Dadgarnia wrote: Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ that satisfy the following conditions: a. $x+f(y+f(x))=y+f(x+f(y)) \quad \forall x,y \in \mathbb{R}$ b. The set $I=\left\{\frac{f(x)-f(y)}{x-y}\mid x,y\in \mathbb{R},x\neq y \right\}$ is an interval. Proposed by Navid Safaei When you write "an interval", do you mean a "bounded interval" (which implies continuity for example)? or is $[0,+\infty)$ for example considered as an "interval" in your problem statement ?

pco wrote: Dadgarnia wrote: Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ that satisfy the following conditions: a. $x+f(y+f(x))=y+f(x+f(y)) \quad \forall x,y \in \mathbb{R}$ b. The set $I=\left\{\frac{f(x)-f(y)}{x-y}\mid x,y\in \mathbb{R},x\neq y \right\}$ is an interval. Proposed by Navid Safaei When you write "an interval", do you mean a "bounded interval" (which implies continuity for example)? or is $[0,+\infty)$ for example considered as an "interval" in your problem statement ? Yes this implies continuity but this is not necessarily a bounded interval.

Dadgarnia wrote: Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ that satisfy the following conditions: a. $x+f(y+f(x))=y+f(x+f(y)) \quad \forall x,y \in \mathbb{R}$ b. The set $I=\left\{\frac{f(x)-f(y)}{x-y}\mid x,y\in \mathbb{R},x\neq y \right\}$ is an interval. Let $P(x,y)$ be the assertion $x+f(y+f(x))=y+f(x+f(y))$ If $1\in I$, then $\exists x\ne y$ such that $f(x)-f(y)=x-y$ And so $y+f(x)=x+f(y)$ and then $P(x,y)$ implies $x=y$, contradiction. So $1\notin I$ It $t\ne 1\in I$, then let $x\ne y$ such that $\frac{f(x)-f(y)}{x-y}=t$ We have $x+f(y)\ne y+f(x)$ and $A=\frac{f(x+f(y))-f(y+f(x))}{(x+f(y))-(y+f(x))}=\frac{x-y}{(x-y)-(f(x)-f(y)}=\frac 1{1-t}$ So $t\in I$ implies $\frac 1{1-t}\in I$ which unfortunately is impossible : If $t>1$, then $\frac 1{1-t}<0$ and we can not have both in I (else $1\in I$) If $t<1$, then we need $\frac 1{1-t}<1$ which means $t<0$ but then $\frac 1{1-t}>0$ and replacing $t$ by $\frac 1{1-t}$ gives contradiction. So no such I. So $\boxed{\text{no such function}}$

Can anyone explain me what does "being interval" mean i.e if some set is interval what do we know about that set? Thanks!

tenplusten wrote: Can anyone explain me what does "being interval" mean i.e if some set is interval what do we know about that set? Thanks! The only property of "interval" I used in my previous post is : $\forall x<y\in I$, then $[x,y]\subseteq I$

pco wrote: $\forall x<y\in I$ Does $x $ also belong to $I $ ,or just $y $? pco wrote: $[x,y]\subseteq I$ Does it mean any number in the interval $[x,y] $ belongs to $I $. Sorry if above questions are stupid. Thanks!

What I wrote : $x,y\in I$ and $x<y$ $\implies$ $z\in I$ $\forall z$ such that $x\le z\le y$

Lol.I now see that first question is quite stupid. Thanks for help.

let $P(x,y)$ be the assertion $$x+f(y+f(x))=y+f(x+f(y)) \quad \forall x,y \in \mathbb{R}$$claim: $f$ is injective proof: note that if $1 \notin I$ we have $$\frac{f(y+f(x))-f(x+f(y))}{y-x+f(x)-f(y)}=\frac{y-x}{y-x+f(x)-f(y)}$$so we have $\frac{y-x}{y-x+f(x)-f(y)} \in I \forall x \neq y$ if $y>x$ then $f(y)>f(x) \forall y>x$ or $f(x)>f(y) \forall y>x$ else we will have $a>1,b<1 : a,b\in I \implies 1 \in I$ contradiction $\blacksquare$ now let $t \in I$ $P(x,ty) \implies x+f(xt+f(y))=yt+f(x+f(yt))$ let it $Q(x,y)$ from $P(x,0)$ and $Q(x,0)$ $$x+f(xt+c)=f(x+c)=x+f(f(x))$$so $f(x)=xt+c$ if we plug it into the original one, we will reach a contradiction so there's no such function and we win