It can be shown that $f(x)=x$ is the only function from $(0,\infty)$ to $(0,\infty)$ with the following properties : (1) $f(x+1)=f(x)+1$ and (2) $f\left(\frac{1}{f(x)}\right)=\frac{1}{x}$. Proof: It is clear that $f(x)=x$ works. Now suppose that $f$ is any function with the desired properties. By (1) $f$ is periodic and by (2) it follows that $f$ is injective and surjective, i.e. bijective. We conclude that $f( (n-1,n] ) = (n-1,n]$ for all natural $n$. Again by (2) we conclude $f(1)=1$, so $f(n)=n$ for natural $n$ and $f(n-1,n)=(n-1,n)$. Now we proceed by induction: Using (2) we can show that for every natural $k$ we have the following: $f( (n-1)/k,n/k ) = ((n-1)/k,n/k)$ and $f(n/k)=n/k$ for all natural $n$. Now it is easy to show that $f(x) = x$ for all positive real $x$. I admit I have left out some details, but I suppose they are easy to fill in.

Let $g(x)=f(x)-x$. From (1) g(x) is periodic (but f is not periodic). From (2) f(x) is bijective. It give f((0,1))=(0,1) and bijective in (0,1). f(1)=1 (f(1)>1 and f(1)<1 give contradition. From (1) $f(n)=n, f((n,n+1))=(n,n+1)$. From (2) $f((\frac{1}{n+1},\frac{1}{n}))=(\frac{1}{n+1},\frac{1}{n}),$ and $f((n+\frac{1}{k+1},n+\frac{1}{k}))=(n+\frac{1}{k+1},n+\frac{1}{k}),f(n+\frac{1}{k})=n+\frac{1}{k}$ and \[f((n_{1}+\frac{1}{n_{2}+\frac{1}{n_{3}}},n_{1}+\frac{1}{n_{2}+\frac{1}{n_{3}+1}}))=(n_{1}+\frac{1}{n_{2}+\frac{1}{n_{3}}},n_{1}+\frac{1}{n_{2}+\frac{1}{n_{3}+1}})\] e.t.c. It give (by continiosly rations) $f(x)=x$.

But $f$ is not contenious

AYMANE wrote: But $f$ is not contenious I don,t say that f is continiosly, this property proved. Let \[x(n_{0},n_{1},...,n_{k})=n_{0}+\frac{1}{n_{1}+\frac{1}{n_{2}+...+\frac{1}{n_{k}}}}.\] We have $f(x(n_{0},n_{1},...,n_{k}))=x(n_{0},n_{1},n_{2},...,n_{k}),n_{0}\ge 0,n_{i}\ge 1,i\ge 1$ Let $I(n_{0},n_{1},...,n_{k})=(x(n_{0},n_{1},...,n_{k},x(n_{0},n_{1},...,n_{k}+1))$ if k is even and $I(n_{0},n_{1},...,n_{k})=(x(n_{0},n_{1},...,n_{k}+1),x(n_{0},n_{1},...,n_{k}))$ if k is odd. We have for any k $if \ x\in I(n_{0},n_{1},n_{2},...,n_{k})$ then $f(x)\in I(n_{0},n_{1},n_{2},...n_{k})$. It give f(x)=x.

solyaris wrote: Now we proceed by induction: Using (2) we can show that for every natural $k$ we have the following: $f( (n-1)/k,n/k ) = ((n-1)/k,n/k)$ and $f(n/k)=n/k$ for all natural $n$. How to do it? I can't see.

Can anyone show me how to do the induction?

Hope this is correct, let's try by induction on the first condition, it's easy to show that, for all $n\in\mathbb N$, $f(x+n)=f(x)+n$. The induction base is the first condition, now let's suppose that $f(x+n)=f(x)+n$ holds, consequently $f(x+n+1)=f((x+n)+1)=f(x+n)+1=f(x)+n+1$ as desired. Now we will prove that $f(x)$ is bijective. Let $h(x)=\frac{1}{x}$. Then $\frac{1}{x}=f(h(f(x)))$, and then the conslusion follows. Let now define the following equality $\frac{f(x+1)}{f(x)}=\frac{1}{f(\frac{x}{x+1})}$ (*) From the first condition we've got, defining $y=\frac{1}{f(x)}$, $f(y+1)=f(y)+1\Rightarrow f(\frac{1+f(x)}{f(x)})=f(\frac{f(x+1)}{f(x)})=f(\frac{1}{f(x)})+1=\frac{x+1}{x}$. From the second we've got $f(\frac{1}{f(\frac{x}{x+1})})=\frac{x+1}{x}$. Comparing the two equalities, and recalling that $f(x)$ is injective, we have the result. Now, let's define $g(x)=f(x)-x$, so $g(x)=g(x+1)$, so $g(x)$ is $1-$periodic, as well as $n\in\mathbb N-$periodic Let's suppose that $g(x)$ is non constant. Now the following holds $g(\frac{1}{x})=g(\frac{1}{x}+1)=g(\frac{x+1}{x})\Rightarrow g(f(\frac{1}{f(x)}))=g(f(\frac{1}{f(\frac{x}{x+1})}))$ (**) Since we supposed that $g(x)$ is non constant, this implies that $f(\frac{1}{f(x)})=f(\frac{1}{f(\frac{x}{x+1})})+n$ Using (*) we get that $f(\frac{1}{f(x)})=f(\frac{f(x+1)}{f(x)})+n=f(\frac{f(x+1)+nf(x)}{f(x)})$, where the last equality of the chain is authorised by the induction on $n$ showed at the beginning. Now, since $f(x)$ is bijective, we've got $\frac{1}{f(x)}=\frac{f(x+1)+nf(x)}{f(x)}\Rightarrow 1=f(x+1)+nf(x)$ (***) Comparing (***) with the first condition, we,ve got $f(x+1)+nf(x)=f(x+1)-f(x)$, and so $f(x)\equiv 0$, absurd. So $g(x)$ is constant, let's say $g(x)=k$, so $f(x)=x+k$. But even if this solutions fits the first condition, when plugged into the second, it leads to $f(\frac{1}{f(x)})=\frac{1}{x+k}+k=\frac{1}{x}$, so $k=0$ and finally we've got our solution, since $\boxed{f(x)=x}$

barasawala wrote: Can anyone show me how to do the induction? OK, I will give some more details. We will show that $f(x)=x$ is the only function from $(0,\infty)$ to $(0,\infty)$ with (1) $f(x+1)=f(x)+1$ and (2) $f\left(\frac{1}{f(x)}\right)=\frac{1}{x}$. It is clear that $f(x)=x$ works. On the other hand let $f$ have the above properties. Then by (2) it follows that $f$ is injective and surjective, i.e. bijective. By induction over $k$ we will show that the following holds: $f( (n-1)/k,n/k ) = ((n-1)/k,n/k)$ and $f(n/k)=n/k$ for all natural $n$. Then we are done as this implies $f(x) = x$ for all rational $x$, and for irrational $x$ this implies $f(x) \in (a,b)$ for all rationals $a,b$ such that $x \in (a,b)$ and thus also $f(x) = x$. For the induction we first do the case $k=1$. By the bijectivity of $f$ the sets $A_{n}: = f(n,n+1] = f(0,1]+n$ (using (1)) have to be disjoint, and the union has to be $(0,\infty)$. Thus we have $A_{0}= (0,1]$. (Indeed for $x \notin A_{0}$ we have $x \in A_{n+1}$ for some $n \ge 0$, so $x-1 \in A_{n}$ by (1), so $x>1$. And for $x>1$ we have $x-1 \in A_{n}$ for some $n \ge 0$, so by (1) we have $x \in A_{n+1}$, so $x \notin A_{0}$.) Now it suffices to show $f(1)=1$. Then the rest of the case $k=1$ follows from (1). $f(1) \le 1$ follows from the above and $f(1)\ge 1$ follows from the following: Let $a = f(1)$, then by (2) we have $f(1/a) = 1$, and thus by the above $1/a \le 1$, so $a = f(1) \ge 1$. For the inductive step we assume the assertion is true for $1,\ldots,k-1$. By (1) it suffices to show that $f( (n-1)/k,n/k ) = ((n-1)/k,n/k)$ for all $1\le n \le k$ and $f(n/k)=n/k$ for all natural $1 \le n < k$. The second assertion follows from $f(n/k) = f(1/(k/n))= f(1/f(k/n)) = 1/(k/n) = n/k,$ where the second step holds by the inductive hypothesis as $n<k$, and the third step holds by (2). Likewise the first assertion follows from $f((n-1)/k,n/k )= \{f(x): (n-1)/k< x< n/k\}$ $= \{ f(1/y): k/n < y < k/(n-1)\}= \{ f(1/f(y)): k/n < y < k/(n-1)\}$ $= \{ 1/y: k/n < y < k/(n-1)\}= ((n-1)/k,n/k),$ where in the first step we substituted $y = 1/x$, in the second step we have used the inductive hypothesis as $n,n-1 < k$, so $f(k/n,(k+1)/n) = (k/n,(k+1)/n)$ and $f((k-1)/(n-1), k/(n-1)) = ((k-1)/(n-1), k/(n-1))$, which inmplies $f(k/n,k/(n-1)) = (k/n,k/(n-1))$, and in the third step we have used (2). I hope this answers you question. (And I am sorry for the delay.)

venatrix wrote: Since we supposed that $g(x)$ is non constant, this implies that $f(\frac{1}{f(x)})=f(\frac{1}{f(\frac{x}{x+1})})+n$ I think this passage is wrong..

solyaris wrote: By (1) $f$ is periodic Rust wrote: From (1) g(x) is periodic (but f is not periodic).

I still do not understand why $f((0,1))=(0,1)$. Can anyone gives an explanation please

For each subset $S\subseteq \mathbb{R}^+$, denote $f(S)=\{ f(x)\mid x\in S\}$. We say that a set $S$ is good if $f(S)=S$. It's obvious that $f$ is bijective. The proof divided into three main steps, I'll show only the detailed prove of the first one: 1. $(0,1]$ is good. Proof: From the first condition, we can easily deduce that $f(x)\geq \lfloor x\rfloor$ for all $x\in \mathbb{R}^+$. For each $r\in (0,1]$, there exists $s=\frac{1}{f(\frac{1}{r})}$ such that $f(s)=r$. We've $\frac{1}{r}>1\implies f(\frac{1}{r})\geq 1\implies s\leq 1$. So, for each $r\in (0,1]$, there exists $s\in (0,1]$ that $f(s)=r$. Suppose there exists $s'\in (0,1]$ that $f(s')>1$. We've proved that there must exists $r'\in (0,1]$ that $f(r')=f(s')-\lceil f(s')\rceil +1$. By the first condition, we get $f(r'+\lceil f(s')\rceil -1)=f(s')$. Injectivity implies $r'+\lceil f(s')\rceil -1 =s'\in (0,1]$, clearly impossible. So, for each $s\in (0,1]$, we get $f(s)\in (0,1]$, this completes the first part. From now, we'll use the (infinite) continued fraction representation of positive real numbers. 2. For any positive integers $n$, the interval set $$\Big([a_0,a_1,a_2,...,a_{n-1},a_n],[a_0,a_1,a_2,...,a_{n-1},a_n+1]\Big]$$is good for all non-negative integers $a_0$ and positive integers $a_1,a_2,...,a_n$. Proof: Induction on $n$. 3. For all positive real numbers $r$ and $\epsilon$, there exists $a,b \in \mathbb{R}^+$ that $a<r<b$ and $|a-r|,|b-r|<\epsilon$ and the set $(a,b)$ is good. After this, if there exists $r\in \mathbb{R}^+$ that $f(r)\neq r$, there exists $\epsilon \in \mathbb{R}^+$ such that for all $a,b\in \mathbb{R}^+$ that $a<r<b$ and $|a-r|,|b-r|<\epsilon$, $$r\in (a,b],f(r)\not\in (a,b].$$The third result gives us the contradiction.