If 0^0=1, then 0^0=0/0

[Andrew26]

If 0^0=1, then 0^-1 equals (0^0)/(0^1)=1/0

and

0^1=(0^0)*(0^1)=1*0=0

Therefore

(1/0)*(0)=0/0 which does not equal just 1.

Showing our original assumption that 0^0 equals 1, and only 1 is a fallacy.

 

[Strange]

0^0 is not well defined, so it is not surprising that you get contradictory results.

Also, one of your derivations uses an undefined value (0^-1) and is therefore meaningless.

[uncool]

Whether 0^0 = 1 is a matter of convention, and depends on the context. For example, if the power is held constant and the base variable (as in the case of Taylor series), the convention is 0^0 = 1. If the base is constant and the power variable (but positive), the convention is 0^0 = 0. 

[Country Boy]

In symbolic logic, any statement that starts "if (a false statement) then …" is true no matter whether the conclusion is true or false.  You have begun several threads in which you start "if" followed by a false statement!

[taeto]

On 11/17/2019 at 7:46 PM, Andrew26 said:

If 0^0=1, then 0^-1 equals (0^0)/(0^1)=1/0

and

0^1=(0^0)*(0^1)=1*0=0

Therefore

(1/0)*(0)=0/0 which does not equal just 1.

Showing our original assumption that 0^0 equals 1, and only 1 is a fallacy.

In combinatorics we often have to look at all functions from a set X to a set Y. For each element x of X there are |Y| possibilities for a y in Y that x can map to, where |Y| is the number of elements in the set Y. It means that the number of such functions is |Y|^|X|, where |X| is the number of elements in X, since for each x we can choose any one of the y for x to be mapped to y by a function. The equality 0^0 = 1 then just follows from the general definition.

Your reasoning is wrong because there is no such thing as 0^-1. It would be a number x for which 0*x = 1. That does not exist.

Edited December 31, 2019 by taeto