I don't believe I ever learned the set theoretic construction of exponentiation for any number set other than than the naturals. That being said, for the sake of this question, assume all arithmetic operations are well defined over the natural numbers. Now lets define the equivalence relation $\sim_Z$ $$\forall a,b,c,d:(a,b,c,d\in\mathbb{N})[(a,b)\sim_Z(c,d)\leftrightarrow a+d=c+b]$$ Then the equivalence class of an ordered pair of natural numbers is an integer: $$\mathbb{Z}=\mathbb{N}^2/\sim_Z$$ I'm really uncertain as to how I should go about constructing exponentiation over this set, since the equivalence class $[(a,b)]_{\sim_Z}$ represents the integer $a-b$, so trying to put one of these equivalence classes to the power of another seems like trying to simplify a binomial to the power of a binomial which sounds like a pain (perhaps I'm lacking some knowledge of elementary number theory that would make this quite obvious). Next, knowing addition and multiplication over the integers, we define $\sim_Q$ $$\forall a,b,c,d:(a,b,c,d\in\mathbb{Z}\wedge b,d\neq0)[(a,b)\sim_R(c,d)\leftrightarrow ad=bc]$$ and the equivalence classes are the rational numbers: $$\mathbb{Q}=\mathbb{Z}\times(\mathbb{Z}-\{0\})/\sim_Q$$ This seems like it would be much more simple but the problem is that the rationals are not closed under exponentiation. If we construct the reals as Cauchy sequences of rationals than I would expect that $$\{a_n\}^{\{b_n\}}=\{a_n^{b_n}\}$$ (Assume the above are equivalence classes of sequences, not just sequences) but the problem is the same: $a_n^{b_n}$ is not necessarily rational. Please help, thank you

  • $\begingroup$ Calculus is set theory too. You can just use any of the ordinary calculus constructions of exponentiation. $\endgroup$
    – Lee Mosher
    Aug 8, 2016 at 2:13

1 Answer 1


Note that every $z \in \mathbb Z$ such that $z \neq [(0,0)]_{\sim_{\mathbb Z}}$ can be uniquely expressed as either

  • $z = [(n,0)]_{\sim_{\mathbb Z}}$ (in this case '$z = n$') or
  • $z = [(0,n)]_{\sim_{\mathbb Z}}$ (in this case '$z = -n$')

for some $n \in \mathbb N$. Let us say that $z \in \mathbb Z^{+}$ iff $z = [(n,0)]_{\sim_{\mathbb Z}}$ for some $n \in \mathbb N^{+}$. Also for $z = [(n,m)]_{\sim_{\mathbb Z}}$ let us write $-z := [(m,n)]_{\sim_{\mathbb Z}}$.

We can now define for $n, m \in \mathbb N$ $$ [(n,0)]_{\sim_{\mathbb Z}}^{[(m,0)]_{\sim_{\mathbb Z}}} := [(n^m,0)]_{\sim_{\mathbb Z}} $$ and $$ [(0,n)]_{\sim_{\mathbb Z}}^{[(m,0)]_{\sim_{\mathbb Z}}} := \begin{cases} [(n^m,0)]_{\sim_{\mathbb Z}} & \text{, if } 2 \mid m \\ [(0,n^m)]_{\sim_{\mathbb Z}} & \text{, if } 2 \not \mid m \end{cases} $$

We can't define $[(0,n)]_{\sim_{\mathbb Z}}^{[(0,m)]_{\sim_{\mathbb Z}}}$ (as elements of $\mathbb Z$ that respect the usual meaning of exponentiation), since $\mathbb Z$ is not closed under exponentiation with negative powers. If you would like to have a total function, maybe just let $$ [(0,n)]_{\sim_{\mathbb Z}}^{[(0,m)]_{\sim_{\mathbb Z}}} := [(0,0)]_{\sim_{\mathbb Z}}^{[(0,m)]_{\sim_{\mathbb Z}}} := [(1,0)]_{\sim_{\mathbb Z}}. $$ Now, for $x = [(p,q)]_{\sim_{\mathbb Q}} \in \mathbb Q$ and $z \in \mathbb Z$ we define $$ \left( [(p,q)]_{\sim_{\mathbb Q}} \right)^z := \begin{cases} [([(0,0)]_{\sim_{\mathbb Z}},[(1,0)]_{\sim_{\mathbb Z}})]_{\sim_{\mathbb Q}} & \text{, if } p = [(0,0)]_{\sim_{\mathbb Z}} \\ [([(1,0)]_{\sim_{\mathbb Z}},[(1,0)]_{\sim_{\mathbb Z}})]_{\sim_{\mathbb Q}} & \text{, if } z = [(0,0)]_{\sim_{\mathbb Z}} \wedge p \neq [(0,0)]_{\sim_{\mathbb Z}}\\ [(p^z,q^z)]_{\sim_{\mathbb Q}} & \text{, if } z \in \mathbb Z^{+} \wedge p \neq [(0,0)]_{\sim_{\mathbb Z}} \\ [(q^{-z}, p^{-z})_{\sim_{\mathbb Q}} & \text{, if } z \in \mathbb Z \setminus \mathbb Z^{+} \wedge z \neq [(0,0)]_{\mathbb Z} \wedge p \neq [(0,0)]_{\sim_{\mathbb Z}} \end{cases} $$

This can easily be generalized to those cases where $x^y \in \mathbb Q$ for $x,y \in \mathbb Q$ as should be apparend from the above.

Now, defining $x^y$ for $x,y \in \mathbb R$ can be done as follows: First suppose that $x,y \in \mathbb Q$ Then $$ \vec{x^y} := \{ q \in \mathbb Q \mid \exists a,b \in \mathbb \colon a^b \in \mathbb Q \wedge q < a^b \}. $$

Choose (here we need some form of choice in order to do that for every $x,y \in \mathbb Q$ simultaneously) a monotone increasing sequence $(q_0, q_1, \ldots)$ of rationals such that $\lim_{n \to \infty} q_n = \sup \vec{x^y}$ and define $$ x^y := (q_0, q_1, \ldots). $$

Now that you've done that you may define $x^y$ for $x \in \mathbb R$ and $y \in \mathbb Q$ just like you suggested in your post and finally, if $y \in \mathbb R^{+}$, fix an increasing sequence $y_n \in \mathbb Q^{+}$ for $n \in \mathbb N$ with limit $y$ and define $$ x^y := \sup \{ x^{y_n} \mid n \in \mathbb N \} $$ and similarly for $x^{-y}, (-x)^{y}, \ldots$. (You may rewrite this using $\vec{x^y}$, if you want to.)

This post should teach you two things:

  1. You can define exponentiation in a set theoretical construction of the reals.
  2. Abstraction is your friend. All of this could've been siginficanlty simplified by adding a layer of abstraction and referring to already developed tools (in this case, some basic analysis which is used - but kept hidden - in my answer).

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