(related to Proposition: General Powers of Positive Numbers)
We have already proven that rational exponents, the power function is identical to the exponential function of general base, i.e. \[a^x=\exp_a(x),\quad\quad x\in\mathbb Q.\] Now, we will prove that the expression \(a^x\) makes perfectly sense and this equation also holds^{1}, if we allow \(x\) to be a general real number. According to the definition of real numbers, a real number \(x\) is a class of all rational Cauchy sequences, which equal each other except a difference, which is a rational Cauchy sequence converging to zero, formally \[x\in\mathbb R\Longleftrightarrow x=(x_n)_{n\in\mathbb N}+I\] where \((x_n)_{n\in\mathbb N}\) is a rational Cauchy sequence representing the real number \(x\), and
\[I:=\{(i_n)_{n\in\mathbb N}~|~i_n\in\mathbb Q,\lim i_n=0\}\] is the set of all rational sequences, which converge to \(0\). Now, we have
\[\begin{array}{rcll} a^x&=&\lim_{n\to\infty}a^{x_n+i_n}&\text{due to the definition of real numbers}\\ &=&\lim_{n\to\infty}\exp_a(x_n+i_n)&\text{definition of rational powers of positive numbers}\\ &=&\lim_{n\to\infty}\exp_a(x_n)\cdot \exp_a(i_n)&\text{functional equation of exponential function of general base}\\ &=&\exp_a(x)\cdot \exp_a(0)&\text{continuity of exponential function of general base}\\ &=&\exp_a(x)\cdot \exp(0\cdot \ln(a))&\text{definition of exponential function of general base}\\ &=&\exp_a(x)\cdot \exp(0)&\text{multiplication by }0\\ &=&\exp_a(x)\cdot 1&\text{proposition stating that} \exp(0)=1\\ &=&\exp_a(x)&\text{multiplication by }1\\ \end{array}\]
Above, we have used, among others, the following propositions: * functional equation of exponential function of general base, * continuity of exponential function of general base, and the * proposition stating that \(\exp(0)=1\).
E.g. a general power \(3^\pi\) makes perfectly sense, although it makes no sense to say that we "multiply \(\pi\) copies of \(3\) together". ↩