Fourth case .
Let
x \displaystyle x x be a non-integer and let
y \displaystyle y y be a non-integer.
If
x = 3.4 \displaystyle x = 3.4 x = 3 . 4 and
y = 5.2 \displaystyle y = 5.2 y = 5 . 2 , then
⌈ 3.4 + 5.2 ⌉ = ⌈ 8.6 ⌉ = 9 \displaystyle \lceil{3.4 + 5.2}\rceil = \lceil{8.6}\rceil = 9 ⌈ 3 . 4 + 5 . 2 ⌉ = ⌈ 8 . 6 ⌉ = 9
⌈ 3.4 ⌉ + ⌊ 5.2 ⌋ = 4 + 5 = 9 \displaystyle \lceil 3.4 \rceil + \lfloor 5.2 \rfloor = 4 + 5 = 9 ⌈ 3 . 4 ⌉ + ⌊ 5 . 2 ⌋ = 4 + 5 = 9
Then,
⌈ 3.4 + 5.2 ⌉ = ⌈ 3.4 ⌉ + ⌊ 5.2 ⌋ \displaystyle \lceil{3.4 + 5.2}\rceil = \lceil 3.4 \rceil + \lfloor 5.2 \rfloor ⌈ 3 . 4 + 5 . 2 ⌉ = ⌈ 3 . 4 ⌉ + ⌊ 5 . 2 ⌋
It worked in the third case
But wait
! ! ! \displaystyle \textcolor{red}{\bold{!!!}} ! ! !
If
x = 3.8 \displaystyle x = 3.8 x = 3 . 8 and
y = 5.3 \displaystyle y = 5.3 y = 5 . 3 , then
⌈ 3.8 + 5.3 ⌉ = ⌈ 9.1 ⌉ = 10 \displaystyle \lceil{3.8 + 5.3}\rceil = \lceil{9.1}\rceil = 10 ⌈ 3 . 8 + 5 . 3 ⌉ = ⌈ 9 . 1 ⌉ = 1 0
⌈ 3.8 ⌉ + ⌊ 5.3 ⌋ = 4 + 5 = 9 \displaystyle \lceil 3.8 \rceil + \lfloor 5.3 \rfloor = 4 + 5 = 9 ⌈ 3 . 8 ⌉ + ⌊ 5 . 3 ⌋ = 4 + 5 = 9
Then,
⌈ 3.8 + 5.3 ⌉ ≠ ⌈ 3.8 ⌉ + ⌊ 5.3 ⌋ \displaystyle \lceil{3.8 + 5.3}\rceil \neq \lceil 3.8 \rceil + \lfloor 5.3 \rfloor ⌈ 3 . 8 + 5 . 3 ⌉ = ⌈ 3 . 8 ⌉ + ⌊ 5 . 3 ⌋
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?
M y C o n c l u s i o n \displaystyle \large \textcolor{blue}{\bold{My \ Conclusion}} M y C o n c l u s i o n
⌈ x + y ⌉ = ⌈ x ⌉ + ⌊ y ⌋ \displaystyle \lceil{x + y}\rceil = \lceil x \rceil + \lfloor y \rfloor ⌈ x + y ⌉ = ⌈ x ⌉ + ⌊ y ⌋ when
1. x , y ∈ Z \displaystyle \bold{1.} \ x,y \in \mathbb{Z} 1 . x , y ∈ Z
2. x ∉ Z , y ∈ Z \displaystyle \bold{2.} \ x \notin \mathbb{Z},y \in \mathbb{Z} 2 . x ∈ / Z , y ∈ Z
3. x , y ∉ Z AND their fractional parts ≤ 1 \displaystyle \bold{3.} \ x,y \notin \mathbb{Z} \ \text{AND their fractional parts} \leq 1 3 . x , y ∈ / Z AND their fractional parts ≤ 1
Last edited: Jul 29, 2025