Notice that both ⌊x/2⌋ and ⌈2/x⌉ are integers, so they must be divisors of 17. But 17 is prime, so either
⌊x/2⌋ = 1 and ⌈2/x⌉ = 17
or
⌊x/2⌋ = 17 and ⌈2/x⌉ = 1
Let x = 2y for some real number y.
• Suppose ⌊x/2⌋ = 1 and ⌈2/x⌉ = 17. Then by definition of floor and ceiling,
⌊x/2⌋ = ⌊2y/2⌋ = ⌊y⌋ = 1 ⇒ 1 ≤ y < 2 ⇒ 2 ≤ x < 4
and
⌈2/x⌉ = ⌈2/(2y)⌉ = ⌈1/y⌉ = 17 ⇒ 16 < 1/y ≤ 17 ⇒ 2/17 ≤ x < 1/8
but this is impossible.
• Suppose ⌊x/2⌋ = 17 and ⌈2/x⌉ = 1. Then
⌊x/2⌋ = ⌊y⌋ = 17 ⇒ 17 ≤ y < 18 ⇒ 34 ≤ x < 38
and
⌈2/x⌉ = ⌈1/y⌉ = 1 ⇒ 0 < 1/y ≤ 1 ⇒ 2 ≤ x
The solution set is the intersection of these two intervals, which is
34 ≤ x < 38
Edit: 17 can also be factorized as (-1) × (-17), so we have two more cases to consider:
• If ⌊x/2⌋ = -1 and ⌈2/x⌉ = -17, then
⌊x/2⌋ = ⌊y⌋ = -1 ⇒ -1 ≤ y < 0 ⇒ -2 ≤ x < 0
⌈2/x⌉ = ⌈1/y⌉ = -17 ⇒ -18 < 1/y ≤ -17 ⇒ -2/17 ≤ x < -1/9
so that -2/17 ≤ x < -1/9 is also a solution.
• If ⌊x/2⌋ = -17 and ⌈2/x⌉ = -1, then
⌊x/2⌋ = ⌊y⌋ = -17 ⇒ -17 ≤ y < -16 ⇒ -34 ≤ x < -32
⌈2/x⌉ = ⌈1/y⌉ = -1 ⇒ -2 < 1/y ≤ -1 ⇒ 2 ≤ x < 4
but this is also impossible.
