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 Post Posted: Sun Dec 04, 2011 3:02 pm 
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Ok, i know this isn't really speculation or directly Erfworld, but give me moment and I'll explain why its here (please move it to a better home if need be!)...

At some point on the boards in response to a recentish page (see, Erfworld link) somebody (probably Kreistor, but I can't actually recall!!) observed that infinity -1 still equals infinity. I agreed at the time, although I did find myself wondering if infinity minus infinity still equalled infinity or if it had to be zero. I digress. I have just moments ago found out the actual equation should read:

Infinity - 1 = 999

My challenge to you all is to explain why this is correct (Obviously there is a trick in this, but I swear I am not making it up!)

Have fun!!

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     Post Posted: Sun Dec 04, 2011 6:51 pm 
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    infinity -1 = 999 -> infinity -1 +1 = 999+ 1 -> infinity = 1000.

    thats not right so infinity - 1 =/= 999

    also if n = infinity, n-n = n(1-1) = n(0) = 0

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     Post Posted: Mon Dec 05, 2011 12:09 am 
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    Infinity is not a number, so standard arithmetical operations cannot be performed on it. Having said that, I'd like to see what 'proof' you offer; I assume it's similar to the proof that 2=1:
    Code:
         a = x            [true for some a's and x's]
       a+a = a+x          [add a to both sides]
        2a = a+x          [a+a = 2a]
     2a-2x = a+x-2x        [subtract 2x from both sides]
    2(a-x) = a+x-2x       [2a-2x = 2(a-x)]
    2(a-x) = a-x          [x-2x = -x]
         2 = 1            [divide both sides by a-x]

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     Post Posted: Mon Dec 05, 2011 5:46 am 
    Here for the 10th Anniversary Has collected at least one unit
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    Nope, none of those and not the 1=2 trick either, I'm afraid!

    I'll give as a hint that the answer is historical rather than mathematical :D

    Hopefully, not too big of a hint though!

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     Post Posted: Mon Dec 05, 2011 9:48 am 
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    Print Book 2 & Draw Book 3 Supporter This user is a Tool! Pin-up Calendar and New Art Team Supporter Here for the 10th Anniversary
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    Welll. It depends what you mean by underlining. If it means 'repeat these digits', then it's pretty clear. The repeated digits were placed to the left of the decimal point.

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     Post Posted: Mon Dec 05, 2011 10:37 am 
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    Ah, cute. :D

    The Roman Numeral representation of 1000 was CIƆ - which was reduced to ↀ and later transformed into infinity by John Wallis.
    So it's just a notational instead of a mathematical trick : "1000" equals "infinity" because they overloaded the symbol.

    http://en.wikipedia.org/wiki/Roman_nume ... ge_numbers

    (Google and Wikipedia has made us all geniuses.)

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     Post Posted: Mon Dec 05, 2011 10:49 am 
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    Aww dang. Well that obviously was the right answer. In any case, great thread btw.

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     Post Posted: Mon Dec 05, 2011 11:59 am 
    Here for the 10th Anniversary Has collected at least one unit
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    Yep, that was it!

    Glad it caused such entertainment. embarrassed that it was one wiki, I had it mentioned to me in passing by a friend who is doing her PhD on philology and medieval manuscripts. I shall tease her now about where she gets her information from!

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     Post Posted: Mon Dec 05, 2011 11:21 pm 
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    MarbitChow wrote:
    Infinity is not a number, so standard arithmetical operations cannot be performed on it. Having said that, I'd like to see what 'proof' you offer; I assume it's similar to the proof that 2=1:
    Code:
         a = x            [true for some a's and x's]
       a+a = a+x          [add a to both sides]
        2a = a+x          [a+a = 2a]
     2a-2x = a+x-2x        [subtract 2x from both sides]
    2(a-x) = a+x-2x       [2a-2x = 2(a-x)]
    2(a-x) = a-x          [x-2x = -x]
         2 = 1            [divide both sides by a-x]


    i hate that proof so much, your not allowed to divide by 0 people "trick" or no

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     Post Posted: Wed Jan 11, 2012 7:23 pm 
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    Thunder wrote:
    MarbitChow wrote:
    Infinity is not a number, so standard arithmetical operations cannot be performed on it. Having said that, I'd like to see what 'proof' you offer; I assume it's similar to the proof that 2=1:
    Code:
         a = x            [true for some a's and x's]
       a+a = a+x          [add a to both sides]
        2a = a+x          [a+a = 2a]
     2a-2x = a+x-2x        [subtract 2x from both sides]
    2(a-x) = a+x-2x       [2a-2x = 2(a-x)]
    2(a-x) = a-x          [x-2x = -x]
         2 = 1            [divide both sides by a-x]


    i hate that proof so much, your not allowed to divide by 0 people "trick" or no

    Yep, this function's domain is all values where a =/= x. Since that was the original function, the whole proof is moot.


    As for "Infinity - Infinity = 0", such subtraction is incorrect (it is called an indeterminant form) because it is unclear which "Infinity" has a higher value. While we can subtract ordinary numerals from infinity (any real number), we cannot subtract Infinity from Infinity since it is unclear which number 'reaches' Infinity first.

    For example:
    Code:
    Limit_x->Infinity [2x - x]  -> 2*Infinity - Infinity -> Infinity - Infinity (Indeterminant)

    BUT!  Technically 2x reaches infinity sooner than x does.  So Infinity - Infinity in this case would be Infinity.  You can evade these problems by  simplifying before evaluating the limit.

    Limit_x->Infinity [2x - x] -> Limit_x->Infinity [x] -> Infinity

    It is the same type of error caused by attempting to divide 0 by 0 (0/0 should be 1, but it is also undefined, and it is also 0, and it is also infinity, so the whole thing is called "indeterminant")

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