(a) Let $m,n,p,q$ be integers, $n,q > 0$ with $m/n=p/q$. Prove$$(b^m)^{1/n}=(b^p)^{1/q}$$Hence it makes sense to define $b^{m/n}=(b^{m})^{1/n}$.
(b) Prove $b^{r+s}=b^rb^s$ for $r,s∈\mathbb{Q}$.
(c) For $x∈\mathbb{R}$, define $B(x) = \{b^t~|~t∈\mathbb{Q},~t≤x\}$. When $r∈\mathbb{Q}$, prove$$b^r=\text{sup }B(r)$$Hence it makes sense to define$$b^x=\text{sup }B(x)$$(d) For $x,y∈\mathbb{R}$, prove $b^{x+y}=b^xb^y$.
Proof: (a) Recall $(z_1z_2)^{1/n}=z_1^{1/n}z_2^{1/n}$ for natural $n$. For natural $n_1,n_2$ it is also clear that $(z^{n_1})^{n_2}=z^{n_1n_2}$, and when $n_1,n_2$ may be negative the equality holds by case analysis together with the definition $z^{-n}=(z^n)^{-1}$. Therefore, we derive$$(b^m)^{1/n}=(b^{1/n})^m=(((b^{1/n})^m)^q)^{1/q}=((b^{1/n})^{mq})^{1/q}=$$$$((b^{1/n})^{pn})^{1/q}=(((b^{1/n})^n)^p)^{1/q}=(b^p)^{1/q}$$ (b) The case is clear for integer $r,s$. As well, we note $z^{1/n_1n_2}=(z^{1/n_1})^{1/n_2}$ as $((z^{1/n_1})^{1/n_2})^{n_1n_2}=(((z^{1/n_1})^{1/n_2})^{n_2})^{n_1}=z$. Otherwise let $r=r_1/r_2$ and $s=s_1/s_2$. We have$$b^{r+s}=b^{r_1/r_2+s_1/s_2}=b^{(r_1s_2+r_2s_1)/(r_2s_2)}=(b^{r_1s_2+r_2s_1})^{1/r_2s_2}=$$$$(b^{r_1s_2}b^{r_2s_1})^{1/r_2s_2}=(((b^{r_1})^{s_2})^{1/s_2})^{1/r_2}(((b^{s_1})^{r_2})^{1/r_2})^{1/s_2}=b^rb^s$$ (c) Lemma 1: When $a = a_1/a_2 > 0$ is rational we have $b^a > 1$. This implies $b^a > b^c$ for rationals $a > c$. Proof: Rewrite $a_2 > 0$ so that $a_1 > 0$ so by induction $b^{a_1} > 1$ and it suffices to prove $b^{1/a_2} > 1$ for $b > 1$. Here it suffices to prove for $c ≤ 1$ then $c^n ≤ 1$ for natural $n$, which is clear by induction $c^n = cc^{n-1} ≤ c ≤ 1$.
Now, for $t≤r$ we show $b^r ≥ b^t ⇔ b^t(b^{r-t}-1) ≥ 0$, so since $b^t$ is nonnegative and by the lemma $b^{r-t}-1$ is nonnegative we have $b^r$ is an upper bound of $B(r)$. As well, if $z < b^r$ then since $b^r∈B(r)$ we have $z$ is not an upper bound of $B(x)$ and thus $b^r = \text{sup }B(r)$.
(d) Lemma 2: Let $X⊆\mathbb{R}$ be bounded above such that when $x∈X$ and $y≤x$, then $y∈X$. We see $\text{sup }X = \text{sup }(X \setminus \{\text{sup }X\})$. Proof: $z = \text{sup }X$ bounds the latter, so assume $y < z$ also bounds. But $y$ doesn't bound $X$ and thus $y < x$ for some $x∈X$ and we may choose some $y < x_1 < x ≤ z$ so $x_1 ∈ X \setminus \{\text{sup }X\}$ yet $y < x_1$ and $z$ is the smallest bound.
Let $B^*(x) = \{v ≤ b^t~|~t∈\mathbb{Q},~t≤x\}$ so that $\text{sup }B^*(x) = \text{sup }B(x)$ and it serves as an equivalent definition for $b^x$. Now if $b^x∈B^*(x)$ then by the above lemma we may remove it, and as well $b^t$ for $t < x$ to define $B^{**}(x)$ with an equivalent supremum. Now we must show that when $z < \text{sup }B^{**}(x)$ then $z < b^v$ for some rational $v < x$. We already see this is the case when for some rational $v ≤ x$, so we must construct rational $v' < v$ such that $z < b^{v'} < b^v$; this may be done by observing lemma 1 and parts (a),(b), and (g) to show $b^{1/n}$ tends to 1 and setting $v' = v - 1/n$ for $1 < b^{1/n} < b^v/z$.
Now, suppose $b^xb^y < b^{x+y}$; then $b^xb^y < b^v$ for some rational $v < x+y$ by the above discussion. Set rational $v - y < v_1 < x$ and $v - v_1 < v_2 < y$ so $v_1+v_2 > v$ and $b^xb^y ≥ b^{v_1}b^{v_2} = b^{v_1+v_2} > b^{v}$.
Suppose $b^{x+y} < b^xb^y$. Then $(b^x)^{-1}b^{x+y} < b^{v_2}$ for some rational $v_2 < y$. Choose rational $v_2 < v_2 < y$ and set $v_4 = v_3 - v_2 > 0$, and then choose $x-v_4 < v_1 < x$. We have $b^{v_2} > (b^x)^{-1}b^{x+y} ≥ (b^x)^{-1}b^{v_1+v_2+v_4}$ implying $1 > (b^x)^{-1}b^{v_1}b^{v_4}$ then $b^x > b^{v_1+v_4}$. But $v_1+v_4 > x$ and choosing rational $x < x' < v_1+v_4$ we see $b^{x'}$ is an upper bound of $b^x$ and $b^{v_1+v_4} > b^{x'} ≥ b^x$ is a contradiction.$~\square$
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