Classic internet argument

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ChaosRocket

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As far as I know, the most classic argument of the internet has never been brought up here, so I think we should definitely debate it.

Does .9999... (repeats forever) equal 1?

I say yes
 
For practical purposes, yes they are equal.

However, mathematically, .9999999999 repeats forever will never equal one.
 
I got into this same debate over at Serebii once. If you use the formula for translating decimal numbers to fractions, it will always come out as one. In every series of numbers where it is the next logical number, it ends up being one instead. I think that .9... is different from 1, though. On the other thread, there was someone that stated some kind of Calculus principle, but I don't know what it was.
 
An infinitely-repeating decimal is shown in fractions as division by nine: .1111~ is 1/9, .2222~ is 2/9, and so forth. 9/9, however, is one. So...
 
The problem is .9999(infinity) is basically *impossible* to round up.
 
The problem is if there is a difference between 1 and .9..., it is a number that defies logic. It would have to be .infinite 0s, ending in 1. In any form of math that we know or common sense, this number cannot exist. How can something be both infinite and finite? It's a contradiction in terms. I only went up to Pre-Cal in math. If there is anyone that knows a lot about Calculus that can prove me wrong, then feel welcome.
 
But 1/3 = .3... right? And 3 times 1/3 = 1, right? And 3 times .3... equals .9... so that means .9... must be the same as 1. Well that's basically what Ket was saying.

>Dumbest topic ever.

Shut up! :p This topic rules


Ooh me and Mozz found a page.

http://mathforum.org/library/drmath/view/53339.html

Ask Dr Math says I'm right.
 
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I'm going to offer the same explanation I gave Misty in the chat.

What you have to understand is that the reals are dense. What it means to be dense is this: given x, y two real numbers, then either x = y or there exists z such that x < z < y. In other words, it is impossible for two different real numbers to be adjacent. There will always be an infinity of real numbers between them, and one of them is their average: (x+y)/2.

Now, consider .999... and 1. Is there any number between them? If there are not, it means either that .999 is not a real number or that they are equal. We know that if they are not equal, then (0.999... + 1) / 2 is a real number that is not 0.999... and not 1 either. If you calculate it, the result is 0.999... ; however, that is not a different number, that is the same one we started with. Therefore, 0.999... = 1, or 0.999... is not a real number (and if it is not a real number, of course, it doesn't behave like one). There are many other arguments that lead to the same conclusion and despite Misty's claim, they're mostly valid.

This is not contradictory or paradoxical. It just means that the syntax for real numbers is ambiguous. Many real numbers have two different decimal expansions: 0.6 = 0.599..., 5 = 4.999..., etc. In binary, 0.11~ = 1. It's not a math problem as much as a quirk about their representation.

Now, there's one thing you -could- defend. Instead of saying they are equal, you could choose the other option I said was possible and say that 0.999... is not a real number. This amounts to posit a model where there is an element that is greater than any number smaller than 1, but is itself smaller than 1. That interpretation might be more intuitive to some, but I think it causes more problems than it solves. It implies that decimal expansions for most rationals don't actually represent real numbers. That's not like saying that 0.999... is an approximation. It's like saying that 0.999... and all its buddies are a totally different clique from the real numbers and don't follow the normal laws of arithmetics. That is... uh... serious business.

Oh and I'd like to point out that you can't use finite induction like "0. followed by any finite sequence of 9s is not 1" and automagically infer that it is true for an infinity of 9s. There is -nothing- in math that allows you to do this, much to the contrary. "Completed infinities" play in their own league.
 
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Archaic said:
Nope. Again, it's an approximation that is made for convenience. It isn't possible to express 1/3 in decimal notation.
Click to expand...

Nope, if the 3s actually go on infinitely, it's exact. As Dr Math also agrees. Read the Dr Math link I posted, it gives several proofs showing that .9... = 1, can you disprove those proofs?
 
I only see two proofs there, and they're not very well explained.

There's another argument that goes in the same way as the one I already presented:

- For any real number x such that 0 <= x < 1, lim n->inf. x^n = 0
- lim n->inf. 0.999... ^ n = 0.999...
- Therefore, it is false that 0 <= 0.999... < 1, or it is false that 0.999... is a real number.

Once again, you're faced with two options: 0.999... = 1 or 0.999... is not a real number.
 
I guess for all practical purposes .9... is 1. You're never going to get the decimal as the result of anything. It always ends up as 1. It may be also be that .9...does not exist, although one would think it should. You can't reach infinity no matter how hard you try.
 
1/9 = 0.1111~
2/9 = 0.2222~
3/9 = 0.3333~
4/9 = 0.4444~
5/9 = 0.5555~
6/9 = 0.6666~
7/9 = 0.7777~
8/9 = 0.8888~
therefore 9/9 = 0.9999~, but by definition if the numerator is equal to the denominator then the fraction equals 1. so 9/9 = 0.9999~ = 1

As far as I'm concerned, this is controversy is most like one of Zeno's paradoxes, where by the magic of infinite division into smaller and smaller units all motion becomes impossible, and a faster runner can never overtake a slower runner that began running first.

It's also a Creationist argument: they will always keep asking for an intermediate.
 
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