Real Exponents

Theorem 17.1

Theorem 17.1. If $x$ is a real number, there exists an increasing rational sequence $\{ r_n \}$ with limit $x$. 

Proof. By Theorem 7.8, $\exists r_1 \in \mathbb{Q}$ such that $x-1 < r_1 < x$. In the same manner, choose $r_2 \in \mathbb{Q}$ such that $\max \left \{ x - \frac{1}{2}, r_1 \right \} < r_2 < x$. In general, having chosen a rational number $r_n < x$, choose $r_{n+1} \in \mathbb{Q}$ such that $\max \left \{ x - \frac{1}{n+1}, r_n \right \} < r_{n+1} < x$. Since $r_n < r_{n+1}$, $\{ r_n \}$ is increasing.
Note that $$x - \frac{1}{n} < r_n < x.$$ By Theorem 12.1, Corollary 12.4, Theorem 14,3, and $\lim_{n \to \infty} \frac{1}{n} = 0$, $$\lim_{n \to \infty} \left( x - \frac{1}{n} \right) = x = \lim_{n \to \infty} x \Longrightarrow \lim_{n \to \infty} r_n = x. \blacksquare$$

Remark

Remark. If $a \geq 1$ and $x$ is a real number, we choose an increasing rational sequence $\{ r_n \}$ such that $\lim_{n \to \infty} r_n = x$. 
Since $r_n \leq r_{n+1}$, $a^{r_n} \leq a^{r_{n+1}}$. Then the sequence $\{ a^{r_n} \}$ is increasing.
Let $R = \{ r_n \, | \, n \in \mathbb{P} \}$. By Theorem 16.2, $R$ is bounded. Then $\exists l = \sup R$ by A13. 
Let $\varepsilon > 0$. Note that $l - \varepsilon$ is not an upper bound for $R$. Then $\exists N \in \mathbb{P}$ such that $l - \varepsilon < r_n, \forall n \geq N$. ($\{ r_n \}$ is increasing.) Since $\{ r_n \}$ converges to $x$, $\exists N' \in \mathbb{P}$ such that $|r_n - x| < \varepsilon, \forall n \geq N'$. Then $$l - \varepsilon < r_n < x + \varepsilon \leq l + \varepsilon, \forall n \geq \max \{ N, N' \} \\ \Longrightarrow |r_n - l| < \varepsilon, \forall n \geq \max \{ N, N' \} \Longrightarrow \lim_{n \to \infty} r_n = l.$$ By Theorem 10.3, $x = l$. Thus $r_n \leq x, \forall n \in \mathbb{P}$. 
Let $r \in \mathbb{Q}$ such that $r > x$. Then $$r_n \leq x < r, \forall n \in \mathbb{P} \Longrightarrow a^{r_n} < a^r.$$ Thus $\{ a^{r_n} \}$ is bounded, and by Theorem 16.2, $\{ a^{r_n} \}$ is convergent. 

위 remark에 의해서 수열 $\{ a^{r_n} \}$은 항상 수렴한다고 말할 수 있으므로, 이와 같이 real exponent를 정의할 수 있다.

Definition 17.2

Definition 17.2. Let $a \geq 1$, and let $x$ be a real number. We define $$a^x = \lim_{n \to \infty} a^{r_n}$$ where $\{r_n\}$ is an increasing rational sequence with limit $x$. If $0 < a < 1$ and $x$ is a real number, we define $$a^x = \left( \frac{1}{1/a} \right)^x = \left( \frac{1}{a} \right)^{-x}.$$

Lemma 17.3

Lemma 17.3. Let $a \geq 1$, and let $x$ be a real number. Let $\{r_n\}$ and $\{s_n\}$ be increasing rational sequences such that $\lim_{n \to \infty} r_n = x = \lim_{n \to \infty} s_n$. Then $$\lim_{n \to \infty} a^{r_n} = \lim_{n \to \infty} a^{s_n}.$$

Proof. Define $$R_n = r_n - \frac{1}{n}, \quad S_n = s_n - \frac{1}{n}, \forall n \in \mathbb{P}.$$ Since $r_n \leq x, s_n \leq x$ and $R_n < r_n, S_n < s_n$ for all $n \in \mathbb{P}$, we have $R_n < x$ and $S_n < x$ for all $n \in \mathbb{P}$. Furthermore, $\{ R_n \}$ and $\{ S_n \}$ is increasing. 
Note that $$\lim_{n \to \infty} a^{R_n} = \lim_{n \to \infty} a^{r_n - \frac{1}{n}} = \lim_{n \to \infty} a^{r_n} \cdot a^{- \frac{1}{n}} = \frac{\lim_{n \to \infty} a^{r_n}}{\lim_{n \to \infty} a^{\frac{1}{n}}} = \lim_{n \to \infty} a^{r_n}$$ by Theorem 12.9 and Theorem 16.4. Similarly, we have $$\lim_{n \to \infty} a^{S_n} = \lim_{n \to \infty} a^{s_n}.$$  
Since $R_1 < x$, we can choose $m_1 \in \mathbb{P}$ such that $R_1 < S_{m_1}$. (If not, then $S_n \leq R_1, \forall n \in \mathbb{P}$, which means that $\{ S_n \}$ is bounded above by $R_1$. By the remark, $x \leq R_1 \bigotimes$.) In the same manner, we can choose $n_2 \in \mathbb{P}$ such that $S_{m_1} < R_{n_2}$ because $S_{m_1} < x$. Continuing in this way, we can choose an increasing sequence $$R_1 = R_{n_1} < S_{m_1} < R_{n_2} < S_{m_2} < \cdots < x.$$ Define the sequence $\{ b_k \}$ by $b_{2k-1} = R_{n_k}$ and $b_{2k} = S_{m_k}$ for all $k \in \mathbb{P}$. Since $\{ b_k \}$ is bounded by $|x|$, $\{ b_k \}$ is convergent by Theorem 16.2. Then we have $$\lim_{n \to \infty} a^{r_n} = \lim_{n \to \infty} a^{R_n} = \lim_{k \to \infty} a^{R_{n_k}} = \lim_{k \to \infty} a^{b_{2k-1}} \\ = \lim_{k \to \infty} a^{b_k} = \lim_{k \to \infty} a^{b_{2k}} = \lim_{k \to \infty} a^{S_{m_k}} = \lim_{n \to \infty} a^{S_n} = \lim_{n \to \infty} a^{s_n}. \blacksquare$$

Theorem 17.4

Theorem 17.4. Let $a$ and $b$ be positive numbers. Then
(i) $a^{x+y} = a^x a^y$ for $x, y \in \mathbb{R}$
(ii) $(a^x)^y = a^{xy}$ for $x, y \in \mathbb{R}$
(iii) $(ab)^x = a^x b^x$ for $x \in \mathbb{R}$
(iv) $a^{-x} = \frac{1}{a^x}$ for $x \in \mathbb{R}$
(v) $\left( \frac{a}{b} \right)^x = \frac{a^x}{b^x}$ for $x \in \mathbb{R}$
(vi) If $a > 1$ and $x < y$, then $a^x < a^y$.
(vii) If $0 < a < 1$ and $x < y$, then $a^x > a^y$.
(viii) If $x > 0$ and $a < b$, then $a^x < b^x$.
(ix) If $x < 0$ and $a < b$, then $a^x > b^x$.

Proof. Let $\{ r_n \}$ and $\{ s_n \}$ be the sequences to be convergent to $x$ and $y$, respectively. 
(i) If $a \geq 1$, then $\lim_{n \to \infty} a^{r_n} = a^x$ and $\lim_{n \to \infty} a^{s_n} = a^y$. Note that $\lim_{n \to \infty} (r_n + s_n) = x + y$ by Theorem 12.2. Thus we have $$\lim_{n \to \infty} a^{x+y} = \lim_{n \to \infty} a^{r_n + s_n}  = \lim_{n \to \infty} a^{r_n} \cdot a^{s_n} = \lim_{n \to \infty} a^{r_n} \lim_{n \to \infty} a^{s_n} = a^x a^y$$ by Theorem 12.6. 
If $0 < a < 1$, then $a^{x+y} = \left( \frac{1}{a} \right)^{-(x+y)} = \left( \frac{1}{a} \right)^{-x - y} = \left( \frac{1}{a} \right)^{-x} \left( \frac{1}{a} \right)^{-y} = a^x a^y$. $\blacksquare$