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.999~ = or != 1
Posted: 2003-11-29 04:32pm
by AnimeJet
So does .9 infinitely equal one or not? Discuss. Unless this has already been discussed.. then tell me what the general concensus is on these boards ._.
EDIT: Now 100% more poll!
![Very Happy :D](./images/smilies/icon_biggrin.gif)
Posted: 2003-11-29 05:03pm
by HemlockGrey
...how can it possibly equal one?
Posted: 2003-11-29 05:23pm
by SyntaxVorlon
.9~/3 = .3~
.3~= 1/3
1/3*3=1
.9~=1
![Laughing :lol:](./images/smilies/icon_lol.gif)
Posted: 2003-11-29 05:25pm
by Mad
.111~ = 1/9
.222~ = 2/9
.333~ = 3/9
...
.888~ = 8/9
All of the above are exact and can easily be proven by doing the division 1/9, 2/9, etc.
Therefore: .999~ = 9/9 = 1
Posted: 2003-11-29 05:38pm
by Crazedwraith
Yes o.9 reaccuring does indeed equal one.
.9~ = x
.99999999999=x
9.9999999999= 10x
9x = 9
x=1
that make it clear?
Re: .999~ = or != 1
Posted: 2003-11-29 05:43pm
by aerius
AnimeJet wrote:So does .9 infinitely equal one or not? Discuss. Unless this has already been discussed.. then tell me what the general concensus is on these boards ._.
It equals 7, but I haven't taken any math courses in a couple years so I could be wrong.
Posted: 2003-11-29 05:59pm
by Darth Wong
It equals 1 when it is taken to a reasonable number of significant figures.
Re: .999~ = or != 1
Posted: 2003-11-29 06:09pm
by Kuroneko
AnimeJet wrote:So does .9 infinitely equal one or not? Discuss. Unless this has already been discussed.. then tell me what the general concensus is on these boards ._.
Of course it does. The decimal representation of a number is simply a series in a particular form: 0.999... = 9/10 + 9/100 + 9/1000 + ... in the exact same way that 295.1 = 2*100 + 9*10 + 5 + 1/10 + 0/100 + ... . This particular series has a sequence partial sums of (.9, .99, .999, ...) that is monotonically increasing and has an upper bound, so it must converge to the lowest upper bound of (.9, .99, .999, ...), which is 1. Therefore, 0.999... = 1.
Posted: 2003-11-29 06:46pm
by HemlockGrey
Could someone explain this slightly better? I mean, won't it always be just short of 1?
Posted: 2003-11-29 06:52pm
by Kuroneko
HemlockGrey wrote:Could someone explain this slightly better? I mean, won't it always be just short of 1?
No. It's a number--it doesn't change. It isn't any of (0.9, 0.99, 0.999, ...); it is the lowest real number greater than or equal to every number found in that sequence. That number is 1.
Posted: 2003-11-29 06:55pm
by CrimsonRaine
Kuroneko wrote:HemlockGrey wrote:Could someone explain this slightly better? I mean, won't it always be just short of 1?
No. It's a number--it doesn't change. It isn't any of (0.9, 0.99, 0.999, ...); it is the lowest real number greater than or equal to every number found in that sequence. That number is 1.
Oh, Christ, I'm still lost. Even though I can see it can be rounded to 1, since it is so close to that number, it still doesn't equal one. Does it?
I'm an English major. I know words, not numbers. Heeeelp!
Crimson Raine
Posted: 2003-11-29 07:03pm
by Mad
HemlockGrey wrote:Could someone explain this slightly better? I mean, won't it always be just short of 1?
For any easy explanation, SyntaxVorlon's is probably the easiest to understand:
1/3 * 3 = 3/3 = 1
.333~ * 3 = .999~
1/3 = .333~, therefore 1 = .999~
As for details, Kureneko's explanation is correct. It follows the same concept that is used very often in calculus: if you have an infinite number of terms to add, with each term smaller than the previous term, then they will all add up to a specific number.
In this case, we're adding up .9 + .09 + .009 + .0009 + ... . If this stretches out to infinity, they will add up to be 1. (The same way the area under a curve can be determined by slicing the problem up into an infinite number of columns.)
You could also look at it by taking the sum of an infinite series, as that'd work, as well.
I guess the easiest way would require a picture of the infinite series being added together, along with the equation and limits and everything.
Posted: 2003-11-29 07:06pm
by Captain Cyran
CrimsonRaine wrote:Kuroneko wrote:HemlockGrey wrote:Could someone explain this slightly better? I mean, won't it always be just short of 1?
No. It's a number--it doesn't change. It isn't any of (0.9, 0.99, 0.999, ...); it is the lowest real number greater than or equal to every number found in that sequence. That number is 1.
Oh, Christ, I'm still lost. Even though I can see it can be rounded to 1, since it is so close to that number, it still doesn't equal one. Does it?
I'm an English major. I know words, not numbers. Heeeelp!
Crimson Raine
Just do what I do. Nod your head, smile, and file it away in that useless information part of your brain for when you want to impress people.
Posted: 2003-11-29 07:06pm
by Darth Wong
Technically, we would say that it is a term which approaches 1 as its terms approach infinity.
Posted: 2003-11-29 07:10pm
by Kuroneko
CrimsonRaine wrote:Oh, Christ, I'm still lost. Even though I can see it can be rounded to 1, since it is so close to that number, it still doesn't equal one. Does it?
Why wouldn't it? Let's call it X = 0.999... .
Pick any number Y between 0 and 1. There is always some number in (0.9, 0.99, 0.999, ...) that's greater than it. For example, suppose I pick Y = 0.931. Then, 0.99 > Y. If I pick Y = 0.9934, then 0.999 > Y. Et cetera.
If X < 1, then there must be a number between them (for example, Y = (1+X)/2). But then by the above demonstration, there is some terminating 0.999...99 > Y. But that cannot be, as X should be greater than any of (0.9, 0.99, 0.999, ...)!
The above doesn't actually strictly
prove the result mathematically, but it should work as demonstration.
Posted: 2003-11-29 07:13pm
by Captain Cyran
Kuroneko wrote:CrimsonRaine wrote:Oh, Christ, I'm still lost. Even though I can see it can be rounded to 1, since it is so close to that number, it still doesn't equal one. Does it?
Why wouldn't it? Let's call it X = 0.999... .
Pick any number Y between 0 and 1. There is always some number in (0.9, 0.99, 0.999, ...) that's greater than it. For example, suppose I pick Y = 0.931. Then, 0.99 > Y. If I pick Y = 0.9934, then 0.999 > Y. Et cetera.
If X < 1, then there must be a number between them (for example, Y = (1+X)/2). But then by the above demonstration, there is some terminating 0.999...99 > Y. But that cannot be, as X should be greater than any of (0.9, 0.99, 0.999, ...)!
The above doesn't actually strictly
prove the result mathematically, but it should work as demonstration.
*blink, blink*
Okay I think I kind of understood that. I'm so glad I'm gonna go for an English Major...
Posted: 2003-11-29 07:31pm
by Jeremy
That's about all I have to say on this matter.
Posted: 2003-11-29 07:34pm
by AnimeJet
Well it seems that there is a general agreement, that .999~ does in fact, equal.. one! Odd, most places dissagree and rip at each other until the topic reaches a limit.. this place scares me.
Posted: 2003-11-29 07:34pm
by Kuroneko
Mad wrote:... if you have an infinite number of terms to add, with each term smaller than the previous term, then they will all add up to a specific number.
Which isn't strictly true--it is so if and only if the sequence of partial sums is Cauchy. As a well-known counterexample, the Harmonic series 1/1+1/2+1/3+1/4+... adds smaller and smaller terms, and yet fails to converge to any real number.
Posted: 2003-11-29 08:12pm
by Master of Ossus
Since the difference between this and one is 1/infinity, it does in fact equal one.
Posted: 2003-11-29 08:59pm
by Andrew J.
This makes so much more sense now that I've studies sequences in math...
Posted: 2003-11-29 09:27pm
by Rye
Guhi hate this...personally i've always thought of it in terms of thirds, the whole 0.3333....equalling a third, times by 3 = 1.
Damn decimals...
Posted: 2003-11-29 09:54pm
by kojikun
Absolutely, no. Its infinitesimally close to but not equal to one. Effectively, and for any reasonable purposes, yes.
Posted: 2003-11-30 12:01am
by Mad
kojikun wrote:Absolutely, no. Its infinitesimally close to but not equal to one. Effectively, and for any reasonable purposes, yes.
You'd be right except for the fact that the fundamental concepts behind calculus beg to differ, as explained earlier in this thread...
Posted: 2003-11-30 01:02am
by kojikun
Mad wrote:You'd be right except for the fact that the fundamental concepts behind calculus beg to differ, as explained earlier in this thread...
shit, right, yeah, sorry. Forgetting calculus = naughty me.