個 = 1 係指無限接近 1 而唔係等於 1 OK？

利申：未學過微分。

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錯concept
當x好接近0而唔等於0時，1-x係好接近1而且唔等於1，但係個limit係等於1而唔係無限接近1

學到嘢，當初自學過下Limit但係學完都唔知做緊咩…

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一舊膠買起你 2017-3-7 23:09:05

Z_p呢
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:^(

:^(

應該係...?
已經還返晒比Professor

柒到盡頭便是型 2017-3-7 23:11:51

[quote]0.999999999999=/=1
但係0.999999999999.........=1
話1-0.99999...=0.000....1個啲不如去讀多次小學
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佢所謂既0.000....1其實係0.1,0.01,0.001...既limit[/quote]
啲人係以為無限個0之後有個1係度

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同埋呢個係咩limit黎

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一舊膠買起你 2017-3-7 23:13:39

[quote]0.999999999999=/=1
但係0.999999999999.........=1
話1-0.99999...=0.000....1個啲不如去讀多次小學
:^(

佢所謂既0.000....1其實係0.1,0.01,0.001...既limit[/quote]
啲人係以為無限個0之後有個1係度
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同埋呢個係咩limit黎
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咪係個sequence既limit
0.1,0.01,0.001...
其實即係0

Andrew_D 2017-3-7 23:17:04

Z_p呢
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:^(

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應該係...?
已經還返晒比Professor

無記錯p-adic入面就唔岩，但係euclidean norm就會岩呢樣野
心諗你地係到講一大論做乜野

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廉政專員 2017-3-7 23:17:10

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米奇老尾 2017-3-7 23:20:42

廢事有盲撚唔識google

If the common ratio r lies between −1 and 1 , we can have the sum of an infinite geometric series.

The sum S of an infinite geometric series with −1<r<1 is given by the formula,

S=a/(1−r), where a=first term, r=common ratio

when a=9/10, r=1/10
0.999999……
=9/10+9/100+9/1000+……+9/10^n (where n=infinity)
=(9/10)/(1-1/10)
=1

https://www.varsitytutors.com/hotmath/hotmath_help/topics/infinite-geometric-series

唔撚識英文呀

咁就去食屎啦

一舊膠買起你 2017-3-7 23:23:31

Z_p呢
:^(

:^(

:^(

:^(

:^(
應該係...?
已經還返晒比Professor

無記錯p-adic入面就唔岩，但係euclidean norm就會岩呢樣野
心諗你地係到講一大論做乜野
:^(

唉
連本NOTES都唔知擺左去邊

:^(

香城獨立 2017-3-7 23:25:50

廢事有盲撚唔識google

If the common ratio r lies between −1 and 1 , we can have the sum of an infinite geometric series.

The sum S of an infinite geometric series with −1<r<1 is given by the formula,

S=a/(1−r), where a=first term, r=common ratio

when a=9/10, r=1/10
0.999999……
=9/10+9/100+9/1000+……+9/10^n (where n=infinity)
=(9/10)/(1-1/10)
=1

https://www.varsitytutors.com/hotmath/hotmath_help/topics/infinite-geometric-series

唔撚識英文呀

咁就去食屎啦

仲爭執緊

:^(

其實唔係讀數嘅人真係唔洗咁執著，想要了解哂成件事就不如自己好好wiki下，睇唔明先上黎，仲會學得更多

:^(

Andrew_D 2017-3-7 23:26:33

廢事有盲撚唔識google

If the common ratio r lies between −1 and 1 , we can have the sum of an infinite geometric series.

The sum S of an infinite geometric series with −1<r<1 is given by the formula,

S=a/(1−r), where a=first term, r=common ratio

when a=9/10, r=1/10
0.999999……
=9/10+9/100+9/1000+……+9/10^n (where n=infinity)
=(9/10)/(1-1/10)
=1

https://www.varsitytutors.com/hotmath/hotmath_help/topics/infinite-geometric-series

唔撚識英文呀

咁就去食屎啦

仲爭執緊
:^(

其實唔係讀數嘅人真係唔洗咁執著，想要了解哂成件事就不如自己好好wiki下，睇唔明先上黎，仲會學得更多
:^(

我都唔明數呢家野已經好好討論，岩就岩錯就錯....
做咩要咁勞氣

:^(

香城獨立 2017-3-7 23:31:39

廢事有盲撚唔識google

If the common ratio r lies between −1 and 1 , we can have the sum of an infinite geometric series.

The sum S of an infinite geometric series with −1<r<1 is given by the formula,

S=a/(1−r), where a=first term, r=common ratio

when a=9/10, r=1/10
0.999999……
=9/10+9/100+9/1000+……+9/10^n (where n=infinity)
=(9/10)/(1-1/10)
=1

https://www.varsitytutors.com/hotmath/hotmath_help/topics/infinite-geometric-series

唔撚識英文呀

咁就去食屎啦

仲爭執緊
:^(

其實唔係讀數嘅人真係唔洗咁執著，想要了解哂成件事就不如自己好好wiki下，睇唔明先上黎，仲會學得更多
:^(

我都唔明數呢家野已經好好討論，岩就岩錯就錯....
做咩要咁勞氣
:^(

:^(

:^(

:^(

:^(

咁係幾難明嘅，你諗下解釋 0-1入面嘅 real number 數目同0-100入面有嘅 real number 數目一樣多俾普通人聽，佢聽完就算明，內心都會抗拒一陣。

而0.9999....呢個問題，比較嚴謹嘅論證我係喺 real analysis 入面學嘅

:^(

我諗唔係讀數嘅人如果只係想靠人地回帖就搞清楚，都唔係一件易事

:^(

狐狸叔叔 2017-3-7 23:40:35

Z_p呢
:^(

:^(

p-adic？

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含吹瀬啜硬膠仔 2017-3-7 23:42:44

廢事有盲撚唔識google

If the common ratio r lies between −1 and 1 , we can have the sum of an infinite geometric series.

The sum S of an infinite geometric series with −1<r<1 is given by the formula,

S=a/(1−r), where a=first term, r=common ratio

when a=9/10, r=1/10
0.999999……
=9/10+9/100+9/1000+……+9/10^n (where n=infinity)
=(9/10)/(1-1/10)
=1

https://www.varsitytutors.com/hotmath/hotmath_help/topics/infinite-geometric-series

唔撚識英文呀

咁就去食屎啦

仲爭執緊
:^(

其實唔係讀數嘅人真係唔洗咁執著，想要了解哂成件事就不如自己好好wiki下，睇唔明先上黎，仲會學得更多
:^(

我都唔明數呢家野已經好好討論，岩就岩錯就錯....
做咩要咁勞氣
:^(

:^(

:^(

:^(

:^(

咁係幾難明嘅，你諗下解釋 0-1入面嘅 real number 數目同0-100入面有嘅 real number 數目一樣多俾普通人聽，佢聽完就算明，內心都會抗拒一陣。

而0.9999....呢個問題，比較嚴謹嘅論證我係喺 real analysis 入面學嘅
:^(
我諗唔係讀數嘅人如果只係想靠人地回帖就搞清楚，都唔係一件易事
:^(

Cardinality 啊嘛中學都有教啦

e^ipi+1=0 2017-3-7 23:46:18 呢個po大概有兩類人
一類係屌樓主up廢話既
另一類就係中學數都未讀過既
不服來辯

濕鳩塞鷗 2017-3-7 23:46:30

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Mr.Lawrence 2017-3-7 23:49:14 假設0.99999999999999.......=1 正確
設x=0.9999999......=1
x-x=0 , 1-0.999......=0.00...01
0=0.0000...01
設k為一常數
做0.000...01*k=1
0*k=0.00.01*k
0=1
0=1+1=2
0=0+0+0=3
1=2=3=............

一蚊都無都可以好滿足就係咁解

一舊膠買起你 2017-3-7 23:50:25

假設0.99999999999999.......=1 正確
設x=0.9999999......=1
x-x=0 , 1-0.999......=0.00...01
0=0.0000...01
設k為一常數
做0.000...01*k=1
0*k=0.00.01*k
0=1
0=1+1=2
0=0+0+0=3
1=2=3=............

一蚊都無都可以好滿足就係咁解

點解呢個k存在?

e^ipi+1=0 2017-3-7 23:52:38

假設0.99999999999999.......=1 正確
設x=0.9999999......=1
x-x=0 , 1-0.999......=0.00...01
0=0.0000...01
設k為一常數
做0.000...01*k=1
0*k=0.00.01*k
0=1
0=1+1=2
0=0+0+0=3
1=2=3=............

一蚊都無都可以好滿足就係咁解

仆街

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睇到中間個prove笑死

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