- #1
why Fenix
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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 xa-a=xax-a as given.
If you have xa-a=xax-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) ...x2, x3... "
"in place of ... xx, xxx, ..."
Let assume (Rudin does not specify that) that x=x1.
From (M5) we get
1=x⋅(1/x) =x1x-1
so
1=x1x-1 ⋅ 1 = x1x-1 ⋅ x1x-1 = (from M3) = x1⋅x1⋅x-1⋅x-1=(from remark 1.13)=
x2⋅x-1⋅x-1.
But we got for x2 from (M5)
1=x2(x2)-1
we get (not proving that if ab=c and ad=c then b=d)
x-1⋅x-1=(x2)-1=(from now on)=x-2 let's call this (equation 1). Now we know what x-n is.
(from remark 1.13) where ∏kX=X⋅X⋅...⋅X k-times
XA+B=∏A+BX=∏AX⋅∏BX=XA⋅XB
so
XA+B=XA⋅XB and A,B ∈ℤ base on (equation 1)
If A=1 and B=-1 we get from above
XA+B=XA⋅XB=X1⋅X-1=(from M5 only for X≠0)=1
and
XA+B=X(1)+(-1)=X0
we get X0=1for X≠0 because of (M5) (we don't know what 0-1 is...)
the end ...
and now just for fun :)
00=(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=0/0
so
00=0/0
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 xa-a=xax-a as given.
If you have xa-a=xax-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) ...x2, x3... "
"in place of ... xx, xxx, ..."
Let assume (Rudin does not specify that) that x=x1.
From (M5) we get
1=x⋅(1/x) =x1x-1
so
1=x1x-1 ⋅ 1 = x1x-1 ⋅ x1x-1 = (from M3) = x1⋅x1⋅x-1⋅x-1=(from remark 1.13)=
x2⋅x-1⋅x-1.
But we got for x2 from (M5)
1=x2(x2)-1
we get (not proving that if ab=c and ad=c then b=d)
x-1⋅x-1=(x2)-1=(from now on)=x-2 let's call this (equation 1). Now we know what x-n is.
(from remark 1.13) where ∏kX=X⋅X⋅...⋅X k-times
XA+B=∏A+BX=∏AX⋅∏BX=XA⋅XB
so
XA+B=XA⋅XB and A,B ∈ℤ base on (equation 1)
If A=1 and B=-1 we get from above
XA+B=XA⋅XB=X1⋅X-1=(from M5 only for X≠0)=1
and
XA+B=X(1)+(-1)=X0
we get X0=1for X≠0 because of (M5) (we don't know what 0-1 is...)
the end ...
and now just for fun :)
00=(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=0/0
so
00=0/0
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