# Having trouble understanding 'definition of nth root'

#### yoscar04

##### Full Member
The n-th root of a can be written as $$\displaystyle \sqrt[n]{a}$$=a1/n. So if you have (-a)n, where n is even, (-a)n=an.
So (an)1/n=|a|. Try and check what happen when n is not even, i.e. ((-3)3)1/3=?
The book gave you an example where (a2)1/2 $$\displaystyle \neq$$ a.

Last edited:

#### Otis

##### Elite Member
What's the point of restating the exact same thing …
Hi stormbytes. The two situations are not exactly the same.

The point they're making is this: When a negative number has been squared, we loose the original sign. In other words, if I give you the number 16, then you can't know what number was squared to obtain that 16.

Here's an example that shows a common beginner's mistake:

Solve the equation x2 = 16

The student takes the square root of each side and writes

x = 4

They missed the solution x = -4 because they incorrectly assumed that the square root of x2 must be x.

The square root of x2 is not simply x. We need to express it as |x| because there's two possibilities.

Therefore, when we take the square root of each side, we write

|x| = 4

That equation means

x = 4 $$\quad$$ OR $$\quad$$ x = -4 • stormbytes

#### stormbytes

##### New member
The n-th root of a can be written as $$\displaystyle \sqrt[n]{a}$$=a1/n. So if you have (-a)n, where n is even, (-a)n=an.
So (an)1/n=|a|. Try and check what happen when n is not even, i.e. ((-3)3)1/3=?
The book gave you an example where (a2)1/2 $$\displaystyle \neq$$ a.
Thank you. I’ll revisit your answer once I’ve done radical exponents. Will have a better sense of what you are illustrating in your examples.

#### stormbytes

##### New member
Hi stormbytes. The two situations are not exactly the same.

The point they're making is this: When a negative number has been squared, we loose the original sign. In other words, if I give you the number 16, then you can't know what number was squared to obtain that 16.

Here's an example that shows a common beginner's mistake:

Solve the equation x2 = 16

The student takes the square root of each side and writes

x = 4

They missed the solution x = -4 because they incorrectly assumed that the square root of x2 must be x.

The square root of x2 is not simply x. We need to express it as |x| because there's two possibilities.

Therefore, when we take the square root of each side, we write

|x| = 4

That equation means

x = 4 $$\quad$$ OR $$\quad$$ x = -4 Thank you. That makes more sense.

#### JeffM

##### Elite Member
The laws of exponents are usually expressed as follows

$$\displaystyle \text{Given } a \text { is a real number } > 0 \text { then stuff}$$

This is done to avoid complications that arise when a is not a positive real number.

One major complication is that if $$\displaystyle a^{2n}$$ is a positive real number, then

$$\displaystyle \exists \ b < 0 \text { such that } b^2 = a^{2n} \text { and } \exists \ c > 0 \text { such that } c^2 = a^{2n}.$$

Of course, it can easily be shown thar b = - c, but still we would like to know whether we are dealing with a negative number or not. Therefore we define

$$\displaystyle \sqrt{u^2} = |u|.$$

How then do we describe the other one?

$$\displaystyle - \sqrt{u^2} = -|u|.$$

Notice that

$$\displaystyle (-|u|)^2 = (-1)^2(|u|)^2 = 1 * (|u|)^2 = (|u|)^2.$$