Exponent Expression Help: Understanding 0^√0 = √0⁰ and Its Truth Value

AI Thread Summary
The expression 0^{\sqrt{0}} = \sqrt{0^{0}} is debated regarding its truth value, largely hinging on the definition of 0^0, which can be considered either 1 or undefined. If 0^0 is deemed undefined, the expression does not hold true. The discussion includes attempts to simplify the expression using logarithms, but it is clarified that log(0) is also undefined, complicating the analysis. Participants emphasize the importance of definitions and assumptions in mathematical proofs. Ultimately, the consensus suggests that without a clear definition of 0^0, the expression cannot be definitively validated.
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Homework Statement


0^{\sqrt{0}}=\sqrt{0^{0}}

Is this expression true?
 
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If you consider "undefined = undefined" to be true, then yes.
 


That depends on your view... 0^0 can be defined as 1, or it can be undefined.

See http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_powerEither way, I think if you define 0^0 as undefined then the expression above does not hold... That would also mean 1/0 = 0^0 which I think doesn't really make sense... I'm not sure on this but I don't think you can say undefined = undefined!
 


Take logarithm and simplify.

root 0 * log 0 = 0 *1= 0

Assumptions= using root 0= 0 on the basis that zero is a real number.
Taking log to the base 10.

R.H.S.> 1/2*0 * log 0
> 0*1
> 0

Assumptions same as above.

The points raised above by honourable members are very meaningful, this is one of the proof methods i learned at the IIT,delhi.
 


physixguru said:
Take logarithm and simplify.

root 0 * log 0 = 0 *1= 0
?? log(0) is NOT 1. It is, one more time, "undefined".
(log(1)= 0, not the other way around.)

Assumptions= using root 0= 0 on the basis that zero is a real number.
Taking log to the base 10.

R.H.S.> 1/2*0 * log 0
> 0*1
> 0

Assumptions same as above.

The points raised above by honourable members are very meaningful, this is one of the proof methods i learned at the IIT,delhi.
 


It was a bad typing error.
 
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