TBC=To be continued...
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.
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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