12 multiple-choice questions, progressively harder.
Solve ⌊x⌋=−5\lfloor x \rfloor = -5⌊x⌋=−5 for all real xxx.
Solution
Correct answer: C
The floor equals −5-5−5 on the step starting at −5-5−5, using n≤x<n+1n \le x < n + 1n≤x<n+1.
−5≤x<−4-5 \le x < -4−5≤x<−4
The left end −5-5−5 is included and the right end −4-4−4 is not.
Solve ⌈x⌉=−4\lceil x \rceil = -4⌈x⌉=−4 for all real xxx.
Correct answer: A
The ceiling equals −4-4−4 on the step ending at −4-4−4, using n−1<x≤nn - 1 < x \le nn−1<x≤n.
−5<x≤−4-5 < x \le -4−5<x≤−4
The left end −5-5−5 is excluded and the right end −4-4−4 is included.
Evaluate ⌈−0.001⌉\lceil -0.001 \rceil⌈−0.001⌉.
Correct answer: B
The input is just below zero, so the least integer at or above it is 000.
−0.001 lies between −1 and 0, so ⌈−0.001⌉=0-0.001 \text{ lies between } -1 \text{ and } 0, \text{ so } \lceil -0.001 \rceil = 0−0.001 lies between −1 and 0, so ⌈−0.001⌉=0
The nearest integer to the right of −0.001-0.001−0.001 is 000.
Compute the fractional part {−2.75}\{-2.75\}{−2.75}, where {x}=x−⌊x⌋\{x\} = x - \lfloor x \rfloor{x}=x−⌊x⌋.
The floor of −2.75-2.75−2.75 rounds down to −3-3−3.
{−2.75}=−2.75−⌊−2.75⌋=−2.75−(−3)=0.25\{-2.75\} = -2.75 - \lfloor -2.75 \rfloor = -2.75 - (-3) = 0.25{−2.75}=−2.75−⌊−2.75⌋=−2.75−(−3)=0.25
The leftover lands in [0,1)[0, 1)[0,1), as it always must.
Solve ⌊x⌋=0\lfloor x \rfloor = 0⌊x⌋=0 for all real xxx.
The floor equals 000 on the step starting at 000, using n≤x<n+1n \le x < n + 1n≤x<n+1.
0≤x<10 \le x < 10≤x<1
Every input from 000 up to (but not including) 111 rounds down to 000.
Rounding 6.56.56.5 to the nearest integer uses ⌊x+12⌋\lfloor x + \tfrac{1}{2} \rfloor⌊x+21⌋. Evaluate ⌊6.5+12⌋\lfloor 6.5 + \tfrac{1}{2} \rfloor⌊6.5+21⌋.
Add one half, then take the floor.
⌊6.5+12⌋=⌊7⌋=7\lfloor 6.5 + \tfrac{1}{2} \rfloor = \lfloor 7 \rfloor = 7⌊6.5+21⌋=⌊7⌋=7
The half-step nudges 6.56.56.5 up to 777 before the floor is applied.
A theater seats 121212 people per row. How many rows are needed to seat 100100100 people?
Correct answer: D
Every leftover person still needs a row, so round up with a ceiling.
⌈100/12⌉=⌈8.33…⌉=9\lceil 100 / 12 \rceil = \lceil 8.33\ldots \rceil = 9⌈100/12⌉=⌈8.33…⌉=9
Eight rows seat 969696 people, so a ninth row is needed for the last 444.
On the graph of y=⌈x⌉y = \lceil x \rceily=⌈x⌉, the step covering 0<x≤10 < x \le 10<x≤1 sits at what height, and where is its closed dot?
On 0<x≤10 < x \le 10<x≤1 the ceiling equals 111, and a ceiling step is closed on its right end.
⌈1⌉=1, so the closed dot is at (1,1)\lceil 1 \rceil = 1, \text{ so the closed dot is at } (1, 1)⌈1⌉=1, so the closed dot is at (1,1)
The left end (0,1)(0, 1)(0,1) is open, since ⌈0⌉=0\lceil 0 \rceil = 0⌈0⌉=0, not 111.
Evaluate ⌊−1.5⌋+⌈−1.5⌉\lfloor -1.5 \rfloor + \lceil -1.5 \rceil⌊−1.5⌋+⌈−1.5⌉.
Round −1.5-1.5−1.5 down for the floor and up for the ceiling, then add.
⌊−1.5⌋+⌈−1.5⌉=−2+(−1)=−3\lfloor -1.5 \rfloor + \lceil -1.5 \rceil = -2 + (-1) = -3⌊−1.5⌋+⌈−1.5⌉=−2+(−1)=−3
The floor is −2-2−2 (nearest integer left) and the ceiling is −1-1−1 (nearest integer right).
The fractional part {x}=x−⌊x⌋\{x\} = x - \lfloor x \rfloor{x}=x−⌊x⌋ equals 000 for which inputs?
The leftover vanishes only when xxx already equals its floor.
{x}=0 ⟺ x=⌊x⌋ ⟺ x is an integer\{x\} = 0 \iff x = \lfloor x \rfloor \iff x \text{ is an integer}{x}=0⟺x=⌊x⌋⟺x is an integer
For every other input the fractional part is strictly between 000 and 111.
For x=3x = 3x=3, how does ⌈x⌉\lceil x \rceil⌈x⌉ compare to ⌊x⌋\lfloor x \rfloor⌊x⌋?
Since 333 is an integer, both round to 333.
⌈3⌉=⌊3⌋=3\lceil 3 \rceil = \lfloor 3 \rfloor = 3⌈3⌉=⌊3⌋=3
The floor and ceiling agree on every integer, so they are equal here.
A 474747 cm ribbon is cut into whole 555 cm pieces. How many complete pieces are there?
Only complete pieces count, so round down with a floor.
⌊47/5⌋=⌊9.4⌋=9\lfloor 47 / 5 \rfloor = \lfloor 9.4 \rfloor = 9⌊47/5⌋=⌊9.4⌋=9
Nine pieces use 454545 cm, and the leftover 222 cm is too short for a tenth piece.
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