Exercise 8.6

Proof. Suppose to the contrary that there is such a function $h$. Then clearly $h(1)=3$ and $h(2)=\sqrt{h(1)-1}=\sqrt{3-1}=\sqrt{2}$. Now, since $1<2<4$, we clearly have $1<\sqrt{2}<\sqrt{4}=2$. Thus $0<\sqrt{2}-1<1$ so that $h(3)=\sqrt{h(2)-1}=\sqrt{\sqrt{2}-1}$ is defined. However, we also have that $0<h(3)=\sqrt{\sqrt{2}-1}<1$ since $0<\sqrt{2}-1<1$, and hence $h(3)-1<0$. We then have that

which is of course impossible since a square is always non-negative. This contradiction shows that such a function $h$ cannot exist. □

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 $f:\left\{1,\dots ,m\right\}\to {ℝ}_{\text{+}}$. If $f(m)>1$ then clearly $f(m)-1>0$ so that $\rho (f)={[f(m)-1]}^{1∕2}$ is defined and positive. If $f(m)\le 1$ then clearly $\rho (f)=5$ is also defined and positive. Since $m$ and $f$ were arbitrary, this shows that $\rho$ is a well-defined function with range ${ℝ}_{\text{+}}$.

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

To see that this $h$ satisfies the recursion formula, clearly $h(1)=3$, and, for $i>1$, we have

as desired.

To show that this function is unique, suppose that $g$ and $h$ both satisfy the recursive formula. We show by induction that $g(n)=h(n)$ for all $n\in {\mathbb{Z}}_{\text{+}}$ so that clearly $g=h$. First, obviously $g(1)=3=h(1)$. Now suppose that $g(n)=h(n)$ for $n\in {\mathbb{Z}}_{\text{+}}$ so that $n+1>1$. Then, if $g(n)=h(n)>1$ then we have $g(n+1)={[g(n)-1]}^{1∕2}={[h(n)-1]}^{1∕2}=h(n+1)$ since $g(n)=h(n)$ and the roots are taken to be positive. Similarly, if $g(n)=h(n)\le 1$, then $g(n+1)=5=h(n+1)$. Thus in either case $g(n+1)=h(n+1)$, which completes the induction. □