0.9_ = 1?

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What he said.

This is a result of fourier series. We consider the fourier series for f(x)=x^2 on [-pi,pi] It can be shown that this is given by
f(x)= (π^2)/3+4Σ[(-1)^n]cos(nx)/n^2
f(π) = π^2 clearly, but the left hand side is
(π^2)/3+4Σ1/n^2 = π^2
Hence,
Σ1/n^2 = (2π^2)/12 = (π^2)/6

DOH!

Put up the wrong image. My original host refused to let me remote link, so I had to save and upload elsewhere. Mustve gotten mixed up. Soz.

All that times (6/pi^2) = 1, then.

But I could easily say that 0.9_ has an infinite number of nines in it and 1 doesn’t instead.



If representations are defined by their numbers then numbers are defined by their representations.



Each representation represents a different thing.



Because Uncyclopedia has a point, and I'm not trolling, a third of people do seem to think that 0.9_ doesn't equal 1. Are they trolling?



But you could say the same thing about 0.9_, Because there are infinite 9's you never get to the last 9 which makes the number 1.

1.0_ has an infinite amount of zeroes.

False

False

Some "last 9" is not necessary, you only need the infinite series.

But doesn't have an infinite amount of nines. 0.9_ has an infinite amount of nines so if 0.9_ is equal to 1 then you can say that 1 has an infinite amount of nines.

Last edited by robo37 on 01 Feb 2009, 3:48 pm, edited 1 time in total.

Not every number has a unique decimal representation. In the case of 1, you can also express it as 0.9_ (as a real number). So even though 0.9_ has an infinite amount of nines and 1 doesn't, there's still no problem with that.

There seems to be confusion regarding the definition of decimal representation. I'll just post this one from Wikipedia:


For 0.9_, a(1) = 0 and a(n) = 9 for all n>1. So this is simply a geometric series, which converges to (9/(1-(1/10)))-9 = 1 using this formula:

it seems that finite decimals do not have unique decimal representation.

That is correct.

Let me put it differently. The sentence '0.9_ has an infinite number of nines' is true. If 0.9_ is equal to 1 then the sentence '1 has an infinite number of nines' would be exactly the same as the sentence '0.9_ has an infinite number of nines'. If the sentence '0.9_ has an infinite number of nines' is true, then the sentence '1 has an infinite number of nines' (being the exact same sentence) is also true. As we all know 1 doesn't have any nines in it meaning that 0.9_ must not equal 1.

Any real number is the limit of an uncountable infinity of Cauchy sequences. There is no unique sequence that converges to a given real number as a limit. Infinite decimal expansions are shorthand for infinite series which in turn are sequences of partial finite sums that converge to a limit (that is what is meant by a convergent series).

ruveyn

The number of 9s is irrelevant, they are two different ways to express the same number. How many 7s are in "Pi?" None, just a "P" and an "I."

lol.

False syllogism is false.

I don't know who can argue against the above, its pretty graphic and easy to understand. It also sneakily helps to avoid the notion of convergence, which I doubt is even relevant in this situation.

The rest of this thread is the natural confusion of people trying to comprehend infinity without accepting that to do so would make it finite and violate its own definition.

I had to dig this thread up because this problem made me lose interest in mathematics. Not due to the sort of argument that has gone on here, but because there are lots of valid proofs for it. At university I came up with one that wasn't in the text book. They looked at it, said I was right, but to do it the way I was told instead. Mathematics was actually pretty exciting for me until that very moment.

Do you know what a convergent series is?

ruveyn

Ok, here's how I understood it. If 0.999... were NOT equal to 1, then there has got to be some other number which you could place between 0.999... and 1. But, since the nines after the decimal place go on forever, you cannot fit in any other number between 0.999... and 1. Therefore, 0.999... IS equal to 1.