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Why is 0^0 undefined? (1 Viewer)

shaon0

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Is it because:
Let b=0
Assume b^b=x
b.ln(b)=ln(x)
0.ln(0)=ln(x)
And ln(0) is undefined since x=0 is an asymptote in a natural logarithmic function.
Thus, 0.Undefined=Undefined.
Therefore, 0^0 is undefined.

---I'm just wondering..
 

IMABOYDAMON!

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shaon0 said:
Is it because:
Let b=0
Assume b^b=x
b.ln(b)=ln(x)
0.ln(0)=ln(x)
And ln(0) is undefined since x=0 is an asymptote in a natural logarithmic function.
Thus, 0.Undefined=Undefined.
Therefore, 0^0 is undefined.

---I'm just wondering..
You officially know you're a maths nerd when you begin to question theorems.
 

shaon0

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IMABOYDAMON! said:
You officially know you're a maths nerd when you begin to question theorems.
lol....but seriously. Is that the reason? Why doesn't the 0.Undefined=0?
 

lyounamu

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shaon0 said:
Is it because:
Let b=0
Assume b^b=x
b.ln(b)=ln(x)
0.ln(0)=ln(x)
And ln(0) is undefined since x=0 is an asymptote in a natural logarithmic function.
Thus, 0.Undefined=Undefined.
Therefore, 0^0 is undefined.

---I'm just wondering..
Let y = 0
Therefore, y^y = x where x is any number greater than 0.
y = lny (x)
Since lny (x) =/= 0, y=/=0, which means that y^y is impossible (or o^o is impossible).
 

shaon0

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lyounamu said:
Let y = 0
Therefore, y^y = x where x is any number greater than 0.
y = lny (x)
Since lny (x) =/= 0, y=/=0, which means that y^y is impossible (or o^o is impossible).
Is the proof i wrote the same thing?
 

Affinity

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These kind of questions are more about what you want to define things as..
you can even define 0^0 = 3.5 if you want to..
 

cooldudehack

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lets no question maths... its gay because its unique!

worry about Qs like that if it interferes when ur paying your bills or buying food from coles
 

Affinity

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cooldudehack said:
lets no question maths... its gay because its unique!

worry about Qs like that if it interferes when ur paying your bills or buying food from coles
I am sure most people here will have no problem with answering those ones..
 

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