1=0.

This is a discussion on 1=0. within the A Brief History of Cprogramming.com forums, part of the Community Boards category; Code: #define 1 0 Haha... HAHAHAHAHA! Take that LOGIC! /dance....

  1. #16
    60% Braindead
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    Code:
    #define 1 0
    Haha... HAHAHAHAHA! Take that LOGIC! /dance.
    Code:
    Error W8057 C:\\Life.cpp: Invalid number of arguments in function run(Brain *)

  2. #17
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    What about the whole .999... = 1 'proof'?

    http://en.wikipedia.org/wiki/0.999...

  3. #18
    CSharpener vart's Avatar
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    sqrt(-1) = i
    sqrt(-1) = +i
    and
    sqrt(-1) = -i
    as well
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  4. #19
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    Quote Originally Posted by vart View Post
    sqrt(-1) = +i
    and
    sqrt(-1) = -i
    as well
    Oh that's right! So in that case:
    Code:
    sqrt( 1 ) = sqrt( -1 ) * sqrt( -1 ) = (+i) * (-i) = 1
    Problem fixed.

  5. #20
    Registered User Stonehambey's Avatar
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    Quote Originally Posted by misterMatt View Post
    What about the whole .999... = 1 'proof'?

    http://en.wikipedia.org/wiki/0.999...
    There is nothing strange about that. 0.999... is indeed equivalent to one. Some people have a hard time accepting it though.

    As for the sqrt one. Remember that "square root" is not a function, since it yields to possibilities (maybe if you do a masters degree in maths you might come across a type of function which gives two answers, but for most people, it is sufficient to define a function as a mapping blah blah which has ONE element in the co-domain ). So we in fact start having trouble on the second line where it says

    sqrt(1) = 1

    what it should say is

    sqrt(1) = 1 <-- AND -1 AS WELL!!


    All the square roots do a good job at confusing you

    Has anyone seen the differential one? It's one of my favourites

    x^2 = x + x + x + x +... (x times)

    (d/dx)(x^2) = (d/dx)(x + x +...)

    2x = 1 + 1 + 1 +... (x time)

    2x = x

    2 = 1
    Last edited by Stonehambey; 08-14-2008 at 10:09 AM.

  6. #21
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    cute. I'm glad I'm not in bed with mathematics.
    Here to Deceive, Inveigle, Obfuscate Since 1945

  7. #22
    CSharpener vart's Avatar
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    you forgot to d/dx of the (x times) part
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  8. #23
    Registered User Stonehambey's Avatar
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    Quote Originally Posted by vart View Post
    you forgot to d/dx of the (x times) part
    Sorry, I just meant the (x times) bracket to mean that x squared is x added to itself x times, it's not actually part of the maths

  9. #24
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    Quote Originally Posted by Stonehambey View Post
    There is nothing strange about that. 0.999... is indeed equivalent to one. Some people have a hard time accepting it though.
    I disagree, it's rightfully hard as no one can say what infinity is and what happens at it, so using it to equate anything with a defined unity may seem quite strange.

    I propose the problem be restated as a limit instead (since limits admit that we cannot define what happens at infinity, but merely what appears to be happening as we approach it) (apologize for on-the-spot math...):

    lim[n->inf] Σ[i=1..n]9/10^i =
    9 * lim[n->inf] Σ[i=1..n](1/10)^i =
    9 * lim[n->inf] ( (1/10) ^ (n+1) - (1/10) ) / ((1/10) - 1) =
    9 * lim[n->inf] ( (1/10) ^ (n+1) - 0.1 ) / -0.9 =
    9 * ( 0 - 0.1 ) / -0.9 = 1

    Therefore it's safe to say that the limit of 0.9999... as the number of 9s grow without bound is 1 (but it is impossible to define its value at infinity as this itself is an undefined value).
    Last edited by @nthony; 08-14-2008 at 10:35 PM.

  10. #25
    CSharpener vart's Avatar
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    it's not actually part of the maths
    yes it is - you have
    Code:
    sum_0^x 1
    so the upper bound of sum depends on x, and as such should be taken in the account when appliyng differential on x
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  11. #26
    Registered User Stonehambey's Avatar
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    Quote Originally Posted by @nthony View Post
    I disagree, it's rightfully hard as no one can say what infinity is and what happens at it, so using it to equate anything with a defined unity may seem quite strange.

    I propose the problem be restated as a limit instead (since limits admit that we cannot define what happens at infinity, but merely what appears to be happening as we approach it) (apologize for on-the-spot math...):

    lim[n->inf] Σ[i=1..n]9/10^i =
    9 * lim[n->inf] Σ[i=1..n](1/10)^i =
    9 * lim[n->inf] ( (1/10) ^ (n+1) - (1/10) ) / ((1/10) - 1) =
    9 * lim[n->inf] ( (1/10) ^ (n+1) - 0.1 ) / -0.9 =
    9 * ( 0 - 0.1 ) / -0.9 = 1

    Therefore it's safe to say that the limit of 0.9999... as the number of 9s grow without bound is 1 (but it is impossible to define its value at infinity as this itself is an undefined value).
    lol, I told you people have a hard time accepting it

    x = 0.9999...

    10x = 9.9999...

    9x = 10x - x = 9.999... - 0.999... = 9

    9x = 9

    x = 1


    yes it is - you have
    Code:

    sum_0^x 1

    so the upper bound of sum depends on x, and as such should be taken in the account when appliyng differential on x
    It was just a comment, honestly. The error is in the very first line, if anyone wants a clue (or you could just look it up on google or something :P )

  12. #27
    Registered User Kernel Sanders's Avatar
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    Quote Originally Posted by Stonehambey View Post
    lol, I told you people have a hard time accepting it

    x = 0.9999...

    10x = 9.9999...

    9x = 10x - x = 9.999... - 0.999... = 9

    9x = 9

    x = 1
    I never liked that one, because 9.999... has one more '9' than 0.999... has. It's confusing because we're dealing with infinity here, but you're still manufacturing an extra decimal place. I'm with @nthony that the limit of 0.999... = 1, but that does not mean that .999... = 1

    **EDIT**
    From the wiki link
    There are many proofs that 0.999… = 1, of varying degrees of mathematical rigour. A short sketch of one rigorous proof can be simply stated as follows. Consider that two real numbers are identical if and only if their difference is equal to zero. Most people would agree that the difference between 0.999… and 1, if it exists at all, must be very small. By considering the convergence of the sequence above, we can show that the magnitude of this difference must be smaller than any positive quantity, and it can be shown (see Archimedean property for details) that the only real number with this property is 0. Since the difference is 0 it follows that the numbers 1 and 0.999… are identical. The same argument also explains why 0.333… = 1⁄3, 0.111… = 1⁄9, etc.
    That does it for me better than anything ever has, but I still don't like it (I never liked dealing with infinity). The difference is 1/inf, which is defined as zero, but to me it's just another limit thing. The limit of 1/x is 0, but it still never gets there


    It was just a comment, honestly. The error is in the very first line, if anyone wants a clue (or you could just look it up on google or something :P )
    Is it because that only holds true for positive numbers?
    Last edited by Kernel Sanders; 08-15-2008 at 04:07 AM.

  13. #28
    Registered User Stonehambey's Avatar
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    Yeah, it's ok for integers, but that's it I think. Saying that x squared is x added to itself x times isn't universally true for all real numbers.

    vart, sorry I see what you're saying now, I initially thought you were implying that I should somehow apply d/dx to a comment ( I took what you said a bit too literally)

    As a passing comment. People usually have no problems whatsoever when they see stuff like 1/9 = 0.111... It's widely accepted that point one recurring is the decimal equivalent of one ninth. However multiply both sides by 9 and you get 0.999... = 1 and something doesn't quite 'click'.

    0.999... recurring is just an infinite sum WHICH IS DEFINED as 1, so it's correct to say that it equals one. Computers don't have a concept of infinity, but mathematics does, trying to picture 0.999... in your head or on a comp screen is trouble because you can't "imagine" what an infinite stream of numbers looks like. That is where the problems lies I think, intuitively 0.999... isn't equal to one because it will always have a little bit left over. But that doesn't make any sense, because this "little bit" (which is the result of the automatic truncation of the number your brain does when you try to picture it) is ALWAYS added on.

    I realise I'm not going to convince some people, this thread and argument could go on forever (oh, the irony!), so I think I'll retire at this point

  14. #29
    Registered User Kernel Sanders's Avatar
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    Quote Originally Posted by Stonehambey View Post
    Yeah, it's ok for integers, but that's it I think. Saying that x squared is x added to itself x times isn't universally true for all real numbers.

    vart, sorry I see what you're saying now, I initially thought you were implying that I should somehow apply d/dx to a comment ( I took what you said a bit too literally)

    As a passing comment. People usually have no problems whatsoever when they see stuff like 1/9 = 0.111... It's widely accepted that point one recurring is the decimal equivalent of one ninth. However multiply both sides by 9 and you get 0.999... = 1 and something doesn't quite 'click'.
    Well, 1/9 isn't actually equal to .111... because 1/9 is impossible to represent using base 10 if you get right down to it

    0.999... recurring is just an infinite sum WHICH IS DEFINED as 1, so it's correct to say that it equals one. Computers don't have a concept of infinity, but mathematics does, trying to picture 0.999... in your head or on a comp screen is trouble because you can't "imagine" what an infinite stream of numbers looks like. That is where the problems lies I think, intuitively 0.999... isn't equal to one because it will always have a little bit left over. But that doesn't make any sense, because this "little bit" (which is the result of the automatic truncation of the number your brain does when you try to picture it) is ALWAYS added on.

    I realise I'm not going to convince some people, this thread and argument could go on forever (oh, the irony!), so I think I'll retire at this point
    Yeah - I can't say I'm convinced, but that doesn't mean I don't believe it's true. A lot of people who are a hell of a lot better at math than I have spent a hell of a lot more time reasoning about it and come to the conclusion that they are equal. I never was very comfortable with infinity and I guess this is just an artifact of that

  15. #30
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    In that case, what is 0.999...8 equal to? i.e. an infinite number of 9's with an 8 at the end. That probably doesn't make much sense mathematically and maybe even logically, but then again, if 0.999... = 1, then that means 1.000...1 = 1 also right? and if you subtract 1 from that it leaves 0.000...1. If you then subtract that from 0.999... you get 0.999...8.
    Just because it can be "proven" mathematically, doesn't necessarily mean it's true. It's just true as far as we can currently prove.

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