I'm wondering if the maths genius' on this board can rip him apart? Thank you
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Anonymous2009-01-24 16:10
lol dongs
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Krieger2009-01-24 19:59
I can't believe anyone who has taken a math class higher than basic algebra still believe 0.999.... is not equal to 1. There are numerous proofs that they are equal. I'll give you a few basic ones to start.
1. Between any two non-equal numbers in the set of rationals, there lies an infinite quantity of numbers. What number lies between 1 and 0.999....?
2. Take 1 - 0.999..... and you'll get that it equals 0.000.....1, which exactly equals 0. So the difference between 1 and 0.999... is exactly equal to 0.
3. 1/3 = 0.333....
multiply both sides by three and you get
3/3 = 0.999....
1 = 0.999....
The whole thing seems obvious, even using 8th-grade level proofs....
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Krieger2009-01-24 20:01
>>42
K, switching between different windows and got my topics mixed up....meant to post in a different forum.
Don't know why you've quoted me in on that one, I agree his proofs are terrible, I'm trying to educate him about the ideas behind the proofs.
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Anonymous2009-01-26 21:45
As a sidelong and unassociated reply to the posts above, attacking and debating the utility of the most abstruse mathematics, without taking any any personal umbrage, and only as it is my pleasure...
Multivariable calculus put men on the moon, compact disc technology was only feasible with complex analysis, and number theory informed RSA encryption, which still plays a role today.
The point: contrary to mathematician G.H. Hardy's noble conceit that pure mathematics exists in a kind of "safe" vacuum, and cannot be applied for any special good or harm (The Bomb, and chemical weapons, derived via physics and chemistry, are the obvious Objects here), the historical trend is that even the most obscure mathematical truths will eventually be applied.
Math is the most useful of all, even if you haven't yet built up the hundred years of military industrial capital.
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Anonymous2009-01-26 23:12
Without computer losers, I would have to leave my apartment to find porn. Therefore programmers > mathematicians.
I don't know why I keep seeing the 1/3=.3333... proof, it assumes something that the opposing side believes to be false.
My own take on the topic is that the value of .9999... is something infinitely close to the number 1. The only thing that can be infinitely close to 1 is 1 itself. Thus, .9999... is just another way of saying 1.
Sage because I know nothing about math, and am just passing by.
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Anonymous2009-01-29 13:42
>>62
Your statement "I know nothing about math" is true. So why would you bother commenting.
The way we define decimal expansions ensure that 0.99... = 1, it's not even a matter of proof it's pretty much a matter of definition.
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Anonymous2009-01-29 22:25
>>62
Can't we just hope the other side dies out in a few years?
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Anonymous2009-01-30 0:06
>>64
I don't really think that's possible though. :/
It's like the gays. Obviously gays aren't having children, so they can't pass the gay gene on, but somehow there are still gays around. So even though all the gays that were around in, say, the 30's have died out by now, there are still gays around.
It's the same with programmers. They don't reproduce, but somehow they're still around.
Just as a little aside, for those who are interested, the troll from the OP is no other than FrozenVoid, who I believe /sci/ should know quite well. http://dis.4chan.org/read/sci/1172747610
So technically, this troll is /sci/'s responsibility. You made him, you clean him up /sci/
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Anonymous2009-01-30 1:11
>>67
Hmmm, so the troll started here, but moved on to more fertile trolling grounds, namely /prog/.
Interesting.
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Anonymous2009-01-30 2:16
>>40
If that was so 1-0.999...=undetermined, but its obviously equals 0
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Anonymous2009-01-30 3:31
>>69
Infinity minus infinity is an indeterminate form. There is no answer to that.
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Krieger2009-01-30 16:18
I'm going to give you several proofs here, and hold on a second, I want you to try something new. Instead of questioning the formality of the proof, explain why you believe it's incorrect.
1.
x = 0.999...
10x = 9.999...
10x-x = 9.999...-0.999...
9x = 9
x = 1
1 = 0.999...
3.
I wouldn't call this a proof, but I'd like you a non-believer to answer this question:
What number lies between 0.999.... and 1?
Seeing as you believe they are distinct, you should be able to answer this fairly simply.
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Anonymous2009-01-30 20:14
1. You haven't properly defined what the symbol 0.99... means, and thus cannot assume that algebraic manipulations (such as addition, multiplication) can be applied to it in any meaningful way.
2. You've defined it as a limit, that's at least a start, however you've applied a theorem on finite geometric series, as a limiting case, to this series.
This works, but you haven't proved that it does, so this one isn't a proof either.
3. As I said last time, this is the closest to a proof.
All that's lacking is you need to state what you're saying in a better way.
The real numbers are a hausdorff space, so given any two distinct points there are two disjoint open sets containing them. However for any open set containing 0.9999... it must have 0.9999.. + e for some e>0, but 0.9999...+ e > 1 for any e >0. Therefore you can't find two such sets, therefore 0.999.... and 1 aren't distinct points.
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Krieger2009-01-31 2:26
>>72
1.
I'm not sure what exactly you're saying here. By 0.999... I mean zero point nine repeating, 0.(9), etc. I'm not sure how else to define it.
2.
If I'm not mistaken, the sum of an infinite geometric series is:
(first term) / [1 - (common ratio)], which is the formula I've used. So I'm not sure what you mean by,"applied a theorem on finite geometric series."
>>73
I'm not sure how this is relevant to anything...
1. Well you have to define somehow, otherwise it's just a symbol, it has to have some sort of mathematical interpretation.
Generally you define an infinite decimal expansion 0.a_0a_1a_2..... to equal the limit of the infinite sum a_i*10^-i.
It's a pretty easy job to show this converges for all value of a_i. during this you'll prove a proof of the form of 2. ie that all numbers which have an infinite string of 9's in their decimal expansion have two equivalent decimal expansions.
The reason your proof of 2 doesn't work is because you're applied a result that is derived to work on finite series, and taken a limiting case.
For example apply your formula to the series 1 + 2 + 4 + 8....
you get the answer -1, nonsense (in a sense anyway). You have to prove that it still holds for any common ratio < 1.
Now once you've prove all this you can start proving that addition and multiplication work for decimal expansions in the way you imagine (ie. that the new series you get still converges and converges to the value you want) and is well defined (ie, if you have A=A', B=B' for equivalent decimal expansions then c*A = c*A' and A+B = A'+B')
Then you can apply your second proof.
tl;dr? Maths is harder then you think.
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Anonymous2009-01-31 9:55
Maybe your third proof attempt can be turned into a proof.
Informal proof sketch/idea for #3.
1. Fact: a real number is uniquely described by a finite sequence of digits, followed by a decimal separator, followed by an infinite sequence of digits
2. Fact: The amount of real numbers is uncountably infinite.
3. We can define the successor function s : R -> R for real numbers as follows: Let "x.y" be a representation of a real number, where x is a finite sequence of digits, and y is an infinite sequence of digits. Then s(x.y) = x'.y', and x'.y' are obtained by incrementing the last digit of y.
(It should be provable that from this definition, for all reals x.y, x.y <= s(x.y); and also any real is s(s(s(...(0.0...)))...), i.e., some above-infinite amount of applications of the successor function)
2. If 0.999... != 1, then by (2) there should be an uncountably infinite amount of numbers between 0.999 and 1.
3. But s(0.999...) = 1.
4. (2) and (3) form a contradiction. Hence 0.999... must be 1.