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It equals 7, but I haven't taken any math courses in a couple years so I could be wrong.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.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 ._.
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.HemlockGrey wrote:Could someone explain this slightly better? I mean, won't it always be just short of 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?Kuroneko wrote: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.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:HemlockGrey wrote:Could someone explain this slightly better? I mean, won't it always be just short of 1?
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.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?Kuroneko wrote: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.HemlockGrey wrote:Could someone explain this slightly better? I mean, won't it always be just short of 1?
I'm an English major. I know words, not numbers. Heeeelp!
Crimson Raine
Why wouldn't it? Let's call it X = 0.999... .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?
*blink, blink*Kuroneko wrote:Why wouldn't it? Let's call it X = 0.999... .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?
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.
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.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.