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by Corry Shores
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[The following is summary. I am not a mathematician, so please consult the original text instead of trusting my summarizations, which are surely mistaken or inelegantly articulated. Bracketed comments and subsection divisions are my own. Proofreading is incomplete, so please forgive my mistakes.]
Summary of
George François Cornelis Griss
(G.F.C. Griss)
“Negationless Intuitionistic Mathematics, I”
0
“Introduction”
Brief summary:
(0.1) Griss will discuss negationless intuitionistic mathematics. (0.2) In intuitionistic mathematics, we have philosophical reasons for needing to reject negation. For, “Proving that something is not right, i.e. proving the incorrectness of a supposition, is no intuitive method. For one cannot have a clear conception of a supposition that eventually proves to be a mistake. Only construction without the use of negation has some sense in intuitionistic mathematics” (1127). (0.3) From this point forward, we no longer consider our philosophical justifications for negationless intuitionistic mathematics and instead we are concerned with the purely mathematical problem of formulating it. (0.4) We begin with some examples. (0.5) Griss gives an example to show two ways to construct proofs. {1} The first one uses negation: it is a reductio argument, so it negates the conclusion. The premise is that we have a triangle ABC whose CA and CB sides are bisected by a fourth line DE such that it brings about the following proportional relation:
CA : CB = CD : CE
The conclusion is that the bisecting line DE is parallel to line AB. We then negate this conclusion and see what follows logically. We then see how the premises plus the negated conclusion yields a contradiction, which proves that the experimentally negated conclusion is false and thus that the originally proposed conclusion is true. {2} The second proof does not use negation. Here we start with the line which fulfills the proportions. Next we construct another line that we know is parallel to the undivided side. Finally we show that this parallel line must necessarily be identical to the line which fulfills the proportions. Thus lines that fulfill the proportions are parallel to the undivided line. (0.6) In the negationless proof, we will need to define our concepts without negation. Parallel lines cannot be defined as “which do not intersect” (for here the “do not intersect” is a negation of “do intersect”). Rather, they must be defined without a negation, as for instance, “parallel lines are such lines, that any point of one of them differs from any point of the other one.” This formulation then requires a positive definition of “difference relative to points.” One reason that negation is used has to do with the triangle figure requiring an additional, different DE line being drawn even though it is identical to that line. (Maybe Griss is saying that this notion of something being both different and identical is counter-intuitive, so people might prefer the reducio proof instead.) Another reason people use negation has to do with how we formulate our mathematical questions. We might take x2 – 2 and form the question, which rational numbers satisfy x2 – 2 = 0, with the answer being the negative “no rational numbers satisfies it.” But we need not think of x2 – 2 in such a formulation that leads us to a negative conceptualization. We can instead say that “x2 – 2 differs positively from zero for every rational number.” (In other words, its value is not seen as not being a rational number when it is equated with zero but rather that given any rational number for x, its value will be another value that is always different from zero.) (0.7) In the second illustration, we wonder if the equation ax + by = 0 has a solution for x and y where x and y are different from zero and the letters represent real numbers? And we will compare the negationless and negative way of answering this question. To do this, we first note the following distinction between real numbers understood as different either positively or negatively: “Two real numbers differ positively, if there can be indicated two approximating intervals which lie outside one another; they differ negatively, if it is impossible that they are equal; you can only divide by a real number if it differs positively from zero.” {1} We begin with the negationless way. We do this first by assuming in one case x has a non-zero value and seeing how that gives a non-zero result for y, secondly we likewise assume that y has a non-zero value and see how that gives a non-zero result for x, and lastly we give both a and b the value zero and see how that yields a non-zero value for both x and y. We conclude from this non-negative approach to the question that “ax + by = 0 has a solution different from zero, if at least one of the coefficients a and b differs from zero or if both are zero.” {2} We next look at how we can use negation to formulate this positive result in a negative way, namely as: “It is impossible that no solution different from zero exists.” We learn this by assuming the only possible solution is zero (“there were no solution different from zero”), which logically yields the contradictory claim that there is such a non-zero solution. Thus it is impossible that there is no solution different from zero. Now, what we learn by comparing the two results is that “The negative formulation is shorter, but distorted, and the details of the positive result are lost. In non-intuitionistic mathematics ax + by = 0 has always a resolution different from zero. In this formulation the positive result has vanished entirely.” (0.8) The third illustration is: “If ax + b ≠ 0 for each value of x, then a = 0.” In our exploration of the proof for it, we make use either of a positive definition for equality or a negative one. {1} The positive definition of equality: “If a differs from c for each value c that differs from b, then a = b.” (In other words, two values are equal if they are both different from all other values.) {2} The negative definition of equality: If a does not differ from b, then a = b. (In the first case, the two equal things share all the same differences to other things. In the second case, they simply are not different to each other). From this Griss concludes that “The positive proposition has to be proved for the different sorts of numbers, to begin with the natural numbers. But therefore again it proves to be necessary to construct the whole of negationless intuitionistic mathematics from the beginning.” (0.9) We can compare a positive formulation, “Two triangles are congruent, if they have equal one side, the angle opposite that side and the sum of the two other sides, while of one of the adjacent angles is known that they are either equal or different” with a negative formulation, “If two triangles have equal one side, the angle opposite that side and the sum of the two other sides, it is impossible, that they are not congruent.” (0.10) Griss now summarizes the results of these illustrations. {1} From example 1 (see section 0.5) we learn that “In some cases it is simpler to avoid the use of the negation.” {2} From example 2 (see section 0.7) we learn that “Positive properties can sometimes be formulated more briefly in a negative way, but details get lost.” {3} From example 4 (see section 0.9) we learn that “The parts of intuitionistic mathematics which in a positive construction are disposed of are less important, for probably examples cannot be constructed for which a negative property could be applied and a corresponding positive property could not.” {4} From examples 1 and 3 (see section 0.5 and section 0.8), we learn that “To construct negationless mathematics one must begin with the elements and a positive definition of difference must be given instead of a negative one.” Moreover, “But even from a general intuitionistic point of view a positive construction of the theory of natural numbers must be given: one cannot define 2 is not equal to 1 (i.e. it is impossible that 2 and 1 are equal), for from this one could never conclude that 2 and 1 differ positively. Conversely one could define in a positive way negation by means of difference, e.g. not equal means different, etc., but, for the present, this seems unfit.” (Perhaps then we might note the following. We need numbers in our negationless mathematics. But to get those numbers, we need more than just inequality (the impossibility of being equal) to tell us that each number is different from the others. We rather need a positive construction of the numbers that does not involve the impossibility of equaling. In section 1 to follow, we learn that there is a notion of distinguishability that grounds inequality.) (0.11) Griss lastly has us “consider the property: If a and b are elements of the set of natural numbers, and if a ≠ b, then a < b or a > b for each element a of the set. If we apply this property to b = 1, we get: For each element a ≠ 1 of the set of natural numbers we have a < 1 or a > 1. a < 1, however, has not any sense in negationless mathematics. If we say: a < b or a > b for each a of a set, we mean 1) that for each a at least one of these conditions is fulfilled, 2) that conversively at least one element fulfils the condition a < b and another one the condition a > b.” We then note that “Negationless intuitionistic logic will differ much from the usual intuitionistic logic by the absence of the negation and the altered meaning of the disjunction” and also that “‘Affirmative’ mathematics is something quite different from the negationless intuitionistic mathematics.”
[Announcing the Topic: Negationless Intuitionistic Mathematics]
[Why We Cannot Have Negation in Intuitionistic Mathematics]
[Turning to the Mathematical Formulation of Negationless Intuitionistic Mathematics]
[Turning to Examples]
[Illustration 1 for Negational vs. Non Negational Mathematical Thinking: Divided Triangle]
[Comments on Illustration 1]
[Illustration 2 for Negational vs. Non-Negational Mathematical Thinking: The Necessity of a Solution vs. The Impossibility of No Solution]
[Illustration 3 for Negational vs. Non-Negational Mathematical Thinking: Equality as Shared Difference to All Others vs. Equality as Lacking Difference between One Another]
[Illustration 4 for Negational vs. Non-Negational Mathematical Thinking: Congruent Triangles]
[Summary of the Results and the Need for a Positive Definition of Difference]
[Final Observations]
Summary
[Announcing the Topic: Negationless Intuitionistic Mathematics]
[Griss will discuss negationless intuitionistic mathematics.]
[ditto]
In these proceedings I already gave a sketch of some parts of negationless intuitionistic mathematics1). In the following I will treat of this subject more fully and systematically.
(1127)
1) Vol. 53 (1944)
[Why We Cannot Have Negation in Intuitionistic Mathematics]
[In intuitionistic mathematics, we have philosophical reasons for needing to reject negation. For, “Proving that something is not right, i.e. proving the incorrectness of a supposition, is no intuitive method. For one cannot have a clear conception of a supposition that eventually proves to be a mistake. Only construction without the use of negation has some sense in intuitionistic mathematics” (1127).]
[ditto]
On philosophic grounds I think the use of the negation in intuitionistic mathematics has to be rejected. Proving that something is not right, i.e. proving the incorrectness of a supposition, is no intuitive method. For one cannot have a clear conception of a supposition that eventually proves to be a mistake. Only construction without the use of negation has some sense in intuitionistic mathematics.
(1127)
[Turning to the Mathematical Formulation of Negationless Intuitionistic Mathematics]
[From this point forward, we no longer consider our philosophical justifications for negationless intuitionistic mathematics and instead we are concerned with the purely mathematical problem of formulating it.]
[ditto]
But I am not going any further into the philosophical side of this question. In intuitionistic mathematics one can consider the construction of negationless intuitionistic mathematics as a pure mathematical problem.
(1127)
[Turning to Examples]
[We begin with some examples.]
[ditto]
To elucidate my intention I’ll first give some examples, which will show what questions arise.
(1127)
[Illustration 1 for Negational vs. Non Negational Mathematical Thinking: Divided Triangle]
[Griss gives an example to show two ways to construct proofs. {1} The first one uses negation: it is a reductio argument, so it negates the conclusion. The premise is that we have a triangle ABC whose CA and CB sides are bisected by a fourth line DE such that it brings about the following proportional relation:
CA : CB = CD : CE
The conclusion is that the bisecting line DE is parallel to line AB. We then negate this conclusion and see what follows logically. We then see how the premises plus the negated conclusion yields a contradiction, which proves that the experimentally negated conclusion is false and thus that the originally proposed conclusion is true. {2} The second proof does not use negation. Here we start with the line which fulfills the proportions. Next we construct another line that we know is parallel to the undivided side. Finally we show that this parallel line must necessarily be identical to the line which fulfills the proportions. Thus lines that fulfill the proportions are parallel to the undivided line.]
[The first example will compare two proofs in geometry, one using negation and one not. (This example seems vaguely similar to Proposition 2 of Book 6 of Euclid’s Elements, but it is not similar enough to help me understand it.) We start with triangle ABC.
We then divide two sides, CA and CB with another line DE such that it creates parts of the sides that are proportional to those sides. (From the way this is worded I would have imagined this means CA is to CD as CB is to CE. The exact wording is, “a line in a triangle cuts off from two sides parts that are proportional to those sides”. But later we see the formulation CA : CB = CD : CE. I think they are equivalent, but the important one is this second one.)
If we do this, then line DE should be parallel with AB. This is what we will prove firstly using negation (through a reductio argument) and secondly without negation. {1} Proof by negation. Since the negational reductio argument will prove that the line which bisects the sides such that it fulfills the proportionality CA : CB = CD : CE is parallel to line AB, then this proof will instead assume that the line which bisects the sides such that it fulfills this proportionality is not parallel. (The reasoning in the next part I do not follow. A consequence of DE not being parallel is that we can draw another line DF such that
CA : CB = CD : CF
But I cannot tell you why this is logically a consequence of DE not being parallel. Are we assuming that the DF line is parallel? And whether it is or is not, how do we know that CA : CB = CD : CF, or at least, what has that to do with DE not being parallel? At any rate, for reasons I do not have:)
If DE was not parallel with AB, we could draw a line, different from DE, that would cut BC in a point F different from E. Then we should have
CA : CB = CD : CF.
Now note that while we have changed E to F, the values for CA, CB, and CD have stayed the same. And since the proportional relation remains intact, that means whatever value they determine for CE in
CA : CB = CD : CE
would have to be the same for CF in
CA : CB = CD : CF
And since CB remains the same length and since CE and CF are the same length and both fall on CB with the same starting point C, that means their other endpoints E and F must fall at the same location. But we previously inferred that point F is different from point E. Thus point F both is and is not point E, which is absurd. Thus it cannot be that DE is not parallel, and therefore it is parallel. {2} Proof without negation. In the proof without negation, it seems we simply construct a parallel line to AB and we show it is the same line as the one made by the proportional relations, thus the bisecting line that fulfills the proportional relations must be parallel. So let us go through the reasoning. We begin with the bisecting line DE, which creates segments with the proportions (as you recall):
CA : CB = CD : CE
We want to know if this line is parallel. We next say, put aside this line for a second, and let us draw another one. We begin at point D, and we draw a line that we know is parallel to line AB. The new point on the other line we will call E′. So now we have this proportional relationship:
CA : CB = CD : CE′
(The reasoning behind this is not clear to me, because we are not stipulating that the line fulfill that proportion but rather that it simply be parallel. So why does that proportion follow from the line being parallel? I do not know. It would seem perhaps that because it is parallel, to whatever degree it portions off the one side it would portion off the other side (so if it halves the one it halves the other, for instance), maybe because they share the same endpoint C. So there is a gap in the reasoning for me here at the moment. At any rate, on account of one or another reason,) the parallel line will bring about the same proportionality. Now, since CA, CB, and CD have not changed lengths, that means the CE′ will be the same length as CE (because in both proportion formulations, the only difference is CE versus CE′, which are determinable by the other three values; and since the other three values are identical, they determine both CE and CE′ as having identical values). With that being the case, E′ will coincide with E (because CE′ and CE have the same lengths, meaning that their endpoints will fall at the same location). Thus line DE coincides with DE′ (as they also have identical endpoints). Since line DE′ is parallel to AB, that means DE is parallel to AB. (What is important here is that instead of making this proof by negating the conclusion, we rather construct the conclusion and show how it follows from the premises. The premise is that there is a line that bisects the sides such that they have that given proportionality. The conclusion is that this line is parallel to the non-bisected line. We then construct a parallel line, and show that it necessarily has that proportionality.)]
1. In elementary geometry there are often given proofs with the aid of the negation, whereas a proof without the use of the negation is simpler, e.g.
If a line in a triangle cuts off from two sides parts that are proportional to those sides, that line is parallel with the third side. Let be given in ∆ABC
CA : CB = CD : CE, then we must prove that DE is parallel with AB.
Proof with negation:
If DE was not parallel with AB, we could draw a line, different from DE, that would cut BC in a point F different from E. Then we should have
CA : CB = CD : CF.
So CE and CF would be equal and E and F would coincide, which is impossible. The supposition, that DE would not be parallel with AB, is absurd, consequently DE ⫽ AB.
Proof without negation:
Draw a line through D parallel with AB, that cuts BC in E′; then
CA : CB = CD : CE′.
So CE = CE′, so that E and E′ coincide. So DE coincides with DE′, whereas DE′ ⫽ AB, so that DE ⫽ AB.
[Comments on Illustration 1]
[In the negationless proof, we will need to define our concepts without negation. Parallel lines cannot be defined as “which do not intersect” (for here the “do not intersect” is a negation of “do intersect”). Rather, they must be defined without a negation, as for instance, “parallel lines are such lines, that any point of one of them differs from any point of the other one.” This formulation then requires a positive definition of “difference relative to points.” One reason that negation is used has to do with the triangle figure requiring an additional, different DE line being drawn even though it is identical to that line. (Maybe Griss is saying that this notion of something being both different and identical is counter-intuitive, so people might prefer the reducio proof instead.) Another reason people use negation has to do with how we formulate our mathematical questions. We might take x2 – 2 and form the question, which rational numbers satisfy x2 – 2 = 0, with the answer being the negative “no rational numbers satisfies it.” But we need not think of x2 – 2 in such a formulation that leads us to a negative conceptualization. We can instead say that “x2 – 2 differs positively from zero for every rational number.” (In other words, its value is not seen as not being a rational number when it is equated with zero but rather that given any rational number for x, its value will be another value that is always different from zero.)]
[ditto]
The second demonstration is really simpler than the first. If, however, one wishes to avoid negation consistently, one must give a positive definition of parallel lines. Instead of saying: parallel lines (in a plane) are lines which do not intersect, one must define: parallel lines are such lines, that any point of one of them differs from any point of the other one. And this, again, presupposes a positive definition (i.e. a definition without negation) of difference relative to points.
If one wants to draw the line DE′ in the second demonstration, one most probably draws a line different from DE, though DE′ and DE prove to coincide. The figure has perhaps been one of the causes of the use of negation.
Another cause is the practice of questioning. One asks for instance: Which rational numbers satisfy x2 – 2 = 0? The answer must be: No rational number satisfies. The question has been put in the wrong way. The fact is that x2 – 2 differs positively from zero for every rational number.
(1128)
[Illustration 2 for Negational vs. Non-Negational Mathematical Thinking: The Necessity of a Solution vs. The Impossibility of No Solution]
[In the second illustration, we wonder if the equation ax + by = 0 has a solution for x and y where x and y are different from zero and the letters represent real numbers? And we will compare the negationless and negative way of answering this question. To do this, we first note the following distinction between real numbers understood as different either positively or negatively: “Two real numbers differ positively, if there can be indicated two approximating intervals which lie outside one another; they differ negatively, if it is impossible that they are equal; you can only divide by a real number if it differs positively from zero.” {1} We begin with the negationless way. We do this first by assuming in one case x has a non-zero value and seeing how that gives a non-zero result for y, secondly we likewise assume that y has a non-zero value and see how that gives a non-zero result for x, and lastly we give both a and b the value zero and see how that yields a non-zero value for both x and y. We conclude from this non-negative approach to the question that “ax + by = 0 has a solution different from zero, if at least one of the coefficients a and b differs from zero or if both are zero.” {2} We next look at how we can use negation to formulate this positive result in a negative way, namely as: “It is impossible that no solution different from zero exists.” We learn this by assuming the only possible solution is zero (“there were no solution different from zero”), which logically yields the contradictory claim that there is such a non-zero solution. Thus it is impossible that there is no solution different from zero. Now, what we learn by comparing the two results is that “The negative formulation is shorter, but distorted, and the details of the positive result are lost. In non-intuitionistic mathematics ax + by = 0 has always a resolution different from zero. In this formulation the positive result has vanished entirely.”] [In the second illustration, we wonder if the equation ax + by = 0 has a solution for x and y where x and y are different from zero and the letters represent real numbers? We next consider an important distinction in intuitionistic mathematics, namely, the difference between real numbers being understood positively as different or negatively as different. “Two real numbers differ positively, if there can be indicated two approximating intervals which lie outside one another; they differ negatively, if it is impossible that they are equal; you can only divide by a real number if it differs positively from zero.” (I do not know the terminology here yet. Let us try to guess through this. “Two real numbers differ positively, if there can be indicated two approximating intervals which lie outside one another”. I do not know yet what an approximating interval is, and I think I cannot even guess it. If two real numbers are different, then for instance their interval to zero will be different. Or maybe the idea has something to do with numbers with non-terminating decimals that need to be approximated. I really cannot say. For now I will just place here the intuition that the “lying outside one another” can be seen somehow as involving different places on a number line. And later I will correct this.) “they differ negatively, if it is impossible that they are equal”. So if you can show that it is impossible for two real numbers to be equal, then of course they are not the same number and must be different numbers. “you can only divide by a real number if it differs positively from zero.” This might be another way, a positive way, of saying that you cannot divide by zero.) We next compare the negationless and negative way for dealing with our above question, namely, does the equation ax + by = 0 has a solution for x and y where x and y are different from zero and the letters represent real numbers? We begin with the negationless way. Here we cannot use the notion of being negatively different, and we can only use positive difference. We first suppose that a is positively different from zero, which means we can divide by a. (In order to get to Griss’ equation, it seems we next subtract by from both sides, and then secondly we divide by a.) This gives us:
x = – b/a y
Next it seems we say that we can determine as one solution that y is equal to one and thus that
x = – b/a
This gives us a solution where the variables are not zero. We next suppose that b is positively different from zero and that x = 1 and so
y = – a/b
This gives us another solution that is different from zero. We next suppose that a and b are both zero. This gives us x = 1 and y = 1, which is yet another solution different from zero.
The result is:
ax + by = 0 has a solution different from zero, if at least one of the coefficients a and b differs from zero or if both are zero.
We next see how we can use negation to formulate this positive result in a negative way: “It is impossible that no solution different from zero exists”. (It seems we arrive upon this by reductio, but I am not sure. The conclusion we are testing for is if the equation has a solution different from zero. For reductio we take the negation of that conclusion, namely we assume that there is no solution different from zero. This implies that any solution would be zero. I do not follow the rest of the reasoning, but let us go through it. Next Griss says that we can conclude that a would need to be zero. His reasoning is that, suppose it is different from zero, then there would be a solution different from zero. By why is that? I have not figured it out yet. As far as I can tell, we assume that x and y are zero, and we are looking for a contradiction. My only guess here is the following. Suppose a is zero. That means you can give any value you want for x, and still get zero. But that is what we want to avoid. The same would go for y. At any rate, somehow we obtain the contradiction that there is no non-zero solution and yet we can infer that in fact there is a non-zero solution. That tells us “it is impossible that no solution different from zero exists.” Then Griss gives us an assessment of the comparison of these two approaches to the question. “The negative formulation is shorter, but distorted, and the details of the positive result are lost. | In non-intuitionistic mathematics ax + by = 0 has always a resolution different from zero. In this formulation the positive result has vanished entirely.” (It seems the idea here is the following. The positive formulation has information that is lacking in the negative formulation. It tells us that the equation will always have a resolution different from zero. But the negative formulation does not have that information. It simply says it is impossible for their to be no solution different from zero. But how is that not the same claim? I can only think that it leaves open the possibility that there is no solution at all, but the positive formulation says there are solutions. Otherwise maybe Griss is saying that the negative formulation is lacking the additional information in the conditional part of this formulation, “ax + by = 0 has a solution different from zero, if at least one of the coefficients a and b differs from zero or if both are zero.” I am not sure. Please consult the quotation below.)]
2. Has the equation ax + by = 0 a solution for x and y, different from zero. i.e. a solution with at least x or y different from zero? The letters represent real numbers.
In intuitionistic mathematics they make a distinction between positively and negatively different with regard to real numbers. Two real numbers differ positively, if there can be indicated two approximating intervals which lie outside one another; they differ negatively, if it is impossible that they are equal; you can only divide by a real number if it differs positively from zero. In negationless mathematics the idea negatively different is, of course, omitted. Therefore we mean henceforth by different positively different.
If now in the equation ax + by = 0 a is different from zero, we can divide by a and find
x = – b/a y
, so among others
x = – b/a
and y = 1 is a solution different from zero. If b is different from zero x = 1 and
y = – a/b
is a solution different from zero. If a and b are both zero, then x = 1 and y = 1 is a solution different from zero. The result is:
ax + by = 0 has a solution different from zero, if at least one of the coefficients a and b differs from zero or if both are zero.
By using the negation this positive result can be formulated negatively: It is impossible that no solution different from zero exists. If namely there were no solution different from zero, a would be zero, for if a differs from zero, there would be a solution different from zero; likewise b would be zero; but then a and b were both zero, so there would still be a solution different from zero. This is impossible according to our supposition. So it is impossible that no solution different from zero exists.
The negative formulation is shorter, but distorted, and the details of the positive result are lost. | In non-intuitionistic mathematics ax + by = 0 has always a resolution different from zero. In this formulation the positive result has vanished entirely.
(1128-1129)
[Illustration 3 for Negational vs. Non-Negational Mathematical Thinking: Equality as Shared Difference to All Others vs. Equality as Lacking Difference between One Another]
[The third illustration is: “If ax + b ≠ 0 for each value of x, then a = 0.” In our exploration of the proof for it, we make use either of a positive definition for equality or a negative one. {1} The positive definition of equality: “If a differs from c for each value c that differs from b, then a = b.” (In other words, two values are equal if they are both different from all other values.) {2} The negative definition of equality: If a does not differ from b, then a = b. (In the first case, the two equal things share all the same differences to other things. In the second case, they simply are not different to each other). From this Griss concludes that “The positive proposition has to be proved for the different sorts of numbers, to begin with the natural numbers. But therefore again it proves to be necessary to construct the whole of negationless intuitionistic mathematics from the beginning.”]
[The third illustration is: “If ax + b ≠ 0 for each value of x, then a = 0.” {1} The proof without negation. (See the text below, as I cannot summarize the reasoning here. Let us go through it. We want to prove “If ax + b ≠ 0 for each value of x, then a = 0.” We begin with c ≠ 0, which means that c is different than zero. We next want to set x1 such that cx1 + b = 0. So we are saying that that we have a c value which is not zero that when multiplied by some x1 value, we get zero. We next know that ax1 + b ≠ 0, but I am not sure why. I suppose it is because we begin with the claim that ax + b ≠ 0, and the x1 here is one such x. At any rate, we next infer from ax1 + b ≠ 0 that (a – c) x1 ≠ 0. I do not know why. From this we can infer that a ≠ c. I also do not know the reasoning for that, either. But we can at least note that if a = c, then subtracting the one from the other would give us 0, which when multiplied by any number would give us 0. But we are saying that it cannot give us zero, so a and c cannot be the same (I realize this however is a negative reductio mode of reasoning. At any rate,) we conclude from this that “So a ≠ c for each value c ≠ 0, so a = 0.” We have this c value which is not zero, but it takes the place of a in the equation, but we also say that it brings about the opposite outcome, namely that it results ultimately in a zero value for the equation. Thus we know that a cannot equal c whenever c is not zero. Or put another way, whatever is in a’s place in the equation cannot be a non-zero value (for then it would give us the opposite conclusion. I am sorry for the negative reasoning here, but it is all I have at the moment.) Thus a must equal 0 in order for ax + b ≠ 0 to hold.) The important point here is that in this proof we use the proposition: If a differs from c for each value c that differs from b, then a = b. (In our example, each value of c differed from 0, and so 0 is the b of this formulation. Since a differs from every value of c and since every value of c differs from b (which is 0 here), that means a equals b (that is, a equals 0 in our example). (In other words, a first thing differs from a second thing if both differ from all the same things.) {2} The negative proof. We do not seem to get the full proof, but we learn that the negative formulation of the above idea is that: If a does not differ from b, then a = b. (In the first case, the two equal things share all the same differences to other things. In the second case, they simply are not different to each other). From this Griss concludes that “The positive proposition has to be proved for the different sorts of numbers, to begin with the natural numbers. But therefore again it proves to be necessary to construct the whole of negationless intuitionistic mathematics from the beginning.”)]
3. If ax + b ≠ 0 for each value of x, then a = 0.
Proof without negation: Take c ≠ 0 (this means c is different from zero) and determine x1, in such a way that cx1 + b = 0.
As ax1 + b ≠ 0, it follows that (a – c) x1 ≠ 0, which gives a ≠ c. So a ≠ c for each value c ≠ 0, so a = 0.
In this proof we used among others the proposition: If a differs from c for each value c that differs from b, then a = b.
This proposition replaces the negative proposition: If a does not differ from b, then a = b.
The positive proposition has to be proved for the different sorts of numbers, to begin with the natural numbers. But therefore again it proves to be necessary to construct the whole of negationless intuitionistic mathematics from the beginning.
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[Illustration 4 for Negational vs. Non-Negational Mathematical Thinking: Congruent Triangles]
[We can compare a positive formulation, “Two triangles are congruent, if they have equal one side, the angle opposite that side and the sum of the two other sides, while of one of the adjacent angles is known that they are either equal or different” with a negative formulation, “If two triangles have equal one side, the angle opposite that side and the sum of the two other sides, it is impossible, that they are not congruent.”]
[(I will not be able to summarize this one, so just see the quotation below. We might simply note the following. We can compare a positive formulation, “Two triangles are congruent, if they have equal one side, the angle opposite that side and the sum of the two other sides, while of one of the adjacent angles is known that they are either equal or different” with a negative formulation, “If two triangles have equal one side, the angle opposite that side and the sum of the two other sides, it is impossible, that they are not congruent.”]
4. We prove first:
Two triangles are congruent, if they have equal two pair of sides and the angles opposite the first pair, while the sum of the angles opposite the other pair differs from 180°.
Proof: Let be given in △ABC and △A′B′C′ : AB = A′B′, BC = B′C′, ∠C = ∠C′ and ∠A + ∠A′ ≠ 180°.
C=LC' and LA+LA':f180o. According to the sinus-rule sin A = sin A′, so that
2 sin 1/2 (A – A′) cos 1/2 (A – A′) = O.
The latter factor differs from zero, so sin 1/2 (A – A′) = o, so that ∠A + ∠A′. So △ABC ≅ △A′B′C′.
I remark. that by using the negation ∠A + ∠A′ ≠ 180° can be replaced by∠A + ∠A′ is not equal to zero. However, no example is known where the second formulation could be applied but not the first, for no real number is known, of which is shown, that it is not possibly equal to zero, whereas it is not shown, that it differs (positively) from zero. It seems as if in intuitionistic mathematics the only way to show that two real numbers are not equal consists in showing that the numbers are (positively) different.
Two triangles are congruent, if they have equal one side, the angle opposite that side and the sum of the two other sides, while of one of the adjacent angles is known that they are either equal or different.
We need only prove the case in which the adjacent angles differ.
Let be given in △ABC and △A′B′C′ : AC = A′C′ ; ∠B′ = ∠B′ AB + BC = A′B′ + B′C′, while A ≠ C′.
Produce AB with BD = BC and A′B′ with B′D′ = B′C′, then △ABC = △A′B′C′ = (∠C + 1/2 ∠B) + (∠C′ +1/2 ∠B′) = 180°. From △ACD ≅ △A′C′D′ it follows immediately△ABC ≅ △A′B′C′.
Bij [sic] using negations one finds:
If two triangles have equal one side, the angle opposite that side and the sum of the two other sides, it is impossible, that they are not congruent.
Compare this with the second example.
[Summary of the Results and the Need for a Positive Definition of Difference]
[Griss now summarizes the results of these illustrations. {1} From example 1 (see section 0.5) we learn that “In some cases it is simpler to avoid the use of the negation.” {2} From example 2 (see section 0.7) we learn that “Positive properties can sometimes be formulated more briefly in a negative way, but details get lost.” {3} From example 4 (see section 0.9) we learn that “The parts of intuitionistic mathematics which in a positive construction are disposed of are less important, for probably examples cannot be constructed for which a negative property could be applied and a corresponding positive property could not.” {4} From examples 1 and 3 (see section 0.5 and section 0.8), we learn that “To construct negationless mathematics one must begin with the elements and a positive definition of difference must be given instead of a negative one.” Moreover, “But even from a general intuitionistic point of view a positive construction of the theory of natural numbers must be given: one cannot define 2 is not equal to 1 (i.e. it is impossible that 2 and 1 are equal), for from this one could never conclude that 2 and 1 differ positively. Conversely one could define in a positive way negation by means of difference, e.g. not equal means different, etc., but, for the present, this seems unfit.” (Perhaps then we might note the following. We need numbers in our negationless mathematics. But to get those numbers, we need more than just inequality (the impossibility of being equal) to tell us that each number is different from the others. We rather need a positive construction of the numbers that does not involve the impossibility of equaling. In section 1 to follow, we learn that there is a notion of distinguishability that grounds inequality.)]
[ditto]
We recapitulate the results we have obtained with these examples. In some cases it is simpler to avoid the use of the negation (ex. 1). Positive properties can sometimes be formulated more briefly in a negative way, but details get lost (ex. 2). The parts of intuitionistic mathematics which in a positive construction are disposed of are less important, for probably examples cannot be constructed for which a negative property could be applied and a corresponding positive property could not (ex. 4). To construct negationless mathematics one must begin with the elements and a positive definition of difference must be given instead of a negative one (ex. 1 and 3).
But even from a general intuitionistic point of view a positive construction of the theory of natural numbers must be given: one cannot define 2 is not equal to 1 (i.e. it is impossible that 2 and 1 are equal), for from this one could never conclude that 2 and 1 differ positively. Conversely one could define in a positive way negation by means of difference, e.g. not equal means different, etc., but, for the present, this seems unfit.
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[Final Observations]
[Griss lastly has us “consider the property: If a and b are elements of the set of natural numbers, and if a ≠ b, then a < b or a > b for each element a of the set. If we apply this property to b = 1, we get: For each element a ≠ 1 of the set of natural numbers we have a < 1 or a > 1. a < 1, however, has not any sense in negationless mathematics. If we say: a < b or a > b for each a of a set, we mean 1) that for each a at least one of these conditions is fulfilled, 2) that conversively at least one element fulfils the condition a < b and another one the condition a > b.” We then note that “Negationless intuitionistic logic will differ much from the usual intuitionistic logic by the absence of the negation and the altered meaning of the disjunction” and also that “‘Affirmative’ mathematics is something quite different from the negationless intuitionistic mathematics.”]
[ditto]
We finally consider the property: If a and b are elements of the set of natural numbers, and if a ≠ b, then a < b or a > b for each element a of the set. If we apply this property to b = 1, we get: For each element a ≠ 1 of the set of natural numbers we have a < 1 or a > 1. a < 1, however, has not any sense in negationless mathematics. If we say: a < b or a > b for each a of a set, we mean 1) that for each a at least one of these conditions is fulfilled, 2) that conversively at least one element fulfils the condition a < b and another one the condition a > b. Properties which do not hold for any element do not occur. The combination of two properties (a natural number is even, a natural number is odd) need not be a property. Suppositions, of which it is not certain that they can be realized by a mathematic system, are not made. In axiomatic mathematics we must also stick to this.
Negationless intuitionistic logic will differ much from the usual intuitionistic logic 2) by the absence of the negation and the altered meaning of the disjunction. Yet I’ll not begin by giving such a system: intuitionistic mathematics has also been studied before a formal system was given.
“Affirmative” mathematics 3) is something quite different from the negationless intuitionistic mathematics I’ll treat of.
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2) A. HEYTING, Die formalen Regeln der intuitionistischen Logik. Preuss. Akad. der Wissenschaften 1930.
3) D. VAN DANTZIG, On the principles of intuitionistic mathematics. Rev. hispan. 1941?
Griss, G.F.C. (1946). “Negationless Intuitionistic Mathematics, I,’’ Proceedings of the Koninklijke Nederlandse Akademie van Wetenschappen, 49, 1127–1133.
Journal PDF here:
http://www.dwc.knaw.nl/DL/publications/PU00014659.pdf
Article PDF here:
http://www.dwc.knaw.nl/DL/publications/PU00018278.pdf
Listing of Griss at this journal:
http://www.dwc.knaw.nl/toegangen/digital-library-knaw/?pagetype=publist&search_author=PE00000531
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