MAIN WRITING PART 3-B

                 MAIN WRITING  PART 3-B


Point 3-A.- Case of a:0 (for a different from zero)


   In general, D:d = q  so  dxq = D. But, in conventional Mathematics, a:0 is equal to no number (a:0 = empty set) for it's considered that zero multiplied by any number (including zero) results in zero and, therefore, doesn't result in a. In other words, it's considered that there isn't any numbers q by which multiplying zero (d), the multiplication results in a. But, if we accept that zero multiplied by zero results in every number EXCEPT zero, we'll be able to see that zero multiplied by zero does result in a. 0x0 = a  (dxq = D) therefore a:0 = 0.
  Reasoning: Absence does not make HABER nor is even a part of it (HABER, the same as SOME?) therefore absence is not contained in HABER. HABER does not contain absence; a does not contain zero nor a part of it, which is zero too. So the number of times that absence (zero) is contained in HABER (a) is zero; a:0 = 0.
                            
                               Problem
  If twelve candies were distributed among a quantity of people giving zero to everyone, how many people would the twelve candies be distributed among?
  
  The procedure to solve the problem takes us to the expression 12:0, which is equal to the number of people among whom the twelve candies would be distributed. In conventional Mathematics, 12:0 is equal to empty set for the reason already expressed at the beginning of this Point 3 but, in spite of that, many people would consider that the answer to the problem is that the number of people among whom the twelve candies would be distributed is infinite because, without mattering how big the number of people to everyone of whom zero candies were given was, there would always be left candies to distribute. I don't know if that kind of reasoning is useful for any kinds of cases. If we accept that zero multiplied by zero results in every number except zero, we should consider 12:0 equal to zero (12:0 = 0) because zero does not constitute twelve (case of a:0) and, based on this, and even if it may seem strange, the answer to the problem is that the twelve candies would be distributed among zero people, which is totally true for, if zero candies were given to everyone, there would be no distribution and the quantity of people among whom the twelve candies would be distributed would be zero or, in other words, there wouldn't be any people among whom the twelve candies would be distributed (for there would be no distribution).
  Note that if we consider that the number of people among whom the twelve candies would be distributed is equal to empty set (based on the result of 12:0 of conventional Mathematics; 12:0 = empty set), we'll be considering that neither would there be distribution nor would there not be distribution, which is not logical.


Point 3-B.- In conventional Mathematics, a/0 (a different from zero) is no number. So a/0 is considered to be equal to empty set. But, as it was seen in the reasoning already done about 0/0, denominator zero implies "parts" equal to zero, and a/0 is constituted by a of those "parts". a/0 is equal, therefore, to zero multiplied by a (zero multiplied by a number different from zero), which is equal to zero.
                       a/0 = 0  (for a different from zero)