If every number to the power of 0 is one, is 0 to the power of 0 also 1?
My teacher doesn't know
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Yeah, it is.
https://de.wikipedia.org/wiki/Null_hoch_null
“in large parts of mathematics”? then
* ln (0) = ln (1) = 0
but the logarithm is not defined for arguments <= 0.
Right, ln(x -> 0) goes against (-infinitely).
I understand. But if 0^0 were defined
it would also be n(0), and that is not it.
0^0 is undefined – although it can be defined as 1, there are problems for this at a different point: if, after the same argument, 0 is equal to 0, then it should be 0^0 = 0. You see the problem?
As already mentioned several times, 0^0 has no clear definition in mathematics. Interesting is, however, if you let the function f(x) = x^x run from right to 0. The limit value is then 1.
The problem is that 0^x for x goes against 0 and x^0 go 1.
No, because you can equally argue 0^k is always 0 and therefore must be 00 0
0 high 0 is not clearly defined in mathematics.
Each positive Number high 0 gives 1.0^0 is not defined and also not for negative figures.
Hello.
There is no general answer to the question, as there are both arguments for it and against it.
LG
It’s not that easy.
https://de.wikipedia.org/wiki/Null_hoch_null
0^0 is undefined.