Re: 0^0
- From: "Androcles" <Androcles@ MyPlace.org>
- Date: Sat, 03 Sep 2005 23:49:28 GMT
"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.
| No, you can't, because division by zero is invalid.
Not at all, division by zero is undefined, which is different from
invalid.
You have incorrectly
| applied the rule of exponents.
Prove it.
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.
Androcles
.
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