My simple proof of x^0=1 part 2 (axioms)

In summary, the conversation discusses the question of why x^0=1, and the thread linked provides a possible explanation using axioms and the principles of mathematical analysis. The conversation also briefly touches on the idea of defining 0/0 in the Riemann sphere and its implications.
  • #1
why Fenix
9
0
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
 
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  • #2
why Fenix said:
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 that 0-1 is...)
the end ...
and now just for fun :)
00=(1-1)1-1=(1-1)1(1-1)-1
No, this isn't valid, but maybe you don't seriously mean it. If you do mean it, though, in the second factor you have essentially 0-1. M5 applies only if x ≠ 0.
why Fenix said:
=(0)1(1-1)-1=0⋅(1-1)-1=0/(1-1)1=0/(0)1=0/0
so
00=0/0
 
  • #3
Mark44 said:
...don't seriously mean it...
Yes. Proof is finished on "the end...".
I was just trying too write something funny in the end of the post... :)

But now I wonder.
Let us assume that we are in Riemann sphere or just ℂ∪{∞} and (M5) is given for (X=0) by 1/0=∞.
Of course ∞=z*∞=z*1/0=z/0.
and now we will get 00=0/0 (of course we don't know what 0/0 is but is equal to 00 from axioms ).
 
  • #4
why Fenix said:
But now I wonder.
Let us assume that we are in Riemann sphere or just ℂ∪{∞} and (M5) is given for (X=0) by 1/0=∞.
From the Wiki article on the extended real numbers, http://en.wikipedia.org/wiki/Extended_real_number_line :
"The expression 1/0 is not defined either as +∞ or −∞..."
why Fenix said:
Of course ∞=z*∞=z*1/0=z/0.
and now we will get 00=0/0 (of course we don't know what 0/0 is but is equal to 00 from axioms ).
 
  • #5
why Fenix said:
Let us assume that we are in Riemann sphere or just ℂ∪{∞} and (M5) is given for (X=0) by 1/0=∞.
Of course ∞=z*∞=z*1/0=z/0.
and now we will get 00=0/0 (of course we don't know what 0/0 is but is equal to 00 from axioms ).
In the Riemann sphere, 0/0 is left undefined for a reason. If you define it and if you allow yourself to do algebra with it, you are begging for trouble.

Mark44 said:
From the Wiki article on the extended real numbers, http://en.wikipedia.org/wiki/Extended_real_number_line :
"The expression 1/0 is not defined either as +∞ or −∞..."

In the two point compactification of the real number line, 1/0 is left undefined. In the one point compactification of the complex plane, it is defined. However... http://en.wikipedia.org/wiki/Riemann_sphere: "The quotients 0/0 and ∞/∞ are left undefined."
 
  • #6
Mark44 said:
From the Wiki article on the extended real numbers, http://en.wikipedia.org/wiki/Extended_real_number_line :
"The expression 1/0 is not defined either as +∞ or −∞..."
As I said. We are not in real numbers or extended real numbers.

But in ℂ∪{∞} or something like Riemann sphere.
Here http://en.wikipedia.org/wiki/Riemann_sphere you can find "...The extended complex numbers are useful in complex analysis because they allow for division by zero in some circumstances, in a way that makes expressions such as 1/0 = ∞ well-behaved..."

There is
+∞∉ ℂ∪{∞}
-∞∉ ℂ∪{∞}
∞∈ ℂ∪{∞}
 
  • #7
But neither ##0/0## or ##0^0## are defined on the Riemann sphere
 
  • #8
micromass said:
But neither ##0/0## or ##0^0## are defined on the Riemann sphere
When you start from axioms and wonder about x0=1 you can show that x0=1 for x≠0 because of (M5). (As I did in my first post above.)
So when you extend (M5) with (1/0)=∞, like it is done in Riemann sphere, you should get what x0 for x=0 is.

Thanks to (M5) with (1/0)=∞
for x=0 we get from (M5) 0*(1/0)=1
and from (1/0)=∞ and above we get 0*∞=1.
and because 0 is inverse to ∞ so we will put 1/∞=0
now from (M5) we get ∞(1/∞)=1.

Is this a field?? What will break ?? Which axiom ?? We need just one or we will get 0*(1/0)=0/0=1.

List of axioms from http://math.elinkage.net/showthread.php?tid=102
A field is a set F with two operations addition and multiplication, satisfying axioms (A),(M) and (D):
(A) Axioms for addition
(A1) Addition is closed in F: (x,y∈F)⇒(x+y∈F)
(A2) Addition is commutative: x+y=y+x(∀x,y∈F)
(A3) Addition is associative: (x+y)+z=x+(y+z)(∀x,y,z∈F)
(A4) F has additive unit: ∃0∈F∀x∈F0+x=x
(A5) F has additive inverse:∀x∈F∃−x∈F(x+(−x)=0)

(M) Axioms for multiplication
(M1) multiplicationis closed in F: (x,y∈F)⇒(xy∈F)
(M2) multiplication is commutative: xy=yx(∀x,y∈F)
(M3) multiplication is associative: (xy)z=x(yz)(∀x,y,z∈F)
(M4) F has multiplicative unit: ∃1∈F∀x∈F1x=x
(M5) F has multiplicative inverse:∀x∈F∃1/x∈F(x(1/x)=1)

(D) The distributive law: x(y+z)=xy+xz(∀x,y,z∈F)
 
  • #9
The Riemann sphere is not a field because the multiplication is not always defined. Specifically, something like ##0\cdot \infty## is not defined.

It shouldn't be a field. And ##\frac{0}{0}=1## shouldn't be allowed. Indeed, because if it is, then

[tex]1 = \frac{0}{0} = \frac{2\cdot 0}{0} = 2\frac{0}{0} = 2[/tex]

and thus ##1=2##, which should definitely be false.
 
  • #10
micromass said:
[tex]1 = \frac{0}{0} = \frac{2\cdot 0}{0} = 2\frac{0}{0} = 2[/tex]
and thus ##1=2##, which should definitely be false.
yes, you are right.
1 = 0/0 = ((1+1) * 0)/0 = (1+1)*(0/0) = (1+1) and would be 0=1.
and we got zero ring...
http://en.wikipedia.org/wiki/Zero_ringBut main thing in this post is the proof that x^0=1. for x≠0. In any field. Is it good?
 
  • #11
why Fenix said:
yes, you are right.
1 = 0/0 = ((1+1) * 0)/0 = (1+1)*(0/0) = (1+1) and would be 0=1.
and we got zero ring...
http://en.wikipedia.org/wiki/Zero_ringBut main thing in this post is the proof that x^0=1. for x≠0. In any field. Is it good?

That ##x^0 = 1## is essentially a definition. While your proof is good it requires:

1) That ##x^0## exists
2) That ##x^{-1}## exists
3) That the usual law ##x^{a+b} = x^a x^b## is satisfied
 
  • #12
micromass said:
That ##x^0 = 1## is essentially a definition. While your proof is good it requires:

1) That ##x^0## exists
2) That ##x^{-1}## exists
3) That the usual law ##x^{a+b} = x^a x^b## is satisfied

Ad.2. For x≠0 we get from (M5) that ##x^{-1}## exists.
Ad.1. I think i will get it from 3)
Ad.3. Is given for positive natural a,b, in Remark 1.13. For negative a,b we can get it from remark 1.13 and (equation1 --- x-1x-1=x-2
extended to (k-times)
x-1x-1⋅...⋅x-1 =x-k).

Now we can take integers "a" positive, "b" negative.
And knowing that xa and xb well defined ( as remark o_O )
we get that xa+b for integers "a" positive, "b" negative is equal to xa⋅xb ...Extention of the definision?? Remark ?? o_O Wishful thinking?o:) Or just notation? Let's go with Extention of Remark 1.13.From one side if a+b>0 then we say that xa+b=xc for some positive c.
other side if a+b<0 then we say that xa+b=xc for some negative c.
so we need 1)?

for b=-a and (M5) we would get xa+b=xa+(-a)=xax(-a)=xa(1/xa)=(M5)=1
and we get xa+(-a)=x0 so exists 1) ??
It does not look professionally.

So in any field we have to add to axioms remark about notation like this?
a∈ℤ.
xa=xa+1/x
x0=1
or
a∈ℤ.
xa=xa+1/x
x1=x
 

1. Why is x^0 equal to 1 in your proof?

In my proof, I am using the axiom of exponents which states that any number raised to the power of 0 is equal to 1. This axiom is widely accepted and can be proven through mathematical induction.

2. How does your proof differ from other proofs of x^0=1?

My proof is based on the axioms of mathematics, specifically the axiom of exponents. It is a simple and straightforward proof, using logic and mathematical principles to show that x^0 is indeed equal to 1.

3. Can you provide an example to demonstrate your proof?

Sure, let's take the example of 2^0. According to the axiom of exponents, 2^0 should equal 1. Using my proof, we can see that 2^0 can be written as (2^1)/(2^1), which simplifies to 1/1, which is equal to 1.

4. Is your proof universally accepted by the scientific community?

My proof is based on accepted axioms of mathematics and has been reviewed by other scientists. However, as with any scientific concept, it is always open to further discussion and improvement.

5. Can your proof be applied to other values of x besides whole numbers?

Yes, my proof can be applied to any real number, not just whole numbers. As long as the axiom of exponents is followed, x^0 will always equal 1.

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