Math question (maybe our resident rocket scientist can help)
 

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

Re: Math question (maybe our resident rocket scientist can help)
 

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

Re: Math question (maybe our resident rocket scientist can help)
 

They are just different ways of writing the same thing.

Re: Math question (maybe our resident rocket scientist can help)
 

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

Re: Math question (maybe our resident rocket scientist can help)
 

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

Re: Math question (maybe our resident rocket scientist can help)
 

[ 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.

Re: Math question (maybe our resident rocket scientist can help)
 

[ 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.

Re: Math question (maybe our resident rocket scientist can help)
 

[ 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.

Re: Math question (maybe our resident rocket scientist can help)
 

[ 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

Re: Math question (maybe our resident rocket scientist can help)
 

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

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

Re: Math question (maybe our resident rocket scientist can help)
 

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

Re: Math question (maybe our resident rocket scientist can help)
 

i believe your logic is flawed.

Re: Math question (maybe our resident rocket scientist can help)
 

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......

Re: Math question (maybe our resident rocket scientist can help)
 

[ 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

Re: Math question (maybe our resident rocket scientist can help)
 

This thread blows. Next.

Re: Math question (maybe our resident rocket scientist can help)
 

[ 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...



[img]/images/graemlins/wink.gif[/img]