What is Zero Raised to Itself?

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Originally posted by jcsd
No, you can only really use the formula to find a Fibonacci number when n is a natural number. But as F2 = 1 and F1 = 1, you can define F0 as 0 from the recurssive formula and this is how it is conventially defined.

0/1 is undefined and it's pretty easy to show that it cannot be a real or a complex number and thus you cannot perform algebraic operations on it.

A good point. Using Fibinocci it can not be proven.

Not using Fibinocci, rather using set theory,
is there a quantity of 1 of the set [undefined]
x a quantity of 1 of the set [undefined]

might that = 1?

If it were so, then 0 x 0 = 1
0 x 1 = 0

Would there be such a thing as -0 ?
If so, what would it be?
 
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S = k log w wrote:
O/1 is infinity, or indeterm., or undefined, or whatever.

and then
jcsd wrote:
0/1 is undefined and it's pretty easy to show that it cannot be a real or a complex number and thus you cannot perform algebraic operations on it.

Am I missing something? I have been suffering under the delusion that 0/1 was equal to 0 for some years! Or is it possible that you meant either 1/0 or 0/0??
 
Oops back to school for a good dose of reading comphrehension for me, I thought it was 1/0 (though why I wrote 0/1 I don't know).
 
This is really interesting.

Look at this:

SQRT AND SQUARE


((+2)+(+2)) = +2^2
((-2)+(-2)) = -2^2
((+2)+(-2)+(+2)+(-2)) = 0^2

0^0 = ?
 
Originally posted by S = k log w
This is really interesting.

Look at this:

SQRT AND SQUARE


((+2)+(+2)) = +2^2
((-2)+(-2)) = -2^2
((+2)+(-2)+(+2)+(-2)) = 0^2

0^0 = ?

Umm, ((-2)+(-2)) = -2^2? Want to rethink?
 
Originally posted by selfAdjoint
Umm, ((-2)+(-2)) = -2^2? Want to rethink?
I think (hope) he meant -(2^2).

Don't forget PEMDAS...

- Warren
 
Originally posted by chroot
I think (hope) he meant -(2^2).

Don't forget PEMDAS...

- Warren

Thank you.