Floor and Ceiling: Free Response

The value $9.4$ lies between $9$ and $10$. The greatest integer at or below it is $9$, and the least integer at or above it is $10$:

$\lfloor 9.4 \rfloor = 9, \qquad \lceil 9.4 \rceil = 10.$

The value $-5.6$ lies between $-6$ and $-5$. Rounding down (toward negative infinity) lands on $-6$, and rounding up (toward positive infinity) lands on $-5$:

$\lfloor -5.6 \rfloor = -6, \qquad \lceil -5.6 \rceil = -5.$

The floor asks for the greatest integer at or below the input. Since $11$ is already an integer, it is at or below itself, and no larger integer is also at or below it, so $\lfloor 11 \rfloor = 11$.

The ceiling asks for the least integer at or above the input. By the same reasoning, $11$ is at or above itself, and no smaller integer is also at or above it, so $\lceil 11 \rceil = 11$:

$\lfloor 11 \rfloor = \lceil 11 \rceil = 11.$

Both definitions point straight back to the input whenever it is already a whole number.

Use the defining inequality $n \le x < n+1$ with $n = -6$:

$-6 \le x < -5.$

Every $x$ from $-6$ up to (but not including) $-5$ has floor $-6$.

Use the defining inequality $n - 1 < x \le n$ with $n = 4$:

$3 < x \le 4.$

Every $x$ above $3$ up to and including $4$ has ceiling $4$.

Substitute $n = -2$ into the defining inequality:

$-2 - 1 < x \le -2, \qquad \text{that is} \qquad -3 < x \le -2.$

The left comparison is strict ($<$), so $x = -3$ itself does not satisfy it and is excluded. The right comparison allows equality ($\le$), so $x = -2$ does satisfy it and is included.

The value $9.6$ lies between $9$ and $10$, so $\lfloor 9.6 \rfloor = 9$. Deleting everything after the decimal point in $9.6$ also leaves $9$:

$\lfloor 9.6 \rfloor = 9 = \text{(truncated } 9.6\text{)}.$

For this positive input, the shortcut matches the floor.

The value $-6.7$ lies between $-7$ and $-6$, and rounding down (toward negative infinity) gives $\lfloor -6.7 \rfloor = -7$. Deleting everything after the decimal point in $-6.7$ instead gives $-6$, since that only removes digits without changing direction:

$\lfloor -6.7 \rfloor = -7 \neq -6 = \text{(truncated } -6.7\text{)}.$

One disagreeing case is enough to refute the claim that the shortcut always matches the floor.

Rounding down and truncating are two different rules about DIRECTION. The floor always moves toward negative infinity, no matter the sign of the input. Truncating always moves toward zero, since it only deletes trailing digits.

For a nonnegative input, both directions point the same way, so the two rules land on the same integer, exactly as in part A. A negative INTEGER has nothing after the decimal point, so neither rule moves it and they agree there too. Only a negative NON-integer splits them: rounding toward negative infinity moves it away from zero while truncation moves it toward zero, so they disagree, exactly as in part B.

$\lfloor x \rfloor = \text{trunc}(x) \iff x \ge 0 \ \text{or} \ x \in \mathbb{Z}.$

Dividing the cans by the crate size gives $138/16 = 8.625$ crates' worth of cans. A fractional crate is not possible, and any leftover cans still need a crate, so round up:

$\lceil 138/16 \rceil = \lceil 8.625 \rceil = 9.$

Nine crates are needed to pack every can.

Only complete crates count, so round down: $\lfloor 138/16 \rfloor = \lfloor 8.625 \rfloor = 8$ full crates. Those $8$ crates hold $16 \times 8 = 128$ cans, so the cans left over are

$138 - 128 = 10.$

Part A rounds the quotient $8.625$ up, and part B rounds the same quotient down, so the two counts are consecutive integers, one apart:

$\lceil 8.625 \rceil = \lfloor 8.625 \rfloor + 1.$

The leftover cans found in part B are precisely the cans that a completely full crate cannot hold; since even one leftover can still needs somewhere to go, it forces exactly one more crate, which is the extra crate part A counted.

Any fraction of a pound still costs a full pound's rate, so the weight is rounded up before multiplying by the rate:

$C(w) = 5 \lceil w \rceil.$

For $w = 6.3$, $\lceil 6.3 \rceil = 7$, so $C(6.3) = 5 \times 7 = 35$ dollars.

Since $12$ is already an integer, it is its own least integer at or above itself, so $\lceil 12 \rceil = 12$. The cost is then

$C(12) = 5 \times 12 = 60 \text{ dollars}.$

The rounding step only ever raises a weight to the next whole pound when there is a fraction to round away; an already-whole weight has nothing to round.

Set the cost expression equal to $50$ and solve for the rounded weight:

$5 \lceil w \rceil = 50 \implies \lceil w \rceil = 10.$

Apply the defining inequality $n - 1 < w \le n$ with $n = 10$:

$9 < w \le 10.$

Postage depends only on $\lceil w \rceil$, not on $w$ itself. On the ceiling's staircase graph, one whole horizontal step covers a full length-one interval of weights, and every weight on that step shares the same height, the same rounded pound count:

$9 < w \le 10 \implies \lceil w \rceil = 10 \implies C(w) = 5\times 10 = 50.$

Since the cost is the rate times that shared height, every weight in the interval you found in part C produces the exact same charge, even though the weights themselves are different real numbers.