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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 doesn't matter, it still isn't EXACTLY equal to 1/3, nothing you can say will change that.
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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.<p style="font-size:1024px">HTML is disabled in signatures
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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, |sj+1 - sj| < epsilon), therefore it's convergent."Beware of the man who works hard to learn something, learns it, and finds himself no wiser than before. He is full of murderous resentment of people who are ignorant without having come by their ignorance the hard way. "
-BokononComment
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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 |sj - 1| < epsilon. And this is trivially true, since for any epsilon > 0, you can find a number in the sequence arbitrarily close to 1."Beware of the man who works hard to learn something, learns it, and finds himself no wiser than before. He is full of murderous resentment of people who are ignorant without having come by their ignorance the hard way. "
-BokononComment
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That's not the condition for Cauchy, Ramo...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.
(and I'm sure of that)
For instance, ln(n) fits that definition...12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio RegumComment
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hmmmm...
And what about Cantor diagonal argument?Comment
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Errr, yeah that's right. j+1 should be j+k for all k>0.
"Beware of the man who works hard to learn something, learns it, and finds himself no wiser than before. He is full of murderous resentment of people who are ignorant without having come by their ignorance the hard way. "
-BokononComment
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Bloody long defintions prevent you from thinking about them when you're writing them down.
"Beware of the man who works hard to learn something, learns it, and finds himself no wiser than before. He is full of murderous resentment of people who are ignorant without having come by their ignorance the hard way. "
-BokononComment
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ignore this
oops I pressed the wrong button ignore this"The first man who, having fenced off a plot of land, thought of saying, 'This is mine' and found people simple enough to believe him was the real founder of civil society. How many crimes, wars, murders, how many miseries and horrors might the human race had been spared by the one who, upon pulling up the stakes or filling in the ditch, had shouted to his fellow men: 'Beware of listening to this imposter; you are lost if you forget the fruits of the earth belong to all and that the earth belongs to no one." - Jean-Jacques RousseauComment
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What about it?Originally posted by carnide_
hmmmm...
And what about Cantor diagonal argument?12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio RegumComment
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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?Comment
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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...12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio RegumComment
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Re: does .9 repeating equal one?
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."He's got the Midas touch.
But he touched it too much!
Hey Goldmember, Hey Goldmember!Comment
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Good question...but really just a quibble.
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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_; September 22, 2002, 09:54.Comment
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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. QEDOriginally posted by carnide_
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Good question...but really just a quibble.
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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.12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio RegumComment
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Damn you Analysis people. You've turned an otherwise esoteric Math thread into a completely illegible Math thread.<p style="font-size:1024px">HTML is disabled in signatures </p>Comment
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