Zero and negative exponents
Dividing powers subtracts their exponents. What do 10^0 and 10^{-3} have to mean so that rule never breaks?
Where you are 3 in Laws of exponents
Why this exists
An exponent counts factors. 10^3 means 10 \cdot 10 \cdot 10, three factors of 10, and that is all exponent notation says. Under that meaning 10^0 and 10^{-3} are nonsense: you cannot write down no factors, and you certainly cannot write down minus three of them.
Yet the rule for dividing powers with the same base produces exactly these symbols. That rule subtracts the exponents, \displaystyle \frac{a^m}{a^n} = a^{m-n}, and it was proved for m larger than n. Use it anyway on two quotients whose values nobody doubts:
The first quotient is 1, because a nonzero number divided by itself is 1. The second is \displaystyle \frac{100}{100\,000} = \frac{1}{1000}. So there are two ways forward. Either the subtraction rule carries a list of cases where it may not be used, or the symbols 10^0 and 10^{-3} are given meanings, and the only meanings that keep the rule true are the values it has just produced.
This lesson takes the second way. The meanings are not found by counting factors, since there are none to count. They are chosen, and the choice is forced: any other choice would break a rule that already works. The pay-off is that powers of ten, and of every other nonzero number, now run in both directions. A millimeter is 10^{-3} meters and a micrometer is 10^{-6} meters, and the same rules that multiply 10^3 by 10^2 handle them.
The lesson is taught over two days. Part 1 is the zero exponent: why a^0 = 1, the tables that agree with it, and why 0^0 is left without a value. Part 2 is negative exponents: what a^{-n} means, how to rewrite an expression with positive exponents, and why the product and power rules still hold.
Definition
Part 1: the zero exponent. For every nonzero number a,
Read a^0 as “a to the power zero”. The base may be any nonzero number, positive, negative or a fraction, or any expression whose value is not zero: (-7)^0 = 1 and \displaystyle \left(\frac{2}{9}\right)^0 = 1. The base 0 is excluded, and the reason is set out under Why it is true.
Part 2: negative exponents. For every nonzero number a and every positive whole number n,
Read a^{-n} as “a to the power negative n”. A negative exponent means the reciprocal of the matching positive power: \displaystyle 5^{-2} = \frac{1}{5^2} = \frac{1}{25}. The minus sign belongs to the exponent, and it says “divide by this many factors”, not “make the number negative”.
With both definitions in place an exponent may be any integer: positive, zero or negative.
As with positive exponents, an exponent belongs only to the base it is written on, the way the order of operations reads powers. In -3^0 the base is 3, so -3^0 = -1; in (-3)^0 the base is -3, so (-3)^0 = 1.
Intuition
Read a column of powers downward. Each step lowers the exponent by 1 and divides the value by the base. Read it upward and each step raises the exponent by 1 and multiplies by the base. That is true on every row where the exponent counts factors, and nothing in it stops at exponent 1: keep dividing and the rows below fill themselves in.
| Power of 5 | Value | Power of 3 | Value |
|---|---|---|---|
| 5^3 | 125 | 3^3 | 27 |
| 5^2 | 25 | 3^2 | 9 |
| 5^1 | 5 | 3^1 | 3 |
| 5^0 | 1 | 3^0 | 1 |
| 5^{-1} | \displaystyle \frac{1}{5} | 3^{-1} | \displaystyle \frac{1}{3} |
| 5^{-2} | \displaystyle \frac{1}{25} | 3^{-2} | \displaystyle \frac{1}{9} |
| 5^{-3} | \displaystyle \frac{1}{125} | 3^{-3} | \displaystyle \frac{1}{27} |
Both columns pass through 1 at exponent 0, and so would a column for any base: the row above is the base itself, and the base divided by itself is 1. Read upward, that same row is where multiplying starts. 5^2 is 1 multiplied by two factors of 5, 5^1 is 1 multiplied by one, and 5^0 is 1 multiplied by none, which leaves the 1. Below the zero row, the rows mirror the ones above: 5^{-3} is the reciprocal of 5^3, and 3^{-2} is the reciprocal of 3^2.
Halving a sheet of paper gives the powers of 2 a picture. Call the whole sheet 1. Each cut halves a piece, so after one cut a piece is \displaystyle \frac{1}{2} = 2^{-1} of the sheet, after two cuts \displaystyle \frac{1}{4} = 2^{-2}, and after n cuts 2^{-n}. Before any cut, the piece is the whole sheet: 2^0 = 1. The exponent counts halvings, and the minus sign says they are halvings rather than doublings.
The sheet also shows the rules surviving. Three cuts followed by two more is five cuts, and \displaystyle 2^{-3} \cdot 2^{-2} = \frac{1}{8} \cdot \frac{1}{4} = \frac{1}{32} = 2^{-5}: the product rule, adding exponents, with both exponents negative.
Results
Zero exponent. For every nonzero a,
Negative exponents. For every nonzero a and b and every positive whole number n,
The rules hold for every integer exponent. For every nonzero a and all integers m and n, positive, negative or zero,
The quotient rule now needs no cases. When the bottom exponent is the larger, the answer simply has a negative exponent, which says the factor belongs in the denominator.
The base 0 is excluded. 0^{-n} would mean \displaystyle \frac{1}{0}, and in this lesson 0^0 is left undefined.
Why it is true
The definitions cannot be proved from counting factors, because a^0 and a^{-n} count none. What can be shown is that each is the only choice that keeps the rules working, and that once made, it keeps all of them working, not just the one it was chosen for.
Part 1: why a^0 has to be 1
Let a be nonzero and n a positive whole number. Then a^n is a product of nonzero numbers, so it is not zero, and a nonzero number divided by itself is 1:
The quotient rule, applied to the same quotient, gives a^{n-n} = a^0. If the rule is to hold here too, a^0 must be 1.
The product rule says the same thing by a different road. It requires a^0 \cdot a^n = a^{0+n} = a^n, so a^0 is a number that multiplies a^n and leaves it unchanged. Dividing both sides by the nonzero a^n gives a^0 = 1. Two different rules force one answer, so the definition makes them agree rather than fight.
Neither argument used which nonzero number a is. So a^0 = 1 for a = 2, for a = -5 and for \displaystyle a = \frac{3}{4} alike.
Why 0^0 is left undefined. Both arguments divide by a^n, and when a = 0 every positive power 0^n is 0, and dividing by 0 has no meaning. So neither argument says anything about 0^0. Patterns pull two ways. The powers 0^3, 0^2, 0^1 are all 0, which suggests 0^0 = 0. The zero powers 3^0, 1^0, \displaystyle \left(\frac{1}{2}\right)^0, \displaystyle \left(\frac{1}{10}\right)^0 are all 1 as the base shrinks toward 0, which suggests 0^0 = 1. Two reasonable patterns give two different answers. Some areas of mathematics, such as counting arrangements and writing out polynomials, agree to take 0^0 = 1 because it makes their formulas work; elsewhere, as in calculus, it is left undefined. The rules of exponents do not force a value, so in this lesson 0^0 is left undefined, in the same way as \displaystyle \frac{0}{0}.
Part 2: why a^{-n} has to be \displaystyle \frac{1}{a^n}
Let a be nonzero and n a positive whole number. If the product rule is to hold for -n,
Two numbers whose product is 1 are reciprocals of each other, so \displaystyle a^{-n} = \frac{1}{a^n}. The quotient rule agrees: \displaystyle \frac{a^0}{a^n} = a^{0-n} = a^{-n}, while \displaystyle \frac{a^0}{a^n} = \frac{1}{a^n} because a^0 = 1. The base 0 is excluded for the same reason as before: 0^{-n} would be \displaystyle \frac{1}{0^n} = \frac{1}{0}.
Two facts follow straight from the definition. First, the product a^{-n} \cdot a^n = 1 says just as well that a^n is the reciprocal of a^{-n}:
Second, the power of a product, with the positive exponent n, gives
so \displaystyle \left(\frac{b}{a}\right)^n is the reciprocal of \displaystyle \left(\frac{a}{b}\right)^n, which by the definition is \displaystyle \left(\frac{a}{b}\right)^{-n}.
The other rules still hold. The definitions were chosen to keep the quotient rule. The product and power rules have to be checked, case by case, with m and n positive whole numbers.
Product, both exponents negative. Multiplying the reciprocals,
Product, one exponent negative. \displaystyle a^m \cdot a^{-n} = \frac{a^m}{a^n}. If m > n this is a^{m-n} by the whole-number quotient rule. If m = n it is 1, which is a^0 = a^{m-n}. If m < n, canceling m factors leaves \displaystyle \frac{1}{a^{n-m}} = a^{-(n-m)} = a^{m-n}. In every case it is a^{m+(-n)}.
Power of a power. Using the definition and the power of a quotient,
and each line is a to the product of the two exponents.
Zero. a^0 \cdot a^n = 1 \cdot a^n = a^{0+n}, and (a^0)^n = 1^n = 1 = a^{0 \cdot n}, and (a^m)^0 = 1 = a^{m \cdot 0}.
Quotient. Now \displaystyle \frac{1}{a^n} = a^{-n} for every integer n (the definition, the case n = 0, and the first fact above), so \displaystyle \frac{a^m}{a^n} = a^m \cdot a^{-n} = a^{m-n} by the product rule just checked.
Worked examples
Examples 1 to 3 go with Part 1, and Examples 4 to 7 with Part 2.
Zero exponents and their bases
Find the value of each.
(a) 9^0 \qquad (b) (-6)^0 \qquad (c) -6^0 \qquad (d) 3 \cdot 8^0 and (3 \cdot 8)^0
(e) For which values of x does (x - 3)^0 equal 1?
Solution(a) The base 9 is nonzero, so 9^0 = 1.
(b) The brackets make the base -6, which is nonzero, so (-6)^0 = 1. A zero exponent gives 1 whatever the sign of the base.
(c) With no brackets the exponent belongs to the 6 alone, and the minus sign is applied afterwards: -6^0 = -(6^0) = -1.
(d) In 3 \cdot 8^0 only the 8 carries the exponent, so 3 \cdot 8^0 = 3 \cdot 1 = 3. In (3 \cdot 8)^0 the whole product is the base, and 24 is nonzero, so (3 \cdot 8)^0 = 1.
(e) The base is x - 3. Whenever it is nonzero the definition gives (x - 3)^0 = 1. It is zero only when x = 3, and then the expression is 0^0, which has no value. So (x - 3)^0 = 1 for every x except x = 3.
A quotient that lands on zero
(a) Find \dfrac{10^5}{10^5} in two ways: from the values, and by subtracting exponents.
(b) Simplify \dfrac{6m^4n^2}{2m^4n}, where m and n are nonzero.
Solution(a) From the values, \displaystyle \frac{100\,000}{100\,000} = 1. By the quotient rule, \displaystyle \frac{10^5}{10^5} = 10^{5-5} = 10^0. Both are the same quotient, and 10^0 = 1, so the two ways agree.
(b) Divide the numbers and each letter separately. The coefficients give \displaystyle \frac{6}{2} = 3. For m, subtract the exponents: 4 - 4 = 0, so the m part is m^0 = 1. For n, remember that n means n^1: 2 - 1 = 1.
The four factors of m on top cancel the four below completely, and m^0 = 1 is the record of that.
Check at m = 2, n = 5. The top is 6 \cdot 16 \cdot 25 = 2400 and the bottom is 2 \cdot 16 \cdot 5 = 160, so the quotient is 2400 \div 160 = 15, and 3n = 15.
Players left in a knockout tournament
Mirela’s tennis club runs a knockout tournament for 64 players. In each round the players still in pair off, each match has one winner, and only the winners go on.
(a) Write the number of players left after r rounds as a power of 2.
(b) How many players are left after 4 rounds?
(c) After how many rounds is one player left, and what does the formula say there?
Solution(a) Each round halves the number of players, so after r rounds there are \displaystyle \frac{64}{2^r} left. Since 64 = 2^6, the quotient rule gives
(b) With r = 4, there are 2^{6-4} = 2^2 = 4 players left: the semi-finalists.
(c) Halving 64 gives 32, 16, 8, 4, 2, 1, so one player is left after 6 rounds. The formula gives 2^{6-6} = 2^0 there, and 2^0 = 1: the champion. This is exactly the case the zero exponent was defined for, top and bottom exponents equal, and the definition gives the answer the tournament does.
The other end works too. Before the first round, r = 0, and \displaystyle \frac{64}{2^0} = \frac{64}{1} = 64, everyone. A formula with a zero exponent at both ends is right at both ends only because 2^0 = 1.
Values of negative powers
Find the value of each.
(a) 10^{-3} \qquad (b) 4^{-3} \qquad (c) (-2)^{-5} \qquad (d) \displaystyle \left(\frac{1}{6}\right)^{-2}
(e) 2^{-1} + 3^{-1}, and compare it with (2 + 3)^{-1}.
Solution(a) By the definition, \displaystyle 10^{-3} = \frac{1}{10^3} = \frac{1}{1000}.
(b) \displaystyle 4^{-3} = \frac{1}{4^3} = \frac{1}{64}. The negative exponent made a fraction, not a negative number.
(c) The base is -2. The definition gives the reciprocal of (-2)^5, and five negative factors make a negative product:
The answer is negative because the base is negative and there are an odd number of factors, not because the exponent is negative.
(d) A fraction to a negative power turns over: \displaystyle \left(\frac{1}{6}\right)^{-2} = 6^2 = 36. From the definition directly, \displaystyle \frac{1}{\left(\frac{1}{6}\right)^2} = \frac{1}{\frac{1}{36}} = 36.
(e) Each term is a reciprocal: \displaystyle 2^{-1} + 3^{-1} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}. But \displaystyle (2 + 3)^{-1} = 5^{-1} = \frac{1}{5}. The two differ: an exponent acts on its base, and the rules for exponents are about products and quotients, never about sums.
Millimeters and micrometers
Odette measures with a ruler marked in millimeters and reads a microscope scale in micrometers. A millimeter is 10^{-3} meters and a micrometer is 10^{-6} meters.
(a) How many micrometers make one millimeter?
(b) A sheet of her notebook paper is about 10^{-1} millimeters thick, a tenth of a millimeter. Write that thickness in meters as a power of ten, and in micrometers.
Solution(a) The question is how many times 10^{-6} goes into 10^{-3}, which is a quotient of powers of 10. Subtract the exponents, minding the signs:
A millimeter is 1000 micrometers. Check with fractions: \displaystyle \frac{1}{1000} \div \frac{1}{1\,000\,000} = \frac{1\,000\,000}{1000} = 1000.
(b) A tenth of a millimeter is 10^{-1} of 10^{-3} meters. Multiply by adding exponents:
which is \displaystyle \frac{1}{10\,000} of a meter. In micrometers, divide by 10^{-6}: \displaystyle \frac{10^{-4}}{10^{-6}} = 10^{2} = 100, so the paper is about 100 micrometers thick.
Clearing negative exponents from an expression
Rewrite each with positive exponents only. Every variable is nonzero.
(a) x^{-6} \qquad (b) 5a^{-2}b \qquad (c) \dfrac{1}{p^{-4}} \qquad (d) \dfrac{x^{-1}}{y^{-3}} \qquad (e) \left(\dfrac{2}{m}\right)^{-3}
Solution(a) By the definition, \displaystyle x^{-6} = \frac{1}{x^6}.
(b) The exponent -2 belongs to a only. Replace a^{-2} by \displaystyle \frac{1}{a^2} and leave the 5 and the b where they are: \displaystyle 5a^{-2}b = 5 \cdot \frac{1}{a^2} \cdot b = \frac{5b}{a^2}.
(c) By the first fact under Why it is true, \displaystyle \frac{1}{p^{-4}} = p^4: the reciprocal of a reciprocal is the power you started from.
(d) Split the fraction into a product, and rewrite each part:
Each factor with a negative exponent has crossed the fraction bar, and its exponent has turned positive.
(e) Turn the fraction over and make the exponent positive, then use the power of a quotient:
The rules with negative exponents
(a) Find 2^{-3} \cdot 2^5 from the definition, and again by the product rule.
(b) Find (2^{-1})^{-3} from the definition, and again by the power-of-a-power rule.
(c) Simplify \dfrac{(r^{-2}s)^3}{r^{-4}s^5}, writing the answer with positive exponents. The variables are nonzero.
Solution(a) From the definition, \displaystyle 2^{-3} \cdot 2^5 = \frac{1}{8} \cdot 32 = 4. By the product rule, add the exponents: 2^{-3+5} = 2^2 = 4. They agree, as Why it is true says they must.
(b) From the definition, \displaystyle (2^{-1})^{-3} = \left(\frac{1}{2}\right)^{-3} = \frac{1}{\frac{1}{8}} = 8. By the power-of-a-power rule, multiply the exponents: 2^{(-1)(-3)} = 2^3 = 8. Again they agree.
(c) First the top. The power of a product sends the 3 to each factor, and the power of a power multiplies -2 by 3:
Now divide, subtracting exponents base by base:
Check at r = 2, s = 3. The top is \displaystyle \left(\frac{3}{4}\right)^3 = \frac{27}{64} and the bottom is \displaystyle \frac{3^5}{2^4} = \frac{243}{16}, so the quotient is \displaystyle \frac{27}{64} \cdot \frac{16}{243} = \frac{1}{36}. The answer gives \displaystyle \frac{1}{4 \cdot 9} = \frac{1}{36}.
Boundaries
The base must not be zero. 0^0 has no value in this lesson, and 0^{-n} would be \displaystyle \frac{1}{0}. So x^0 = 1 and \displaystyle x^{-2} = \frac{1}{x^2} are statements about nonzero x only, and an expression such as (x - 3)^0 equals 1 everywhere except where its base is zero.
An exponent belongs to its own base. 3 \cdot 8^0 = 3 but (3 \cdot 8)^0 = 1; -6^0 = -1 but (-6)^0 = 1; and \displaystyle 5a^{-2} = \frac{5}{a^2} while \displaystyle (5a)^{-2} = \frac{1}{25a^2}.
A negative exponent does not make a number negative. For a positive base, every power is positive: \displaystyle 4^{-3} = \frac{1}{64}. The sign of a power comes from the base, and a negative base gives a negative value only when the number of factors is odd.
The rules are about products and quotients, not sums. A factor may cross a fraction bar and change the sign of its exponent, but a term in a sum may not. For instance \displaystyle \frac{1}{2^{-1} + 3^{-1}} = \frac{1}{\frac{5}{6}} = \frac{6}{5}, which is not 2 + 3.
The exponents here are integers. A definition for an exponent such as \displaystyle \frac{1}{2} is a further choice, made by the same principle of keeping the rules: see fractional exponents.
Where this goes next
With exponents allowed to be any integer, the rules for exponents stop needing cases, and monomials and the exponent rules puts them to work on terms with a number in front and several letters at once, where a negative exponent is simply another way of saying a factor belongs on the bottom.
The column of powers of ten that now runs down past 10^0 is what lets a very small quantity be written as an ordinary number times 10^{-n}; that is writing numbers in scientific notation.
Halving again and again is the simplest case of decay, and 2^{-n} is the size after n halvings. Exponential growth and decay uses exactly these powers, with a negative exponent meaning a step backwards in time. Further on, the same question this lesson asked, “what must the symbol mean so the rules keep working?”, decides what an exponent of \displaystyle \frac{1}{2} or \displaystyle \frac{1}{3} means, in fractional exponents.
Exercises
Part 1: zero exponents
Find the value of each.
- 13^0
- -9^0 and (-9)^0
- 6^0 + 6^1
Simplify. Every variable is nonzero.
- \dfrac{15a^3b^6}{5a^3b^2}
- \dfrac{y^2 \cdot y^7}{y^9}
Part 1: using the zero exponent
- For which values of t is (2t + 8)^0 equal to 1? Explain the value that is left out.
- Rasmus-Ole runs a knockout chess tournament for 128 players, and each round halves the number still in. Write the number left after r rounds as a power of 2, and find after how many rounds one player is left.
Part 2: negative exponents as numbers
Find the value of each.
- 7^{-2}
- 10^{-5}
- (-3)^{-3}
- \displaystyle \left(\frac{1}{4}\right)^{-3}
- \displaystyle \left(\frac{5}{2}\right)^{-2}
- 2^{-2} + 2^{-3}
Part 2: writing with positive exponents
Rewrite with positive exponents only. Every variable is nonzero.
- b^{-9}
- 8c^{-1}d^2
- \dfrac{3}{h^{-5}}
- \dfrac{w^{-4}z}{6}
Part 2: the rules with negative exponents
Simplify, writing the answer with positive exponents. Every variable is nonzero.
- k^{-8} \cdot k^3
- \dfrac{n^{-3}}{n^4}
- (x^{-2})^{-4}
- \dfrac{(2t^{-1})^3}{t^{-5}}
- Ines has a full jug of juice. Each time she pours, she pours out half of what is left in the jug. She pours 3 times, and later 5 more times. What part of the full jug is still in it? Give it as a power of 2 and as a fraction.
Reasoning
- Use the product rule a^m \cdot a^n = a^{m+n}, and not the quotient rule, to explain why 2^0 has to be 1.
- Is 5^{-2} positive or negative? Explain, and say when a power with a negative exponent can be negative.
- Show from the definition that a^{-2} \cdot a^{-3} = a^{-5} for every nonzero a, and explain why a = 0 has to be left out.
Answers
- 1. The base 13 is nonzero.
- -9^0 = -1 and (-9)^0 = 1. Without brackets the exponent belongs to the 9 alone, so -9^0 = -(9^0) = -1; with brackets the base is -9.
- 7. 6^0 + 6^1 = 1 + 6.
- 3b^4. The coefficients give \displaystyle \frac{15}{5} = 3; for a, 3 - 3 = 0 and a^0 = 1; for b, 6 - 2 = 4.
- 1. The top is y^{2+7} = y^9, and \displaystyle \frac{y^9}{y^9} = y^{9-9} = y^0 = 1.
- Every t except t = -4. The base 2t + 8 is zero only when t = -4, and there the expression is 0^0, which has no value; for every other t the base is nonzero and the power is 1.
- 2^{7-r} players, and one is left after 7 rounds. Since 128 = 2^7, after r rounds there are \displaystyle \frac{2^7}{2^r} = 2^{7-r} left, and 2^{7-7} = 2^0 = 1.
- \displaystyle \frac{1}{49}. \displaystyle 7^{-2} = \frac{1}{7^2}.
- \displaystyle \frac{1}{100\,000}. \displaystyle 10^{-5} = \frac{1}{10^5}.
- \displaystyle -\frac{1}{27}. \displaystyle (-3)^{-3} = \frac{1}{(-3)^3} = \frac{1}{-27}.
- 64. The fraction turns over: \displaystyle \left(\frac{1}{4}\right)^{-3} = 4^3 = 64.
- \displaystyle \frac{4}{25}. \displaystyle \left(\frac{5}{2}\right)^{-2} = \left(\frac{2}{5}\right)^2 = \frac{4}{25}.
- \displaystyle \frac{3}{8}. \displaystyle \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8}.
- \displaystyle \frac{1}{b^9}, by the definition.
- \displaystyle \frac{8d^2}{c}. Only c carries the negative exponent, and \displaystyle c^{-1} = \frac{1}{c}.
- 3h^5. \displaystyle \frac{1}{h^{-5}} = h^5, so \displaystyle \frac{3}{h^{-5}} = 3 \cdot h^5.
- \displaystyle \frac{z}{6w^4}. \displaystyle w^{-4} = \frac{1}{w^4}, so the w^4 joins the 6 on the bottom.
- \displaystyle \frac{1}{k^5}. Add the exponents: -8 + 3 = -5, and \displaystyle k^{-5} = \frac{1}{k^5}.
- \displaystyle \frac{1}{n^7}. Subtract the exponents: -3 - 4 = -7.
- x^8. Multiply the exponents: (-2)(-4) = 8.
- 8t^2. The top is 2^3t^{-3} = 8t^{-3}, and dividing by t^{-5} gives 8t^{-3-(-5)} = 8t^2.
- \displaystyle 2^{-8} = \frac{1}{256} of the jug. Each pour leaves half of what was there, so 3 pours leave 2^{-3} of the jug and 5 more leave 2^{-5} of that: 2^{-3} \cdot 2^{-5} = 2^{-8}.
- The product rule asks that 2^0 \cdot 2^3 = 2^{0+3} = 2^3 = 8. So 2^0 is a number that multiplies 8 and leaves 8 unchanged, and the only such number is 1. The same works with any power of 2 in place of 2^3.
- Positive: \displaystyle 5^{-2} = \frac{1}{25}. A negative exponent gives the reciprocal of a positive power, and the reciprocal of a positive number is positive. A power can be negative only when the base is negative and the number of factors is odd, as in \displaystyle (-2)^{-3} = \frac{1}{-8} = -\frac{1}{8}.
- \displaystyle a^{-2} \cdot a^{-3} = \frac{1}{a^2} \cdot \frac{1}{a^3} = \frac{1}{a^5} = a^{-5}, using the product rule for the positive powers on the bottom. With a = 0 the first step would need \displaystyle \frac{1}{0^2} = \frac{1}{0}, which has no value, so the definition does not apply.