anonymous
  • anonymous
why x raised to power 0 equals one?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
this is another definition
anonymous
  • anonymous
however, be careful 0^0 is undefined i believe
anonymous
  • anonymous
\[x^n \div x^n = x^{n-n} = x^0 = 1 \]

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anonymous
  • 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
  • anonymous
Of course this doesn't hold if x = 0, as it is division by zero. But that is whole different topic.
anonymous
  • anonymous
newton, but thats not a proof
anonymous
  • anonymous
excellent newton
anonymous
  • anonymous
It's a way to explain it - they asked 'why', not for a proof.
anonymous
  • anonymous
ok
anonymous
  • 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
  • anonymous
THATS the reason. i dont want to hear anything else
anonymous
  • 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
  • anonymous
no it isnt
anonymous
  • anonymous
its a definition, thats all. but there is a reason for the definition
anonymous
  • anonymous
ok prove it, show me the deductive steps , axioms, etc
anonymous
  • anonymous
sorry my style of learning is argumentative, dont take it personally
anonymous
  • anonymous
but ive thought this through, so im fairly confident
anonymous
  • 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
  • anonymous
well why is x^n/x^n = x ^ (n-n) ?
anonymous
  • anonymous
i would say thats false when n=0, and say x^0 is undefined
anonymous
  • 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
  • anonymous
so thats the weakest link here
anonymous
  • anonymous
also you have to assume x is not zero
anonymous
  • 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
  • anonymous
\[x^a \div x^b = x^{a-b}\ \forall \mathbb{R};\ x \not= 0\]
anonymous
  • anonymous
Yes, assuming \[x \neq 0\] would solve that problem
anonymous
  • anonymous
and 0^0 is indeterminate, well when we are dealing with functions
anonymous
  • anonymous
You're looking for a problem which isn't there.
anonymous
  • anonymous
hmmm, that does look like a proof,
anonymous
  • anonymous
ie, as a counsequence of a previous definition
anonymous
  • 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
  • anonymous
err, you know what i mean
anonymous
  • anonymous
It's a consequence of the rule \[x^a \div x^b = x^{a-b} ... \]
anonymous
  • anonymous
exactly
anonymous
  • anonymous
now ...
anonymous
  • anonymous
Do you prove every rule you use in your proofs?
anonymous
  • anonymous
you can use definitions, thats fine
anonymous
  • anonymous
youre free to define as long as the definition is consistent
anonymous
  • anonymous
yes i try to
anonymous
  • anonymous
Then your proofs must be very boring.
anonymous
  • anonymous
well i dont go back to ZFC , if thats what youre asking
anonymous
  • anonymous
zermelo frankel cohen set theory (where we assume the continuum hypothesis is true)
anonymous
  • anonymous
well one man's boredom is another man's delight. one man's garbaege is another man's delight
anonymous
  • anonymous
one man's cow of a wife, is another man's wet dream
anonymous
  • anonymous
so on and so forth
anonymous
  • anonymous
shall i continue
anonymous
  • anonymous
Topic has been exhausted. I'm out! Hope some help can be found in here lol. :)
anonymous
  • anonymous
No, no - I'm pretty sure anyone would find a proof that went back to the axioms for evey last thing boring.
anonymous
  • anonymous
yes i know i am pedantic. thats the way my brain works
anonymous
  • anonymous
but he asked an axiomatic question
anonymous
  • 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
  • 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
  • anonymous
for example, you can define the real numbers as a complete ordered field. but thats inelegant
anonymous
  • 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
  • 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
  • anonymous
hmmm
anonymous
  • anonymous
you mean if x = 0 ?
anonymous
  • anonymous
This is what you said: i would say thats false when n=0, and say x^0 is undefined
anonymous
  • anonymous
well say x = 0 , and say n = 3. so 0^3 / 0^3 is undefined
anonymous
  • anonymous
OK so it was a type-o
anonymous
  • anonymous
no its fine if n=0, but x != 0
anonymous
  • anonymous
Just wondering :) See ya!
anonymous
  • anonymous
i have a contradiction
anonymous
  • anonymous
0^2 = 0^4 / 0^2 which is undefined, but 0^2 = 0*0

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