00 9 an Algebraic Approach to the Set of Intervals

In this paper we present the set of intervals as a normed vector space. We define also a four-dimensional associative algebra whose product gives the product of intervals in any cases. This approach allows to give a notion of divisibility and in some cases an euclidian division. 1. Intervals and generalized intervals An interval is a connected closed subset of R. The classical arithmetic operations on intervals are defined such that the result of the corresponding operation on elements belonging to operand intervals belongs to the resulting interval. That is, if ⋄ denotes one of the classical operation +,−, ∗, we have [x, x] ⋄ [y, y] = {x ⋄ y / x ∈ [x, x], y ∈ [y, y]} . (1) In particular we have { [x, x] + [y, y] = [x + y, x + y], [x, x]− [y, y] = [x − y, x − y] and [x, x]− [x, x] = [x − x, x − x] 6= 0. Let IR be the set of intervals. It is in one to one correspondance with the half plane of R: P1 = {(a, b), a ≤ b}. This set is closed for the addition and P1 is endowed with a regular semigroup structure. Let P2 be the half plane symmetric to P1 with respect to the first bisector ∆ of equation y − x = 0. The substraction on IR, which is not the symmetric operation of +, corresponds to the following operation on P1: (a, b)− (c, d) = (a, b) + s∆ ◦ s0(c, d), where s0 is the symmetry with respect to 0, and s∆ with respect to ∆. The multiplication ∗ is not globally defined. Consider the following subset of P1:    P1,1 = {(a, b) ∈ P1, a ≥ 0, b ≥ 0}, P1,2 = {(a, b) ∈ P1, a ≤ 0, b ≥ 0}, P1,3 = {(a, b) ∈ P1, a ≤ 0, b ≤ 0}. We have the following cases: 1) If (a, b), (c, d) ∈ P1,1 the product is written (a, b) ∗ (c, d) = (ac, bd). This work is supported by ’System’Virtual Prototyping S.A.S, a scientific service company involved in ’complex systems design’ and ’Functional Virtual Prototyping’ (www.systemvip.com) . 1 2 NICOLAS GOZE, ELISABETH REMM Then if e1 = (1, 1) and e2 = (0, 1), these ”vectors” generate P1,1 : ∀(x, y) ∈ P1,1 then (x, y) = xe1 + (y − x)e2, x > 0, y − x > 0. The multiplication corresponds in this case to the following associative commutative algebra: { e1e1 = e1, e1e2 = e2e1 = e2e2 = e2. 2) Assume that (a, b) ∈ P1,1 and (c, d) ∈ P1,2 so c ≤ 0 and d ≥ 0. Thus we obtain (a, b)∗(c, d) = (bc, bd) and this product does not depend of a. Then we obtain the same result for any a < b. Let e1 = (0, 1) and e2 = (−1, 0). Any interval of P1,1 is written ae1 + be2 with b < 0 and any interval of P1,2, ce1 + de2 with c, d > 0. We have no associative multiplication between (e1, e2) which describes the product. We have to add a formal dimension to obtain a 3-dimensional associative algebra and the product appears as the projection in the plane (e1, e2) of this associative algebra. Here if we consider the following associative commutative algebra    e1e1 = e1, e1e2 = e2, e1e3 = −e2, e2e3 = −e1, e2e2 = e1, e3e3 = e3. then (αe1+βe2+βe3)(γe1+ δe2) = αγe1+αδe2. As (a, b) = be1−ae2 and (c, d) = de1− ce2, we obtain the expected product. 3) If (a, b) ∈ P1,1 and (c, d) ∈ P1,3 then a ≥ 0, b ≥ 0 and c ≤ 0, d ≤ 0 and we have (a, b)∗ (c, d) = (bd, ac). Let e1 = (1, 1), e2 = (0, 1). This product corresponds to the following associative algebra:    e1e1 = e1, e1e2 = e1 − e2, e2e2 = e1 − e2. We have similar results for the cases (P1,2,P1,2), (P1,2,P1,3) and (P1,3,P1,3). All this shows that the set IR is not algebraically structured. Let us describe a vectorial structure on IR using the previous geometrical interpretation of IR with P1. First we extend P1 to R 2 and we obtain an extended set IR which corresponds to the classical interval [a, b] and ”generalized intervals” [a, b] with a > b. Of course using the addition of R, we obtain on IR a structure of abelian group and the symmetric of [a, b] ∈ IR is [−a,−b] ∈ IR \ IR. In this context [a, b] + [−a,−b] = 0. This aspect as been developed in [5]. We have a group homomorphism φ on IR given by φ : IR −→ IR (a, b) −→ (b, a). This map is called dual and we denote by dual (a, b) the generalized interval (b, a). The corresponding arithmetic has been developped by Kaucher [4] and is naturally called the Kaucher arithmetic. In the following we recall how to complete the semigroup IR to obtain a natural vetorial structure on IR. AN ALGEBRAIC APPROACH TO THE SET OF INTERVALS. 3 2. The real vector space IR 2.1. The semigroup (IR,+). Consider x = [x, x] and y = [y, y] two elements of IR. From (1) we get the addition x+ y = [x + y, x + y]. This operation is commutative, associative and has an unit [0, 0] simply denoted by 0. Theorem 1. The semigroup (IR,+) is commutative and regular. Proof. We recall that a semigroup is a nonempty set with an associative unitary operation +. It is regular if it satisfies x+ z = x+ y =⇒ z = y, for all x, y, z . The semigroup (IR,+) is regular. In fact x+ z = x+ y =⇒ [x + z, x + z] = [x + y, x + y] which gives z = y and z = y, that is z = y. 2.2. The group (IR,+). The goal is to define a substraction corresponding to an inverse of the addition. For that we build the symmetrized of the semigroup (IR,+). We consider on the set IR× IR the equivalence relation: (x, y) ∼ (z, t) ⇐⇒ x+ t = y + z, for all x, y, z, t ∈ IR. The quotient set is denoted by IR. The addition of intervals is compatible with this equivalence relation: (x, y) + (z, t) = (x+ z, y + t) where (x, y) is the equivalence class of (x, y). The unit is 0 = {(x, x), x ∈ IR} and each element (x, y) has an inverse r(x, y) = (y, x). Then (IR,+) is a commutative group. For all x = [x, x] ∈ IR, we denote by l(x) his lenght, so l(x) = x − x, and by c(x) his center, so c(x) = x + x 2 . Proposition 2. Let X = (x, y) be in IR. Thus • if l(y) < l(x), there is an unique A ∈ IR \ R such that X = (A, 0), • if l(y) > l(x), there is an unique A ∈ IR \ R such that X = (0, A) = r(A, 0), • if l(y) = l(x), there is an unique A = α ∈ R such that X = (α, 0) = (0,−α). Proof. It is based on the following lemmas: Lemma 3. Consider (x, y) ∈ IR with l(x) < l(y). Then (x, y) ∼ (0, [y − x, y − x]). 4 NICOLAS GOZE, ELISABETH REMM Proof. For all x, y, z, t ∈ IR, we have (x, y) ∼ (z, t) ⇐⇒ { z + y = x + t, z + y = x + t. If we put z = z = 0, then { t = y − x, t = y − x. with the necessary condition t > t. We obtain l(x) < l(y). So we have (x, y) = (0, [y − x, y − x]). Lemma 4. Consider (x, y) ∈ IR with l(y) < l(x), then (x, y) ∼ ([x − y, x − y], 0). Proof. For all x, y, z ∈ IR, we have (x, y) ∼ (z, 0) ⇐⇒ { z + y = x z + y = x . or { z = x − y z = x − y with the condition z > z which gives l(y) < l(x). So (x, y) = ([x − y, x − y], 0). Lemma 5. Consider (x, y) ∈ IR with l(x) = l(y), then