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 nb 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 3b the real cube root of b. Notice that 1b = b, because b1 = b, so we usually just call it b.)

What is the sign of nb?

If b is negative: If b is zero: If b is positive:
nb is negative; nb is zero; nb 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 nb 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:

(We usually write 2b as simply √b and call it the principal square root of b.)

What is the sign of nb?

If b is negative: If b is zero: If b is positive:
nb is undefined (or imaginary); nb is zero; nb is positive.
We will look at imaginary numbers later on. These allow us to make sense of nb 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 nb. We can generalize this to any rational number m/n in lowest terms:
bm/n = nbm.
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 364.
64 = 43, so 364 = 4.
Find 3−27.
−27 = (−3)3, so 3−27 = −3.
Find √25.
25 means 225, 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 −364.
364 = 4, so −364 = −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 364 = 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 3x3.
3x3 = x.
Find 3x6.
x6 = (−x2)3, so 3x6 = −x2.
Find √x2.
x2 = (x)2 and x2 = (−x)2; either way, x2 = |x|2 and |x| ≥ 0, so √x2 = |x|.
Find 416x8y4.
16x8y4 = (2x2|y|)4 and 2x2|y| ≥ 0, so 416x8y4 = 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.
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