Roots (§9.2)
Just as subtraction reverses addition and division reverses multiplication,
so taking roots reverses raising to powers.
By considering how raising to the power of a natural number
affects whether a number is positive or negative,
we can see under what conditions and how much this operation may be reversed.
Raising to the power of an odd number
What is the sign of an
when n is an odd natural number?
| If a
is negative: |
If a is zero: |
If a is positive: |
| an is negative; |
an is zero; |
an is positive. |
How many real solutions are there
to xn = b?
| If b
is negative: |
If b is zero: |
If b is positive: |
There's one real solution,
which is negative; |
There's one real solution,
which is zero; |
There's one real solution,
which
is positive. |
Define
n√b
to be this solution,
called the real nth root of b
(or the real root of b of index n).
In other words,
these two statements mean the same thing
when n is an odd natural number:
(We usually call
3√b
the real cube root of b.
Notice that
1√b =
b,
because b1 = b,
so we usually just call it b.)
What is the sign
of
n√b?
| If b
is negative: |
If b is zero: |
If b is positive: |
| n√b
is negative; |
n√b
is zero; |
n√b
is positive. |
Raising to the power of an even number
What is the sign of an
when n is an even natural number?
| If a
is negative: |
If a is zero: |
If a is positive: |
| an is positive; |
an is zero; |
an is positive. |
How many real solutions are there
to xn = b?
| If b
is negative: |
If b is zero: |
If b is positive: |
There's no real solution;
|
There's one real solution,
which is zero; |
There are two real solutions,
one negative and one positive. |
Define
n√b
to be the non-negative solution,
called the principal nth root of b
(or the principal root of b of index n),
if such a solution exists.
In other words,
these two statements mean the same thing
when n is an even natural number:
- n√b =
a;
- an = b,
and a ≥ 0.
(We usually write
2√b
as simply √b
and call it the principal square root of b.)
What is the sign
of
n√b?
| If b
is negative: |
If b is zero: |
If b is positive: |
| n√b
is undefined (or imaginary); |
n√b
is zero; |
n√b
is positive. |
We will look at imaginary numbers later on.
These allow us to make sense of
n√b
when b is negative and n is even,
although we will only consider the case when n = 2 (square roots)
in this course.
Fractional exponents
Because it makes most of the rules of exponents continue to work,
we define b1/n
to mean
n√b.
We can generalize this
to any rational number m/n in lowest terms:
bm/n =
n√bm.
If b is positive, then this always exists (and is positive).
If b is zero,
then this is zero if m is positive,
and undefined (or infinite)
if m is negative.*
(Since m/n is in lowest terms, n must be positive.)
If b is negative,
then this is negative if m and n are both odd,
positive if m is even and n is odd,
and undefined (or imaginary) if m is odd and n is even.
(Since m/n is in lowest terms,
m and n cannot both be even.)
When b is positive,
it's possible to define bx
(as another positive number)
even when x is irrational,
but we won't pursue that in this course.
(If b is negative and x is irrational,
then bx is imaginary.
If b is zero and x is irrational,
then the result is the same as when x is rational:
zero when x is positive, infinite when x is negative.)
Examples
- Find
3√64.
- 64 = 43,
so 3√64 =
4.
- Find
3√−27.
- −27 = (−3)3,
so
3√−27 =
−3.
- Find √25.
- √25
means 2√25,
25 = 52, and 5 ≥ 0,
so √25 = 5.
- Find
4√−81.
- −81 is negative and 4 is even,
so 4√−81
is undefined (or imaginary).
- Find
−3√64.
- 3√64 =
4,
so
−3√64 =
−4.
- Find
−3√−27.
- 3√−27 =
−3,
so
−3√−27 =
3.
- Find −√25.
- √25 = 5,
so −√25 =
−5.
- Find
−4√−81.
- 4√−81
is undefined (or imaginary),
so
−4√−81
is also undefined (or imaginary).
- Find (−8)2/3.
- (−8)2/3
means
3√(−8)2,
(−8)2 = 64,
and 3√64 =
4,
so (−8)2/3 = 4.
- Find (−27)2/6.
- 2/6 = 1/3 in lowest terms,
(−27)1/3
means
3√−27,
and
3√−27 =
−3,
so (−27)2/6 = −3.
- Find 251/2.
- 251/2
means √25,
and √25 = 5,
so 251/2 = 5.
- Find (−81)3/12.
- 3/12 = 1/4 in lowest terms,
(−81)1/4
means
4√−81,
and 4√−81
is undefined (or imaginary),
so (−81)3/12 is undefined (or imaginary).
- Find
3√x3.
- 3√x3 =
x.
- Find
3√−x6.
- −x6 =
(−x2)3,
so
3√−x6 =
−x2.
- Find
√x2.
- x2 = (x)2
and x2 = (−x)2;
either way, x2 = |x|2
and |x| ≥ 0,
so
√x2 =
|x|.
- Find
4√16x8y4.
- 16x8y4 =
(2x2|y|)4
and 2x2|y| ≥ 0,
so
4√16x8y4 =
2x2|y|.
*
In the special case where b and m are both zero,
modern mathematics defines 00 = 1.
However, our textbook
takes the old-fashioned view that 00 is undefined.
To avoid confusion, I will never test you on 00.
Go back to the course homepage.
This web page was written by Toby Bartels, last edited on 2024 August 15.
Toby reserves no legal rights to it.
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