Math question (maybe our resident rocket scientist can help)

[elwoodblues]

This has always bothered me. I think it is just trying to wrap your head around infinity, but any help from anyone would be appreciated.

Does .999999(repeating) really = 1? I think it's just really, really (repeating) close, but the below suggests otherwise:

1/9 = .11111(repeating)
1/9 x 9 = 1
.11111(repeating) x 9 = .99999(repeating)
Therefore .99999(repeating) = 1

[ThaSaltCracka]

I think a really anal math teacher will tell you .9999999 does not actually equal one.

[Tron]

They are just different ways of writing the same thing.

[GuyOnTilt]

Any middle school level math text would be able to tell you this. I'm sure you've read it in reliable sources before, so why are you still asking? The answer is still yes.

GoT

[Mano]

That is correct, there is no difference between .9999... and 1 ( i.e. 1 - .9999... = 0).

[elwoodblues]

[ QUOTE ]
I'm sure you've read it in reliable sources before, so why are you still asking? The answer is still yes.


[/ QUOTE ]

I "know" the answer is yes because people have told me and I can "prove." I just have a hard time wrapping my head around it. My mind tells me that they aren't equal and that they have a difference of .0000(repeating)1.

[TimM]

[ QUOTE ]
I "know" the answer is yes because people have told me and I can "prove." I just have a hard time wrapping my head around it. My mind tells me that they aren't equal and that they have a difference of .0000(repeating)1.

[/ QUOTE ]

The repeating zeros are infinite in number, so you never get the 1.

[junkmail3]

[ QUOTE ]
[ QUOTE ]
I "know" the answer is yes because people have told me and I can "prove." I just have a hard time wrapping my head around it. My mind tells me that they aren't equal and that they have a difference of .0000(repeating)1.

[/ QUOTE ]

The repeating zeros are infinite in number, so you never get the 1.

[/ QUOTE ]

This is correct. Otherwise, what would be the point of haveing a number that is .9999999999999999... and a number that is 1.

if they are the same thing, call them the same thing. You will approach 1 (and for many applications can call it one) but it is not turely 1. Ever.

[anisotropy]

[ QUOTE ]
1/9 = .11111(repeating)

[/ QUOTE ]
incorrect. The decimal form is an approximation of an irregular number.
[ QUOTE ]
1/9 x 9 = 1

[/ QUOTE ]
correct
[ QUOTE ]
.11111(repeating) x 9 = .99999(repeating)

[/ QUOTE ]
correct
[ QUOTE ]
Therefore .99999(repeating) = 1

[/ QUOTE ]
incorrect

[BeerMoney]

Let
n=.9999999999999999...
10n=9.9999999999999999.....
10n-n = 9
9n = 9
n = 1

It is wierd. I guess our number system is messed up.

[Patrick del Poker Grande]

Let's not forget this little gem:

a = x
a+a = a+x
2a = a+x
2a-2x = a+x-2x
2(a-x) = a+x-2x
2(a-x) = a-x
2 = 1

[Sporky]

i believe your logic is flawed.

[DMBFan23]

what we're actually saying is that 1 - [1 - {1/(10*n)}] tends to 0 as n tends to infinity.

1 - 0.9
1 - 0.99
1 - 0.999


...


1- 0.9999999999999

etc.

so since A minus the limit of B equals 0, the limit of B = A. but limit of B = 0.99999999999999......

[BeerMoney]

[ QUOTE ]
i believe your logic is flawed.

[/ QUOTE ]
There's nothing wrong with it at all.. ITs the way you convert a repeating decimal to fraction..

eg..

n = .27272727......
100n =27.2727272727

100n -n = 27
99n = 27
n=27/99

27/99 = .2727272727272727

Or, how about an infinite geometric sum..

.999999999999= .9+.09+.009 +.0009 etc..
= sum .9*.1^n let n -> infinity= sum a*r^n where a = .9, r = .1.. = a/(1-r) = .9/(1-.1) = .9/.9 = 1

[Patrick del Poker Grande]

This thread blows. Next.

[pudley4]

[ QUOTE ]
Let's not forget this little gem:

a = x
a+a = a+x
2a = a+x
2a-2x = a+x-2x
2(a-x) = a+x-2x
2(a-x) = a-x
2 = 1

[/ QUOTE ]

Can't divide by zero (a-x = 0)

Sheesh, they don't make rocket scientists like they used to...



/images/graemlins/wink.gif