By Barnabas Bede
This publication offers a mathematically-based creation into the attention-grabbing subject of Fuzzy units and Fuzzy good judgment and can be used as textbook at either undergraduate and graduate degrees and likewise as reference consultant for mathematician, scientists or engineers who want to get an perception into Fuzzy good judgment. Fuzzy units were brought through Lotfi Zadeh in 1965 and because then, they've been utilized in many functions. for this reason, there's a monstrous literature at the functional purposes of fuzzy units, whereas conception has a extra modest assurance. the most function of the current ebook is to minimize this hole via delivering a theoretical creation into Fuzzy units in accordance with Mathematical research and Approximation concept. famous functions, as for instance fuzzy keep watch over, also are mentioned during this publication and put on new flooring, a theoretical beginning. furthermore, a couple of complicated chapters and a number of other new effects are incorporated. those include, between others, a brand new systematic and confident technique for fuzzy inference structures of Mamdani and Takagi-Sugeno kinds, that investigates their approximation power through supplying new mistakes estimates.
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Extra resources for Mathematics of Fuzzy Sets and Fuzzy Logic
30. (see Acz´el ) The only continuous solution F : R → R of Cauchy’s functional equation F (x + y) = F (x) + F (y) is F (x) = cx, for some constant c ∈ R. Proof. We have F (0) = 2F (0), and so F (0) = 0. Then, F (n) = nF (1). Let m F (1) = c. , F (r) = cr, ∀r ∈ Q. Then we extend F by continuity and we get F (x) = cx, for any x ∈ R. 31. g. , there is a continuous strictly decreasing t : [0, 1] → [0, ∞], t(1) = 0, which is uniquely determined up to a multiplicative constant such that for all x, y ∈ [0, 1] we have T (x, y) = t(−1) (t(x) + t(y)).
Proof. (i) Case 1. If x ∨ y ≤ z then (x ∨ y) → z = 1 and also we have x ≤ z and y ≤ z. Then (x → z) ∧ (y → z) = 1 ∧ 1 = 1. Case 2. If x ∨ y > z then (x ∨ y) → z = z and x ≥ z or ,y ≥ z so we have (x → z) = z or (y → z) = z and then (x → z) ∧ (y → z) = z. (ii) Case 1. If x ∧ y ≤ z then (x ∧ y) → z = 1. Also, we have either x ≤ z or y ≤ z in which case either x → z = 1 or y → z = 1 and so (x → z) ∨ (y → z) = 1. Case 2. If x ∧ y > z, then (x ∧ y) → z = z. Also, if (x ∧ y) > z then both x > z and y > z and x → z = z and y → z = z.
23. The following inequalities hold true: (i) P ≤ R−1 (R ◦ P ); (ii) R ◦ (R−1 Q) ≤ Q; (iii) R ≤ (P (R ◦ P )−1 )−1 ; (iv) (P Q−1 )−1 ◦ P ≤ Q. Proof. 18, (vi) we have for every y ∈ Y and z ∈ Z, R−1 (R ◦ P )(y, z) = R−1 (y, x) → (R ◦ P )(x, z) x∈X R−1 (y, x) → = x∈X R(x, t) ∧ P (t, z) t∈Y R(x, y) → R(x, y) ∧ P (y, z) ≥ ≥ x∈X P (y, z) = P (y, z). 18, (v) we obtain for every x ∈ X and z ∈ Z R ◦ (R−1 Q)(x, z) = R(x, y) ∧ (R−1 Q)(y, z) y∈Y R(x, y) ∧ ( = y∈Y R−1 (y, t) → Q(t, z)) t∈X R(x, y) ∧ (R(x, y) → Q(x, z)) ≤ ≤ y∈Y Q(x, z) = Q(x, z).