Relations and Functions: Free Response

Take the declarations one at a time, because each one asks a different question of the same four points.

Declare the first coordinate the input. The inputs are then

$-4,\ 1,\ 5,\ 9$

and these are four different numbers, so no input is used twice and none can carry two outputs. Under this declaration $R$ is a function. The repeated $6$ among the second coordinates is no obstacle: the definition restricts inputs, and says nothing about how often an output may be reused.

Now declare the second coordinate the input instead. The same four points are read as

$6 \mapsto -4, \quad -2 \mapsto 1, \quad 6 \mapsto 5, \quad 0 \mapsto 9$

and the input $6$ appears twice, carrying $-4$ once and $5$ the other time. One input with two different outputs is exactly what a function forbids, so under this declaration $R$ is not a function. The pairs responsible are $(-4, 6)$ and $(5, 6)$.

Work backwards from what each verdict needs. Declaring the second coordinate the input turns the second coordinates into inputs, so that declaration succeeds exactly when the three second coordinates are all different. Declaring the first coordinate the input fails exactly when two pairs share a first coordinate. So build a set with a repeated first coordinate and three distinct second coordinates.

$\{(3, -1),\ (3, 8),\ (-6, 4)\}$

Under the first declaration the input $3$ sits in two pairs and carries the two different outputs $-1$ and $8$, so it is not a function. Under the second declaration the inputs are $-1$, $8$ and $4$, three different numbers, so no input repeats and it is a function.

The construction is never blocked, and one small fact is why: two different pairs sharing a first coordinate must differ in the second, since pairs agreeing in both coordinates would be the same pair. A repeated first coordinate therefore costs the other declaration nothing.

The objection has a real point buried in it. A relation is a completed object: the pairs are what they are, and declaring one coordinate or the other to be the input adds no point and removes none. In that sense nothing about the set changes when the declaration does.

What does not follow is that every question about the set has a single answer. Being a function is not a property the points carry by themselves, because the definition speaks of inputs and outputs while a bare pair says nothing about which of its two entries is which.

$(a, b): \quad a \mapsto b \quad \text{or} \quad b \mapsto a$

Once a declaration is supplied the question is perfectly sharp and has exactly one answer. Before it is supplied the question is incomplete, and four points are enough to show that the two completions can disagree, as parts A and B between them do.

So 'a function of which variable' is not a trick. It is the half of the question the objection left out, and it is the half that decides the verdict.

All $30$ metres are used and the two widths take $2w$ of them, so the other pair of sides shares the remainder equally.

$\frac{30 - 2w}{2} = 15 - w$

That is the length of each of those sides, and multiplying it by the width gives the rule in the stem.

Now read the domain off the situation rather than off the formula. The expression $w(15 - w)$ accepts $w = -4$ or $w = 20$ without complaint and returns a number in each case, so the formula alone rules nothing out at all. The pen does. A width is a positive length, which requires $w > 0$, and each of the other two sides is also a positive length, which requires $15 - w > 0$, that is $w < 15$.

$0 < w < 15$

At $w = 15$ the rule still returns the perfectly good number $0$, so nothing about the arithmetic breaks there. It is the pen, not the formula, that refuses that input, and the domain records the refusal.

Look for two inputs the rule cannot tell apart. Swapping a width for the other side length leaves the product alone.

$A(15 - w) = (15 - w)w = A(w)$

So any allowed width $w$ and its partner $15 - w$ enclose the same area. Taking $w = 4$ gives the partner $11$, and both lie strictly between $0$ and $15$, so both are allowed.

$A(4) = 4(11) = 44, \qquad A(11) = 11(4) = 44$

Each encloses $44$ square metres. What the two widths have in common is that they add to $15$, and the reason is worth naming: they describe the same rectangle twice, once with the $4$ metre side called the width and once with the $11$ metre side called the width.

Same rule, two domains, so the question is whether a domain is part of a function or merely a window it is viewed through. Test a property on each.

On the wider domain the function is not one-to-one. Whenever $w \ne 7.5$, the width $w$ and its partner $15 - w$ are two different numbers, both allowed, and they enclose the same area.

On $0 < w \le 7.5$ that partner is usually gone. Suppose two allowed widths $a$ and $b$ enclose the same area, so $a(15 - a) = b(15 - b)$. Move everything to one side.

$15a - a^2 - 15b + b^2 = 0$

Group the two differences, using $b^2 - a^2 = -(a - b)(a + b)$.

$(a - b)\bigl(15 - (a + b)\bigr) = 0$

So either $a = b$, or $a + b = 15$. The second can only happen when $a$ and $b$ are both exactly $7.5$, since each is at most $7.5$ and anything smaller would drop the sum below $15$. That branch also gives $a = b$. Either way equal areas force equal widths, and the restricted function is one-to-one.

One of the two functions has a property the other lacks, so they are not the same function. The rule never changed. The domain did, and that was enough.

Take any allowed width $w$, so $0 < w < 15$, and pair it with $15 - w$, which is also strictly between $0$ and $15$ and so also allowed. The two enclose the same area.

$A(15 - w) = (15 - w)w = A(w)$

At least one of $w$ and $15 - w$ is at most $7.5$, since their sum is exactly $15$ and two numbers both larger than $7.5$ would sum to more. The smaller of the pair therefore lies in $0 < w \le 7.5$ and encloses the very area we started from. Every area available on the wider domain is available on the smaller one, so the two functions produce exactly the same collection of outputs.

That settles nothing about whether they are the same function, because a function is not its list of outputs. It is the pairing: which input goes with which output. On the wider domain the area $4(11) = 44$ arrives twice, from the width $4$ and from the width $11$; on the smaller domain it arrives once. So the pair $(11, 44)$ belongs to the first collection of ordered pairs and not to the second, the two collections differ, and the functions differ with them.

Test the conclusion by trying to build a relation that defeats it, one that fails under both declarations. Failing the first needs two pairs sharing a first coordinate; failing the second needs two pairs sharing a second coordinate. Nothing prevents a three-pair set from doing both at once, provided the two sharings use different pairs.

$\{(-2, 5),\ (-2, 0),\ (4, 5)\}$

Read the first coordinate as the input. The input $-2$ sits in two pairs and carries the outputs $5$ and $0$, so this declaration fails.

Read the second coordinate as the input. The inputs are then $5$, $0$ and $5$, so the input $5$ sits in two pairs and carries the outputs $-2$ and $4$. This declaration fails as well.

One relation is a function under neither declaration, so the conclusion is false. Notice how little it took: three points, arranged so that the repeat in one column and the repeat in the other use different pairs.

Separate the two claims and judge them one at a time.

The first is true. If $(a, c)$ and $(a, d)$ are two different pairs of a relation, they cannot agree in the second coordinate either, because two pairs agreeing in both coordinates are one and the same pair. Among the pairs sitting above a repeated input, the second coordinates really are all different.

$(a, c) \ne (a, d) \ \Longrightarrow \ c \ne d$

The second does not follow, because that guarantee is local. It constrains the pairs above one repeated input and nothing else, and a relation has other pairs. A second coordinate that is fresh within its own group can still be repeated by a pair from a different group, and one repeat anywhere in the column is all the other declaration needs in order to fail.

So the true sentence controls one column within one group of pairs. The conclusion needs control of that column across the whole relation, and nothing in the argument has supplied it.

The first declaration fails exactly when a value repeats among the first coordinates, and the second fails exactly when a value repeats among the second coordinates. Those are conditions on two different columns of the same list of pairs, and a repeat in one column places no requirement on the other beyond the local fact of part B.

So no implication should be expected either way, and the way to prove none exists is to realise every combination.

$\{(1, 2),\ (5, 9)\}$

Neither column repeats, so this is a function under both declarations.

$\{(1, 2),\ (5, 2)\}$

The first column is clean and the second repeats $2$, so this is a function under the first declaration only.

$\{(1, 2),\ (1, 9)\}$

Now the first column repeats $1$ and the second is clean, so this is a function under the second declaration only.

$\{(1, 2),\ (1, 9),\ (5, 2)\}$

Both columns repeat, so this is a function under neither. Four relations, four outcomes: the two verdicts vary independently, and knowing one of them leaves the other exactly as open as it was before.

The numerator is the constant $9$, and there is no root anywhere in the rule, so the only way it can fail is a denominator of zero. Find the inputs that make the denominator vanish.

$|x - 20| - 6 = 0 \ \Longrightarrow \ |x - 20| = 6$

An absolute value equals $6$ exactly when the quantity inside is $6$ or $-6$, which gives two equations.

$x - 20 = 6 \quad \text{or} \quad x - 20 = -6$

These give $x = 26$ and $x = 14$. At every other real input the denominator is some nonzero number and the division goes through, returning a real output, so the natural domain is every real number except $26$ and $14$.

A function must assign an output to every input in its domain, so a declared domain is legitimate exactly when the rule delivers a value at each of its members.

Take $20 \le x \le 23$ first. Across that interval $x - 20$ runs from $0$ to $3$, so $|x - 20|$ is at most $3$ and the denominator is at most $-3$.

$|x - 20| - 6 \le 3 - 6 = -3$

It is never zero, so every input from $20$ to $23$ receives exactly one output and this declaration produces a genuine function. It is a smaller function than the default, in the sense that the inputs outside the interval are simply not in it, and nothing is wrong with that: the declaration is data, and shrinking is its privilege.

Now take all real numbers. That declaration includes $x = 26$, where the rule reads

$\frac{9}{|26 - 20| - 6} = \frac{9}{0},$

which is not a number at all. The input $26$ would have to receive an output and receives none, so this declared set is not the domain of any function built from this rule, and $x = 14$ fails in the same way.

The asymmetry is the point. A declaration may shrink the default domain as far as it likes, and may not stretch it beyond the inputs at which the rule delivers.

The two share every letter of their rule, so any difference has to come from the domain. Hunt for a property that separates them.

On the natural domain the inputs $18$ and $22$ are both allowed, and both sit two units away from $20$, so the absolute value cannot tell them apart.

$v(18) = \frac{9}{2 - 6} = -\frac{9}{4} = v(22)$

Two different inputs share an output, so this function is not one-to-one.

On the declared domain $20 \le x \le 23$ the input $18$ is gone. Every allowed input has $x - 20 \ge 0$, so $|x - 20| = x - 20$ and the rule can be rewritten without the bars.

$v(x) = \frac{9}{x - 26}$

Suppose two allowed inputs $a$ and $b$ satisfy $v(a) = v(b)$. Neither denominator is zero, so $\dfrac{9}{a - 26} = \dfrac{9}{b - 26}$ gives $a - 26 = b - 26$, hence $a = b$. No two allowed inputs share an output, and this function is one-to-one.

One of the two has a property the other lacks, so they are not the same function. The rule never changed; the domain did, and the domain is part of what a function is.

Two obligations hide inside the word function, and both have to be met: no input may carry two different outputs, and every input of the claimed domain must carry one.

For the first, suppose $(a, c)$ and $(a, d)$ both belong to $R_E$. Every pair of $R_E$ is a pair of $R$ by construction, so both belong to $R$, and $R$ is a function, so the input $a$ carries a single output there.

$(a, c), (a, d) \in R \ \Longrightarrow \ c = d$

No input of $R_E$ carries two different outputs.

For the second, take any $a$ in $E$. Since $E$ is contained in $D$, this $a$ lies in the domain of $R$, so $R$ contains a pair whose first coordinate is $a$. That pair has its first coordinate in $E$, so it survives the cut into $R_E$. Every input of $E$ therefore carries exactly one output, and the domain of $R_E$ is exactly $E$.

The argument used nothing about $R$ beyond its being a function, and nothing about $E$ beyond its sitting inside $D$, so it holds for every restriction: cutting a domain down can never destroy the function property.

Take the surviving direction first. Suppose no two inputs of $D$ share an output. Any two inputs of $E$ are inputs of $D$, since $E$ lies inside $D$, so they cannot share an output either. One-to-oneness therefore survives every cut, without any calculation.

The reverse direction claims that the smaller set can report on the larger one, and four points defeat it.

$R = \{(1, 8),\ (4, 3),\ (6, 8),\ (7, 1)\}$

Read the first coordinate as the input. The four inputs $1$, $4$, $6$, $7$ are all different, so $R$ is a function with domain $D = \{1, 4, 6, 7\}$. It is not one-to-one, because the inputs $1$ and $6$ both carry the output $8$.

Now cut the domain down to $E = \{4, 6, 7\}$.

$R_E = \{(4, 3),\ (6, 8),\ (7, 1)\}$

Its outputs are $3$, $8$ and $1$, three different numbers, so no two inputs of $E$ share an output and $R_E$ is one-to-one. The cut gained a property the original never had, so one-to-oneness on a smaller domain reports nothing about the larger one.

Part A says the function property is untouchable: whichever inputs are kept, what remains is a function on exactly those inputs. Part B says one-to-oneness moves one way only: a one-to-one function stays one-to-one on every smaller domain, while a function that is not one-to-one may become one. Together, cutting a declared domain down is always safe and is sometimes a genuine gain.

That is exactly why the two objects cannot be the same function once $E$ omits an input of $D$. If a restriction were only a change of viewpoint, then every property of the original would still be there to be seen, and the property gained in part B would have to hold of the original too. It does not, so something more than the viewpoint changed.

$R_E \subseteq R, \qquad R_E \ne R \text{ when } E \ne D$

Underneath that sits the plainer reason: a function is its pairing, the collection of ordered pairs, together with the codomain that never changed here, and a restriction that drops even one input is a smaller collection. Two different collections are two different functions, however similar the formula that described them. A domain is not the scenery a function is viewed against. It is part of the function.