Fourth case.
Let \(\displaystyle x\) be a non-integer and let \(\displaystyle y\) be a non-integer.
If \(\displaystyle x = 3.4\) and \(\displaystyle y = 5.2\), then
\(\displaystyle \lceil{3.4 + 5.2}\rceil = \lceil{8.6}\rceil = 9\)
\(\displaystyle \lceil 3.4 \rceil + \lfloor 5.2 \rfloor = 4 + 5 = 9\)
Then,
\(\displaystyle \lceil{3.4 + 5.2}\rceil = \lceil 3.4 \rceil + \lfloor 5.2 \rfloor\)
It worked in the third case
But wait\(\displaystyle \textcolor{red}{\bold{!!!}}\)
If \(\displaystyle x = 3.8\) and \(\displaystyle y = 5.3\), then
\(\displaystyle \lceil{3.8 + 5.3}\rceil = \lceil{9.1}\rceil = 10\)
\(\displaystyle \lceil 3.8 \rceil + \lfloor 5.3 \rfloor = 4 + 5 = 9\)
Then,
\(\displaystyle \lceil{3.8 + 5.3}\rceil \neq \lceil 3.8 \rceil + \lfloor 5.3 \rfloor\)
It didn't work in the third case
@fresh_42
What should I say when the fourth case is partially satisfied? And how to solve this problem (the whole problem) by using arbitrary constants?
\(\displaystyle \large \textcolor{blue}{\bold{My \ Conclusion}}\)
\(\displaystyle \lceil{x + y}\rceil = \lceil x \rceil + \lfloor y \rfloor\) when
\(\displaystyle \bold{1.} \ x,y \in \mathbb{Z}\)
\(\displaystyle \bold{2.} \ x \notin \mathbb{Z},y \in \mathbb{Z}\)
\(\displaystyle \bold{3.} \ x,y \notin \mathbb{Z} \ \text{AND their fractional parts} \leq 1 \)