0.9_ = 1?

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Shiggily
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27 Jan 2009, 3:55 pm

twoshots wrote:

Quote:

you could probably disprove the .999 repeating decimal is 1 assumption in any number of mathematical ways starting with set theory and the fact that 1 is a natural number and an integer, while .9999... is not. 1 is rational and .999 is not. Identity properties of 1, etc.

Let's see you try.

I might look at it later.

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kalantir
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27 Jan 2009, 4:09 pm

1/3 = 0.33333_
2/3 = 0.66666_
1 = 3/3 = 0.99999_

Thats my take on it...

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twoshots
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27 Jan 2009, 4:14 pm

Shiggily wrote:

ruveyn wrote:

Shiggily wrote:

Wrong. One cannot disprove what has been correctly proven.

ruveyn

you have correctly proven that .9 repeating is 1?

lolz

Yes. A repeating decimal is defined as the limit of a series. For a similar case

Quote:

(b) Show that the real number 1/3 can be represented in a ternary expansion as 0.10 and 0.02.

Kosmala, Witold A.J., A Friendly Introduction to Analysis, p333,
Where I have underlined the part of the expansion which is repeated.

My number theory book likewise defines a base b expansion of numbers between 0 and 1 in terms of a series in giving a proof of the uniqueness of the the expansion where the expansion is not allowed to be Σ(b-1)/b^k at the end. This is quite well known stuff; the central confusion is that people are unaware of the rigorous definition of a base b expansion of numbers betwixt 0 and 1.

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robo37
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27 Jan 2009, 4:56 pm

If 0.0_1=0 then 1/infinity=0, so if 0*infinity is 0 then infinity=0. (because X/Y*Y=X) Since infinity isn't equal to zero 0.0_1 must be a real number, and 1-0.9_ is 0.0_1 and 0.9_ is not equal to 1.

twoshots
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27 Jan 2009, 4:58 pm

Infinity isn't a number.

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RockDrummer616
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27 Jan 2009, 5:54 pm

robo37 wrote:

If 0.0_1=0 then 1/infinity=0, so if 0*infinity is 0 then infinity=0. (because X/Y*Y=X) Since infinity isn't equal to zero 0.0_1 must be a real number, and 1-0.9_ is 0.0_1 and 0.9_ is not equal to 1.

Well, as limits are considered, 1/infinity is 0. In other words, as the x value in the function y=1/x approaches infinity, the y value approaches 0. Therefore the limit (the y value at x=infinity) is extrapolated to be equal to 0. And also, as I already said, there is no such number as 0.0_1, since there are infinite 0s, you will never get to the 1.

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27 Jan 2009, 6:04 pm

.999... plus the number in my avatar = 1.

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Hector
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27 Jan 2009, 6:06 pm

There's an "extended real line" with infinity and -infinity considered, but infinity/infinity is undefined.

http://en.wikipedia.org/wiki/Extended_real_number_line

Hector
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27 Jan 2009, 6:07 pm

Also, guys, this is what a decimal representation of a real number is: http://en.wikipedia.org/wiki/Decimal_representation

twoshots
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27 Jan 2009, 6:37 pm

RockDrummer616 wrote:

robo37 wrote:

If 0.0_1=0 then 1/infinity=0, so if 0*infinity is 0 then infinity=0. (because X/Y*Y=X) Since infinity isn't equal to zero 0.0_1 must be a real number, and 1-0.9_ is 0.0_1 and 0.9_ is not equal to 1.

Well, you *could* define it as the limit of the sequence (.01,.001,.0001,...) but obviously this is zero. Infinities in general are frowned upon by many mathematicians, so it's best just not to think of something like 0.0_1 as having an infinite number of 0s (although I'm very pro-infinity myself).

Hector wrote:

There's an "extended real line" with infinity and -infinity considered, but infinity/infinity is undefined.

Well, yes, to be more precise RU{+∞,-∞} isn't a field, and when most people want to deal with numbers they want to deal with a field.

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dannit
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27 Jan 2009, 8:10 pm

Just out of curiousity how many of you 'anti-completeness of metric space' individuals also believe that there does not exist a number whose square is minus 1?

dannit
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27 Jan 2009, 8:14 pm

twoshots wrote:

RockDrummer616 wrote:

robo37 wrote:

If 0.0_1=0 then 1/infinity=0, so if 0*infinity is 0 then infinity=0. (because X/Y*Y=X) Since infinity isn't equal to zero 0.0_1 must be a real number, and 1-0.9_ is 0.0_1 and 0.9_ is not equal to 1.

Hector wrote:

There's an "extended real line" with infinity and -infinity considered, but infinity/infinity is undefined.

Well, yes, to be more precise RU{+∞,-∞} isn't a field, and when most people want to deal with numbers they want to deal with a field.

You can actually include infinity if you like - you just have to change your notion of distance. The usual distance between numbers is |x-y|, but this is just the Euclidean distance from geometry. If we were to say that the distance was given as |(1\x) - (1\y)| then in fact the sequence 1,2,3,4,5,... is convergent (it is called a Cauchy sequence - one whose terms get arbitrarily close together as you progress to infinity), it's up to you whether you call the limit of this sequence infinity, banana or gandaulf (It exists regardless of whether or not you have a name for it). Look up things like the p-adic numbers for more examples of number systems that include infinity and do not break the usual 'nice' algebraic rules

twoshots
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27 Jan 2009, 11:26 pm

Hmm. That hadn't occurred to me. Neat (:

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Death_of_Pathos
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28 Jan 2009, 12:31 am

I figured he wasn't trolling, but when in doubt its funnier the call someone a troll when they are harmlessly mislead instead of intentionally baiting an argument (because the only difference in the result is intent).

Also, Ill just leave this here:

robo37
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28 Jan 2009, 12:40 pm

Divergent series dont obey the ordinary laws of arithmetic, and Cramer's rule can only be applied to systems with a unique solution.

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28 Jan 2009, 3:33 pm

I wish people would discuss other mathematical problems with as much interest and skepticism.
0.9999999999...=1 is getting old.
This so reminds me of Zenos Paradox: The runner covers 90% of the remaining distance with every step. So he will never reach his destination. Or will he?

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