<s>[INST] Floor function of a real number The floor of a real number $x$, denoted by $⌊x⌋$, is the largest integer that is less than or equal to $x$. The ceiling of a real number x, denoted by $⌈x⌉$, is the smallest integer that is greater than or equal to $x$. Let $x$ and $y$ be any real numbers. Let $n, m$ and $k$ be any integers.


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*$⌊x⌋ \le x < ⌊x⌋+1$, equivalently $x-1 < ⌊x⌋ \le x$.

*$⌈x⌉-1 < x \le ⌈x⌉$, equivalently $x \le ⌈x⌉ < x+1$.

*If $n \le x < n+1$ (or $x-1 < n \le x$), then $n=⌊x⌋$; similarly, if $n-1 < x \le n$ (or $x \le n < x+1$), then $n=⌈x⌉$.

*If $x \le y$, then $⌊x⌋ \le ⌊y⌋$ and $⌈x⌉ \le ⌈y⌉$. Similarly for $x \ge y$.

*If $n \le x$, then $n \le ⌊x⌋$. If $n \ge x$, then $n \ge ⌈x⌉$.

*If $x$ has an integer value, then $⌊x⌋=x=⌈x⌉$. If x has a non-integer value, then $⌈x⌉=⌊x⌋+1$.
Furthermore, we have the well-known inequality: $ \frac {x-1}{x} \le \ln x \le \frac {x^2 -1}{2x} \le x-1$
I am trying to find all pairs of positive real numbers $(x,y)$ such that $ \ln⁡(⌊x+y⌋ )=⌊y⌋+⌊x⌋,$ but the above identities are hard to follow. Thank you
 [/INST] 
The following does not use the fact that $e$ is transcendental.

First , note that $$\tag 1\lfloor x+y\rfloor \le\lfloor x\rfloor +\lfloor y\rfloor +1$$ 
for all real $x,y$.
Take the exponential of both sides of teh target equation to arrive at the equivalent condition
$$ \lfloor x+y\rfloor =e^{\lfloor x\rfloor}e^{\lfloor y\rfloor}$$
Using the mother of all $e$-inequalities, 
$$\tag{$\star$} e^t\ge 1+t\qquad\text{with equality iff }t=0,$$
we find (as $x,y$ are positive by assumption) 
$$\tag2 \begin{align}\lfloor x\rfloor+\lfloor y\rfloor +1&\ge \lfloor x+y\rfloor \\&=e^{\lfloor x\rfloor} e^{\lfloor y\rfloor}\\&\ge(1+\lfloor x\rfloor)(1+\lfloor y\rfloor)\\&=1+\lfloor x\rfloor+\lfloor y\rfloor +\lfloor x\rfloor\lfloor y\rfloor\end{align}$$
so that we must have $\lfloor x\rfloor\lfloor y\rfloor =0$ and the inequality in $(2)$ is sharp. by the sharpness condition in $(\star)$, we conclude that $$\lfloor x\rfloor =\lfloor y\rfloor=0$$
and by exploiting $(2)$, now an equality,
$$ \lfloor x+y\rfloor =\lfloor x\rfloor+\lfloor  y\rfloor +1=1.$$
hence a necessary condition is that $$\fbox{$\quad0\le x<1, \quad 0\le y<1, \quad1\le x+y.\quad\vphantom\int$}$$
As this implies $x+y<2$, we see immediately that these are also sufficient as we compute $\ln\lfloor x+y\rfloor=\ln 1=0$ and $\lfloor x\rfloor+\lfloor y\rfloor = 0+0=0$.
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