Negative and Zero Exponents: Free Response

Each power comes straight from the definition, as a string of factors of the base:

$6^{3} = 6 \times 6 \times 6 = 216, \qquad 6^{2} = 6 \times 6 = 36, \qquad 6^{1} = 6.$

With the exponent falling, the row reads $216$, $36$, $6$. Now compare each entry with the one on its right. From $216$ to $36$ the entry has been divided by $6$, and from $36$ to $6$ it has been divided by $6$ again.

That is no coincidence. Dropping the exponent by one removes one factor of the base, and removing a factor is exactly dividing by it. So one step to the right means one division by $6$, and the entry after $6^{1}$ is found the same way as every entry before it:

$6 \div 6 = 1, \qquad \text{so} \qquad 6^{0} = 1.$

Nothing had to be assumed about what a zero exponent means. The row was simply continued by the operation it was already running on.

The operation does not change at the new steps, so keep dividing by $6$:

$1 \div 6 = \frac{1}{6}, \qquad \frac{1}{6} \div 6 = \frac{1}{36}.$

Those two entries sit under the exponents $-1$ and $-2$:

$6^{-1} = \frac{1}{6}, \qquad 6^{-2} = \frac{1}{36}.$

The row now reads $216$, $36$, $6$, $1$, $\tfrac{1}{6}$, $\tfrac{1}{36}$. Every entry is positive. Dividing a positive number by $6$ makes it smaller, but it can never carry the result down past zero, so the entries close in on zero from above and never cross it. A negative exponent is producing a small positive fraction here, not a negative number, and it is the divide step that guarantees it.

The row runs on one rule, and that rule has two halves which always move together: one step to the right lowers the exponent by $1$, and one step to the right divides the value by $6$.

$6^{n} \div 6 = 6^{n-1}.$

On the left-hand part of the row this is not a guess. Each of $216$, $36$ and $6$ can be multiplied out straight from the definition of a power, and the divide-by-$6$ step is then observed to hold between them. So the rule is not being invented for the new entries; it is already in force where it can be tested.

Now suppose someone hands $6^{0}$ a value other than $1$. The exponent half of the rule still applies, since the exponent does drop from $1$ to $0$. It is the value half that would have to be abandoned, at exactly that step, and abandoned again at every step after it. The cost is a second rule, taking over precisely where the multiplying out runs out, with nothing to recommend it except that the checking stopped there.

So the classmate has the situation backwards. Keeping one rule for the whole row leaves exactly one entry available at each new step, and those entries are $1$, then $\tfrac{1}{6}$, then $\tfrac{1}{36}$. What the classmate is right about is that nothing below $6^{1}$ can be multiplied out. That is the point: there is nothing there to multiply out, which is why the rule has to settle the value instead.

First the ordinary fraction. Since $5^{4} = 625$, the top and the bottom are the same number, and a nonzero number divided by itself is $1$:

$\frac{5^{4}}{5^{4}} = \frac{625}{625} = 1.$

Notice that this reading uses nothing from this lesson or the last one. It is division, and it would have been available long before powers were introduced.

Now the same quotient through the rule, which subtracts the exponent underneath from the exponent on top:

$\frac{5^{4}}{5^{4}} = 5^{4-4} = 5^{0}.$

The two readings start from identical left-hand sides, so their right-hand sides describe the same number:

$5^{0} = 1.$

There are three factors of $5$ on top and six underneath, so three pairs cancel:

$\frac{5^{3}}{5^{6}} = \frac{5 \times 5 \times 5}{5 \times 5 \times 5 \times 5 \times 5 \times 5} = \frac{1}{5 \times 5 \times 5} = \frac{1}{5^{3}} = \frac{1}{125}.$

When every factor on top has been cancelled, what is left there is $1$, not nothing: cancelling replaces each pair by $1$, and the top is a product of those.

The rule reads the very same quotient by subtracting the exponents, and this time the subtraction runs past zero:

$\frac{5^{3}}{5^{6}} = 5^{3-6} = 5^{-3}.$

Setting the two results beside each other, and remembering they are two descriptions of one quotient,

$5^{-3} = \frac{1}{5^{3}} = \frac{1}{125}.$

So the negative exponent has turned out to mean the reciprocal of the matching positive power. It was not decided; it was read off.

Why the shape of the argument is binding. Each part wrote down one expression, $\dfrac{5^{4}}{5^{4}}$ or $\dfrac{5^{3}}{5^{6}}$, and then described it twice. An expression like that stands for a single number, so two correct descriptions of it cannot name different numbers. That is what turns a pair of readings into an equation:

$\frac{5^{3}}{5^{6}} = \frac{1}{125} \qquad \text{and} \qquad \frac{5^{3}}{5^{6}} = 5^{-3}.$

Which reading does the pinning. Not both equally. The fraction reading is entitled to settle the matter because it uses only cancelling and ordinary division, neither of which mentions a zero or negative exponent. It was available before the question was even asked. The rule's reading is the one on trial: it produces a symbol nobody has yet given a meaning to. So the fraction supplies the value and the rule supplies the name, and the definition is what pins the name to the value. Had the fraction reading itself relied on the meaning of $5^{-3}$, the argument would have been circular and would have proved nothing.

Where a base of $0$ breaks it. In part A the fraction reading turns on the fact that a number divided by itself is $1$. With base $0$ that step reads

$\frac{0^{4}}{0^{4}} = \frac{0}{0},$

and $\tfrac{0}{0}$ is not $1$. It is not any number at all: it would have to be a number that multiplies $0$ back up to $0$, and every number does that, so nothing picks one out. With nothing definite standing on the left, there is nothing to pin the name to, which is exactly why $0^{0}$ is left undefined while every other base gets a value.

Handle each expression by asking first what the base is, then doing the reciprocal and the power as separate steps.

In $7^{-2}$ the base is $7$. The negative exponent calls for the reciprocal of the matching positive power:

$7^{-2} = \frac{1}{7^{2}} = \frac{1}{49}.$

In $(-7)^{-2}$ the parentheses put the minus sign inside the base, so the base is $-7$ and the whole of it is used twice. Two negative factors multiply to a positive:

$(-7)^{-2} = \frac{1}{(-7)^{2}} = \frac{1}{(-7) \times (-7)} = \frac{1}{49}.$

In $-7^{-2}$ there are no parentheses, so the exponent is attached to the $7$ alone and the leading minus sign waits outside for the power to be finished:

$-7^{-2} = -\left(7^{-2}\right) = -\frac{1}{49}.$

So the three values are $\tfrac{1}{49}$, $\tfrac{1}{49}$ and $-\tfrac{1}{49}$. The only thing that changed between the second and the third was a pair of parentheses, and it changed which number was being raised to the power.

Take the four in turn, deciding on sight.

$7^{-2}$: the base is $7$, a positive number, so every factor is positive and so is the reciprocal of the result. Positive.

$(-7)^{-2}$: the base is $-7$ and the exponent is even. Negative factors pair off, and each pair multiplies to a positive:

$(-7) \times (-7) = 49, \qquad (-7) \times (-7) \times (-7) = -343.$

So an even count of negative factors leaves nothing negative behind. Taking the reciprocal of a positive number keeps it positive. Positive.

$-7^{-2}$: the power itself is $7^{-2}$, which is positive, and the minus sign outside then negates it. Negative.

$(-7)^{-3}$: the base is $-7$ and the exponent is odd, so one minus sign is left over after the pairing, as the second display above shows. The reciprocal of a negative number is negative, since the number and its reciprocal multiply to a positive $1$. Negative.

Comparing the two kinds of minus sign, they do completely different jobs. The one in the exponent decides only whether the power is flipped over into a reciprocal, and flipping a number over never changes its sign. The one belonging to the base decides whether the factors being multiplied are negative, and then the parity of the exponent decides whether those minus signs pair off. A minus sign left outside the power is a third thing again: it is not part of the power at all, and it simply reverses the sign of whatever the power comes to.

First counterexample: the minus sign is only in the exponent. Take $7^{-2}$. A minus sign certainly appears in it, so the rule predicts a negative value. But

$7^{-2} = \frac{1}{7^{2}} = \frac{1}{49},$

which is positive. One counterexample already retires the rule.

Second counterexample, of a different kind: the base carries a minus sign too. Take $(-7)^{-2}$. Now there are two minus signs, so on the revision sheet's reading the case ought to be even clearer. Yet

$(-7)^{-2} = \frac{1}{(-7)^{2}} = \frac{1}{49},$

also positive. This one matters because it blocks the obvious patch. Somebody might try to save the rule by saying a minus sign counts only when it is on the base; this case has one there, and the value is still positive.

A corrected rule. What actually decides the sign is which number is being multiplied and how many times, so state it that way. For a nonzero base written in parentheses, the power is negative exactly when the base is negative and the exponent is odd.

Check it both ways round, since a rule of that form claims two things. If the base is negative and the exponent is odd, one factor is left unpaired and the result is negative, as in

$(-7)^{-3} = \frac{1}{(-7)^{3}} = \frac{1}{-343} = -\frac{1}{343}.$

And if a power is negative, the base cannot be positive, since positive factors only ever make positive products, and the exponent cannot be even, since the minus signs would pair off. Now run the counterexamples past it: $7^{-2}$ has a positive base, so the corrected rule predicts positive, and it is; $(-7)^{-2}$ has a negative base but an even exponent, so again the corrected rule predicts positive, and it is. Neither breaks it.

Zoom In clicks count as positive and Zoom Out clicks as negative, so the two settings are $n = 2$ and $n = -3$. The first multiplier needs no new machinery:

$4^{2} = 16.$

The second has a negative exponent, so take the reciprocal of the matching positive power and then work that power out:

$4^{-3} = \frac{1}{4^{3}} = \frac{1}{64}.$

Read back into the picture, a multiplier of $16$ means a line drawn $3$ units long in the opening view is drawn $48$ units long after two clicks in, since $3 \times 16 = 48$. A multiplier of $\tfrac{1}{64}$ means the same line is drawn $\tfrac{3}{64}$ of a unit long after three clicks out, a great deal shorter than it started. The second multiplier is a small positive fraction, which is what shrinking looks like. It is not a negative length, and nothing in the picture is turned round or reflected; the negative exponent only records that the clicks went the other way.

Each burst of clicking contributes its own multiplier, and doing one after the other multiplies them:

$4^{-3} \times 4^{5}.$

The product rule adds the exponents, and it does that whatever their signs, so the negative one is added exactly as it stands:

$4^{-3} \times 4^{5} = 4^{-3+5} = 4^{2} = 16.$

Now check it against the buttons rather than against the algebra. Three clicks out followed by five clicks in: each click out is undone by a click in, so three of the five clicks in are spent cancelling the three clicks out, and two clicks in are left over. Two clicks in is a multiplier of $4^{2} = 16$, the same answer.

The check is worth pausing on. The addition of the exponents and the cancelling of the clicks are the same bookkeeping: a click out is a click in run backwards, which is why one law with a negative exponent in it can describe both buttons at once. The alternative, arithmetic all the way, gives the same thing but tells you less:

$\frac{1}{64} \times 1024 = \frac{1024}{64} = 16.$

What the user would see. The multiplier is applied to every length in the drawing, so with a multiplier of $0$ a line $12$ units long in the opening view would be drawn

$12 \times 0 = 0$

units long, and so would every other line. The drawing would open as an empty screen, and no amount of zooming afterwards would help, since $0$ multiplied by anything is still $0$. Worse, the manual would be describing the opening view itself, the one view where nothing has been done to the drawing at all.

What it has to be instead. The opening view is the view all the multipliers are measured against. After no clicks nothing has happened, so the multiplier has to leave every length exactly as it was:

$12 \times 1 = 12.$

The number that leaves a multiplication unchanged is $1$, and it is the only one, so the multiplier after $0$ clicks is $1$, giving $4^{0} = 1$.

Notice which way the reasoning ran. Nothing was quoted from a rule about exponents here; the story fixed the value on its own, because a multiplier that changes nothing has only one possible size. That the definition of a zero exponent agrees with it is the whole point of choosing the definition the way it was chosen.

The product rule, if it is to go on holding when one of the exponents is zero, adds the exponents like any other pair:

$9^{3} \times 9^{0} = 9^{3+0} = 9^{3} = 729.$

Now read the same product with the rival value put in. The rival says $9^{0} = 0$, so

$9^{3} \times 9^{0} = 729 \times 0 = 0.$

The two readings begin from the same product, so between them they claim that $729 = 0$, which is false. One of the two has to be given up.

Which one is not a close call. The product rule was proved for exponents of $1$ or more before any of this came up, and protecting it is the entire reason for defining the new symbols at all. The rival value is the newcomer, so it is the newcomer that goes.

It is worth running the same test on the value actually adopted, to see that the test is a real one and not rigged. With $9^{0} = 1$ the second reading becomes $729 \times 1 = 729$, which agrees with the first, and no contradiction appears.

Start with the requirement, before any value is chosen for the negative power. The product rule adds the exponents, and the two here cancel:

$9^{2} \times 9^{-2} = 9^{2 + (-2)} = 9^{0}.$

So whatever $9^{-2}$ turns out to mean, multiplying it by $81$ has to give $9^{0}$.

Now the rival meaning. It says $9^{-2} = -9^{2} = -81$, so the product is

$81 \times (-81) = -6561.$

Compare that with the requirement. Under the definitions this lesson adopted, $9^{0} = 1$, and $-6561$ is not $1$. But the rival does not even survive on its own terms: its own value for $9^{0}$ is $0$, and $-6561$ is not $0$ either. So the rival's two halves disagree with each other, quite apart from disagreeing with anything else.

Finally, find what the rule does leave available. Write $x$ for the value of $9^{-2}$. The requirement is $81 \times x = 9^{0} = 1$, and only one number multiplies $81$ up to $1$:

$81 \times x = 1, \qquad \text{so} \qquad x = \frac{1}{81} = \frac{1}{9^{2}}.$

That is the reciprocal of the matching positive power, which is exactly the definition the lesson adopted. There was no room to prefer anything else: one equation with one unknown had one solution.

The standard a definition is held to. The symbols $a^{0}$ and $a^{-n}$ meant nothing under the old definition of a power, so anyone may write down a meaning for them. What is not free is the company they must keep. A body of results about exponents had already been proved for the cases the old definition covered, and a new meaning earns its place only if those results survive it. Otherwise the notation splits in two, a region where a law holds and a region where it does not, with a boundary to be remembered at every calculation. The student's argument, that the rival pair keeps the minus signs visible, is an argument about how the notation looks, and looks are not the standard.

Why only one pair survives. Hold each symbol up to the product rule. For the zero power the rule demands $9^{3} \times 9^{0} = 9^{3}$, so $9^{0}$ must be a number that leaves $729$ unchanged under multiplication:

$729 \times x = 729, \qquad \text{so} \qquad x = 1.$

For the negative power the rule demands $9^{2} \times 9^{-2} = 9^{0} = 1$, so $9^{-2}$ must be the number that multiplies $81$ up to $1$:

$81 \times x = 1, \qquad \text{so} \qquad x = \frac{1}{81} = \frac{1}{9^{2}}.$

Neither step offered a choice: each is one equation in one unknown with one solution, and neither used anything about the base except that it was not $0$. So the adopted pair is not a convention that happened to win a vote. It is the only pair the already proved laws permit, and the rival pair is not an alternative convention but a contradiction.

Why $0^{0}$ is still left out. Run the very same test with the base $0$. The rule demands $0^{3} \times 0^{0} = 0^{3}$, which reads

$0 \times x = 0,$

and that is true for every number $x$. This test does not force a value, so this course leaves $0^{0}$ undefined: the test that named a single value for every other base names all of them at once here. Reading a quotient instead gives the same verdict from the other side, since the direct calculation there would be $\tfrac{0}{0}$, which is no number at all. These exponent laws force a definition only when they leave exactly one value available. At $0^{0}$ they leave either every value or none, so nothing is forced and the symbol is left with no value.