Everyone knows that most truths in the world can be proven from the Pythagorean Theorem. For example, Einstein's Theory of Relativity. And existence of G-d. Amongst other things.
2. the limit of 0.999... is 1. that doesn't mean 0.999... is 1 mamosh anymore than 0sin(1/0) is equal to 0 mamosh. (I know that the math establishment is on your side, but i'm still sticking with my opinion until someone explains to me otherwise.)
3. There's controversy how do interpret 0^0. In school they taught us that it's "indeterminate."
4. The way you have the sigma thing written, you're supposed to go from infinty to one, which is nonsense.
you should have n=1 below the sigma and infinity on top of the sigma.
1. I got paranoid for a second that in order to prove that area of the triangle is half its base times height you need P.t., but after I ran to my geometry textbook, I realized that you don’t need that. It’s actually a very neat proof (for the area of the triangle).
2. Pomoyemu, when we say 1/3 is equal to 0.333... we're implicitly talking about a limit. 0.333... is as close to 1/3 as you please. But it is not equal mamosh to 1/3
When used to specify a recurring decimal, "…" means that some infinite portion is left unstated, which can only be interpreted as a number by using the mathematical concept of limits. As a result, in conventional mathematical usage, the value assigned to the notation "0.999…" is defined to be the real number which is the limit of the convergent sequence (0.9, 0.99, 0.999, 0.9999, …).
Here is an explanation by Ilya Birman. 0.999... cannot “approach” infinity. It’s not a function. It’s a number. And in that that number, the number of nines is not approaching infinity, but is mamosh infinite. So, the number 0.999... in which the number of 9’s is infinite is equal to 1.
If one wants, one can look at .999... as a function. Consider the sum from i=1 until n of 9*(1/10)^i. This can be viewed as a function of n. The real number to which this series converges is the limit as n approaches infinity of f(n).
Wikipedia brings up a good point. The unenlightened (i.e. those like me) think that a limit is a process rather than a fixed value: you give me a delta, and I give you an epsilon. You can keep on giving me smaller deltas, and i'll keep giving you smaller epsilons. But you can never give me a delta of 0. So I can never give you an epsilon of 0. and .999... will never reach 1.
the problem is that infinity is not a real number. So the only way you can understand infinity without introducing non-real numbers is in terms of a limit.
I don’t understand anymore where Birman is coming from, when he says 0^0 = 1. And he doesn’t really prove it or explain it anywhere, as far as I see, so it just seems to me he is ripping the piss out of people.
The spambots are evolving! First they evolved altruism and spread good cheer throughout the blogosphere. Now they have evolved existential angst and are wondering if spreading viruses is such a good idea. I see great potential in the future of spambothood!
20 comments:
1. that is a neat proof.
2. the limit of 0.999... is 1. that doesn't mean 0.999... is 1 mamosh anymore than 0sin(1/0) is equal to 0 mamosh. (I know that the math establishment is on your side, but i'm still sticking with my opinion until someone explains to me otherwise.)
3. There's controversy how do interpret 0^0. In school they taught us that it's "indeterminate."
4. The way you have the sigma thing written, you're supposed to go from infinty to one, which is nonsense.
you should have n=1 below the sigma and infinity on top of the sigma.
But all in all, these are nice tibits.
1. I got paranoid for a second that in order to prove that area of the triangle is half its base times height you need P.t., but after I ran to my geometry textbook, I realized that you don’t need that. It’s actually a very neat proof (for the area of the triangle).
2. 1/3 = 0.333...
(1/3) 3 = (0.333...) 3
1 = 0.999...
Look it up.
3. Your teachers were wrong. According to this guy. I am happy to say it’s a machloikes.
4. I noticed that too. It’s not mine.
2. Pomoyemu, when we say 1/3 is equal to 0.333... we're implicitly talking about a limit. 0.333... is as close to 1/3 as you please. But it is not equal mamosh to 1/3
Are you sure? I remember reading about it on Wikipedia.
http://en.wikipedia.org/wiki/0.999_%3D_1
When used to specify a recurring decimal, "…" means that some infinite portion is left unstated, which can only be interpreted as a number by using the mathematical concept of limits. As a result, in conventional mathematical usage, the value assigned to the notation "0.999…" is defined to be the real number which is the limit of the convergent sequence (0.9, 0.99, 0.999, 0.9999, …).
(emphasis added)
So, what’s the difference? 0.333... = 1/3, no? Its limit as the number of numbers approaches infinity is 1/3.
(The limit as x approaches c of f(x) = L) does not imply (f(c) = L).
Similarly, (the limit as the number of digits approaches infinity = 1/3) does not imply (.33333... = 1/3)
Doesn’t imply, but in this case it is.
But we are just talking about notation, aren’t we?
Also: the real number that is the limit of the converging sequence (0.3, 0.33, 0.333, 0.3333, …) is 1/3, isn’t it?
Here is an explanation by Ilya Birman. 0.999... cannot “approach” infinity. It’s not a function. It’s a number. And in that that number, the number of nines is not approaching infinity, but is mamosh infinite. So, the number 0.999... in which the number of 9’s is infinite is equal to 1.
If one wants, one can look at .999... as a function. Consider the sum from i=1 until n of 9*(1/10)^i. This can be viewed as a function of n. The real number to which this series converges is the limit as n approaches infinity of f(n).
Wikipedia brings up a good point. The unenlightened (i.e. those like me) think that a limit is a process rather than a fixed value: you give me a delta, and I give you an epsilon. You can keep on giving me smaller deltas, and i'll keep giving you smaller epsilons. But you can never give me a delta of 0. So I can never give you an epsilon of 0. and .999... will never reach 1.
lim (x --> 5) of f(x), where f(x) = x is?
Why don’t I look at 5 as a function instead of a number?
the problem is that infinity is not a real number. So the only way you can understand infinity without introducing non-real numbers is in terms of a limit.
I am actually more interested in what 0^0 is.
I don’t understand anymore where Birman is coming from, when he says 0^0 = 1. And he doesn’t really prove it or explain it anywhere, as far as I see, so it just seems to me he is ripping the piss out of people.
But the answer to a limit can be a real number.
By the way, is 0 an even number?
http://ilyabirman.ru/meanwhile/2006/07/22/1/comments
1. http://en.wikipedia.org/wiki/0%5E0#Zero_to_the_zero_power
2. A limit is a real number, meaning it's the real number which you approach but never reach.
3. 0 is of the form 2n, so it is even. In addition, a function of the form f(x) = ax^0 is an even function.
4. My russian isn't that good.
regarding 0^0: see http://www.askamathematician.com/?p=4524
regarding .999999:
I've now grown older and wiser and have come to agree that
0.9999999999... = 1.0000000000...
The spambots are evolving! First they evolved altruism and spread good cheer throughout the blogosphere. Now they have evolved existential angst and are wondering if spreading viruses is such a good idea. I see great potential in the future of spambothood!
http://xkcd.com/810/
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