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I was looking for explanation why x^0=1.

thread

https://www.physicsforums.com/threads/my-simple-proof-of-x-0-1.172073/

is locked and i did't found solution in it from axioms. People using exp(x) and log(x) and x

If you have x

But from axioms ... maybe something like this???...:

Starting from "Principles of mathematical analysis. Walter Rudin" page 5 and 6.

For any field F we get axioms:

...

"(M3) Multiplication is commutative: xy=yx for all x,y ∈ F"

...

"(M5) If x ∈ F and x ≠ 0 then there exists an element 1/x ∈ F such that

x⋅(1/x) =1." Let assume (Rudin does not specify that) that we call this element x

On next page (6) Rudin in Remark 1.13 states:

"One usually writes (in any field) .....x

"in place of ... xx, xxx, ..."

Let assume (Rudin does not specify that) that x=x

From (M5) we get

1=x⋅(1/x) =x

so

1=x

x

But we got for x

1=x

we get (not proving that if ab=c and ad=c then b=d)

x

(from remark 1.13) where ∏

X

so

X

If A=1 and B=-1 we get from above

X

and

X

we get X

the end ...

and now just for fun :)

0

so

0

thread

https://www.physicsforums.com/threads/my-simple-proof-of-x-0-1.172073/

is locked and i did't found solution in it from axioms. People using exp(x) and log(x) and x

^{a-a}=x^{a}x^{-a}as given.If you have x

^{a-a}=x^{a}x^{-a}for a∈ℤ and x∈ℕ_{+}then there is no question... (thank to M5 below...)But from axioms ... maybe something like this???...:

Starting from "Principles of mathematical analysis. Walter Rudin" page 5 and 6.

For any field F we get axioms:

...

"(M3) Multiplication is commutative: xy=yx for all x,y ∈ F"

...

"(M5) If x ∈ F and x ≠ 0 then there exists an element 1/x ∈ F such that

x⋅(1/x) =1." Let assume (Rudin does not specify that) that we call this element x

^{-1}.On next page (6) Rudin in Remark 1.13 states:

"One usually writes (in any field) .....x

^{2}, x^{3}... ""in place of ... xx, xxx, ..."

Let assume (Rudin does not specify that) that x=x

^{1}.From (M5) we get

1=x⋅(1/x) =x

^{1}x^{-1}so

1=x

^{1}x^{-1}⋅ 1 = x^{1}x^{-1}⋅ x^{1}x^{-1}= (from M3) = x^{1}⋅x^{1}⋅x^{-1}⋅x^{-1}=(from remark 1.13)=x

^{2}⋅x^{-1}⋅x^{-1}.But we got for x

^{2}from (M5)1=x

^{2}(x^{2})^{-1}we get (not proving that if ab=c and ad=c then b=d)

x

^{-1}⋅x^{-1}=(x^{2})^{-1}=(from now on)=x^{-2}let's call this (equation 1). Now we know what x^{-n}is.(from remark 1.13) where ∏

_{k}X=X⋅X⋅...⋅X k-timesX

^{A+B}=∏_{A+B}X=∏_{A}X⋅∏_{B}X=X^{A}⋅X^{B}so

X

^{A+B}=X^{A}⋅X^{B}and A,B ∈ℤ base on (equation 1)If A=1 and B=-1 we get from above

X

^{A+B}=X^{A}⋅X^{B}=X^{1}⋅X^{-1}=(from M5 only for X≠0)=1and

X

^{A+B}=X^{(1)+(-1)}=X^{0}we get X

^{0}=1for X≠0 because of (M5) (we don't know what 0^{-1}is...)the end ...

and now just for fun :)

0

^{0}=(1-1)^{1-1}=(1-1)^{1}(1-1)^{-1}=(0)^{1}(1-1)^{-1}=0⋅(1-1)^{-1}=0/(1-1)^{1}=0/(0)^{1}=0/0so

0

^{0}=0/0
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