Which of these statements about the floor and ceiling functions are correct ?…
2021
Which of these statements about the floor and ceiling functions are correct ?
Statement I : \(\lfloor 2x \rfloor = \lfloor x \rfloor + \lfloor x + (1/2) \rfloor \) for all real numbers x.
Statement II : \(\lceil x + y \rceil = \lceil x \rceil + \lceil y \rceil \) for all real numbers \(x\) and \(y\).
Answer: C. Statement I is true but Statement II is false — Answer: Statement I is true; Statement II is false. Proof that the first identity holds: Write x = n + f where n = floor(x) and 0 ≤ f < 1. Then 2x = 2n + 2f,…
- A.
Both Statement I and Statement II are true
- B.
Both Statement I and Statement II are false
- C.
Statement I is true but Statement II is false
- D.
Statement I is false but Statement II is true
Attempted by 86 students.
Show answer & explanation
Correct answer: C
Answer: Statement I is true; Statement II is false.
Proof that the first identity holds:
Write x = n + f where n = floor(x) and 0 ≤ f < 1.
Then 2x = 2n + 2f, so floor(2x) = 2n + floor(2f).
Also floor(x) + floor(x + 1/2) = n + floor(n + f + 1/2) = 2n + floor(f + 1/2).
For 0 ≤ f < 1 we have floor(2f) = 0 when 0 ≤ f < 1/2 and = 1 when 1/2 ≤ f < 1; similarly floor(f + 1/2) has the same values. Therefore floor(2f) = floor(f + 1/2), so the two sides are equal for all real x.
Counterexample showing the second identity is false:
Take x = 0.3 and y = 0.4. Then ceil(x + y) = ceil(0.7) = 1, but ceil(x) + ceil(y) = 1 + 1 = 2, so the equality fails.
Thus Statement I holds for all real x, and Statement II is not true in general.