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It's more of an open question than your reply would seem to indicate: http://mathforum.org/dr.math/faq/faq.0.to.0.power.html http://mathforum.org/dr.math/faq/f

by lutusp 12y ago

It's more of an open question than your reply would seem to indicate: http://mathforum.org/dr.math/faq/faq.0.to.0.power.html http://mathforum.org/dr.math/faq/faq.0.to.0.power.html It's less controversial when the terms are integers, but even there an argument can be made for indeterminacy. As the linked article says, "There is no one definition that always works well for 0^0" I can use 0^0 = 1 to prove that 1 = 2, but I'm sure you can anticipate the argument's form.

akater 12y ago
> "There is no one definition that always works well for 0^0" Yes there is. :-) 0^0 = 1. Actually, the only case for claiming it to be an “indeterminate”, comes from so-called “continuous exponents”. Which, arguably, something that never occurs in reality — only in exam sheets by lazy calculus teachers. Whenever you meet an algebraic equation with sum over 0 <= k <= n… and there's some m^n or m^k in it, it's always only true when 0^0 = 1. I don't claim I've seen them all but really, try to find a counterexample. What exactly was that argument for indeterminacy you were talking about? Those equations come from reasoning about meaningful entities, not chimeras of “x^x”, or “x^y”, or worse. Again, try to find, say, a physics paper with x^y in it. I haven't read much physics papers but I'm pretty sure you'll find precisely 0. As one mathematician I knew put it, “hard analysis is the primary source of all obscurantism in mathematics out there”. ;-)
- lutusp 12y ago
  > What exactly was that argument for indeterminacy you were talking about? http://www.math.vanderbilt.edu/~schectex/commerrs/#Infinity http://www.math.vanderbilt.edu/~schectex/commerrs/#Infinity From the bookmark given above, search for the phrase "That reminds me of a related question that seems to bother many students", followed by a brief and instructive exposition on the indeterminacy of 0^0. Also http://mathforum.org/dr.math/faq/faq.0.to.0.power.html http://mathforum.org/dr.math/faq/faq.0.to.0.power.html, already given. It's nice to be so sure of oneself, but in this case it's misleading.
  - lutusp 12y ago
    Ingenious solution -- faced with contradicting evidence, and rather than debate the topic on its merits, just downvote the post and walk away.
  - akater 12y ago
    I'm sorry but your links lead to the same continuity example that has nothing to do with real world (and real mathematics, as well). Can you prove (0^0 = 1) -> (1 = 2), as you claimed above? And no, we don't need “temporary definitions”. For any combinatorial investigation imaginable leads unambiguously to 0^0 = 1: https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zero https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zer... The only concept a general expression of the form x^y could possibly represent is the space of mappings y -> x. There's one and only one such mapping when x and y represent empty set. Arguing this to be wrong as in “but it can't be 1 because there are no mappings to empty space from non-empty sets!” ( = “0^x = 0”) is plain ridiculous. This definition simply does not fall apart. 0^0 is not a special case for it in any way. Enumerating numbers one can represent with 0 digits put in a string of length 0 leads to the same conclusion. It's clear that there is one and only one string of length 0, unless you demand it to contain more than 0 digits, in which case there are none. [However, this combinatorial problem is not an independent one: it's equivalent to enumerating mappings from space of strings to space of chars.] In other words, not only there are definitions that work well for all cases, including 0^0, there's actually only one such definition.
    - lutusp 12y ago
      > I'm sorry but your links lead to the same continuity example that has nothing to do with real world (and real mathematics, as well). I guess that would explain why it's located in a litany of common student errors compiled by math educators, as well as the other reference I provided. But you know what? I'm not interested in posting to a thread that downvotes posts with a probability proportional to their accuracy and relevance. From: http://www.wolframalpha.com/input/?i=0%5E0 http://www.wolframalpha.com/input/?i=0%5E0 Result: 0^0 = indeterminate

by lutusp 12y ago

akater 12y ago

> "There is no one definition that always works well for 0^0" Yes there is. :-) 0^0 = 1. Actually, the only case for claiming it to be an “indeterminate”, comes from so-called “continuous exponents”. Which, arguably, something that never occurs in reality — only in exam sheets by lazy calculus teachers. Whenever you meet an algebraic equation with sum over 0 <= k <= n… and there's some m^n or m^k in it, it's always only true when 0^0 = 1. I don't claim I've seen them all but really, try to find a counterexample. What exactly was that argument for indeterminacy you were talking about? Those equations come from reasoning about meaningful entities, not chimeras of “x^x”, or “x^y”, or worse. Again, try to find, say, a physics paper with x^y in it. I haven't read much physics papers but I'm pretty sure you'll find precisely 0. As one mathematician I knew put it, “hard analysis is the primary source of all obscurantism in mathematics out there”. ;-)

lutusp 12y ago
> What exactly was that argument for indeterminacy you were talking about? http://www.math.vanderbilt.edu/~schectex/commerrs/#Infinity http://www.math.vanderbilt.edu/~schectex/commerrs/#Infinity From the bookmark given above, search for the phrase "That reminds me of a related question that seems to bother many students", followed by a brief and instructive exposition on the indeterminacy of 0^0. Also http://mathforum.org/dr.math/faq/faq.0.to.0.power.html http://mathforum.org/dr.math/faq/faq.0.to.0.power.html, already given. It's nice to be so sure of oneself, but in this case it's misleading.
- lutusp 12y ago
  Ingenious solution -- faced with contradicting evidence, and rather than debate the topic on its merits, just downvote the post and walk away.
- akater 12y ago
  I'm sorry but your links lead to the same continuity example that has nothing to do with real world (and real mathematics, as well). Can you prove (0^0 = 1) -> (1 = 2), as you claimed above? And no, we don't need “temporary definitions”. For any combinatorial investigation imaginable leads unambiguously to 0^0 = 1: https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zero https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zer... The only concept a general expression of the form x^y could possibly represent is the space of mappings y -> x. There's one and only one such mapping when x and y represent empty set. Arguing this to be wrong as in “but it can't be 1 because there are no mappings to empty space from non-empty sets!” ( = “0^x = 0”) is plain ridiculous. This definition simply does not fall apart. 0^0 is not a special case for it in any way. Enumerating numbers one can represent with 0 digits put in a string of length 0 leads to the same conclusion. It's clear that there is one and only one string of length 0, unless you demand it to contain more than 0 digits, in which case there are none. [However, this combinatorial problem is not an independent one: it's equivalent to enumerating mappings from space of strings to space of chars.] In other words, not only there are definitions that work well for all cases, including 0^0, there's actually only one such definition.
  - lutusp 12y ago
    > I'm sorry but your links lead to the same continuity example that has nothing to do with real world (and real mathematics, as well). I guess that would explain why it's located in a litany of common student errors compiled by math educators, as well as the other reference I provided. But you know what? I'm not interested in posting to a thread that downvotes posts with a probability proportional to their accuracy and relevance. From: http://www.wolframalpha.com/input/?i=0%5E0 http://www.wolframalpha.com/input/?i=0%5E0 Result: 0^0 = indeterminate