| 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_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. |
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
Result: 0^0 = indeterminate