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Oh, the ambiguity!

October 9, 2013

Oh, the ambiguity!

Extensive revisions: November 12, 2019

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  1. Manfred von Willich permalink

    Your argument for leaving 0^0 undefined seems to be based on the non-uniqueness of a value that would satisfy a particular power law. However, it would be reasonable to require all the power laws to hold, such as a^(b+c)=(a^b)(a^c) and a^-b=1/a^b (the latter at least on any ring in where the power law is extended to negative exponents through division). The latter law restricts the value of 0^0 to a unique value: 1; and even where this latter law does not hold, the former allows only two possible values: 0 and 1. It would seem that your statement “infinitely many functions on N will satisfy the above requirements for exponentiation” needs revision if any of the other power laws are to remain valid, and the uniqueness argument fails in any context where negative exponents are permitted. My conclusion: in most contexts, only two options present themselves: undefined and 1, but either way your argument fails.

    • To summarize my position: If the ONLY requirements for the exponentiation function ^ on N are such that

      (1) for all n in N, n^2=n*n
      (2) for all n, m in N, n^(m+1) = n^m * n

      then there are infinitely many possibilities for ^ on ALL of N.

      If you add other conditions, you may well be able to narrow down the possibilities. Interestingly though, ^ is uniquely determined on ALL of N by only these two conditions, except for the value assigned 0^0 for which any value in N will work. It may then make sense to represent ^ as a partial function on N, leaving only 0^0 undefined. Using such a partial function, we can derive the usual Laws of Exponents with certain restrictions to avoid a zero base and exponent — just like we learned in high school. Nothing new or radical here. A novel argument for an old idea perhaps, but nothing more.

      Note that we can develop addition and multiplication on N in similar ways:

      The addition function + on N is uniquely determined by the conditions for repeated succession (not covered in my blog posting):
      (1) For all n in N, n+2=S(S(n))
      (2) For all n, m in N, n+S(m)=S(n+m).

      Likewise, the multiplication function * on N is uniquely determined by the conditions for repeated addition:
      (1) For all n in N, n*2=n+n
      (2) For all n, m in N, n*(m+1) = n*m + n.

      Similarly, the exponentiation function ^ on N is uniquely determined, except for 0^0, by the conditions for repeated multiplication:
      (1) For all n in N, n^2=n*n
      (2) for all n, m in N, n^(m+1) = n*m * n.


  2. Mike Housky permalink

    If you look at n^m as the product of a sequence of m numbers, each equal to n, then the value of 1 for 0^0 makes sense, much as 0! = 1 does. It’s long been accepted practice (in discrete math applications, anyway) that the product of an empty list is 1–the identity element for multiplication, just as the sum of an empty sequence is the identity element for addition.

    For nonzero n, this is consistent with the n^(m+0) = (n^m)(n^0) property that gets n^0 = 1 for nonzero n. Since an empty sequence of zeroes is indistinguishable from an empty sequence of nonzero values, this interpretation seems to demand that 0^0 be equal to 1 as well.

    • It is certainly tempting to put the matter to rest by drawing analogies to various non-arithmetic conventions that suggest that 0^0 might equal 1, e.g. the convention on products of empty lists. Based on purely arithmetic considerations, however, I really can’t justify any particular value for 0^0.

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