Question

why x raised to power 0 equals one?

Answer 1

this is another definition

Answer 2

however, be careful 0^0 is undefined i believe

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

newton, but thats not a proof

Answer 7

excellent newton

Answer 8

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

Answer 9

ok

Answer 10

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

Answer 11

THATS the reason. i dont want to hear anything else

Answer 12

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.

Answer 13

no it isnt

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

but ive thought this through, so im fairly confident

Answer 18

"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 ;)

Answer 19

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

Answer 20

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

Answer 21

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...

Answer 22

so thats the weakest link here

Answer 23

also you have to assume x is not zero

Answer 24

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.

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

hmmm, that does look like a proof,

Answer 30

ie, as a counsequence of a previous definition

Answer 31

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

Answer 32

err, you know what i mean

Answer 33

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

Answer 34

exactly

Answer 35

now ...

Answer 36

Do you prove every rule you use in your proofs?

Answer 37

you can use definitions, thats fine

Answer 38

youre free to define as long as the definition is consistent

Answer 39

yes i try to

Answer 40

Then your proofs must be very boring.

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

so on and so forth

Answer 46

shall i continue

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

but he asked an axiomatic question

Answer 51

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

Answer 52

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)

Answer 53

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

Answer 54

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...?

Answer 55

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

Answer 56

hmmm

Answer 57

you mean if x = 0 ?

Answer 58

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

Answer 59

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

Answer 60

OK so it was a type-o

Answer 61

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

Answer 62

Just wondering :) See ya!

Answer 63

i have a contradiction

Answer 64

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