Press space
If you believe this is wrong, do say so, explaining why. @Jamulus proved it wrong: Assign 2 numbers a and b such that a=b therefore a-b=0 Assuming 0/0=0, (a-b)/(a-b)=a-b Multiply by a-b to get a-b=(a-b)(a-b) a-b=0 so n(a-b)=a-b for all n so n(a-b)=(a-b)(a-b) Subtracting n(a-b) gives 0=(a-b)(a-b)-n(a-b) This factorises to 0=(a-b)(a-b-n) For this to equal 0, a can equal b or b+n, but a=b therefore a=a (which it does) but can also equal a+n for all n (which it does not) Which means that if 0=0/0 then any number equals any other number, therefore 0 cannot equal 0/0, qed.
