MAIN WRITING PART 1
Does zero multiplied by zero result in zero? DOES ABSENCE OF ABSENCE RESULT IN ABSENCE? Of course NOT!! I have the conviction that zero multiplied by zero results in every number EXCEPT zero. Absence of absence implies what, in spanish, I call HABER (existence, for real things, and its equivalent -or counterpart- for imaginary things. Something like SOME?), which is what constitutes units and fractions. ZERO MULTIPLIED BY ZERO IS A DOUBLE NEGATION. Zero multiplied by zero results in every number EXCEPT zero and, therefore, is equal to the set of all numbers EXCEPT zero.
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Read Main Writing Part 3-A
Saturday, August 27, 2011
MAIN WRITING PART 2
MAIN WRITING PART 2
Additional Sustaining Reasoning
Due to a certain practical reason, the expression 0:0 is considered indeterminate (0:0 = ind). But, in conventional Mathematics, 0:0 is equal to an infinite quantity of numbers, including zero,* for zero multiplied by any number is supposed to result in zero (axb = c so c:a = b) or, in other words, zero would be contained in zero any quantity of times. Therefore, IN CONVENTIONAL MATHEMATICS, though 0:0 is usually considered indeterminate, 0:0 is equal to the set of all numbers (including zero). This must be kept in mind for the following.
The "fraction" 0/0 is considered to be equal to the expression 0:0 (in general, a:b = a/b). It's considered that 0:0 is equal to an infinite quantity of numbers (including zero) and is equal to the set of all numbers. Because of that, without enough reflection, it's considered that 0/0 is an infinite quantity of numbers too, equal to the set of all numbers.
a) But, if we analyze 0/0 considering, wrongly, that zero multiplied by zero results in zero, we'll get to the WRONG conclusion that 0/0 is no number. 0/0 would equal empty set (absence of any numbers). So 0/0 would be opposite 0:0, which would equal the set of all numbers (including zero). In the general expression a/b, the denominator (b) is a number of equal parts that constitute a unit (the whole b-parts constitutes a unit and is equal to it). One part (p) multiplied by b results in one. But if the denominator (b) is zero, what can a "part" (p) be equal to, if every number multiplied by zero is supposed to result in zero? In this case, the "part" would be NO number (p = empty set). If a = 0 and b = 0 then a/b is 0/0. The number of "parts" which would constitute the number 0/0, which is the numerator (a), is zero then 0/0 WOULD BE equal to empty set multiplied by zero (pxa), which results in empty set.*** Therefore, if zero multiplied by zero resulted in zero, the expression 0:0 wouldn't be equal to the "fraction" 0/0 (0:0 different from 0/0), and a:b = a/b wouldn't be a general equality.
b) BUT if we consider that zero multiplied by zero results in every number except zero, 0:0 is equal to the "fraction" 0/0. The expression 0:0 is equal to every number except zero because the divisor zero multiplied by the quotient, which is every number of an infinite quantity of numbers of every value different from zero, results in the dividend zero (dxq = D so D:d = q)**. The "fraction" 0/0 is zero of zero "parts" that constitute a unit, "parts" (p) which are equal to zero because zero (p) multiplied by zero (b), resulting in every number except zero, does result in one (among an infinite quantity of numbers different from zero). So the "fraction" 0/0 is equal to zero multiplied by zero (pxa), which results in every number except zero. The expression 0:0 and the "fraction" 0/0 are both equal to the set of all numbers except zero. THEREFORE 0:0 = 0/0 .
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Read Main Writing Part 3-B
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*In the condition of being a particular set of times zero, 0:0 is equal to every number equal to that particular set. For example, in the particular condition of being a set of times zero whose value is 7, 0:0 is equal to 7.
a) IN CONVENTIONAL MATHEMATICS, 0:0 is equal to every number so, in general, it's equal to the set of all numbers.
** b) If 0x0 results in every number EXCEPT zero, 0:0 is equal to every number EXCEPT zero so, in general, it's equal to the set of all numbers EXCEPT zero.
*** The product of empty set multiplied by zero results in empty set because, there not being zero as a multiplicand, it's not possible to obtain any numbers different from zero as a product and, there not being any numbers different from zero as a multiplicand, it's not possible to obtain zero as a product. Therefore the multiplication empty set multiplied by zero does not result in any numbers; it results in empty set. BUT the reasoning with which we got to the FALSE conclusion that 0:0 was different to 0/0 is based on the false assumption that zero multiplied by zero results in zero so I'll prove that, even on that false assumption, empty set multiplied by zero would result in empty set too.
I) In order that empty set multiplied by zero resulted in any number different from zero, the multiplicand should be zero (zero multiplied by zero is absence of absence). But:
a) this contradicts the assumption that zero multiplied by zero results in zero.
b) the multiplicand is empty set, absence of any numbers; there is no zero as the multiplicand.
So no number different from zero would be obtained.
II) In order that empty set multiplied by zero resulted in zero, the multiplicand should be a number different from zero, with which the multiplication would be absence of HABER (absence of SOME?), or the multiplicand should be zero, wrongly considered by conventional Mathematics; numbers that there are not as the multiplicand for this is empty set. So zero would not be obtained either.
Therefore, even on the false assumption that zero multiplied by zero results in zero, empty set multiplied by zero would not result in zero nor any numbers different from zero. Empty set multiplied by zero would result in empty set.
Some people may not be satisfied with the part II of this reasoning. All right!! If you want, consider that empty set multiplied by zero results in zero. If zero multiplied by zero resulted in zero, 0/0, which would be equal to empty set multiplied by zero (pxa), would be equal to zero so, even in the case that empty set multiplied by zero resulted in zero, 0/0 would be different from 0:0, which, in conventional Mathematics, is equal to the set of all numbers.
Additional Sustaining Reasoning
Due to a certain practical reason, the expression 0:0 is considered indeterminate (0:0 = ind). But, in conventional Mathematics, 0:0 is equal to an infinite quantity of numbers, including zero,* for zero multiplied by any number is supposed to result in zero (axb = c so c:a = b) or, in other words, zero would be contained in zero any quantity of times. Therefore, IN CONVENTIONAL MATHEMATICS, though 0:0 is usually considered indeterminate, 0:0 is equal to the set of all numbers (including zero). This must be kept in mind for the following.
The "fraction" 0/0 is considered to be equal to the expression 0:0 (in general, a:b = a/b). It's considered that 0:0 is equal to an infinite quantity of numbers (including zero) and is equal to the set of all numbers. Because of that, without enough reflection, it's considered that 0/0 is an infinite quantity of numbers too, equal to the set of all numbers.
a) But, if we analyze 0/0 considering, wrongly, that zero multiplied by zero results in zero, we'll get to the WRONG conclusion that 0/0 is no number. 0/0 would equal empty set (absence of any numbers). So 0/0 would be opposite 0:0, which would equal the set of all numbers (including zero). In the general expression a/b, the denominator (b) is a number of equal parts that constitute a unit (the whole b-parts constitutes a unit and is equal to it). One part (p) multiplied by b results in one. But if the denominator (b) is zero, what can a "part" (p) be equal to, if every number multiplied by zero is supposed to result in zero? In this case, the "part" would be NO number (p = empty set). If a = 0 and b = 0 then a/b is 0/0. The number of "parts" which would constitute the number 0/0, which is the numerator (a), is zero then 0/0 WOULD BE equal to empty set multiplied by zero (pxa), which results in empty set.*** Therefore, if zero multiplied by zero resulted in zero, the expression 0:0 wouldn't be equal to the "fraction" 0/0 (0:0 different from 0/0), and a:b = a/b wouldn't be a general equality.
b) BUT if we consider that zero multiplied by zero results in every number except zero, 0:0 is equal to the "fraction" 0/0. The expression 0:0 is equal to every number except zero because the divisor zero multiplied by the quotient, which is every number of an infinite quantity of numbers of every value different from zero, results in the dividend zero (dxq = D so D:d = q)**. The "fraction" 0/0 is zero of zero "parts" that constitute a unit, "parts" (p) which are equal to zero because zero (p) multiplied by zero (b), resulting in every number except zero, does result in one (among an infinite quantity of numbers different from zero). So the "fraction" 0/0 is equal to zero multiplied by zero (pxa), which results in every number except zero. The expression 0:0 and the "fraction" 0/0 are both equal to the set of all numbers except zero. THEREFORE 0:0 = 0/0 .
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Read Main Writing Part 3-B
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*In the condition of being a particular set of times zero, 0:0 is equal to every number equal to that particular set. For example, in the particular condition of being a set of times zero whose value is 7, 0:0 is equal to 7.
a) IN CONVENTIONAL MATHEMATICS, 0:0 is equal to every number so, in general, it's equal to the set of all numbers.
** b) If 0x0 results in every number EXCEPT zero, 0:0 is equal to every number EXCEPT zero so, in general, it's equal to the set of all numbers EXCEPT zero.
*** The product of empty set multiplied by zero results in empty set because, there not being zero as a multiplicand, it's not possible to obtain any numbers different from zero as a product and, there not being any numbers different from zero as a multiplicand, it's not possible to obtain zero as a product. Therefore the multiplication empty set multiplied by zero does not result in any numbers; it results in empty set. BUT the reasoning with which we got to the FALSE conclusion that 0:0 was different to 0/0 is based on the false assumption that zero multiplied by zero results in zero so I'll prove that, even on that false assumption, empty set multiplied by zero would result in empty set too.
I) In order that empty set multiplied by zero resulted in any number different from zero, the multiplicand should be zero (zero multiplied by zero is absence of absence). But:
a) this contradicts the assumption that zero multiplied by zero results in zero.
b) the multiplicand is empty set, absence of any numbers; there is no zero as the multiplicand.
So no number different from zero would be obtained.
II) In order that empty set multiplied by zero resulted in zero, the multiplicand should be a number different from zero, with which the multiplication would be absence of HABER (absence of SOME?), or the multiplicand should be zero, wrongly considered by conventional Mathematics; numbers that there are not as the multiplicand for this is empty set. So zero would not be obtained either.
Therefore, even on the false assumption that zero multiplied by zero results in zero, empty set multiplied by zero would not result in zero nor any numbers different from zero. Empty set multiplied by zero would result in empty set.
Some people may not be satisfied with the part II of this reasoning. All right!! If you want, consider that empty set multiplied by zero results in zero. If zero multiplied by zero resulted in zero, 0/0, which would be equal to empty set multiplied by zero (pxa), would be equal to zero so, even in the case that empty set multiplied by zero resulted in zero, 0/0 would be different from 0:0, which, in conventional Mathematics, is equal to the set of all numbers.
MAIN WRITING PART 3-A
MAIN WRITING PART 3-A
Some Implications and Considerations
Point 1-A.- Case of multiplications of the same type as 20x10
The regular procedure is such as it's shown here:
20
x10
00
20
200
which is an abbreviation of this:
(20 + 0)
x(10 + 0)
(0x0)_ _ _ _ _ _ 0
(20x0)_ _ _ _ _ 00
(0x10)_ _ _ _ _ _ 0
(20x10)_ _ _ _ _ 200
200
To consider that zero multiplied by zero results in zero doesn't hinder us from getting the searched result ALTHOGH THAT CONSIDERATION IS FALSE. This last thing takes us to the issue of the consideration of exceptions where there weren't considered to be.
The axiom (a + b)x(c + d) = ac + bc + ad + bd, AXIOM WHICH MULTIPLICATION PROCEDURES ARE USUALLY RELATED TO, is an example. The axiom is true EXCEPT for (a = 0 or b = 0) and (c = 0 or d = 0).
For the multiplication 20x10, we should proceed just like this: 20x10 = 200, equality which comes from (20 + 0)x(10 + 0) = 20x10 + 0x10 + 20x0 = 200 + 0 + 0 = 200.
In general, (a + 0)x(c + 0) is equal to axc + 0xc + ax0 (or axc + 0xc + ax0 + 0) BUT IS NOT EQUAL TO axc + 0xc + ax0 + 0x0 (for a different from zero and c different from zero).
Point 1-B.- Another important case I'd like to mention here is ax0 = a-a. In this axiom, a is every number except zero for 0x0 results in every number except zero whereas 0-0 results in zero.
Point 2.- MATHEMATICS IS NOT CONDITIONED BY WHAT OCCURES IN THE FIELD OF FACTUAL SCIENCES. IT IS AN INSTRUMENT OF THEM BUT IT IS NOT CONDITIONED BY THEM. Let's imagine three magnitudes A, B and C related between them, ALMOST always, like this: A = a, B = b, C = c, and axb = c. Note A is not exactly the same as a, B is not exactly the same as b, and C is not exactly the same as c. A, B and C are particular magnitudes whereas a is a multiplicand, b is a multiplier and c is a product. NOTHING should say that if A, B and C are equal to such a special number as zero (A = 0, B = 0 and C = 0) then a, b and c have to be equal to zero too.
To consider that zero multiplied by zero results in zero may usually be very useful IN SPITE OF BEING THE CONSIDERATION OF A FALSITY.
Point 1-A.- Case of multiplications of the same type as 20x10
The regular procedure is such as it's shown here:
20
x10
00
20
200
which is an abbreviation of this:
(20 + 0)
x(10 + 0)
(0x0)_ _ _ _ _ _ 0
(20x0)_ _ _ _ _ 00
(0x10)_ _ _ _ _ _ 0
(20x10)_ _ _ _ _ 200
200
To consider that zero multiplied by zero results in zero doesn't hinder us from getting the searched result ALTHOGH THAT CONSIDERATION IS FALSE. This last thing takes us to the issue of the consideration of exceptions where there weren't considered to be.
The axiom (a + b)x(c + d) = ac + bc + ad + bd, AXIOM WHICH MULTIPLICATION PROCEDURES ARE USUALLY RELATED TO, is an example. The axiom is true EXCEPT for (a = 0 or b = 0) and (c = 0 or d = 0).
For the multiplication 20x10, we should proceed just like this: 20x10 = 200, equality which comes from (20 + 0)x(10 + 0) = 20x10 + 0x10 + 20x0 = 200 + 0 + 0 = 200.
In general, (a + 0)x(c + 0) is equal to axc + 0xc + ax0 (or axc + 0xc + ax0 + 0) BUT IS NOT EQUAL TO axc + 0xc + ax0 + 0x0 (for a different from zero and c different from zero).
Point 1-B.- Another important case I'd like to mention here is ax0 = a-a. In this axiom, a is every number except zero for 0x0 results in every number except zero whereas 0-0 results in zero.
Point 2.- MATHEMATICS IS NOT CONDITIONED BY WHAT OCCURES IN THE FIELD OF FACTUAL SCIENCES. IT IS AN INSTRUMENT OF THEM BUT IT IS NOT CONDITIONED BY THEM. Let's imagine three magnitudes A, B and C related between them, ALMOST always, like this: A = a, B = b, C = c, and axb = c. Note A is not exactly the same as a, B is not exactly the same as b, and C is not exactly the same as c. A, B and C are particular magnitudes whereas a is a multiplicand, b is a multiplier and c is a product. NOTHING should say that if A, B and C are equal to such a special number as zero (A = 0, B = 0 and C = 0) then a, b and c have to be equal to zero too.
To consider that zero multiplied by zero results in zero may usually be very useful IN SPITE OF BEING THE CONSIDERATION OF A FALSITY.
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)
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)
MAIN WRITING PART 3-C
MAIN WRITING PART 3-C
Point 4.- Case of Roots
An implication that zero multiplied by zero does not result in zero is that zero does not have any numbers as a square root for there is no number with which, multiplied by another number equal to it, the multiplication results in zero. Zero does not have any numbers as a square root, as a fourth root, as a sixth root, etc, but does have a number as a cube root, as a fifth root, as a seventh root, etc for an odd quantity of factors zero does result in zero whereas an even quantity of factors zero results in every number except zero.
0x0x0x0 =
= The set of all numbers except zero.
0x0x0 =
= 0
0x0x0x0 = (0x0)x(0x0) = the set of all numbers except zero multiplied by the set of all numbers except zero = the set of all numbers except zero
OR
0x0x0x0 = (0x0x0)x0 = 0x0 = the set of all numbers except zero.
0x0x0 = (0x0)x0 = (the set of all numbers except zero)x0 = 0
Point 5.- Another important exception
= 
(for b or c different from zero).
=
but is different from 
.
Point 6.- Reasoning
= 
= 0
= 
= 0x0 = the set of all numbers except zero.
1/3 = 2/6
2/3 = 4/6 = 1/3+2/6
so a) = 
= 
= 
. As we've already seen, zero does not have any numbers as a square root, as a fourth root, as a sixth root, etc. 
= 
= empty set. = = = = empty set multiplied by empty set. (
symbol of empty set) So
results in , which is equal to zero. = . So = 0.
b)==
=
=
=
=
=.
So=
== 0x0 = the set of all numbers except zero.
So= the set of all numbers different from zero.
c)
=
=
=
=.
=
=
=
So=.
THEREFORE empty set multiplied by empty set results in empty set as well as it results in every number, INCLUDING zero.
ASTONISHED?? I think you shouldn't. This reasoning is only a theory for part b is based on the assumption that the associative law is applicable to factors which are empty set BUT there is a fact that seems to support this part (note that parts a and c do not need any additional support). In conventional Mathematics itself, there are the so called Imaginary Numbers. The product of two real numbers of the same sign is a positive number (+2 x +3 = +6, -2 x -3 = +6) but the product of two imaginary numbers of the same sign is a negative one (+2i x +3i = -6). These so called imaginary numbers are not real numbers (so they are not numbers at all). Therefore, couldn't they be some particular manifestations of empty set?
Point 4.- Case of Roots
An implication that zero multiplied by zero does not result in zero is that zero does not have any numbers as a square root for there is no number with which, multiplied by another number equal to it, the multiplication results in zero. Zero does not have any numbers as a square root, as a fourth root, as a sixth root, etc, but does have a number as a cube root, as a fifth root, as a seventh root, etc for an odd quantity of factors zero does result in zero whereas an even quantity of factors zero results in every number except zero.
0x0x0x0 =
= The set of all numbers except zero.0x0x0 =
= 00x0x0x0 = (0x0)x(0x0) = the set of all numbers except zero multiplied by the set of all numbers except zero = the set of all numbers except zero
OR
0x0x0x0 = (0x0x0)x0 = 0x0 = the set of all numbers except zero.
0x0x0 = (0x0)x0 = (the set of all numbers except zero)x0 = 0
Point 5.- Another important exception
= (for b or c different from zero). = but is different from .Point 6.- Reasoning
= = 0 = = 0x0 = the set of all numbers except zero.1/3 = 2/6
2/3 = 4/6 = 1/3+2/6
so a)
= = = . As we've already seen, zero does not have any numbers as a square root, as a fourth root, as a sixth root, etc. = = empty set. = = = = empty set multiplied by empty set. ( symbol of empty set) So results in , which is equal to zero. = . So = 0.b)
=======.So
=== 0x0 = the set of all numbers except zero.So
= the set of all numbers different from zero.c)
====.===So
=.THEREFORE empty set multiplied by empty set results in empty set as well as it results in every number, INCLUDING zero.
ASTONISHED?? I think you shouldn't. This reasoning is only a theory for part b is based on the assumption that the associative law is applicable to factors which are empty set BUT there is a fact that seems to support this part (note that parts a and c do not need any additional support). In conventional Mathematics itself, there are the so called Imaginary Numbers. The product of two real numbers of the same sign is a positive number (+2 x +3 = +6, -2 x -3 = +6) but the product of two imaginary numbers of the same sign is a negative one (+2i x +3i = -6). These so called imaginary numbers are not real numbers (so they are not numbers at all). Therefore, couldn't they be some particular manifestations of empty set?
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