TBC=To be continued...

From previous threads, we learnt that defining division by zero by adding an extra axiom which state that

$$q \times 0=0 \times q=1$$ -----(1)

does not work when all axioms of reals are present as one will always end up with the following result

$$0=1=2=3...$$

Which is the trivial ring {0,+,*}

In addition, attempt to remove the above contradictory result by removing the assumption of associativity and distributivitiy end up with

$$q^{-1}=1$$

which contradict with (1)

Recent detailed reading of Wheel theory came across the following result

$$ \frac0 0 +x=\frac0 0$$ --------(2)

In light of this, this might gave us a way to set up another number system where division by zero can be included

Our aims is as follows. Given a set of axioms, letting u=0/0, we need to prove that

$$q={1} \times q^{-1}$$

and

$$0 \times q=q \times 0$$

and

$$There.are.no.contradictory.results$$

and

$$With.the.exception.of.the.axiom.given,there.should.not.be.any.results.derived.in.the.table.of.this.link$$ http://en.wikipedia.org/wiki/James_Anderson_(computer_scientist)

Axioms (Those which are not sure marked with ?)

?1. x+y=y+x

2. x*(y+z)=xy+xz

?3. For x=/=u -x+x=0 (Note I have not mention that x+(-x)=0)

?4. x*1=x (Note I have not mention that 1*x=x)

?5. x+0=x (Note I have not mention that 0+x=x)

6. x*x[sup]-1[/sup]=1 (Note I have not mention that x[sup]-1[/sup]*x=1)

7.0=/=1

8.0*q=u

9.u+x=u

10. 2 is a successor of 1 such that 1+1=2

?11. x(yz)=(xy)z

Derivations (Note that every occurence of x is assumed to be =/=u):

1.

**Warning: Pseudoscientific and cranky nonsense, ignore if necessary****(This serves as a drawing board for yet another (possibly failed attempt) in defining division by zero). Constructive criticism if the following actually make sense, ignore if otherwise.**From previous threads, we learnt that defining division by zero by adding an extra axiom which state that

$$q \times 0=0 \times q=1$$ -----(1)

does not work when all axioms of reals are present as one will always end up with the following result

$$0=1=2=3...$$

Which is the trivial ring {0,+,*}

In addition, attempt to remove the above contradictory result by removing the assumption of associativity and distributivitiy end up with

$$q^{-1}=1$$

which contradict with (1)

Recent detailed reading of Wheel theory came across the following result

$$ \frac0 0 +x=\frac0 0$$ --------(2)

In light of this, this might gave us a way to set up another number system where division by zero can be included

Our aims is as follows. Given a set of axioms, letting u=0/0, we need to prove that

$$q={1} \times q^{-1}$$

and

$$0 \times q=q \times 0$$

and

$$There.are.no.contradictory.results$$

and

$$With.the.exception.of.the.axiom.given,there.should.not.be.any.results.derived.in.the.table.of.this.link$$ http://en.wikipedia.org/wiki/James_Anderson_(computer_scientist)

Axioms (Those which are not sure marked with ?)

?1. x+y=y+x

2. x*(y+z)=xy+xz

?3. For x=/=u -x+x=0 (Note I have not mention that x+(-x)=0)

?4. x*1=x (Note I have not mention that 1*x=x)

?5. x+0=x (Note I have not mention that 0+x=x)

6. x*x[sup]-1[/sup]=1 (Note I have not mention that x[sup]-1[/sup]*x=1)

7.0=/=1

8.0*q=u

9.u+x=u

10. 2 is a successor of 1 such that 1+1=2

?11. x(yz)=(xy)z

Derivations (Note that every occurence of x is assumed to be =/=u):

1.

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