# Logic behind negative exponents

1. Jan 10, 2010

### seiche

I know that if you have x-2, that's the same thing as saying 1/x2. But I'm just wondering what is the mathematical reasoning for why that's true?

Thanks!

2. Jan 10, 2010

### DaveC426913

I looked around a bit.

http://en.wikipedia.org/wiki/Exponentiation#Negative_integer_exponents" both say it is defined that way (for convenience and consistency (i.e. there is no mathematical basis for it)).

Last edited by a moderator: Apr 24, 2017
3. Jan 10, 2010

### seiche

Hmmm.... A little dissappointing.

But thanks

4. Jan 10, 2010

### DaveC426913

Please don't take my word for it though; my answers are simply from Googling. There are some math heavy-weights around here who can answer with authority.

5. Jan 10, 2010

### seiche

Alright I will continue my search for the truth!

6. Jan 11, 2010

### rochfor1

For a != 0, we reserve the notation $$a^{-1}$$, for the multiplicative inverse of a. That is, the number such that $$a\ a^{-1}=1=a^{-1}\ a$$. It is clear then that when a is a real number, $$a^{-1} = 1/a$$ because a (1/a) = 1 = (1/a) a. From there, we want our notation to play well with the rules of exponents, so we must have $$a^{-n} = ( a^{-1} )^n = \left ( \frac{1}{a} \right )^n = \frac{1}{a^n}$$. Hope this answers your questions a bit.

7. Jan 11, 2010

### HallsofIvy

Staff Emeritus
While we are free to define things as we please, there is some logic involved in how we want to define things.

If n is a positive integer, then an is defined as "a multiplied by itself n times".

From that, we have anam= [a*a*a*... *a(n times)][a*a*a*a...*a(m times)]. It is easy to see that there are a total of n+ m "a"s there. In other words, for m and n positive integers, anam= an+m, a very useful formula.

Now, how should we define a0? It certainly is NOT "a multiplied by itself '0' times"! Since anam is so useful, how should we define a0 so that is still true when n= 0? Well, we want a0am to be equal to a0+ m= am since 0+ m= 0. That is, we want a0am= am and, as long as a is not 0 am is not 0 either so we can divide both sides by am to get a0= 1.

In order that anam= an+m be true even if m or n are 0, we define a0= 1. (But notice we have to restrict a to be non-zero to do this. 00 is undefined.)

Now what about a-n where n is a positive integer? We still want to have ana-n= an+ (-n) and that, since n+(-n)= 0, is the same as ana-n= a0= 1. Dividing both sides by an (again, a is not 0 so an is not 0) we must have a-n= 1/an.

Yes, that is a definition, but there is 'logic' even behind definitions.

(Since the original post was a question rather than "learning materials", I am moving this to "General Math".)

8. Jan 11, 2010

### Tac-Tics

Halls has it. It's a convenient definition, nothing more. It introduces a lot of symmetry into our equations.

You'll find definitions chosen like that in various places in math. Why is 1 not a prime number? Some people mumble b.s. about how it's a "unit" or something, but the real reason is that if 1 was prime, we'd have to change 99% of the theorems about prime numbers to start with "Let p be a non-zero prime number." It'd just be a pain.

Similarly, 0^0 is often defined as 1 for the integers. You'll often see expressions like $$\Sigma_i x^i$$. If x = 0, though, the formula still works as long as you let 0^0 = 1.

On the other hand, 0^0 is left undefined when talking about reals or complex numbers. The reason is because f(x, y) = x^y is a continuous everywhere except at (0, 0). If you leave out that one point, you get a continuous function defined on almost all of R (or C).

9. Jan 11, 2010

### JSuarez

HallsofIvy is correct in saying that this is a definition, but not a completely arbitrary one; it is the best that it's compatible with the law of exponents.

But let me make a few remarks concerning the previous post:

(1) 1 is not considered a prime because allowing units (and 1 is a unit in the ring of integers) would ruin the unique factorization theorem. It's not just a matter of being a pain: many theorems would be false.

(2) There is reason for $$0^0 = 1$$ that it's more than just a definition. This is a particular (and extreme) case of a combinatorial equality. Remember that, given sets $$A$$ and $$B$$, the set of functions $$f:A\rightarrow B$$ is denoted by $$A^B$$ and its cardinality is
$$\left|A^B\right|=\left|A\right|^\left|B\right|$$
Now consider the case where $$A=B=\emptyset$$; then $$\left|A\right|=\left|B\right|=0$$ and the only function $$f:\emptyset \rightarrow \emptyset$$ is the empty function, so:

$$0^0 = \left|\emptyset\right|^{\left|\emptyset\right|}=\left|\emptyset^\emptyset\right| = 1$$

(3) In the context of Analysis, $$0^0$$ is considered undefined because it stand for a shorthand for a limit of the form:

$$lim_{x\rightarrow a}f(x)^{g(x)}$$

And $$lim_{x\rightarrow a}f(x) = lim_{x\rightarrow a}g(x) = 0$$ and this cannot be calculated without knowing more about the local behaviour of $$f$$ and $$g$$ close to $$a$$. It doesn't have directly to do with the continuity $$x^y$$.

10. Jan 11, 2010

### Tac-Tics

[/quote]

"All integers have a factorization which is unique up to the number of unit divisors." The theorem isn't ruined. It just puts on some weight.

It's still rather vacuous. How many ways are there to spell a word with no letters? The answer is "1" if you're a mathematician and "what?" if you're not.

The 0^0 = 1 convention is even useful when the base is a real number. In power series, the x^0 term is reduced to 1 even though x might take on any real value, including 0. Since we're dealing with an arbitrary real, the combinatoric example isn't always applicable.

I guess that's the more traditional way of looking at it, but I think the continuity result is simpler. You're treating exponentiation as a regular real-valued function on R^2 instead of a shorthand for some awkward limit.