# Why does inverting the base of a negative exponent cancel the negative?

1. Aug 24, 2012

### bcheck

So I understand the math, that is, how to solve these problems. But how does inverting the base cancel out the negative in the exponent? I worked out inversion to simply be a setback of 2 decimal places, e.g. 12.5% of 8 = 1, so 1/8 is .125 % of 1. But why does this nullify the negative?

2. Aug 24, 2012

### chiro

Hey bcheck and welcome to the forums.

I think you are talking about why a-x = $\frac{1}{a^x}$.

The best way to understand this is through the following identities:

ab*ac = ab+c. Now consider b = -c and consider
$\frac{a^c}{a^b} = 1$

3. Aug 24, 2012

### Old Wolf

Hopefully this is a typo and you meant to write "1/8 is .125 of 1". I think you are mixing up two different concepts here though:

The "setback of 2 decimal places" is what "%" means. "per cent" literally means "for each hundred".

"12.5%" just means ".125" , dividing 12.5 by 100. As you know, dividing a number by 100 sets the decimal point back 2 places. It's nothing to do with "inversion".

Returning to your original example, if we write it without involving per cents, here are some different ways of saying the same thing (Bearing in mind that "of" just means multiplication):

1 / 8 = .125
1 / 8 = .125 * 1
1 / 8 = .125 of 1
.125 * 8 = 1
.125 of 8 = 1

No mystery?

4. Aug 25, 2012

### bcheck

Thanks man.

5. Aug 25, 2012

### bcheck

Yeah the percent thing was just something I discovered for myself while doing this. I never thought of what "per cent" literally meant, and it what was really cool for me when I found that the reciprocal of every whole number as a decimal had a relationship that got me back to .01 every time (I could have definitely worded that better, but hopefully you got it). And yes, I did mean "1/8 is .125 of 1", but you got the gist of it. Anyway, I was really just wondering why this operation works in mathematics, but I more or less get it now with chiro's equation, along with the last post of this thread: https://www.physicsforums.com/showthread.php?t=254455 Thanks to both of you.