$\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers of $x$ and $y$

Question

Give a convincing argument that $\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers $x$ and $y$. Could someone please explain how to prove this? I attempted to say that the largest values that could be added to $x$ and $y$ is $0.99$ and that doing so still made $\lfloor x\rfloor + \lfloor y\rfloor < \lfloor x+y\rfloor$. However, my answer was not accepted.

Answer 1

Let $x = \lfloor x \rfloor + \epsilon_1$ and $y = \lfloor y \rfloor + \epsilon_2$.

Then $\lfloor x + y \rfloor = \lfloor\lfloor x \rfloor + \epsilon_1 + \lfloor y \rfloor + \epsilon_2\rfloor = \lfloor x \rfloor + \lfloor y \rfloor + \lfloor \epsilon_1 + \epsilon_2 \rfloor \ge \lfloor x \rfloor + \lfloor y \rfloor$.

Answer 2

If $\left \lfloor x \right \rfloor = n$ then $x = n + r$ where $0 \leq r < 1$ and If $\left \lfloor y \right \rfloor = n'$ then $y = n' + r'$ where $0 \leq r' < 1$. Then $$x+y = n+n'+r+r'\geq n + n' = \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor $$.

Answer 3

Let $x = a + m$ and $y = b + n$ where $a, b$ are integers and $m,n$ are real numbers such that $0 \le m,n < 1$. Then,

$$\lfloor x \rfloor + \lfloor y \rfloor = a + b$$

while

$$\begin{align}\lfloor x + y \rfloor &= \lfloor a + b + m + n \rfloor\\ &= a + b + \lfloor m + n \rfloor\\ & \ge a + b\\ &= \lfloor x \rfloor + \lfloor y \rfloor\end{align}$$

Equality occurs when the fractional part of $x$ and $y$ sum up to be less than $1$, in which case $\lfloor m + n \rfloor = 0$. For example, $x = 0.5$ and $y = 1.1$.

Answer 4

There are additional answers at the following duplicate(s):

How to prove that $\lfloor a\rfloor+\lfloor b\rfloor\leq \lfloor a+b\rfloor$

(I'm making this community wiki so that anyone can add to it in case other instances come to light.)