Assume there exists function $f$ satisfies given conditions. Let $P(x,y)$ be the assertion. $P(0;0)$ gives $f(0) \in \{0;-1\}$ and since $\lfloor f(0) \rfloor = 0 $ we get $f(0) = 0$. $P(x;-x)$ gives $f(x) + f(-x) \in \{0;-1\}$ $\forall x \in \mathbb R$, yeilds $f(1) + f(-1) \in \{0;-1\}$. If $f(1) + f(-1) = -1$, $f(1) = f(-1) +1 \in [0;1)$ since $\lfloor f(-1) \rfloor = -1 $ and this is a contradiction. Therefore $f(-1) + f(1) = 0$.Let $f(1) = 1+c$ with $c\ge0$ ($\lfloor f(1) \rfloor =1 $) then $f(-1) = -1 -c$, yeilds $c=0$ and $f(1) =1, f(-1) = -1$. Using induction we get $f(x) = x$ $\forall x \in \mathbb Z$. We have $\lfloor f(x) \rfloor = 0 $ $\forall x \in [0;1)$. Note that $f(x) \in [0;1)$ $\forall x \in [0;1)$, combine with $P(x;y)$ we get $f(x+y) = f(x) + f(y)$ $\forall x, y \in [0;1]$. From this we determine $f(x) = x$ $\forall x \in [0;1]$. Now using induction we get $f(x) = x$ $\forall x \in \mathbb R$.

Typo: Must be $\left\{ 0,1\right\} $ isntedd of $\left\{ 0,-1\right\} $ I don't see why it is not possible $f(x+y)=f(x)+f(y)+1$?

finn123 wrote: Assume there exists function $f$ satisfies given conditions. Let $P(x,y)$ be the assertion. $P(0;0)$ gives $f(0) \in \{0;-1\}$ and since $\lfloor f(0) \rfloor = 0 $ we get $f(0) = 0$. $P(x;-x)$ gives $f(x) + f(-x) \in \{0;-1\}$ $\forall x \in \mathbb R$, yeilds $f(1) + f(-1) \in \{0;-1\}$. If $f(1) + f(-1) = -1$, $f(1) = f(-1) +1 \in [0;1)$ since $\lfloor f(-1) \rfloor = -1 $ and this is a contradiction. Therefore $f(-1) + f(1) = 0$.Let $f(1) = 1+c$ with $c\ge0$ ($\lfloor f(1) \rfloor =1 $) then $f(-1) = -1 -c$, yeilds $c=0$ and $f(1) =1, f(-1) = -1$. Using induction we get $f(x) = x$ $\forall x \in \mathbb Z$. We have $\lfloor f(x) \rfloor = 0 $ $\forall x \in [0;1)$. Note that $f(x) \in [0;1)$ $\forall x \in [0;1)$, combine with $P(x;y)$ we get $f(x+y) = f(x) + f(y)$ $\forall x, y \in [0;1]$. From this we determine $f(x) = x$ $\forall x \in [0;1]$. Now using induction we get $f(x) = x$ $\forall x \in \mathbb R$. Why $f(x)=x$ $\forall x \in [0;1]$? It's wrong. For example, function $f(x)=\lfloor x \rfloor$ satysfying all conditions of problem.

richrow12 wrote: finn123 wrote: Assume there exists function $f$ satisfies given conditions. Let $P(x,y)$ be the assertion. $P(0;0)$ gives $f(0) \in \{0;-1\}$ and since $\lfloor f(0) \rfloor = 0 $ we get $f(0) = 0$. $P(x;-x)$ gives $f(x) + f(-x) \in \{0;-1\}$ $\forall x \in \mathbb R$, yeilds $f(1) + f(-1) \in \{0;-1\}$. If $f(1) + f(-1) = -1$, $f(1) = f(-1) +1 \in [0;1)$ since $\lfloor f(-1) \rfloor = -1 $ and this is a contradiction. Therefore $f(-1) + f(1) = 0$.Let $f(1) = 1+c$ with $c\ge0$ ($\lfloor f(1) \rfloor =1 $) then $f(-1) = -1 -c$, yeilds $c=0$ and $f(1) =1, f(-1) = -1$. Using induction we get $f(x) = x$ $\forall x \in \mathbb Z$. We have $\lfloor f(x) \rfloor = 0 $ $\forall x \in [0;1)$. Note that $f(x) \in [0;1)$ $\forall x \in [0;1)$, combine with $P(x;y)$ we get $f(x+y) = f(x) + f(y)$ $\forall x, y \in [0;1]$. From this we determine $f(x) = x$ $\forall x \in [0;1]$. Now using induction we get $f(x) = x$ $\forall x \in \mathbb R$. Why $f(x)=x$ $\forall x \in [0;1]$? It's wrong. For example, function $f(x)=\lfloor x \rfloor$ satysfying all conditions of problem. Yeah you are right!. From $f(x+y) = f(x) + f(y)$ $\forall x, y \in [0;1)$ and $f$ is not decreasing in $[0;1)$, $f(x) \in [0;1)$ $\forall x \in [0;1)$, using the result of Cauchy's functional equation we get $f(x) = cx$ $\forall x \in [0;1)$ and $0 \le c \le1$ and from this using induction we can find all $f(x)$ with $x \in \mathbb R$.

Generalization by educator: What about $f(x+y)-f(x)-f(y) \in \{0, 1, 1/2, 1/3, ....\}$ (@below.. Sorry. Fixed)

You need to include backslash before every brace in order to make the brace shown.

Maybe there would be a lot of answers.(Also educator didn't solve it to us.) Can anyone try this one?

lminsl wrote: Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfying the following conditions : 1) $f(x+y)-f(x)-f(y) \in \{0,1\} $ for all $x,y \in \mathbb{R}$ 2) $\lfloor f(x) \rfloor = \lfloor x \rfloor $ for all real $x$. Let $P(x,y)$ be the assertion. $P(0;0)$ gives $f(0) \in \{0;-1\}$ and since $\lfloor f(0) \rfloor = 0 $ we get $f(0) = 0$. Let $f(x)=g(x)+\lfloor x \rfloor$ we find $1>g(x)\geq0$. We have $g(x+y)-g(x)-g(y)+\lfloor x+y \rfloor-\lfloor x \rfloor-\lfloor y \rfloor \in \{0,1\} $ we know that $\lfloor x+y \rfloor-\lfloor x \rfloor-\lfloor y \rfloor=\lfloor\left\{ x\right\}+\left\{y\right\} \rfloor$. If $\lfloor\left\{ x\right\}+\left\{y\right\} \rfloor=0$ we find that $g(x+y)=g(x)+g(y)$ its easy to see $g(1)=0$ and $g(x+1)=g(x)$ for all real $x$.For all $0<x<1$, $0<y<1$ and $0<x+y<1$ we have $g(x+y)=g(x)+g(y)$ where $1>g(x)\geq0$.Thus we have $g(x)=cx$ for all $0<x<1$ and for some $1\geq c\geq0$.But other hand we have $g(x+1)=g(x)$ for all real $x$.Thus we have $g(x)=c\left\{ x\right\}$ and $f(x)=c\left\{ x\right\}+\lfloor x \rfloor$ for all real numbers $x$.Its easy to find $c=1$ or $c=0$.Thus solutions are $$f(x)=\left\{ x\right\}+\lfloor x \rfloor$$and $$f(x)=\lfloor x \rfloor$$.

How do you guys use Cauchy here? I thought Cauchy over $\mathbb{R}$ requires continuity?

@above No, if we know that $f(x)$ is bounded ($0\leq f(x) <1$).

CEH wrote: Generalization by educator: What about $f(x+y)-f(x)-f(y) \in \{0, 1, 1/2, 1/3, ....\}$ (@below.. Sorry. Fixed) Any ideas for this?

CEH wrote: Generalization by educator: What about $f(x+y)-f(x)-f(y) \in \{0, 1, 1/2, 1/3, ....\}$ (@below.. Sorry. Fixed) Bump! (I think after showing $f(x)\le x$, the next step should be $f(x+1)=f(x)+1$ for all $x\in\mathbb{Q}$.) and btw, some weird solutions I've found were $$f(x)=\frac{1}{6}\lfloor 3x\rfloor +\frac{1}{12}\lfloor 6x\rfloor$$

Bump the generalization once more

Bump the generalization part.