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    what is the answer?

    0 or 1?
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    (Original post by MC To Tha Zee)
    what is the answer?

    0 or 1?

    Probably 1.
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    (Original post by MC To Tha Zee)
    what is the answer?

    0 or 1?
    On a sunny day it can be 3, on a rainy day, it might be five. If you were to talk to a philosopher, he/she might ask you how many times can you take nothing out of nothing? answer, infinite times

    Note: 0^0 is essentially 0^1/0^1 (0,1,^,0,1,^,/ in reverse polish).
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    Anything to the power of 0 is 1, so I assume it is the same in this case.
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    (Original post by byb3)
    Anything to the power of 0 is 1, so I assume it is the same in this case.
    Most mathematicians tend to agree 0^0 = 1 although strictly it's undefined. See: http://www.faqs.org/faqs/sci-math-fa...lnumbers/0to0/
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    (Original post by byb3)
    Anything to the power of 0 is 1, so I assume it is the same in this case.
    As camford implied 0^0 must be indeterminate as it is equivalent to 0/0 which you can pretty much make eqeual anything - 0/0 = k (some integer) => 0 = k.0 <=> 0=0. See.
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    (Original post by It'sPhil...)
    As camford implied 0^0 must be indeterminate as it is equivalent to 0/0 which you can pretty much make eqeual anything - 0/0 = k (some integer) => 0 = k.0 <=> 0=0. See.
    Divison by nought is undefined and as much is meaningless - you can not attribute a value to the result when dividing by 0, because the process itself is impossible.
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    its 0
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    (Original post by theone)
    Divison by nought is undefined and as much is meaningless - you can not attribute a value to the result when dividing by 0, because the process itself is impossible.
    that was my point
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    (Original post by It'sPhil...)
    that was my point
    I was just saying that your statement 0/0 = k where k can be anything you want is really not true because you can not have x/0 in the first place, whatever x is. Although I agree that it is indeterminate.
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    I can prove that 0^0 is undefined:

    let y = 0^0
    take natural logs of each side : ln (y) = 0ln0
    ln0 is undefined. This means that the whole of the rhs is undefined. Since RHS = LHS, ln(y) is undefined, meaning y is undefined. QED.
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    oh your 'QED' bits at the end of all your mathematical genious posts, make me giggle!
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    It is more philosophical than you think...
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    (Original post by Camford)
    It is more philosophical than you think...
    Is it? I always assosciated it mathematical proofs...
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    (Original post by theone)
    Is it? I always assosciated it mathematical proofs...
    Now, consider the situation where you have an empty room. How many times can you take nothing out of the empty room? Once, twice? Or may be more times?
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    (Original post by Camford)
    Now, consider the situation where you have an empty room. How many times can you take nothing out of the empty room? Once, twice? Or may be more times?
    Hehe, I thought you meant the use of QED
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    (Original post by chrisbphd)
    I can prove that 0^0 is undefined:

    let y = 0^0
    take natural logs of each side : ln (y) = 0ln0
    ln0 is undefined. This means that the whole of the rhs is undefined. Since RHS = LHS, ln(y) is undefined, meaning y is undefined. QED.
    Does it matter that ln0 is undefined? sqrt(-1) is undefined but that doesn't stop people using it, ln(-1) is undefined but that doesn't stop people saying that ln(-1) = (2n-1)pi*i.

    Just because ln0 is undefined does not mean that 0*ln0 is not equal to 0. Would you not also say that 0*arcsin(2) = 0 ? So, in that case we have:

    lny = 0ln0 = 0
    => y = e^0 = 1

    -

    Couldn't you also say this:

    0^0 = 0^(1-1) = 0^(1+(-1)) = 0^1 + 0^(-1)
    = 0 + 1/0 = infinity

    ?
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    (Original post by mikesgt2)
    couldn't you also say this:

    0^0 = 0^(1-1) = 0^(1+(-1)) = 0^1 + 0^(-1)
    = 0 + 1/0 = infinity

    ?
    no - x^(a+b) = x^a.x^b not x^a + x^b
 
 
 
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