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^{-2}, that's the same thing as saying 1/x

^{2}. But I'm just wondering what is the mathematical reasoning for why that's true?

Thanks!

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Thanks!

- #2

DaveC426913

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I looked around a bit.^{-2}, that's the same thing as saying 1/x^{2}. But I'm just wondering what is the mathematical reasoning for why that's true?

Thanks!

http://en.wikipedia.org/wiki/Exponentiation#Negative_integer_exponents" both say it is

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Hmmm.... A little dissappointing.

But thanks

But thanks

- #4

DaveC426913

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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.Hmmm.... A little dissappointing.

But thanks

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Alright I will continue my search for the truth!

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- #7

HallsofIvy

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If n is a positive integer, then a

From that, we have a

Now, how should we define a

In order that a

Now what about a

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

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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

Similarly, 0^0 is often defined as 1 for the integers. You'll often see expressions like [tex]\Sigma_i x^i[/tex]. 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

- #9

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

(1)

(2) There is reason for [tex]0^0 = 1[/tex] that it's more than just a definition. This is a particular (and extreme) case of a combinatorial equality. Remember that, given sets [tex]A[/tex] and [tex]B[/tex], the set of functions [tex]f:A\rightarrow B[/tex] is denoted by [tex]A^B[/tex] and its cardinality is

[tex]

\left|A^B\right|=\left|A\right|^\left|B\right|

[/tex]

Now consider the case where [tex]A=B=\emptyset[/tex]; then [tex]\left|A\right|=\left|B\right|=0[/tex] and the only function [tex]f:\emptyset \rightarrow \emptyset[/tex] is the empty function, so:

[tex]

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

[/tex]

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

[tex]

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

[/tex]

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

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

"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.There is reason for [tex]0^0 = 1[/tex] that it's more than just a definition. This is a particular (and extreme) case of a combinatorial equality.

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

[tex]

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

[/tex]

.... It doesn't have directly to do with the continuity [tex]x^y[/tex].

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