- anonymous

why x raised to power 0 equals one?

- jamiebookeater

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- anonymous

this is another definition

- anonymous

however, be careful 0^0 is undefined i believe

- anonymous

\[x^n \div x^n = x^{n-n} = x^0 = 1 \]

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## More answers

- anonymous

we want that x^m / x^n = x^(m-n) , so what if m = n , then you have
x^(0) , which should equal 1

- anonymous

Of course this doesn't hold if x = 0, as it is division by zero. But that is whole different topic.

- anonymous

newton, but thats not a proof

- anonymous

excellent newton

- anonymous

It's a way to explain it - they asked 'why', not for a proof.

- anonymous

ok

- anonymous

by defining x^0 = 1, we are allowed to use the x^m/x^n = x^(m-n) rule without qualifications or caveats

- anonymous

THATS the reason. i dont want to hear anything else

- anonymous

If 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

no it isnt

- anonymous

its a definition, thats all. but there is a reason for the definition

- anonymous

ok prove it, show me the deductive steps , axioms, etc

- anonymous

sorry my style of learning is argumentative, dont take it personally

- anonymous

but ive thought this through, so im fairly confident

- anonymous

"In mathematics, a proof is a convincing demonstration (within the accepted standards of the field) that some mathematical statement is necessarily true."
\[x ^{0} = x ^{n-n} = x ^{n} / x ^{n} = 1\]
unless I have the definition of "definition" wrong ;)

- anonymous

well why is x^n/x^n = x ^ (n-n) ?

- anonymous

i would say thats false when n=0, and say x^0 is undefined

- anonymous

Personally, I think it's a proof (more so than 0!).
You use the definition
\[x^n \div x^m = x^{n-m} \]
Not every proof have to prove everything right down to the axioms, you know...

- anonymous

so thats the weakest link here

- anonymous

also you have to assume x is not zero

- anonymous

Maybe 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

\[x^a \div x^b = x^{a-b}\ \forall \mathbb{R};\ x \not= 0\]

- anonymous

Yes, assuming \[x \neq 0\] would solve that problem

- anonymous

and 0^0 is indeterminate, well when we are dealing with functions

- anonymous

You're looking for a problem which isn't there.

- anonymous

hmmm, that does look like a proof,

- anonymous

ie, as a counsequence of a previous definition

- anonymous

i think thats what my point was. its a proof if you first accept the definition x^m / x^n = x^(n-n)

- anonymous

err, you know what i mean

- anonymous

It's a consequence of the rule \[x^a \div x^b = x^{a-b} ... \]

- anonymous

exactly

- anonymous

now ...

- anonymous

Do you prove every rule you use in your proofs?

- anonymous

you can use definitions, thats fine

- anonymous

youre free to define as long as the definition is consistent

- anonymous

yes i try to

- anonymous

Then your proofs must be very boring.

- anonymous

well i dont go back to ZFC , if thats what youre asking

- anonymous

zermelo frankel cohen set theory (where we assume the continuum hypothesis is true)

- anonymous

well one man's boredom is another man's delight. one man's garbaege is another man's delight

- anonymous

one man's cow of a wife, is another man's wet dream

- anonymous

so on and so forth

- anonymous

shall i continue

- anonymous

Topic has been exhausted. I'm out! Hope some help can be found in here lol. :)

- anonymous

No, no - I'm pretty sure anyone would find a proof that went back to the axioms for evey last thing boring.

- anonymous

yes i know i am pedantic. thats the way my brain works

- anonymous

but he asked an axiomatic question

- anonymous

anyways, you answered fine. i was just saying, merely calling it a definition is a bit unneceesary. it has a more fundamental reason

- anonymous

if 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

for example, you can define the real numbers as a complete ordered field. but thats inelegant

- anonymous

Wait, 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

id rather say, the real numbers is an example of a complete ordered field, and complete ordered fields are unique up to isomorphism

- anonymous

hmmm

- anonymous

you mean if x = 0 ?

- anonymous

This is what you said:
i would say thats false when n=0, and say x^0 is undefined

- anonymous

well say x = 0 , and say n = 3. so 0^3 / 0^3 is undefined

- anonymous

OK so it was a type-o

- anonymous

no its fine if n=0, but x != 0

- anonymous

Just wondering :) See ya!

- anonymous

i have a contradiction

- anonymous

0^2 = 0^4 / 0^2 which is undefined, but 0^2 = 0*0

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