21-09-2002 10:16									#61
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From Asher

quote: Sounds like you're more into the theory aspect more? The math behind everything?

Funny thing to say that I am not even sure! Perhaps yes, but in the other hand I also like to see the application of it.

	21-09-2002 10:16									#62
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And I also was remembering the arrow going the wrong way on compactness and completeness.

Compactness->completeness, but completeness does not require compactness...

	21-09-2002 10:22									#63
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Quoted from Frogger:

quote: Yup... Mixed up convergence and uniform convergence... I'm not on the ball tonight.

I am the one who is to blame about. I started all this completeness stuff when I (incorrectly) stated that the decimal representable numbers were complete...

You are not alone!

	21-09-2002 17:40									#64
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No, as close as it gets, it never is 1.

Take any number and divide it in half an infinite amount of times, do you ever get to 0?

	21-09-2002 18:02									#65
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Ever heard of limits?

Zero is not just nothing, but also an infinitely small number.

Last edited by Dauphin on 21-09-2002 at 18:23

	21-09-2002 20:22									#66
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quote: Originally posted by Sava No, as close as it gets, it never is 1.

Did you just not bother reading the thread or something?

Anyway, here's the way I learned it back in Calc:

.999... == x, so 10 * .999... = 10x = 9.999...
9.999... - .999... = 9 = 9x
so x == 1, but since .999... == x, then .999... == 1.

Quick and dirty way to find the fractional representation of any rational number.

	21-09-2002 20:26									#67
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Which was pretty much what was posted in the first post.

	21-09-2002 20:27									#68
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loinburger, 1/3 does not equal .333 repeator, it is just a numerical approximation...

for all intents and purposes, .999 repeator is close to the value of 1 to say that substituting one for the other would be acceptable. But they are by no means equal.

	21-09-2002 20:30									#69
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quote: Originally posted by Sava loinburger, 1/3 does not equal .333 repeator, it is just a numerical approximation...

It's only a numerical approximation if you truncate .333... somewhere along the lines, like you'd have to do when using a computer or something. But then you don't have .333... anymore.

.333... == 1/3, exactly.

	21-09-2002 20:33									#70
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quote: Originally posted by Sagacious Dolphin Which was pretty much what was posted in the first post.

Methinks that all that talk about bijection and smegma confused the issue to holy hell.

	21-09-2002 20:37									#71
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If you told a computer to find the value for 1/3 without limiting a decimal point, the computer would never complete the operation.

1/3 does not equal .333 repeator exactly

	21-09-2002 20:39									#72
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quote: Originally posted by Sava If you told a computer to find the value for 1/3 without limiting a decimal point, the computer would never complete the operation.

So what?

quote: 1/3 does not equal .333 repeator exactly

Yes it does. I'm still waiting for you to point out the error in my first post on this thread. Gainsaying is not a valid counterargument. You're just being contrary.

	21-09-2002 20:45									#73
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Okay, here's another way to look at the problem. .999... is really just the sum of an infinite geometric series, with a0 = .9 and r = .1. The sum of an infinite geometric series (where -1 < r < 1) is a0 / (1-r), which is 1, thus .999... == 1. Substitute a0 = .3, and you get that .333... == 1/3.

	21-09-2002 20:48									#74
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it doesn't matter, it still isn't EXACTLY equal to 1/3, nothing you can say will change that.

	21-09-2002 20:51									#75
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The way I was informed that .999999 recurring is equal to 1 was by proving that 0.xxxx recuring in any base is identically equal to 0.yyyyy recuring in any other base. (x and y being the base number less 1.)

The only occasion on which such an identity would be true is if they are all equal to 1.

Not sure how accurate that is, as I've never seen it proven, but I'll take their word for it.

	21-09-2002 20:52									#76
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quote: Originally posted by Sava it doesn't matter, it still isn't EXACTLY equal to 1/3, nothing you can say will change that.

It's unfortunately that mathematical reasoning is completely lost on you. You can say that black is white and 1 is 0 and equal is not-equal as much as you want, but it doesn't make it any less false than the first time you said it.

	22-09-2002 03:42									#77
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It is one.

As, I believe has already been pointed out, the sequence {.9, .99, .999, .999....} is cauchy (that is, for all epsilons greater than 0, there exist an N such that for all j>N, |s_j+1 - s_j| < epsilon), therefore it's convergent.

	22-09-2002 03:52									#78
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Alternatively, you can use the standard definition of convergence to prove this; that is for all epsilon > 0, there exists a N such that for all j > n |s_j - 1| < epsilon. And this is trivially true, since for any epsilon > 0, you can find a number in the sequence arbitrarily close to 1.

	22-09-2002 06:26									#79
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quote: Originally posted by Ramo It is one. As, I believe has already been pointed out, the sequence {.9, .99, .999, .999....} is cauchy (that is, for all epsilons greater than 0, there exist an N such that for all j>N, |sj+1 - sj| < epsilon), therefore it's convergent.

That's not the condition for Cauchy, Ramo...

(and I'm sure of that)

For instance, ln(n) fits that definition...

	22-09-2002 09:32									#80
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hmmmm...
And what about Cantor diagonal argument?

	22-09-2002 09:35									#81
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Errr, yeah that's right. j+1 should be j+k for all k>0.

	22-09-2002 09:37									#82
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Bloody long defintions prevent you from thinking about them when you're writing them down.

	22-09-2002 10:36									#83
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oops I pressed the wrong button ignore this

	22-09-2002 10:50									#84
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quote: Originally posted by carnide_ hmmmm... And what about Cantor diagonal argument?

What about it?

	22-09-2002 11:01									#85
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Cantor diagonal argument deals with the representation of numbers. Now, if a number has 2 representations (like 0.99999... == 1), how can we believe that the diagonal is a unique number, not present elsewhere?

	22-09-2002 11:10									#86
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Good question...but really just a quibble.

The decimal system provides unique representations for all reals with the sole exception of infinite trailing nines. It is very possible to construct cantor's diagonal argument such the the resultant diagonal number does not have a single 9 in it...

	22-09-2002 14:36									#87
	Re: does .9 repeating equal one? Get a bigger avatar today!

More or less. Or maybe just less. Or maybe not.

PHOTOJOURNALIST
"Do you know what the man is saying? Do you? This is dialectics. It's very simple dialectics. One through nine, no maybes, no supposes, no fractions -- you can't travel in space, you can't go out into space, you know, without, like, you know, with fractions -- what are you going to land on, one quarter, three-eighths -- what are you going to do when you go from here to Venus or something -- that's dialectic physics, OK? Dialectic logic is there's only love and hate, you either love somebody or you hate them."

	22-09-2002 18:41									#88
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----------------------------
Good question...but really just a quibble.
----------------------------
Not so sure about that. An algorithm must produce its result in a finite number of steps. We write 1, not 1.000000000000...
We can write all numbers as that, 2 being 2.00000000... , and so on. But then, all our algorithms would fail.
Just try to find 1.0000000...+2.0000000... . With the standard algorithm, designed to work with finite string of digits, you will be unable to produce a result. You will have to start at the infinite summing zeros, and will never reach the left end, where the action is. Unless you modify the algorithm somehow, to avoid starting with the zeros at the infinite. But that will be an "ad hoc" procedure.
To ilustrate the "ad hoc" in it, its only necessary to realize that the same number can be writen in many bases: 7.0000000.... (Base 10) can be writen 111.000000.... (Base 2) or 21.000000.... (base 3). In all previous example, the base has been an integer. But the base can be any number.
Take Pi for example. 10 (base Pi) will mean Pi, 100 (base Pi) will mean 9.8696044010893... (base 10), Pi^2.
With this base, a 4 (base 10) will have to be written
as an infinite string of digits. But we cannot assume repeating digits until the infinite. As a matter of fact, we need to know all the digits until infinite, to know for sure that we are talking about 4 (base 10).
Now try to sum 4 (base 10) + 5 (base 10), but working on base Pi.
You will be unable, now matter what "ad hoc" procedure you figure out.
Bottom line: You cannot work with infinite strings of digits using standard algoritms and "ad hoc" procedures.
So, its not just a quibble.

Last edited by carnide_ on 22-09-2002 at 18:54

	23-09-2002 08:25									#89
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quote: Originally posted by carnide_ ---------------------------- Good question...but really just a quibble. ---------------------------- Not so sure about that. An algorithm must produce its result in a finite number of steps. We write 1, not 1.000000000000... We can write all numbers as that, 2 being 2.00000000... , and so on. But then, all our algorithms would fail. Just try to find 1.0000000...+2.0000000... . With the standard algorithm, designed to work with finite string of digits, you will be unable to produce a result. You will have to start at the infinite summing zeros, and will never reach the left end, where the action is. Unless you modify the algorithm somehow, to avoid starting with the zeros at the infinite. But that will be an "ad hoc" procedure.

This is a bunch of crud, since we assume omega-consistency (an axiom when dealing with standard real analysis). Construct a pyramidal proof that bundles the infinite trailing zeros. QED

	23-09-2002 08:40									#90
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Damn you Analysis people. You've turned an otherwise esoteric Math thread into a completely illegible Math thread.