IRL_ABooklet_2023

We first note that if $f$ is a semi-reciprocal function and $f(1)=a\in S$, then $f(a)=f(f(1))=1$. Hence, $a=f(1)=f(f(a))=1/a$, and so $a^2=1$. Since $S$ does not contain negative numbers, we must have $a=1$, i.e. $f(1)=1$ for each semi-reciprocal function $f$.

If $a\in S$ and $b=f(a)$, the condition $f(f(x))=1/x$ implies $$f(b)=f(f(a))=1/a,f(1/a)=f(f(b))=1/b,f(1/b)=f(f(1/a))=a,$$ i.e., the function $f$ cyclically permutes the four numbers $a,b,1/a,1/b$: $$a\mapsto b\mapsto \frac{1}{a}\mapsto \frac{1}{b}\mapsto a.(5)$$

To solve (a) we construct a semi-reciprocal function $f$ by defining $$\begin{array}{rlrlrl}f(1)& =1& f(2n)& =2n+1& f(2n+1)& =\frac{1}{2n}\\ f\left(\frac{1}{2n}\right)& =\frac{1}{2n+1}& f\left(\frac{1}{2n+1}\right)& =2n.& & \end{array}$$ for all integers $n\ge 1$. This settles part (a). More generally, such functions can be obtained from a partition of the positive integers greater than 1 into ordered pairs $(a,b)$ by applying (5). In the example given above, the pairs $(2n,2n+1)$ are used.

To finish part (b), we first note that we have seen above that $f(p)=p$ has the solution $p=1$. We need to show that there is no other solution.

Suppose $p\in S$ satisfies $f(p)=p$. Because $f$ is semi-reciprocal, we obtain $$p=f(p)=f(f(p))=1/p,$$ hence $p^2=1$ and therefore $p=1$, since $p\in S$ is positive.