If one defines the integers as equivalence classes of pairs of natural numbers, there is a (canonical?) way to define addition and multiplication for the integers based on addition and multiplication for the natural numbers, but I’m not sure if the same may be said for exponentiation.

One could define exponentiation for the integers similar to how it can be defined for the natural numbers, by applying the recursion principle to multiplication, but I’m wondering if there is an approach that “agrees in spirit” with how addition and multiplication on $\mathbf{Z}$ can be defined in terms of those operations in $\mathbf{N}$.

  • $\begingroup$ of course you could. use the biomial theorem. $\endgroup$
    – miracle173
    Jul 1, 2015 at 2:40
  • $\begingroup$ I'm not sure I follow. $\endgroup$
    – user235033
    Jul 1, 2015 at 2:44

1 Answer 1


It follows from the binomia theorem, how exponentiation has to be declared for a pair: $$(a,b)^n=(a-b)^n=\sum(-1)^k \binom{n}{k} a^kb^{n-k}=\sum_{k \; \text{even}}\binom{n}{k} a^kb^{n-k} - \sum_{k \;\text{odd}}\binom{n}{k}a^kb^{n-k}=(\sum_{k \; \text{even}} \binom{n}{k}a^kb^{n-k}, \sum_{k \;\text{odd}}\binom{n}{k}a^kb^{n-k})$$ This is exponentiation for natural number exponents.

For nonnatural exponents exponentiation does not always result in a natural number. But one can define $(a,b)^{(c,d)}$ is the solution $x$ of $$(a,b)^c=x(a,b)^d$$ if one exists.

  • $\begingroup$ Did you forget the binomial coefficients, there? $\endgroup$ Jul 1, 2015 at 3:00
  • $\begingroup$ Also, this doesn't address how to deal with nonnatural exponents. $\endgroup$ Jul 1, 2015 at 3:02
  • $\begingroup$ @Cameron Buie: yes, I forgot them. Thx $\endgroup$
    – miracle173
    Jul 1, 2015 at 3:18
  • $\begingroup$ Of course! sorry for the trouble. $\endgroup$
    – user235033
    Jul 1, 2015 at 3:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .