joy Posted February 14, 2016 Report Posted February 14, 2016 Proof by geometric series The number "0.9999..." can be "expanded" as: 0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ... In other words, each term in this endless summation will have a "9" preceded by some number of zeroes. This may also be written as: 0.999... = 9/10 + (9/10)(1/10)^1 + (9/10)(1/10)^2 + (9/10)(1/10)^3 + ... That is, this is an infinite geometric series with first term a = 9/10 and common ratio r = 1/10. Since the size of the common ratio r is less than 1, we can use the infinite-sum formula to find the value: 0.999... = (9/10)[1/(1 - 1/10)] = (9/10)(10/9) = 1 So the formula proves that 0.9999... = 1. Note: Technically, the above proof requires that some fairly advanced concepts be taken on faith. If you study "foundations" or mathematical philosophy (way after calculus), you may encounter the requisite theoretical constructs. Copyright © Elizabeth Stapel 2006-2011 All Rights Reserved Other pre-calculus arguments Argument from precedence: If you haven't already learned that 1/3 = 0.333... in decimal form, you can prove this easily by doing the long division: long division of 1 by 3, showing quotient of 0.3333.... ...and so forth, ad infinitum. So 1/3 + 1/3 + 1/3 = 3( 1/3 ) = 1. Reasonably then, 0.333... + 0.333... + 0.333... = 3(0.333...) should also equal 1. But 3(0.333...) = 0.999.... Then 0.999... must equal 1.
Silver_mani Posted February 14, 2016 Report Posted February 14, 2016 Proof by geometric series The number "0.9999..." can be "expanded" as: 0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ... In other words, each term in this endless summation will have a "9" preceded by some number of zeroes. This may also be written as: 0.999... = 9/10 + (9/10)(1/10)^1 + (9/10)(1/10)^2 + (9/10)(1/10)^3 + ... That is, this is an infinite geometric series with first term a = 9/10 and common ratio r = 1/10. Since the size of the common ratio r is less than 1, we can use the infinite-sum formula to find the value: 0.999... = (9/10)[1/(1 - 1/10)] = (9/10)(10/9) = 1 So the formula proves that 0.9999... = 1. Note: Technically, the above proof requires that some fairly advanced concepts be taken on faith. If you study "foundations" or mathematical philosophy (way after calculus), you may encounter the requisite theoretical constructs. Copyright © Elizabeth Stapel 2006-2011 All Rights Reserved Other pre-calculus arguments Argument from precedence: If you haven't already learned that 1/3 = 0.333... in decimal form, you can prove this easily by doing the long division: long division of 1 by 3, showing quotient of 0.3333.... ...and so forth, ad infinitum. So 1/3 + 1/3 + 1/3 = 3( 1/3 ) = 1. Reasonably then, 0.333... + 0.333... + 0.333... = 3(0.333...) should also equal 1. But 3(0.333...) = 0.999.... Then 0.999... must equal 1.
Roger_that Posted February 14, 2016 Report Posted February 14, 2016 Proof by geometric series The number "0.9999..." can be "expanded" as: 0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ... In other words, each term in this endless summation will have a "9" preceded by some number of zeroes. This may also be written as: 0.999... = 9/10 + (9/10)(1/10)^1 + (9/10)(1/10)^2 + (9/10)(1/10)^3 + ... That is, this is an infinite geometric series with first term a = 9/10 and common ratio r = 1/10. Since the size of the common ratio r is less than 1, we can use the infinite-sum formula to find the value: 0.999... = (9/10)[1/(1 - 1/10)] = (9/10)(10/9) = 1 So the formula proves that 0.9999... = 1. Note: Technically, the above proof requires that some fairly advanced concepts be taken on faith. If you study "foundations" or mathematical philosophy (way after calculus), you may encounter the requisite theoretical constructs. Copyright © Elizabeth Stapel 2006-2011 All Rights Reserved Other pre-calculus arguments Argument from precedence: If you haven't already learned that 1/3 = 0.333... in decimal form, you can prove this easily by doing the long division: long division of 1 by 3, showing quotient of 0.3333.... ...and so forth, ad infinitum. So 1/3 + 1/3 + 1/3 = 3( 1/3 ) = 1. Reasonably then, 0.333... + 0.333... + 0.333... = 3(0.333...) should also equal 1. But 3(0.333...) = 0.999.... Then 0.999... must equal 1. orei google search nuvvu lekapothe emi ayipoyevallo ee kalam kurrallu
ramudu3 Posted February 14, 2016 Report Posted February 14, 2016 bhayaa oka doubt ... apatlo meeru full telangana/KCR/TRS anti post lu vese varu gaaaa ... ippudu suddengaaa CBN ki anti post lu why???? orei google search nuvvu lekapothe emi ayipoyevallo ee kalam kurrallu
Roger_that Posted February 14, 2016 Report Posted February 14, 2016 bhayaa oka doubt ... apatlo meeru full telangana/KCR/TRS anti post lu vese varu gaaaa ... ippudu suddengaaa CBN ki anti post lu why???? @3$%
ramudu3 Posted February 14, 2016 Report Posted February 14, 2016 idhi okateee smily .. eee post lo chusinaa eee smiley vesi janalani sava 10geevaru bhayaaa.... but sudden change???? @3$%
Roger_that Posted February 14, 2016 Report Posted February 14, 2016 idhi okateee smily .. eee post lo chusinaa eee smiley vesi janalani sava 10geevaru bhayaaa.... but sudden change???? KCR chese manchi panulaki nenu KCR fan ni ayipoya ba.. ade nakka ni chudu lol nakka... waste product @3$%
kakatiya Posted February 14, 2016 Author Report Posted February 14, 2016 posani != posani_ bhayaa oka doubt ... apatlo meeru full telangana/KCR/TRS anti post lu vese varu gaaaa ... ippudu suddengaaa CBN ki anti post lu why????
ramudu3 Posted February 14, 2016 Report Posted February 14, 2016 is it :3D_Smiles: .. ohh my bala posani != posani_
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