Case 1) [([4x] + 2)/3] = 1 and [(2x + 1)/3] = 0
then ([4x] + 2)/3 is in the range [1,2)
([4x] + 2) is in the range [3,6)
[4x] is in the range [1,4)
4x is in the range [1,4)
x is in the range [1/4,1)
[(2x + 1)/3] = 0
(2x + 1)/3 is in the range [0,1)
(2x + 1) is in the range [0,3)
2x is in the range [-1,2)
x is in the range [-1/2,1)
The intersection of these 2 ranges is [1/4,1)
Case 2) [([4x] + 2)/3] = 0 and [(2x + 1)/3] = 1
then ([4x] + 2)/3 is in the range [0,1)
([4x] + 2) is in the range [0,3)
[4x] is in the range [-2,1)
4x is in the range [-2,1)
x is in the range [-1/2,1/4)
[(2x + 1)/3] = 1
(2x + 1)/3 is in the range [1,2)
(2x + 1) is in the range [3,6)
2x is in the range [2,5)
x is in the range [1,5/2)
The intersection of these 2 ranges is null
No need to do more cases.
It is pretty clear that if x >= 1, [(2x + 1)/3] >=1 and [([4x] + 2)/3] >= 2
Also, if x < 1/4, then [(2x + 1)/3] <= 0 and [([4x] + 2)/3] <= 0
So the only values of x that work are 1/4 <= x < 1
Edited on September 13, 2024, 7:14 am
Doing the math (spoiler)
