# Negative exponents

1. Sep 7, 2008

### StupidGenius

Recently in class (11 math) we explored exponents in negatives. My teacher placed a question on the board and it said

7*1/7^-7
( 7 times 1 over 7 to the power of -7 )

She said that in order to get rid of a negative, we must imagine that there is an imaginary "1" multipliying the fraction.

1/7^-7 *1

Then we must "flip" the fraction around to eliminate the negative exponent. so it will be

7*1*7^7/1

I am a type of person to want to to know why and how stuff are done, especially in the subject of math. I want to know the reasoning behind. Unfortunately i couldn't ask her that day, and then the weekend came and the problem still remains.
So now i know how to get ride of the negative exponent, but will someone explain to me where the imaginary"1" came from and will i be able to do this in any equation where a negative exponent is present?

2. Sep 7, 2008

### wind522

Personally, I have never heard of the "imaginary one" but from how your teacher explained it, I think it makes the whole problem a lot harder than necessary. So here's my explanation: whenever a number is to a negative exponent, all you have to do is move it to the other side of the fraction to get rid of the negative. If the negative exponent is on the denominator, move it to the numerator and vice versa.
Ex. (2^5)/(3^-4) = (2^5)(3^4)
All I did was move 3^-4 from the denominator to the numerator, thus making it's negative exponent positive.

3. Sep 7, 2008

### StupidGenius

Makes sense, but that "1" still exists in the fraction if (2^5)(3^4)/1. But I see how it is sort of irrelevant. Thank you, now what if the number was not in a fraction and was just simply;
(7^-3)
will this be correct? 1/(7^3)

4. Sep 7, 2008

### CRGreathouse

7^-3 = 1/7^3, yes.

5. Sep 9, 2008

### DrGreg

The basic idea behind this is the equation

$$7^a 7^b = 7^{a+b}$$​

We know this is true for a and b both positive integers, so we'd like to make it true for other values too.

Put b=0 and the only solution is $7^0 = 1$. Then put b=-a to see that

$$7^{-a} = \frac{7^0}{7^a} = \frac{1}{7^a}$$​