True Or False ?

[kakatiya]

true or false ?

 

1 > 0.999999

 

 

[ramudu3]

false

[ramudu3]

[kakatiya]

show some other proof 

[joy]

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]

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]

 

 

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]

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

 

[joy]

[Roger_that]

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]

idhi okateee smily .. eee post lo chusinaa eee smiley vesi janalani sava 10geevaru bhayaaa.... but sudden change????

@3$%

 

[Roger_that]

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]

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????

 

[Roger_that]

posani != posani_

@3$%

[ramudu3]

is it  :3D_Smiles: .. ohh my bala 

posani != posani_