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.

Sorry, please merge this thread and the triplicate ones, my comp had screw up and result in triplicate, thanks!

Trial 2 Axioms (Those which are not sure marked with ? and are assumed to not hold unless followed by "assume") 1. For x=/=u? x+(-x)=0 (Note I have not mention that (-x)+x=0) 2. x*1=x (Note I have not mention that 1*x=x) 3. x+0=x (Note I have not mention that 0+x=x) 4. x*x[sup]-1[/sup]=1 (Note I have not mention that x-1*x=1) x=/=u? 5. 0=/=1 6. 0*q=u 7. u+x=u, x=/=u? 8. 2 is a successor of 1 such that 1+1=2 ?9. x(yz)=(xy)z 10. x+y=y+x 11. xy=yx 12. x*(y+z)=xy+xz (Note I said nothing about right distributivity) Step 1 Use 6 and 11 0*q=q*0=u ----(1) Use 2,3 resp with 11 x*1=1*x=x x=/=u ----(2) x+0=0+x=x ----(3) x*x[sup]-1[/sup]=x[sup]-1[/sup]*x=1----(3a) Step 1.5, assume x=/=u=/=q x=x both sides *0 x*0=x*0 +x both sides x*0+x=x*0+x Use 12 x*0+x=x*(0+1) Use (3) x*0+x=x*1 Use 2 x*0+x=x Use 1 x*0=0 Use 11 x*0=0*x=0----(O) Step 2 Use 7, sub x=q u+q=u 0* both sides and use 12 0*(u+q)=0*u 0*u+0*q=0*u Use 6 0*u+u=0*u ----(4.1) u+q=u both sides *0 and use 12 (u+q)*0=u*0 ----(4.2) Now ASSUME(1): (x+y)*z=xz+yz is true, and see where it will lead us to, and use (4.1), (4.2) (4.1) 0*u+u=0*u Use ASSUME(1) (0+1)*u=0*u Use (3) 1*u=0*u Use (2) 0*u=u (Given ASSUME(1)=TRUE)----(4.1a) (4.2) (u+q)*0=u*0 Use ASSUME(1) u*0+q*0=u*0 Use (1) u*0+u=u*0 Use 12 u*(0+1)=u*0 Use (3) u*1=u*0 u*0=u (Given ASSUME(1)=TRUE)----(4.2a) Therefore u*0=0*u=u when ASSUME(1) is true Let's call this result u*0=0*u=u (Given ASSUME(1)=TRUE) ----(A) Step 3 u+u Use 12 =u*(1+1) Use 8 =u*2 ----(5.1) u+u Use ASSUME(1) =(1+1)*u =2*u ----(5.2) Therefore u+u=u*2 ----(5.1a) OR u+u=2*u ---(5.2a) (Given ASSUME(1)=TRUE) 0* both sides and use (5.1a) 0*(u+u)=0*(u*2) Use 12 but leave RHS for now 0*u+0*u=0*(u*2) Use (A) u+u=0*(u*2) ----(5.1b) both sides *0 and use (5.1a) (u+u)*0=(u*2)*0 Use 12 but leave RHS for now u*0+u*0=(u*2)*0 Use (A) u+u=(u*2)*0 ----(5.1c) Therefore u+u=0*(u*2)=(u*2)*0 ----(5.1*) Similarly for 5.2a, we get u+u=0*(2*u)=(2*u)*0 (Given ASSUME(1)=TRUE)----(5.2*) Therefore u+u=0*(u*2)=(u*2)*0=0*(2*u)=(2*u)*0 (Given ASSUME(1)=TRUE)----(5.3) Now ASSUME 9 holds and use 5.3 u+u=u*2=u=u=2*u ----(B) (Given 9 AND ASSUME(1)=TRUE) Step 4 u+2*u Use 12 =u*(1+2) =u*3 So u+2*u=u*3 ----(6) 0* both sides (6) 0*(u+(2*u))=0*(u*3) Use 12 0*u+0*(2*u)=0*(u*3) ----(6.1a) both sides (6) *0 gives u*0+(2*u)*0=(u*3)*0 ----(6.1b) Similarly for the expressions u+u*2, u*2+u, 2*u+u we get u+u*2=u*3----(6.1c) AND u*2+u=2*u+u=3*u ----(6.1d) (Given ASSUME(1)=TRUE) which after 0* both sides, then becomes 0*u+0*(u*2)=0*(u*3) ----(6.1e) AND 0*(u*2)+0*u=0*(2*u)+0*u=0*(3*u) ----(6.1f) (Given ASSUME(1)=TRUE) both sides (6.1c),(6.1d) *0 gives u*0+(u*2)*0=(u*3)*0 ----(6.1g) AND (u*2)*0+u*0=(2*u)*0+u*0=(3*u)*0 ----(6.1h) (Given ASSUME(1)=TRUE) Use (B) on (6.1a),(6.1b) gives 0*u+0*u=u*3 u+u=u*3 u=u*3 (Given 9 AND ASSUME(1)=TRUE) AND u*0+u*0=u*0 u+u=u u=u (Given 9 AND ASSUME(1)=TRUE) Use (B) on (6.1e-h) gives 0*u+0*u=u*3 u+u=u*3 u=u*3 (Given 9 AND ASSUME(1)=TRUE) 0*u+0*u=0*u u+u=u (Given 9 AND ASSUME(1)=TRUE) u*0+u*0=u*0 u+u=u (Given 9 AND ASSUME(1)=TRUE) u*0+u*0=3*u u+u=3*u u=3*u (Given 9 AND ASSUME(1)=TRUE) Therefore u+u=u=3*u=u*3 u=2*u=3*u ----(6.2) Generalise this by induction (detail omitted else too many cases to fill in this post) you will obtain (1+x)*u=u*(x+1)=u*x=x*u=u (for x>=0)----(C.1) (Given 9 AND ASSUME(1)=TRUE) WARNING this does not mean 1=2=3=4=... as we currently have not found out what u[sup]-1[/sup] is, let alone whether it exist, this will be dealt with later Step 5 Using (A) 0*u=u*0=u (Given ASSUME(1)=TRUE) Use 1 (1+(-1))*u=u*(1+(-1))=u (Given ASSUME(1)=TRUE) Use 12 and ASSUME(1) 1*u+(-1)*u=u*1+u*(-1)=u (Given ASSUME(1)=TRUE) Use (2) u+(-1)*u=u+u*(-1)=u (Given ASSUME(1)=TRUE) Use (A) 0*u+(-1)*u=0*u+u*(-1)=u (Given ASSUME(1)=TRUE) Use 12 and ASSUME(1) (0+(-1))*u=u*(0+(-1))=u Use (3) (-1)*u=u*(-1)=u Using similar methods as Step 4, one can show that (1+x)*u=u*(x+1)=u*x=x*u=u (for x<0) ----(C.2) (Given 9 AND ASSUME(1)=TRUE) Using (B) u+u=u Assume there exist a y such that u+y=0 +y both sides 0+u=0 Use 3 or 7 u=0 which is a contradiction since this will make 0*q=0 thus violate axiom 6 Therefore y (i.e. additive inverse of u: -u) does not exist, as expected Step 6 Use 6 set x=q q*q[sup]-1[/sup]=1 0* both sides 0*q*q[sup]-1[/sup]=0*1 Use (O) 0*q*q[sup]-1[/sup]=0 Use 6 u*q[sup]-1[/sup]=0 Step 7 Assume there exist r such that r*u=u*r=1 0* both sides 0*(r*u)=0*(u*r)=0*1 Use (A) r*u=u*r=0 (Given 9 AND ASSUME(1)=TRUE) Which is a contradiction Therefore either r (i.e. u[sup]-1[/sup]) does not exist OR 9 is false To Be Continued... P.S. Sorry but it prevent me from editing the OP due to the edit there has expired