The definition of exponents given at the beginning of this section—
106106 size 12{"10" rSup { size 8{6} } } {}means
10⋅10⋅10⋅10⋅10⋅1010⋅10⋅10⋅10⋅10⋅10 size 12{"10" cdot "10" cdot "10" cdot "10" cdot "10" cdot "10"} {}—does not enable us to answer questions such as:
4
0
=
?
4
0
=
?
size 12{4 rSup { size 8{0} } =?} {}
(1)
5
−
4
=
?
5
−
4
=
?
size 12{5 rSup { size 8{ - 4} } =?} {}
(2)
9
1
2
=
?
9
1
2
=
?
size 12{9 rSup { size 8{ { { size 6{1} } over { size 6{2} } } } } =?} {}
(3)
You can’t “multiply 9 by itself half a time” or “multiply 5 by itself
−4−4 size 12{ - 4} {} times.” In general, the original definition only applies if the exponent is a positive integer.
It’s very important to understand this point. The question is not “What answer does our original definition give in these cases?” The original definition does not give any answer in these cases. If we want to include these numbers, we need a whole new definition of what an exponent is.
In principle, many such definitions are possible. We could define
5−45−4 size 12{5 rSup { size 8{ - 4} } } {} as
5/5/5/55/5/5/5 size 12{5/5/5/5} {}: in other words, divide four times instead of multiplying four times. Or we could define
5−45−4 size 12{5 rSup { size 8{ - 4} } } {}as
5−45−4 size 12{5 rSup { size 8{ - 4} } } {}: take 5 to the fourth power, and then multiply it by
−1−1 size 12{ - 1} {}. Or we could define
5−45−4 size 12{5 rSup { size 8{ - 4} } } {} as
−(5)4−(5)4 size 12{ - \( 5 \) rSup { size 8{4} } } {}: take
−5−5 size 12{ - 5} {} to the fourth power (which gives a different answer from the previous definition). It seems at first that we are at liberty to choose any definition we want.
Given that degree of freedom, you may be very surprised at the definitions that are actually used: they seem far more arbitrary and complicated than some others you could come up with.
Table 1: Definitions: When the exponent is not a positive integer
|
Zero exponents
|
Negative exponents
|
Fractional exponents
(numerator = 1)
|
Fractional exponents
(numerator ≠ 1)
|
| Always 1 |
Go in the denominator |
Act as roots |
The numerator is an exponent
The denominator is a root |
7
0
=
1
7
0
=1
9
0
=
1
9
0
=1
x
0
=
1
x
0
=1
|
7
-3
=
1
7
3
=
1
343
7
-3
=
1
7
3
=
1
343
x
-5
=
1
x
5
x
-5
=
1
x
5
1
5
-3
=
5
3
1
5
-3
=
5
3
|
9
1
2
=
9
=
3
9
1
2
=
9
=3
2
1
2
=
2
2
1
2
=
2
8
1
3
=
8
3
=
2
8
1
3
=
8
3
=2
x
1
4
=
x
4
x
1
4
=
x
4
|
8
2
3
=
8
2
3
8
2
3
=
8
2
3
or
(
8
3
)
2
(
8
3
)
2
Order doesn't matter!
8
2
3
=
64
3
=
4
8
2
3
=
64
3
=4
or
(
8
3
)
2
=
2
2
=
4
(
8
3
)
2
=
2
2
=4
832
=
8
3
8
3
2
=
8
3
or
(
8
)
3
(
8
)
3
|
Note that you can combine these definitions. For instance,
8−238−23 size 12{8 rSup { size 8{ - { { size 6{2} } over { size 6{3} } } } } } {} is a negative, fractional exponent. The negative exponent means, as always, “put me in the denominator.” So we can write:
8
−
2
3
=
1
8
2
3
=
1
8
2
3
=
1
4
8
−
2
3
=
1
8
2
3
=
1
8
2
3
=
1
4
size 12{8 rSup { size 8{ - { { size 6{2} } over { size 6{3} } } } } = { {1} over {8 rSup { { { size 6{2} } over { size 6{3} } } } } } size 12{ {}= { {1} over { nroot {3} {8 rSup {2} } } } } size 12{ {}= { {1} over {4} } }} {}
(4)
These are obviously not chosen to be the simplest possible definitions. But they are chosen to be consistent with the behavior of positive-integer exponents.
One way to see that consistency is to consider the following progression:
19
4
=
19
⋅
19
⋅
19
⋅
19
19
4
=
19
⋅
19
⋅
19
⋅
19
size 12{"19" rSup { size 8{4} } ="19" cdot "19" cdot "19" cdot "19"} {}
(5)
19
3
=
19
⋅
19
⋅
19
19
3
=
19
⋅
19
⋅
19
size 12{"19" rSup { size 8{3} } ="19" cdot "19" cdot "19"} {}
(6)
19
2
=
19
⋅
19
19
2
=
19
⋅
19
size 12{"19" rSup { size 8{2} } ="19" cdot "19"} {}
(7)
19
1
=
19
19
1
=
19
size 12{"19" rSup { size 8{1} } ="19"} {}
(8)
What happens each time we decrease the exponent by 1? Your first response might be “we have one less 19.” But what is really happening, mathematically, to the numbers on the right? The answer is that, with each step, they are dividing by 19. If you take
19⋅19⋅19⋅1919⋅19⋅19⋅19 size 12{"19" cdot "19" cdot "19" cdot "19"} {}, and divide it by 19, you get
19⋅19⋅1919⋅19⋅19 size 12{"19" cdot "19" cdot "19"} {}. Divide that by 19 again, and you get
19⋅1919⋅19 size 12{"19" cdot "19"} {}...and so on. From this we can formulate the following principle for the powers of 19:
Whenever you subtract 1 from the exponent, you divide the answer by 19.
As I said earlier, we want the behavior of our new exponents to be consistent with the behavior of the old (positive-integer) exponents. So we can continue this progression as follows:
19
0
=
19
19
=
1
19
0
=
19
19
=
1
size 12{"19" rSup { size 8{0} } = { {"19"} over {"19"} } =1} {}
(9)
19
−
1
=
1
19
19
−
1
=
1
19
size 12{"19" rSup { size 8{ - 1} } = { {1} over {"19"} } } {}
(10)
19
−
2
=
1
19
19
=
1
19
2
19
−
2
=
1
19
19
=
1
19
2
size 12{"19" rSup { size 8{ - 2} } = { { { { size 8{1} } over { size 8{"19"} } } } over {"19"} } = { {1} over {"19" rSup { size 8{2} } } } } {}
(11)
...and so on. We can arrive at our definitions
anything0=1anything0=1 size 12{"anything" rSup { size 8{0} } =1} {} and negative exponents go in the denominator by simply requiring this progression to be consistent.
More rigorously, we can find all our exponent definitions by using the laws of exponents. For instance, what is
4040 size 12{4 rSup { size 8{0} } } {}? We can approach this question indirectly by asking: what is
42424242 size 12{ { {4 rSup { size 8{2} } } over {4 rSup { size 8{2} } } } } {}?
- The second law of exponents tells us that
4242=42−24242=42−2 size 12{ { {4 rSup { size 8{2} } } over {4 rSup { size 8{2} } } } =4 rSup { size 8{2 - 2} } } {}, which is of course
4040 size 12{4 rSup { size 8{0} } } {}.
- But of course,
42424242 size 12{ { {4 rSup { size 8{2} } } over {4 rSup { size 8{2} } } } } {} is just
16161616 size 12{ { {"16"} over {"16"} } } {}, or 1.
- Since
42424242 size 12{ { {4 rSup { size 8{2} } } over {4 rSup { size 8{2} } } } } {} is both
4040 size 12{4 rSup { size 8{0} } } {}and 1,
4040 size 12{4 rSup { size 8{0} } } {} and 1 must be the same thing!
The proofs given below all follow this pattern. They use the laws of exponents to rewrite expressions such as
42424242 size 12{ { {4 rSup { size 8{2} } } over {4 rSup { size 8{2} } } } } {}, and go on to show how zero, negative, and fractional exponents must be defined. We started with the definition of an exponent for a positive integer,
106=10⋅10⋅10⋅10⋅10⋅10106=10⋅10⋅10⋅10⋅10⋅10 size 12{"10" rSup { size 8{6} } ="10" cdot "10" cdot "10" cdot "10" cdot "10" cdot "10"} {}. From there, we developed the laws of exponents. Now we find that, if we want those same laws to apply to other kinds of exponents, there is only one correct way to define those other kinds of exponents.
Table 2: Proofs: When the exponent is not a positive integer
|
Zero exponents
|
Negative exponents
|
Fractional exponents
(numerator = 1)
|
Fractional exponents
(numerator ≠ 1)
|
| Always 1 |
Go in the denominator |
Act as roots |
The numerator is an exponent
The denominator is a root |
4
2
4
2
=
4
2
-
2
=
4
0
4
2
4
2
=
4
2
-
2
=
4
0
but
4
2
4
2
=
16
16
=
1
4
2
4
2
=
16
16
=1
so
4
0
4
0
must be 1!
|
10
1
10
3
=
10
1
-
3
=
10
-2
10
1
10
3
=
10
1
-
3
=
10
-2
but
10
1
10
3
=
10
10
·
10
·
10
=
1
10
·
10
10
1
10
3
=
10
10
·
10
·
10
=
1
10
·
10
so
10
-2
10
-2
must be
1
10
2
1
10
2
|
(
9
1
2
)
2
=
9
1
2
·
2
=
9
1
=
9
(
9
1
2
)
2
=
9
1
2
·
2
=
9
1
=9
So what is
9
1
2
9
1
2
?
Well, when you square it, you get 9.
So it must be
9
9
, or 3!
|
8
2
3
=
(
8
1
3
)
2
=
(
8
3
)
2
8
2
3
=
(
8
1
3
)
2
=
(
8
3
)
2
or
8
2
3
=
(
8
2
)
1
3
=
8
2
3
8
2
3
=
(
8
2
)
1
3
=
8
2
3
|
You may want to experiment with making these proofs more general and more rigorous by using letters instead of numbers. For instance, in the third case, we could write:
(
x
1
a
)
a
=
x
1
a
(
a
)
=
x
1
(
x
1
a
)
a
=
x
1
a
(
a
)
=
x
1
size 12{ \( x rSup { size 8{ { { size 6{1} } over { size 6{a} } } } } \) rSup {a} size 12{ {}=x rSup { left ( { { size 6{1} } over { size 6{a} } } right ) \( a \) } } size 12{ {}=x rSup {1} }} {}
(12)
(
x
1
a
)
a
=
x
(
x
1
a
)
a
=
x
size 12{ \( x rSup { size 8{ { { size 6{1} } over { size 6{a} } } } } \) rSup {a} size 12{ {}=x}} {}
(13)
x
a
a
=
x
a
x
a
a
=
x
a
size 12{ nroot { size 8{a} } { left (x rSup { size 8{ {1} wideslash {a} } } right ) rSup { size 8{a} } } = nroot { size 8{a} } {x} } {}
(14)
x
1
a
=
x
a
x
1
a
=
x
a
size 12{x rSup { size 8{ { { size 6{1} } over { size 6{a} } } } } = nroot {a} {x} } {}
(15)
"DAISY and BRF versions of this collection are available."