Floor function of a real number

Question

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.

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

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

3. 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 $n=⌈x⌉$.

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

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

6. 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

Answer 1

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

Answer 2

Consider two cases for positive reals $x$ and $y$. Note that we require $x + y \geq 1$ for the left side of the equation to be defined, so we make this assumption.

• If $x,y < 1$, then the equation holds, as both sides are equal to zero.
• If $x \geq 1$, then suppose the equality holds, so that $e^{\lfloor x \rfloor}e^{\lfloor y \rfloor} =\lfloor x + y \rfloor$. Since $x \geq 1$, $e^{\lfloor x \rfloor}$ is not an integer, hence the left side is not an integer, a contradiction since the right side is. Similar when $y \geq 1$.

Thus, the only positive real solutions $(x,y)$ are characterized by the first case.

Answer 3

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 $\lfloor x\rfloor+\lfloor y\rfloor = 0+0=0$.