Let $g: \Z \to \R$ be the mapping defined as: :$(1): \quad \map g n = \map H {\dfrac 1 n, \dfrac 1 n, \dotsc, \dfrac 1 n}$ Let $k \in \Z_{>0}$. We have: {{begin-eqn}} {{eqn | l = \map g {n^k}       | r = \map g {n \times n^{k - 1} }       | c =  }} {{eqn | r = \map g n + \map g {n^{k - 1} }       | c = Axiom 7 }} {{eqn | r = \map g n + \map g n + \map g {n^{k - 2} }       | c = Axiom 7 again }} {{eqn | r = \underbrace {\map g n + \map g n + \dotsb + \map g n}_{\text {$k - 1$ times} } + \map g {n^{k - \paren {k - 1} } }       | c = after $k - 1$ applications of Axiom 7 }} {{eqn | r = \underbrace {\map g n + \map g n + \dotsb + \map g n}_{\text {$k - 1$ times} } + \map g n       | c =  }} {{eqn | n = 2       | r = k \map g n       | c =  }} {{end-eqn}} Let $r, s \in \R$ and $n \in \Z_{>0}$. Let $m$ satisfy: :$(3): \quad r^m \le s^n \le r^{m + 1}$ From $(2)$ and Axiom 5: :$\map g {r^m} \le \map g {s^n} \le \map g {r^{m + 1} }$ and so: :$m \map g r \le n \map g s \le \paren {m + 1} \map g r$ Taking logarithms of $(3)$: :$m \ln r \le n \ln s \le \paren {m + 1} \ln r$ which holds because Logarithm is Strictly Increasing. Hence: :$\size {\dfrac {\map g s} {\map g r} - \dfrac {\map \ln s} {\map \ln r} } \le \dfrac 1 n$ We have that $n$ is an arbitrary positive integer. Hence the {{RHS}} of the above can be made arbitrarily small. Hence: :$\size {\dfrac {\map g s} {\map g r} - \dfrac {\map \ln s} {\map \ln r} } = 0$ and so: :$(4): \quad \dfrac {\map g s} {\map \ln s} = \dfrac {\map g r} {\map \ln r} = A$ for some constant $A$. That is: :$(5): \quad \map g s = A \map \ln s$ for $s \in \Z$. Let $p$ be a rational number such that $0 < p < 1$. That is: :$p = \dfrac t n$ for some integers $t$ and $n$. Thus we can set: :$q = 1 = p = \dfrac {n - t} n$ Axiom 8 gives: :$\map g n = \map H {\dfrac 1 n, \dfrac 1 n, \dotsc, \dfrac 1 n} = \map H {\dfrac 1 n, \dfrac {n - t} n} + \dfrac t n \map g t + \dfrac {n - t} n \map g {n - t}$ Using $(5)$ and gathering up terms: :$\map H {\dfrac 1 n, \dfrac {n - t} n} = A \paren {\dfrac t n} \ln \dfrac t n + A \paren {\dfrac {n - t} n} \ln \dfrac {n - t} n$ which gives: :$(6): \quad \map H {p, 1 - p} = A p \ln p + A \paren {1 - p} \, \map \ln {1 - p}$ for rational $p$. Axiom 6 gives that $H$ is continuous. Thus $(6)$ extends to all real $p$ such that $0 < p < 1$. It remains to be demonstrated that: :$(7): \quad \map H {p_1, p_2, \ldots, p_N} = A \displaystyle \sum_{i \mathop = 1}^N p_i \ln p_i$ where: :$0 < p_i < 1$ for all $p_i$ :$p_1 + p_2 + \dotsb + p_N$ for all $N \in \Z_{\ge 2}$. We have demonstrated that $(7)$ is true for $N = 2$. Assume the induction hypothesis that $(7)$ holds for $N = k$. Consider: :$\map H {p_1, p_2, \ldots, p_k, p_{k + 1} }$ Let: :$p = p_1 + p_2 + \dotsb + p_k$ :$q = p_{k + 1}$ and apply Axiom 8: {{begin-eqn}} {{eqn | l = \map H {p_1, p_2, \ldots, p_k, p_{k + 1} }       | r = \map H {p, q} + p \map H {\dfrac {p_1} p, \dfrac {p_2} p, \dotsc, \dfrac {p_k} p} + q \map H 1       | c =  }} {{eqn | r = A p \ln p + A q \ln q + p A \sum_{i \mathop = 1}^k \dfrac {p_i} p \ln \dfrac {p_i} p       | c = from the induction hypothesis }} {{eqn | r = A p \ln p + A p_{k + 1} \ln p_{k + 1} + A \sum_{i \mathop = 1}^k p_i \paren {\ln p_i - \ln p}       | c = Difference of Logarithms }} {{eqn | r = A p \ln p + A p_{k + 1} \ln p_{k + 1} + A \sum_{i \mathop = 1}^k p_i \ln p_i - A p \ln p       | c = as $A \displaystyle \sum_{i \mathop = 1}^k p_i = p$ }} {{eqn | ll= \leadsto       | l = \map H {p_1, p_2, \ldots, p_k, p_{k + 1} }       | r = A \sum_{i \mathop = 1}^{k + 1} p_i \ln p_i       | c =  }} {{end-eqn}} Hence the result from the Principle of Mathematical Induction. {{qed}}