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

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1. Feb 23, 2015

### why Fenix

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

Last edited: Feb 23, 2015
2. Feb 23, 2015

### Staff: Mentor

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.

3. Feb 24, 2015

### why Fenix

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. Feb 24, 2015

### Staff: Mentor

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 −∞..."

5. Feb 24, 2015

### jbriggs444

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.

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. Feb 24, 2015

### why Fenix

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. Feb 24, 2015

### micromass

Staff Emeritus
But neither $0/0$ or $0^0$ are defined on the Riemann sphere

8. Feb 24, 2015

### why Fenix

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.

A field is a set F with two operations addition and multiplication, satisfying axioms (A),(M) and (D):
(A1) Addition is closed in F: (x,y∈F)⇒(x+y∈F)
(A4) F has additive unit: ∃0∈F∀x∈F0+x=x

(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. Feb 24, 2015

### micromass

Staff Emeritus
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

$$1 = \frac{0}{0} = \frac{2\cdot 0}{0} = 2\frac{0}{0} = 2$$

and thus $1=2$, which should definitely be false.

10. Feb 24, 2015

### why Fenix

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_ring

But main thing in this post is the proof that x^0=1. for x≠0. In any field. Is it good?

11. Feb 24, 2015

### micromass

Staff Emeritus
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. Feb 24, 2015

### why Fenix

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 )
we get that xa+b for integers "a" positive, "b" negative is equal to xa⋅xb .....Extention of the definision?? Remark ?? Wishful thinking? Or just notation? Lets 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