# Verifying the Existence of Real Valued Functions

How can we be certain that expressions involving real numbers exist? For example, how do we know the functions \begin{align*} f(x)&=x^\pi,\\ g(x)&=\pi^x \end{align*} are well defined for all entries $x\in\mathbb{R}^+\cup\{0\}$, and $x\in\mathbb{R}$ respectively? I'm willing to assume the real number system is a field, so multiplication and addition of real numbers are well defined. As such, is it sufficient to show there's a convergent power series expansion for the function? Or, alternatively, could you use the fact that there is a sequence of rationals $q_n\to \pi$, giving rise to a sequence of functions \begin{align*} f_n=x^{q_n}\\ g_n={q_n}^x \end{align*} and prove that both of these will converge pointwise for all $x\in\mathbb{R}$?

• Does it help to write these as $e^{\pi\log x}$ and $e^{x\log \pi}$? In other words, do you believe that $\log x$ and $\log\pi$ are real? – MPW Jun 3 '17 at 3:55
• Actually $x^\pi$ is not defined for all $x\in \mathbb R.$ – zhw. Jun 3 '17 at 4:07
• I suppose it does in this context, and yes I do believe those functions are real, but I'm mainly interested in the general setting. – user238841 Jun 3 '17 at 4:07
• @zhw good point, that's easy to overlook. Will update accordingly – user238841 Jun 3 '17 at 4:13

Suppose we define $\mathbb R$ by means of equivalence classes of $\mathbb Q$-Cauchy sequences (...That is, sequences $(q_n)_{n \in \mathbb N}$ in $\mathbb Q$ such that $\forall q\in \mathbb Q^+\;\exists n_0\in \mathbb N \;\forall m,n\geq n_0\;(q>|q_m-q_n|)$...).
We show that Cauchy sequences in $\mathbb R$ converge to members of $\mathbb R.$
We show that whenever $x\in \mathbb R^+$ and $a,b\in \mathbb N$ there is a unique $y\in \mathbb R^+$ such that $x^a=y^b.$ We define $x^{a/b}=y$.
We then show that if $x,y\in \mathbb R^+$ and if $(q_n)_n, (q'_n)_n$ are any $\mathbb Q$-Cauchy sequences converging to $y,$ then the sequences $(x^{q_n})_n,\; (x^{q'_n})_n$ are Cauchy sequences in $\mathbb R,$ both converging to the same value $z\in \mathbb R$. We define $x^y=z.$