<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] Note that the right side is an integer so the left side must be, too.  That means $\lfloor x+y\rfloor$ is an integer power of $e$ but it is also an integer, so it must be $1$ and the $\log$ is $0$.  This gives
$$\lfloor x+y\rfloor=1\\ \lfloor x \rfloor + \lfloor y \rfloor=0$$
If either $x$ or $y$ is greater than or equal to $1$ the second will fail, so our final solution is $$0 \lt x \lt 1\\ 1-x \lt y \lt 1$$
or the part of the unit square above $x+y=1$  It is the region strictly inside the triangle below

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