When Is the Product of Intervals Also an Interval?

When is the product of intervals also an interval?

Olga Kosheleva1 and Piet G. Vroegindeweij2

1Department of Electrical and Computer Engineering University of Texas at El Paso El Paso, TX 79968, USA email [email protected]

2Koenestraat 10 NL 3958 XG Amerongen, the Netherlands

Abstract Interval arithmetic is based on the fact that for intervals on the real line, the element-wise product of two intervals is also an interval. This property is not always true: e.g., it is not true if we consider intervals on the set of integers instead of intervals on the set of real numbers. When is an element-wise product or a sum of two intervals always an interval? In this paper, we analyze this problem in a general algebraic setting: we need the corresponding algebraic structures to have (related) addition, multiplication, and order; thus, we consider (consistently) ordered rings. We describe all consistently ordered rings for which the element-wise product and sum of two intervals are always intervals.

1 Introduction

Intervals are important. Measurements are never 100% precise. As a result, if the measurement result is, say, 100, this does not mean that the actual value is exactly 100, it rather means that the actual value must be within an interval [100 − ∆, 100 + ∆], where ∆ is an upper bound on the measurement error (this upper bound is usually supplied by the manufacturer of the measuring instrument). For example, if ∆ = 1, then the actual value can be equal to any number within the interval [99, 101]. Similarly, if we, say, measure the total charge to determine the total amount of electrons in a given beam, and the measurements are not accurate enough, we do not get the exact number of electrons, but only the interval of possible values, such as [99, 101], meaning that there can be either 99, or 100, or 101 electrons (i.e., that [99, 101] = {99, 100, 101}).

1 In both cases, after the measurement, we get not the exact value of the measured quantity, but only the interval of its possible values:

a = [a, a] = {a ∈ X | a ≤ a ≤ a}. (1)

In the ﬁrst example, the set X of possible actual values of the measured quantity is a set R of all real numbers; in the second example, X coincides with the set of all integers. Intervals on the real line: Traditional interval arithmetic. Often, we need to process the measured data. If the measured quantities are real numbers, then the measurement results can be expressed by intervals on the real line. To process such “real-interval” data, we can use interval arithmetic originally proposed by R. E. Moore (for recent surveys, see, e.g., [2, 3]). In general, in data processing, we are interested in the quantity b = f(a1, . . . , an) that is related, in a known way, with directly measured quantities a1, . . . , an. To describe possible values of a, we measure the quantities a1, . . . , an, and use the resulting intervals a1 = [a1, a1],..., an = [an, an] of possible values of ai to describe the set B of possible values of b as

B = {b | b = f(a1, . . . , an) for some a1 ∈ [a1, a1], . . . , an ∈ [an, an]}. (2)

The resulting set B is called the range of the function f(a1, . . . , an) on the intervals a1,..., an, or the result of applying the function f(a1, . . . , an) to the intervals a1,..., an. In particular, if the data processing operation f(a1, . . . , an) consists of simply applying an arithmetic operation + or ∗, i.e., if f(a1, a2) = a1 + a2 or f(a1 ∗ a2) = a1 ∗ a2, then we get the element-wise deﬁnitions of the sum and of the product of two intervals:

[a1, a1]+[a2, a2] = {b | b = a1 +a2 for some a1 ∈ [a1, a1] and a2 ∈ [a2, a2]}; (3)

[a1, a1] ∗ [a2, a2] = {b | b = a1 ∗ a2 for some a1 ∈ [a1, a1] and a2 ∈ [a2, a2]}. (4) For intervals on the real line, thus defined element-wise sum and product are also intervals (and the formulas for computing the endpoints of these intervals form the basis of interval arithmetic). Intervals on the set of integers are more difficult to handle than intervals on the real line. For intervals on the set of all integers, we can also define element-wise addition and multiplication by using formulas (3) and (4), but these integer-intervals are more computationally difficult to handle because the product of two integer-intervals is not always an interval: e.g., the element- wise product of two integer-intervals [1, 2] = {1, 2} and [1, 3] = {1, 2, 3} is equal to [1, 2] ∗ [1, 3] = {1, 2, 3, 4, 6}; this product is not an integer-interval.

2 A natural question. A natural question is:

How can we describe all algebraic structures for which the product of two intervals is always an interval? To formulate this problem in precise terms, we need to have an algebraic structure with addition, multiplication, and order that are (in some reasonable way) consistent. In the next section, we show that natural consistency demands lead to a notion of a consistently ordered ring. Thus, in algebraic terms, we get the following problem:

Given an arbitrary consistently ordered ring K, we can deﬁne another structure of interval objects of the form {a ∈ K | a ≤ a ≤ a}, with addition and multiplication deﬁned element-wise. We want to describe all the rings for which this new interval structure is closed under addition and multiplication.

2 Motivation for the Following Deﬁnitions

We need two consistent operations and one relation. We want to formulate our problem in general algebraic terms. Since our question is about addition and multiplication of intervals, this structure must contain addition, multiplication, and the order (so that we will be able to define intervals); we must also require that these operations be consistent with each other. Addition. First, we need an operation called addition. We will assume that addition + is commutative and associative, that there exists an element 0 for which 0 + a = a for all a, and that for every element a, there is an inverse element −a for which a + (−a) = 0. In other words, we will assume that our structure K is an Abelian (commutative) group under addition. Multiplication. Second, we need an operation called multiplication. It is natural to require that multiplication is associative and distributive with respect to addition (i.e., (a + b) ∗ c = a ∗ c + b ∗ c and a ∗ (b + c) = a ∗ c + b ∗ c). These properties define a ring. So, we will consider rings. Comment. Multiplication of real numbers has an additional property: it is commutative (a∗b = b∗a). However, we do not want to impose this restriction on the general structure, because there are meaningful computation-related examples of non-commutative ordered rings: e.g., a ring of all n × n real-valued square matrices over A (with A ≥ 0 iff A is non-negative definite) is non-commutative.

Order. In order to deﬁne intervals on a ring, we must have an order. The above matrix example shows that the order need not be total (linear); therefore, in this paper, we will consider the general case of a (partial) order. This order

3 must be consistent with + and ∗. There are several reasonable requirements that express this consistency: • First, there must exist a positive element a > 0. • Second, the order must be shift-invariant: if a < b, then for every c, we must have a + c < b + c. • Third, the sum and the product of two positive numbers must be positive. • Fourth, if a ∗ b is positive and a is positive, then b must be positive too, and if both a ∗ b and b are positive, then a must be positive. There is a fifth natural property that does not need to be specifically postulated for linear orders, but which is needed for partial orders: the product of two non- zero numbers must be non-zero. (If we have a linear order, then every non-zero number is either positive or negative, and this property simply follows from the third one.) These five properties are true for all three examples that we had considered so far: for the ring of all real numbers, for the ring of all integers (with natural order in both examples), and for the ring of all n × n square matrices. The notions described by these properties are well known in algebra: an object that satisfies the first three properties is called an ordered ring; an ordered ring that satisfies the fourth property as well is called a consistently ordered ring, and a ring that satisfies the additional fifth property is called a ring with no divisors of zero. In these terms, in the following text, we will consider consistently ordered rings with no divisors of 0. We are now ready to introduce formal definitions:

3 Definitions and the Main Result 3.1 Definitions of the algebraic structure: consistently ordered rings, intervals, operations with intervals Definition 1. (see, e.g., [4]) • A ring K is a set of elements with two binary operations + and ∗ (called addition and multiplication) that satisfies the following three properties: – K is an Abelian (commutative) group under addition; – multiplication is associative; – the right- and left-distributive laws hold, i.e., a ∗ (b + c) = a ∗ b + a ∗ c and (a + b) ∗ c = a ∗ c + b ∗ c. • We say that a ring has no (proper) divisors of zero if from a ∗ b = 0, it follows that a = 0 or b = 0.

4 Deﬁnition 2. (see, e.g., [1, 8, 10])

• By an ordered ring, we mean a ring K with a partial order < such that:

– there exists an element a for which a > 0; – if a < b, then for every c ∈ K, we have a + c < b + c; – if a > 0 and b > 0, then a ∗ b > 0;

• An order is called consistent if the following two properties hold: – if a > 0 and a ∗ b > 0 then b > 0; – if b > 0 and a ∗ b > 0 then a > 0. • An order is called linear or total if for every two elements a and b, either a < b, or a = b, or a > b. Comment. In the following text, we will only consider consistently ordered rings with no divisors of zero. Deﬁnition 3. Let K be an ordered ring with no divisors of 0. By an interval, we mean a set [a, a] = {a ∈ K | a ≤ a ≤ a}, where a ≤ a. The element a is called the lower endpoint of this interval, and the element a is called its upper endpoint. The sum and product of two intervals are deﬁned element-wise (i.e., by formulas (3) and (4)).

3.2 Ring properties that are “equivalent” to interval arithmetic Comment. We want to reformulate the desired property — that the element- wise sum and element-wise product of two intervals are always intervals — in purely algebraic terms. This property is true for the ring of real numbers and it is not true for the ring of integers. From the algebraic viewpoint, the main diﬀerence between these two rings is as follows: • in the ring of real numbers, multiplication is invertible in the sense that for every two real numbers x 6= 0 and a, there exists a number q (q = a/x) for which x ∗ q = a;

• on the other hand, multiplication in the ring of all integers is not invertible: for example, 2 6= 0, but there exists no integer q for which 2 ∗ q = 3. In view of this diﬀerence, it is natural to conjecture that the desired interval property is equivalent to invertibility of multiplication. This ﬁrst guess turns out to be wrong: there are examples of ordered rings with the above interval property in which multiplication is not invertible. These example enable to

5 correct the original conjecture into an exact theorem, according to which the interval property is equivalent to a special property called almost invertibility. To introduce this new notion, let us first recall the formal definition of invertibility: Definition 4. Let K be a ring. We say that multiplication is invertible if for every two elements a and x 6= 0, there exist:

• an element ql for which ql ∗ x = a; and

• an element qr for which x ∗ qr = a. Comments. • If a ring has a unit element e (i.e., with an element for which e∗a = a∗e = a for all a), then a ring is invertible iﬀ it is a ﬁeld (commutative or non- commutative)1.

• Not every ring is invertible (for example, as we have already mentioned, the ring of all integers is not invertible). There can be two reasons for that:

– In some cases, multiplication is not invertible in the original ring, but we can still easily add intermediate elements and make it invertible. For example, in the ring of integers, we do not have a number q for which 2 ∗ q = 3, but we have 2 ∗ 1 < 3 and 2 ∗ 2 > 3, so, we can add an extra element 3/2 for which 1 < 3/2 < 2. If we add all such elements, we will thus expand the ring of integers to the field of all rational numbers. – It can also be that an element x is “infinitely smaller” than a in the sense that whatever element q we take, we always have x ∗ q < a. Rings in which this second reason is the only obstacle to invertibility will be called almost invertible: Definition 5. Let K be an ordered ring.

• We say that an element x 6= 0 is left-inﬁnitely smaller than an element a (and denote it by x ¿l a), if y ∗ x < a for all y ∈ K. • We say that an element x 6= 0 is right-inﬁnitely smaller than an element a (and denote it by x ¿r a), if x ∗ y < a for all y ∈ K.

1In some algebra textbooks, a field is defined as a commutative ring with a property that non-zero elements form a group. In such books, a non-commutative ring with this property is called a skew field, or a s-field.

6 Deﬁnition 6. Let K be an ordered ring. We say that multiplication is almost invertible (and, respectively, that the ring K is almost invertible) if for every a and x 6= 0, the following two properties hold:

• if x is not left-inﬁnitely smaller than a, then there exists an element q for which q ∗ x = a.

• if x is not right-inﬁnitely smaller than a, then there exists an element q for which x ∗ q = a.

Example. To illustrate this notion, let us give a computationally meaningful example of an ordered ring that is almost invertible, but not invertible: the ring 2 n K of all formal infinite series f(x) = a0 + a1 · x + a2 · x + ... + an · x + ... with real coefficients, with normal algebraic sum and product and the following 2 lexicographic order: if f(x) = a0 + a1 · x + a2 · x + ... and g(x) = b0 + b1 · x + 2 b2 · x + ..., then f < g if and only if either a0 < b0, or a0 = b0 and a1 < b1, or a0 = b0, a1 = b1, and a2 < b2, etc. One can easily check that this is a consistently ordered commutative ring with no divisors of zero. In particular, in this order, f(x) = x < g(x) = 1, because f(x) = 0 + 1 · x, g(x) = 1+0·x, and a0 = 0 < b0 = 1. Similarly, for every series h(x), the product x · h(x) starts with a 0 term x · h(x) = 0 + ..., and therefore, x · h(x) < 1. Due to commutativity, we also have h(x) · x < 1 for all series h(x). Thus, in terms of our definition, this means that x is both left- and right-infinitely smaller than 1 (x ¿l 1 and x ¿r 1). This ring is not invertible, because x · h(x) < 1 and therefore, there is no inverse element for series q = x and a = 1. One can see, however, that this ring is almost invertible: indeed, for elements of this ring, f(x) ¿l g(x) means that n the lowest non-zero element an · x in the expansion of f(x) is of larger degree m than the first non-zero element bm · x in the expansion of g(x)(n > m). If n ≤ m, then we can formally divide two series and get the desired element q for which q ∗ x = x ∗ q = a (see, e.g., [4]). Now, we are ready to formulate our main result: Theorem. • For every consistently ordered ring K with no divisors of 0, the following two conditions are equivalent to each other:

i) The product of every two intervals is also an interval. ii) The ring K is totally (linearly) ordered and almost invertible. • If one of these conditions is satisﬁed, then the sum of every two intervals is also an interval.

7 Comments.

• All three above examples are well consistent with this theorem:

– For real numbers, multiplication is invertible (and therefore, almost invertible), and the product of two intervals is indeed an interval. – For integers, multiplication is not invertible and not even almost invertible and indeed, the product of two integer-intervals is not nec- essarily an interval. – For matrices, the order is not linear, and indeed, the product of matrix-intervals is not always a matrix-interval. • For reader’s convenience, all the proofs are placed in a special (last) section. • The proof of the Theorem is somewhat similar to the proofs from [9].

4 Corollaries of the Main Result

For the rings in which there are no infinitely smaller elements, this result can be further simplified: Definition 7. We say that a ring is weakly Archimedean if none of its elements is left- or right-infinitely smaller than any other. Comments.

• Many rings are weakly Archimedean, including the ring of all real numbers, the ring of all integers, and many others.

• A weakly Archimedean ring is almost invertible if and only if it is invertible. Hence, we get the following corollary: Corollary 1. • For every weakly Archimedean consistently ordered ring K with no divisors of 0, the following two conditions are equivalent to each other:

i) The product of every two intervals is also an interval. ii) The ring K is linearly ordered and invertible.

• If one of these conditions is satisﬁed, then the sum of every two intervals is also an interval.

We have already mentioned that a ring with a unit element is invertible if and only if it is a ﬁeld. Hence, we get the following second corollary:

8 Corollary 2.

• For every weakly Archimedean consistently ordered ring K with a unit element e and with no divisors of 0, the following two conditions are equivalent to each other:

i) The product of every two intervals is also an interval. ii) The ordered ring K is a linearly ordered ﬁeld.

• If one of these conditions is satisfied, then the sum of every two intervals is also an interval. Comments. In short, this result says that for interval arithmetic to be possible in a ring, this ring has to be a field. It is worth mentioning that, somewhat in contrast to this result, the intervals themselves do not form a field; for example, the set of intervals does not even have a distributivity property (it was noticed already by Moore [7]) with respect to element-wise addition and multiplication. However, intervals can be extended to a structure that satisfies several important algebraic properties of a field; see, e.g., Markov [5, 6].

5 Auxiliary Problem

We have found the conditions under which both the product and the sum of the two intervals form an interval. It may be interesting to consider the case when only addition is defined, and find out when, in this case, the sum of any two intervals is also an interval. Definition 8. By an ordered (Abelian) group, we mean an Abelian group G with a partial order < such that: • there exists an element a for which a > 0; • if a < b, then for every c ∈ K, we have a + c < b + c. To formulate our result, we will need the following property: Definition 9. We say that an order is 2-2-separating if for every four elements a, b, a, b, if each of the two elements a and b is smaller than or equal to each of the elements a, b (i.e., if a ≤ a, a ≤ b, b ≤ a, and b ≤ b), then there exist an element c that lies in between, i.e., for which a ≤ c ≤ a and b ≤ c ≤ b. Comments. • This definition is close to the notion of TR(2,2) (Riesz) ordered groups. • Every group in which an order forms a lattice has this property.

9 • If a group G is a compact topological group in which the order is (in some reasonable sense) consistent with topology, then this property is equivalent to G being a lattice. Proposition. For every ordered Abelian group G, the following two conditions are equivalent to each other: i) The sum of every two intervals is also an interval. ii) The order on the group G is 2-2-separating.

6 Proofs 6.1 Proof of the Theorem Proof of i)→ii). Let us first prove that from i), it follows that the ring K is linearly ordered. Since addition forms an Abelian group, for every q ∈ K, there exists an element −q for which q + (−q) = 0. As usual, we will denote p + (−q) by p − q. Linearly ordered means that for every p and q, either p < q, or p > q, or p = q. By definition of an ordered ring, these conditions are equivalent, correspondingly, to c < 0 (where we denoted c = p − q), c > 0, and c = 0. Therefore, to prove that the ring K is linearly ordered, it is sufficient to show that for every element c ∈ K, either c < 0, or c > 0, or c = 0. To prove it, let us take an arbitrary element c ∈ K, c 6= 0, and show that either c > 0, or c < 0. According to the definition of an ordered ring, there exist an element a > 0. Due to a > 0, the set [0, a] is an interval. Now, the set [c, c] is also an interval (that consists of a single element c). According to i), the product [0, a] ∗ [c, c] of these two intervals is also an interval. Let us denote the lower and the upper endpoints of this interval by l and u. By definition of the product of two intervals, the interval [l, u] = [0, a] ∗ [c, c] contains an element 0 ∗ c = 0. Therefore, by definition of an interval, l ≤ 0 ≤ u. The elements l and u cannot be both equal to 0. Indeed, in this case, the entire interval [l, u] = [0, a] ∗ [c, c] would consist of the only element 0, while in reality, it contains an element a ∗ c which is non-zero because a 6= 0, c 6= 0, and K has no proper divisors of 0. So, either l, or u are different from 0. Without losing generality, we can assume that u 6= 0, i.e., u > 0. The element u belongs to the interval [l, u] = [0, a] ∗ [c, c]. Be definition of a product of two intervals, it means that there exist elements a0 ∈ [0, a] and c0 ∈ [c, c] for which u = a0 ∗ c0. From c0 ∈ [c, c], it follows that c0 = c, so, a0 ∗ c = u for some a0 ∈ [0, a]. By definition of an interval, from the fact that a0 ∈ [0, a], it follows that a0 ≥ 0. We cannot have a0 = 0, because then, we would have a0 ∗ c = 0 6= c. Therefore, a0 > 0. So, a0 ∗ c = u > 0 and a0 > 0. By definition of a consistently ordered ring, we can now conclude that c > 0. Similarly, if l 6= 0, we conclude that c < 0.

10 Let us now prove that the ring K is almost invertible. Indeed, let x 6= 0 and a are two elements of this ring, for which x is not right-inﬁnitely smaller than a, i.e., for which x ∗ y ≥ a for some y. Let us show that there exists an element q for which x ∗ q = a. Without losing generality, we can assume that x > 0 and a > 0 (otherwise, we could take −x and −a respectively). Then, from x ∗ y ≥ a > 0, we conclude that x∗y > 0 and hence, taking into consideration that x > 0, we conclude that y > 0. So, the set [0, y] is an interval, and therefore, due to i), the product of the two intervals [x, x] and [0, y] is also an interval. This interval contains the points x ∗ 0 = 0 and x ∗ y ≥ a and must, therefore, contain all intermediate points, including the point a. Thence, there exists a q ∈ [0, y] for which x ∗ q = a. A similar result can be proven for a pair in which x is not left-inﬁnitely smaller than a. The property ii) is proven. Proof of ii)→i). We will show if ii) is true, then that the product of two intervals [a, a] and [b, b] is an interval with the endpoints determined by the formulas of standard interval arithmetic [l, u], where

l = min(a ∗ b, a ∗ b, a ∗ b, a ∗ b);

u = max(a ∗ b, a ∗ b, a ∗ b, a ∗ b). (Minimum and maximum exist because the ring is linearly ordered.) By enumerating all possible cases of signs, it is easy to show that if a ≤ a ≤ a and b ≤ b ≤ b, then l ≤ a ∗ b ≤ u. To complete the proof, we must show that the inverse is also true, i.e., that every element c ∈ [l, u] can be represented as product c = a∗b for some a ∈ [a, a] and b ∈ [b, b]. It is suﬃcient to prove this result for the case when both intervals are non- negative; indeed: • If one of the intervals is non-positive, we can reduce this case to the non- negative one by simply changing a sign (or signs). • If one of the intervals contains 0, e.g., if a < 0 < a, then, since the ordering is total, we can represent this interval as a union [a, 0] ∪ [0, a] of a non- positive interval [a, 0] and a non-negative interval [0, a]. Then, the desired result follows from the similar results for the corresponding sub-intervals. So, let us assume that 0 ≤ a < a and 0 ≤ b < b. To describe the value of a for a general ring, let us ﬁrst describe what happens in the case when K is the set of all real numbers. In this case, we must satisfy the inequalities a ≤ a ≤ a and b ≤ c/a ≤ b. For real numbers, we can use division /, and therefore, the second set of inequalities is equivalent to c/b ≤ a ≤ c/b. Thus, the requirements on a can be formulated as follows: max(a, c/b) ≤ a ≤ min(a, c/b). In particular, we can take a = max(a, c/b). Let us show that a similar expression can help in the general case.

11 In the general case, l = a ∗ b and u = a ∗ b. From the fact that c ∈ [l, u] it follows that c ≤ u = a∗b and therefore, that b is not left-inﬁnitely small relative to c. Thus, since the ring is almost invertible, we can conclude that there exists an elementa ˜ such thata ˜ ∗b = c. From c =a ˜ ∗b ≤ a∗b it follows thata ˜ ≤ a. We then take a = max(a, a˜). To complete the proof, we will consider the following two cases:

• If a ≤ a˜, this means that a ≤ a ≤ a. In this case, c = a ∗ b for b = b.

• If a > a˜, then, multiplying both sides of this inequality by b, we conclude that a ∗ b > a˜ ∗ b = c. Thus, a is not right-infinitely small relative to c and therefore, there exists a b such that a ∗ b = c. From a ∗ b = c < a ∗ b, it follows that b < b. Similarly, from the fact that a ∗ b = c ≥ l = a ∗ b, it follows that b ≥ b. So, c = a ∗ b for some b ∈ [b, b]. In both cases, we have c = a ∗ b for some a ∈ [a, a] and b ∈ [b, b]. The result is proven. Proof that the sum of two intervals is also an interval. It is easy to see that if a ≤ a ≤ a and b ≤ b ≤ b, then a + b ≤ a + b ≤ a + b, i.e., that [a, a] + [b, b] ⊆ [a + b, a + b]. So, to prove this part, it is sufficient to show that every element c ∈ [a, a, b + b] can be represented as a sum of some elements a ∈ [a, a] and b ∈ [b, b]. The requirements on a are: a ≤ a ≤ a from the first condition, and b ≤ c − a ≤ b, i.e., c − b ≤ a ≤ c − b from the second condition. Hence, we can, e.g., take a = max(a, c − b) and b = c − a. The theorem is proven. 2.

6.2 Proof of the Proposition Proof of i)→ii). Let a, a, b, b be as in Definition 9. Then, we can form two intervals [a, a] and [−b, −b]. Due to i), the sum of these two intervals is also an interval. The smallest element in this sum is the sum of the smallest elements in each interval, i.e., of a and −b. Similarly, we can find the largest possible element in the sum. Therefore, [a, a] + [−b, −b] = [a − b, a − b]. Since a ≤ b and b ≤ a, we conclude that a − b ≤ 0 ≤ a − b. Thus, 0 belongs to the interval [a − b, a − b]. Hence, 0 must belong to the sum of the two intervals, i.e., we must be able to represent 0 as 0 = a + b, where a ∈ [a, a] and b ∈ [−b, −b]. The first of these two inclusions means that a ≤ a ≤ a; the second means that −b ≤ b ≤ −b. From a + b = 0, it follows that b = −a. Hence, these inequalities are equivalent to b ≤ a ≤ b. Thus, a is the desired separating element. Proof of ii)→i). Let [p, p] and [q, q] be two intervals. It is easy to see that [p, p] + [−q, −q] ⊆ [p − q, p − q]. So, to conclude our proof, we must show that every element r ∈ [p−q, p−q] can be represented as r = p+q for some p ∈ [p, p]

12 and q ∈ [q, q]. The desired conditions on p and q are p ≤ p ≤ p and q ≤ q ≤ q. Since q = r − p, the second condition can be reformulated as r − q ≤ p ≤ r − q. So, here we have four elements a = p, a = p, b = r − q, and b = r − q. We already know that a ≤ a and b ≤ b. The fact that a ≤ b, i.e., that p ≤ r − q, follows from the inequality p+q ≤ r. Similarly, from r ≤ p+q, we conclude that b ≤ a. Hence, due to the 2-2-separation property, we have the desired element p. The proposition is proven. 2. Acknowledgments. This research was partly supported by the National Sci- ence Foundation via a Model Institutions for Excellence grant. The authors are thankful to Svetoslav Markov and Vladik Kreinovich whose email discus- sion on the intervals as an algebraic structure stimulated this result, to Piotr Wojciechowski for valuable discussions, and to the anonymous referees for their great help.

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