Exercise 8.5

Proof. First, for any function $f:\left\{1,\dots ,m\right\}\to {ℝ}_{\text{+}}$, define

Consider any $m\in {\mathbb{Z}}_{\text{+}}$ and any function $f:\left\{1,\dots ,m\right\}\to {ℝ}_{\text{+}}$. Since $f(m)\in {ℝ}_{\text{+}}$, it follows that $f(m)+1\in {ℝ}_{\text{+}}$ also so that $\rho (f)={[f(m)+1]}^{1∕2}$ is defined and is positive. Hence $\rho$ is a well-defined function with range ${ℝ}_{\text{+}}$ since $m$ and $f$ were arbitrary. It then follows from the principle of recursive definition (Theorem 8.4) that there is a unique function $h:{\mathbb{Z}}_{\text{+}}\to {ℝ}_{\text{+}}$ such that

It is easy to see that this $h$ satisfies the required property since $h(1)=3$ and

for $i>1$ as desired.

Now we show that such a function is unique. Suppose that $g$ and $h$ both satisfy the inductive formula. We show by induction that $g(i)=h(i)$ for all $i\in {\mathbb{Z}}_{\text{+}}$, from which it clearly follows that $g=h$. First, we clearly have $g(1)=3=h(1)$. Now suppose that $g(i)=h(i)$ for $i\in {\mathbb{Z}}_{\text{+}}$. Then we have that $i+1>1$ so that $g(i+1)={[g(i)+1]}^{1∕2}={[h(i)+1]}^{1∕2}=h(i+1)$ since $g(i)=h(i)$ and we are taking the positive root. This completes the induction. □