Logic behind negative exponents

[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!

 

[DaveC426913]

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!

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)).

 

[seiche]

Hmmm... A little dissappointing.

But thanks

 

[DaveC426913]

Hmmm... A little dissappointing.

But thanks

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.

 

[seiche]

Alright I will continue my search for the truth!

 

[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.

 

[HallsofIvy]

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 aⁿa^m= [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, aⁿa^m= a^n+m, a very useful formula.

Now, how should we define a⁰? It certainly is NOT "a multiplied by itself '0' times"! Since aⁿa^m is so useful, how should we define a⁰ so that is still true when n= 0? Well, we want a⁰a^m to be equal to a^{0+ m}= a^m since 0+ m= 0. That is, we want a⁰a^m= a^m and, as long as a is not 0 a^m is not 0 either so we can divide both sides by a^m to get a⁰= 1.

In order that aⁿa^m= a^n+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= a^{n+ (-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".)

 

[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).

 

[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
<br /> \left|A^B\right|=\left|A\right|^\left|B\right|<br />
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:

<br /> 0^0 = \left|\emptyset\right|^{\left|\emptyset\right|}=\left|\emptyset^\emptyset\right| = 1<br />

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

<br /> lim_{x\rightarrow a}f(x)^{g(x)}<br />

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.

 

[Tac-Tics]

1 is not considered a prime because allowing units would ruin the unique factorization theorem.

[/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.

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.

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.

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

<br /> lim_{x\rightarrow a}f(x)^{g(x)}<br />
... It doesn't have directly to do with the continuity x^y.

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.