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Layra-chan
Crew

PostPosted: Sun Mar 18, 2007 12:57 am


super.funk
Morberticus

There was a thread a while back on gaia and the debating just kept going and going and going. bleh


i see what you mean! i brought this up to my friend, and he refused to buy it. we argued for a long long time.

this argument went all over the place. we even talked about singularities in the universe!

we're actually still arguing about it lol.


somehow... we just brought the argument into linguistics...

this is a dangerous topic.

haha


Oh, God. Please stop. Do not mix mathematics and linguistics, because you will end up in etymology, and then you will go to ontology and epistemology, and by then you're in philosophy, and philosophy is basically the antithesis of math.
PostPosted: Mon Mar 19, 2007 12:21 pm


MyOwnBestCritic
Yes, .9999... is equal to one. That's why we round whee

lolz, one of the most simple explanations yet. cool

`Fenix


Layra-chan
Crew

PostPosted: Mon Mar 19, 2007 10:22 pm


The Sexy Lady
MyOwnBestCritic
Yes, .9999... is equal to one. That's why we round whee

lolz, one of the most simple explanations yet. cool


Yet I'm not sure if that's quite correct. We round because we're willing to deal with positive error. But .9999... equals 1 exactly, with no error at all; the trick is proving that to stubborn people.
For instance, if you're willing to have an error of .001, then .999~1. This is an instance of rounding. If you're willing to have an error of .000001, then .999999~1, another instance of rounding.

But .99999.... is exactly 1, with 0 error; this is not rounding.
PostPosted: Fri Apr 20, 2007 4:24 pm


Yes, it is correct.
Here are two other ways to prove this correct.
I. 1/3=.333 repeating
2/3=.666 repeating
3/3 must equal .999 repeating
3/3=1

II. Try doing 108/9 using long division. 9 goes into 10 one time. Say that 9 goes into 18 only 1 time. Continue dividing. Even though 9 goes into 90 ten times, you can only use 9, given that it is the largest 1-digit number. Continue dividing like this. You will realize that the division will continue forever. Your answer should be 11.999 repeating. We know that 108/9=12. Since they are the answer to the same problem, they must equal each other.

Hope this helps! biggrin biggrin biggrin

Jathak


Ash Engel

PostPosted: Wed Apr 25, 2007 4:11 am


So the relation = really means is functionally indistinguishable from?
PostPosted: Wed May 02, 2007 10:53 am


Ash Engel
So the relation = really means is functionally indistinguishable from?


No, it means "is the same as." There is a relation that means "functionally (for certain functions) indistinguishable from," otherwise known as isomorphic, but there is a slight difference between isomorphic and equal in that two objects can be isomorphic but not equal in that isomorphic doesn't necessarily mean indistinguishable for all functions, whereas if two objects were "functionally indistinguishable from" for all functions, we would say that they are in fact the same object mathematically.

A fundamental problem here is that it is hard to distinguish between symbol and object; the symbols "1" and ".999..." are functionally equivalent, but "=" refers to a relation on the objects that "1" and ".999..." refer to.

The statement "1 = .999..." can mean that "1" and ".999..." refer to the same object, which would be a statement about the functional equivalence of the labels, but it is taken to mean that the object entitled "1" and the object entitled ".999..." are the same object, which is a statement about the equality of the objects, a slightly different statement from the one about labels.

Layra-chan
Crew


Dewdew

PostPosted: Wed Jun 27, 2007 9:51 pm


Layra-chan
The Sexy Lady
MyOwnBestCritic
Yes, .9999... is equal to one. That's why we round whee

lolz, one of the most simple explanations yet. cool


Yet I'm not sure if that's quite correct. We round because we're willing to deal with positive error. But .9999... equals 1 exactly, with no error at all; the trick is proving that to stubborn people.
For instance, if you're willing to have an error of .001, then .999~1. This is an instance of rounding. If you're willing to have an error of .000001, then .999999~1, another instance of rounding.

But .99999.... is exactly 1, with 0 error; this is not rounding.


How small is nothing?
PostPosted: Thu Jun 28, 2007 10:58 pm


Dewdew
Layra-chan
The Sexy Lady
MyOwnBestCritic
Yes, .9999... is equal to one. That's why we round whee

lolz, one of the most simple explanations yet. cool


Yet I'm not sure if that's quite correct. We round because we're willing to deal with positive error. But .9999... equals 1 exactly, with no error at all; the trick is proving that to stubborn people.
For instance, if you're willing to have an error of .001, then .999~1. This is an instance of rounding. If you're willing to have an error of .000001, then .999999~1, another instance of rounding.

But .99999.... is exactly 1, with 0 error; this is not rounding.


How small is nothing?


Exactly what kind of nothing are we talking about? It's generally of size/measure/volume 0, but the exact wording depends on the kind of nothing.

Layra-chan
Crew


Dewdew

PostPosted: Mon Jul 02, 2007 8:24 pm


Layra-chan
Dewdew
Layra-chan
The Sexy Lady
MyOwnBestCritic
Yes, .9999... is equal to one. That's why we round whee

lolz, one of the most simple explanations yet. cool


Yet I'm not sure if that's quite correct. We round because we're willing to deal with positive error. But .9999... equals 1 exactly, with no error at all; the trick is proving that to stubborn people.
For instance, if you're willing to have an error of .001, then .999~1. This is an instance of rounding. If you're willing to have an error of .000001, then .999999~1, another instance of rounding.

But .99999.... is exactly 1, with 0 error; this is not rounding.


How small is nothing?


Exactly what kind of nothing are we talking about? It's generally of size/measure/volume 0, but the exact wording depends on the kind of nothing.


Well it's just if .999999=1 then the difference between them must be nothing. So is the difference really nothing? or is it just so small that it can be considered nothing?
PostPosted: Mon Jul 02, 2007 8:42 pm


The there is no difference. 0.999… can be represented as the sum of a geometric series which converges to one.

i.e.

0.999…
= 9/10 + 9/10² + 9/10³ + higher terms
= ∑_{n=1}^{∞} 9/(10^n) = 1

Exactly.

Wait, Moberticus has already mentioned this one.

I'm not that well versed on this, but I'm sure Layra or similar can set the foundations for why the series does converge exactly [some blag to do with Cauchy].

A Lost Iguana
Crew

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Layra-chan
Crew

PostPosted: Wed Jul 04, 2007 11:29 pm


Quick abusive version of why ∑_{n=1}^{∞} 9/(10^n) =1:

Let ∑_{n=1}^{∞} 9/(10^n) = S.
.9+.09+.009... = .9 + .1(.9+.09+.009...) <= This works but only very carefully
Thus S = 9/10+S/10.
Multiply both sides by 10 to get: 10S = 9+S
Subtract S from both sides to get: 9S = 9
Divide both sides by 9 to get: S = 1.
biggrin
Actually building the reals topologically is unhappy.
It starts with the reals being Hausdorff and ends in tears.

[Warning: lots o' text]

The reals are a field, which means that you can take any two real numbers and multiply them together to get another real number. It also means that if you take two numbers a and b, and b is not 1 or 0, then ab is not equal to a.

So, suppose you have two numbers a and b. We can denote the distance between these numbers as d(a, b) = |a - b|, i.e. the absolute difference between them. Note that |a-b| is a real number, so we can multiply this value by 1/2, thus getting a different real number |a-b|/2.

Suppose that a does not equal b. In that case, we can construct a pair of sets such that set A contains a, set B contains b, the two sets are disjoint (i.e. share no numbers) and both are of positive size (in this case length).

What we do is we make set A all numbers c such that |a-c| < |a-b|/2, and similarly set B is all numbers d such that |b-d|<|a-b|/2.
|a-a| = 0<|a-b|/2, and |b-b|=0<|a-b|/2, so a is in A and b is in B.
For c in A, |a-c| < |a-b|/2, and by the triangle inequality |c-b|+|a-c| > |a-b|, |c-b| > |a-b|/2, so c is not in B. Similarly, if d is in B, d is not in A. Thus A and B are disjoint.
Finally, set A has length |a-b|, as does set B.

So now we have two sets that fit the criteria we want.
Since we can construct these two sets, we call the set of real numbers "Hausdorff".

Now, suppose that we had a sequence that converged to a. What does this mean? This means that if we denote the sequence as a_1, a_2, a_3... then for any positive real number e, there exists an N such that for all n > N, |a_n-a| < e. So the sequence gets as close to a as we want it to, no matter how small the distance (as long as the distance is greater than 0).

But suppose that we tried to make a sequence that converged to both a and b.
Suppose that we've chosen e < |a-b|/2. Then for some N, for all n > N,
|a_n-a| N. But then what about b? For all n > N, |a_n - b| > |a-b|/2 by definition, and thus we can't get it any closer to b. Therefore it can't converge to b; it can only converge to one number.

So what does all of this mean?

Well, look at the sequence a_1=.9, a_2=.99, a_3=.999, etc. It obviously converges to .9999.... since the difference between a_n and .9999... is one tenth of that for a_{n-1}, so going far enough along will give us a difference of any smallness we want.

But it also converges to 1, since |a_n-1| = .1^n, and thus we also have the difference being divided by 10 each time.

So therefore, since it converges to both 1 and .999..., and we just showed that any sequence can only converge to one thing, we get that 1 and .999... must be equal.
PostPosted: Sat Jul 14, 2007 6:07 pm


.999... (repeating forever) does not have a number between it and 1.


How's that?

So they must be equal.

AirisMagik


Layra-chan
Crew

PostPosted: Sun Jul 15, 2007 12:35 pm


And before anyone says anything, yes, I should have showed that ∑_{n=1}^{∞} 9/(10^n) is a meaningful number, and a real number, and included something about Cauchy sequences, but feh on you. I don't have to worry about Cauchy because ∑_{n=1}^{∞} 9/(10^n) is rational, and I'm not going to bother showing that either.
PostPosted: Tue Aug 21, 2007 9:47 pm


AirisMagik
.999... (repeating forever) does not have a number between it and 1.


How's that?

So they must be equal.


That is probably the dumbest proof possible. If one were to define a system in which only integers existed, your proof would require every number equal the other.

Since one and two have no number between them, they are equal. Since this is true of two and three, and one and two are equal, one and three are the same!

Besides, there are numbers between the two, if you follow a number system outside the "norm." The simplest one would be the Hackenstrings number:

BR(B)B

Which happens to be analogous to .999... 5.

zz1000zz


zz1000zz

PostPosted: Tue Aug 21, 2007 10:20 pm


In this post, i will attempt to explain the "problem" with proofs stating .999... = 1. If this debate has shown anything, it has shown people do not understand the nature of mathematics.

The most common proof i see is .333... x 3 = .999..., and we know 1/3 x 3 = 1. To be fair, i think the proof should be changed to a number less optically pleasing, such as 0.14285714... × 7 = 0.999.... In either event, the result is meaningless, as provided by A. N. Walker:

Hackentstrings FAQ
Conventionally, we multiply two finite strings of digits together from the right, propagating `carries' to the left as necessary. If we work from the left, then we must always be prepared to change the `result-so-far' in the light of later carries. For finite strings, the final results are equivalent, but in working from the left arbitrarily many digits may be only provisionally known at any one time, so that (for example) the computation cannot be performed by a finite-state machine even if the multiplier is specified in advance. So much the worse if either or both strings are infinite. Thus, if we multiply 0.333...3d... by 3, where the next digit, d, is thus far unknown, then all we know is that the answer is no less than 0.999...90 and no more than 1.000...20. If d turns out to be less than 3, then the start of the result is confirmed to be 0.999..., if greater than 3 then 1.000...; but if d is 3, then the decision is postponed. If the 3's persist indefinitely, then so does the postponement of the decision. Of course, if we know that the multiplicand is 1/3, then we know already that the 3's will persist indefinitely, and we can make out a special case for known rational numbers; but the theory required to do this will be entirely equivalent to the formal analytic proof that 0.999... = 1, and there is no didactic gain.


In a similar manner, the proof in which we multiply .999... by ten is visually attractive, but is justified only by making special cases. At which point there is no "proof." The other common example was explained in my post immediately prior to this one.

Every proof offered to prove .999... = 1 is nothing more than an optical delusion. They do well to convince high school students, but nothing more. Each is inherently self-referential to the system for which is defined.

This is not to say the two numbers are different. In any system following Archimedean mathematics the two numbers will be the same. The reason is in standard analysis infinitesimals, infinitely small numbers, do not exist. This is axiomatic to the system, meaning it is accepted as true without proof. A different system may use different axioms, which makes it quite possible for .999... to be different than 1. Such systems do exist, with the most known being surreal numbers.

This debate is not one of proofs, but one of mathematical systems. In the system taught in elementary school, the two numbers are equal. Because this system is so ingrained into the mind of society, it is treated as the the only truth. This problem is so predominant those who teach mathematics and science often do not know better. When students encounter this problem, they are given these "proofs" which do not answer anything. This is a grave injustice, which can easily be avoided by encouraging an understanding of non-standard mathematics.

(Other proofs exist, using Cauchy Sequences and Dedekind Cuts. Both methods derive their validity directly from the "standard" system, making them just as tautological as the three proof i already mentioned.)

Edit: If you want information on Xeno's paradoxes, i would be glad to do so. However, i think it would be best done in another topic.
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Mathematics

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