# Roots

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:

• nb = a;
• an = b.
(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: 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 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:

• nb = a;
• an = b, and a ≥ 0.
(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: 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 nb when b is negative and n is even, although we will only consider the case when n = 2 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 (not even imaginary, but completely undefined) 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, completely undefined 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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