Re: 0^0


"Dave Seaman" <dseaman@xxxxxxxxxxxx> wrote in message
news:dfdehb$34t$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| On Sat, 03 Sep 2005 23:49:28 GMT, Androcles wrote:
|
| > "Dave Seaman" <dseaman@xxxxxxxxxxxx> wrote in message
| > news:dfd8lt$vek$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| >| On Sat, 03 Sep 2005 22:27:48 GMT, Androcles wrote:
| >|
| >| > "Christian Bau" <christian.bau@xxxxxxxxxxxxxxxxxxxx> wrote in
| > message
| >| >
news:christian.bau-F5D920.22281703092005@xxxxxxxxxxxxxxxxxxxxxxxxxxx
| >| >| I could also define that 0/0 = 1.
| >| > You did. 0^(1-1) = 0 * 1/0 = 0/0.
| >|
| >| Can you prove that 0^(1-1) = 0 * 1/0?
|
| > It's by definition of x^1 = x and x^-1 = 1/x for all x, and the
addition
| > of exponents, I don't need to prove it.
|
| Yes, you do need to prove it.

No I don't, it's a definition and well-defined.
You need to prove 0^0 = 1 which is not well-defined as I've shown.

I've glanced at the rest of your post and I do not see your proof,
so I'll end here.
Androcles.

The additive rule for exponents certainly
| applies to cardinal numbers, because there is a natural bijection
between the
| sets of mappings that define a^b*a^c and a^(b+c).
|
| Unfortunately, -1 is not a cardinal. The rule can be extended to
cover
| negative exponents in the case where the base is nonzero, but that
doesn't help
| your argument.
|
|
| >| No, you can't, because division by zero is invalid.
|
| > Not at all, division by zero is undefined, which is different from
| > invalid.
|
| I wasn't talking about the operation of division by zero. I was
talking about
| the argument that says 0^(1-1) = 0^1 * 0^(-1). This is an invalid
argument,
| precisely because division by zero is undefined.
|
| > You have incorrectly
| >| applied the rule of exponents.
|
| > Prove it.
|
| See above.
|
|
| > By the same reasoning,
| >|
| >| 0^1 = 0^(2-1) = 0^2/0^1 = 0/0, which "proves" that 0^1 is
undefined.
|
| > Not at all, it shows that x/0^1 is undefined as we agree,
| > it doesn't prove 0^1 is undefined, that's defined as 0, as we agree.
|
| And by the same token, your argument doesn't show 0^0 is undefined.
The
| reasoning is identical.
|
|
| --
| Dave Seaman
| Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
| <http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>

.

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