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anonymous
 5 years ago
why x raised to power 0 equals one?
anonymous
 5 years ago
why x raised to power 0 equals one?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is another definition

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0however, be careful 0^0 is undefined i believe

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^n \div x^n = x^{nn} = x^0 = 1 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we want that x^m / x^n = x^(mn) , so what if m = n , then you have x^(0) , which should equal 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Of course this doesn't hold if x = 0, as it is division by zero. But that is whole different topic.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0newton, but thats not a proof

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's a way to explain it  they asked 'why', not for a proof.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0by defining x^0 = 1, we are allowed to use the x^m/x^n = x^(mn) rule without qualifications or caveats

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0THATS the reason. i dont want to hear anything else

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you write it the other way around it is a proof. View \[x ^{n} \div x ^{n}\] as a fraction. Any number divided by itself is 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its a definition, thats all. but there is a reason for the definition

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok prove it, show me the deductive steps , axioms, etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry my style of learning is argumentative, dont take it personally

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but ive thought this through, so im fairly confident

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"In mathematics, a proof is a convincing demonstration (within the accepted standards of the field) that some mathematical statement is necessarily true." \[x ^{0} = x ^{nn} = x ^{n} / x ^{n} = 1\] unless I have the definition of "definition" wrong ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well why is x^n/x^n = x ^ (nn) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i would say thats false when n=0, and say x^0 is undefined

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Personally, I think it's a proof (more so than 0!). You use the definition \[x^n \div x^m = x^{nm} \] Not every proof have to prove everything right down to the axioms, you know...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so thats the weakest link here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also you have to assume x is not zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Maybe it needs to have limits set on it? I think the basic question has been answered by INewton though, this is becoming a whole different topic.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^a \div x^b = x^{ab}\ \forall \mathbb{R};\ x \not= 0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, assuming \[x \neq 0\] would solve that problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 0^0 is indeterminate, well when we are dealing with functions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're looking for a problem which isn't there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm, that does look like a proof,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ie, as a counsequence of a previous definition

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think thats what my point was. its a proof if you first accept the definition x^m / x^n = x^(nn)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0err, you know what i mean

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's a consequence of the rule \[x^a \div x^b = x^{ab} ... \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you prove every rule you use in your proofs?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can use definitions, thats fine

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0youre free to define as long as the definition is consistent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then your proofs must be very boring.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well i dont go back to ZFC , if thats what youre asking

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0zermelo frankel cohen set theory (where we assume the continuum hypothesis is true)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well one man's boredom is another man's delight. one man's garbaege is another man's delight

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one man's cow of a wife, is another man's wet dream

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Topic has been exhausted. I'm out! Hope some help can be found in here lol. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, no  I'm pretty sure anyone would find a proof that went back to the axioms for evey last thing boring.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i know i am pedantic. thats the way my brain works

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but he asked an axiomatic question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0anyways, you answered fine. i was just saying, merely calling it a definition is a bit unneceesary. it has a more fundamental reason

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if say , because i defined it that way. ummm, thats not necessary. but its true. you can get away with many theorems by just calling it a definition (in some cases)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for example, you can define the real numbers as a complete ordered field. but thats inelegant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wait, now that I'm looking at it  why would it be indeterminate when n = 0? That doesn't put a 0 in the denominator...?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0id rather say, the real numbers is an example of a complete ordered field, and complete ordered fields are unique up to isomorphism

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is what you said: i would say thats false when n=0, and say x^0 is undefined

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well say x = 0 , and say n = 3. so 0^3 / 0^3 is undefined

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK so it was a typeo

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no its fine if n=0, but x != 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just wondering :) See ya!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have a contradiction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.00^2 = 0^4 / 0^2 which is undefined, but 0^2 = 0*0
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