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nice profile pic. :)
same to you
correct un defined
i like the new pic better lol
i need to get to next class the verification for correcting the teachers error
No its defined as 1
a^0=1 for all real a not equal to 0 a^0 is undefined when a=0 @alekos
No, that's incorrect. It is definitely mathematically defined as 1
Yes, that's wonderful and I know all of this, but in the end it's been defined mathematically and the modern convention is to define 0^0=1, for good reason. Why? Because it lets us manipulate exponentials without adding special cases. Donald Knuth set things straight in 1992, Donald Knuth is a professor Emeritus at Stanford University.
I think it is more of a convenience to define \(0^0\) to be 1. The limit of \(x^y\) does not exist as (x,y) to (0,0). Different direction of approach will yield different results.
there's a difference between pure maths and what's regarded as a definition. this is one of those cases. kind of similar to 0!=1
You could argue 0!=1 as (n-1)!=n!/n so 0!=1!/1=1. For 0^0 you can't really make such argument and it is for convenience only IMO.
Yes, I don't disagree with you. But that is the way it has been defined, and it has to be treated in that regard.