Question

Solve for

Answer 1

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.