Mathematics / Fuzzy Mathematics Fuzzy Sets, Logic & Relational Systems 100% Free Open Access
Chapter 5 • Theory & Derivations

Unit 5: Fuzzy Arithmetic & Solution of Fuzzy Equations

Comprehensive theory of arithmetic on fuzzy numbers and the solution of fuzzy algebraic equations: addition, subtraction, multiplication, and division defined through Zadeh's Extension Principle and verified through interval α-cuts, shape alterations (e.g. non-triangular product of two TFNs), MIN and MAX operations on fuzzy numbers, the fundamental non-invertibility problem (why B - A does NOT solve A + X = B), solvability criteria and closed-form solutions for linear fuzzy equations A + X = B and multiplicative equations A · X = B.

§5.1 Addition and Subtraction of Fuzzy Numbers via the Extension Principle & $\alpha$-Cuts

1. Addition of Fuzzy Numbers ($A + B$)

Let $A$ and $B$ be two fuzzy numbers on $\mathbb{R}$. By Zadeh's Extension Principle, their sum $C = A + B$ has membership function:

$$\mu_{A + B}(z) = \sup_{z = x + y} \min(\mu_A(x), \mu_B(y))$$
Verification via $\alpha$-Cuts:

For every $\alpha \in (0, 1]$, let $A_\alpha = [a_1(\alpha), a_2(\alpha)]$ and $B_\alpha = [b_1(\alpha), b_2(\alpha)]$. By Nguyen's theorem:

$$[A + B]_\alpha = A_\alpha + B_\alpha = [a_1(\alpha) + b_1(\alpha), \; a_2(\alpha) + b_2(\alpha)]$$
Addition of Triangular Fuzzy Numbers (TFNs):

If $A = (a_1, a_2, a_3)$ and $B = (b_1, b_2, b_3)$ are TFNs:

$$a_1(\alpha) + b_1(\alpha) = (a_1 + b_1) + \alpha[(a_2 + b_2) - (a_1 + b_1)]$$
$$a_2(\alpha) + b_2(\alpha) = (a_3 + b_3) - \alpha[(a_3 + b_3) - (a_2 + b_2)]$$

Both endpoints remain linear in $\alpha$. Therefore:

Theorem 5.1 (TFN Addition): The sum of two Triangular Fuzzy Numbers is strictly a Triangular Fuzzy Number:

$$(a_1, a_2, a_3) + (b_1, b_2, b_3) = (a_1 + b_1, \; a_2 + b_2, \; a_3 + b_3)$$

2. Subtraction of Fuzzy Numbers ($A - B$)

By the Extension Principle, $C = A - B$ has membership function:

$$\mu_{A - B}(z) = \sup_{z = x - y} \min(\mu_A(x), \mu_B(y))$$

In terms of $\alpha$-cuts:

$$[A - B]_\alpha = A_\alpha - B_\alpha = [a_1(\alpha) - b_2(\alpha), \; a_2(\alpha) - b_1(\alpha)]$$
Subtraction of TFNs:

For $A = (a_1, a_2, a_3)$ and $B = (b_1, b_2, b_3)$:

$$[A - B]_\alpha = [(a_1 + \alpha(a_2 - a_1)) - (b_3 - \alpha(b_3 - b_2)), \; (a_3 - \alpha(a_3 - a_2)) - (b_1 + \alpha(b_2 - b_1))]$$
$$= [(a_1 - b_3) + \alpha((a_2 - b_2) - (a_1 - b_3)), \; (a_3 - b_1) - \alpha((a_3 - b_1) - (a_2 - b_2))]$$

Therefore:

Theorem 5.2 (TFN Subtraction):

$$(a_1, a_2, a_3) - (b_1, b_2, b_3) = (a_1 - b_3, \; a_2 - b_2, \; a_3 - b_1)$$

Notice that the left bound is $a_1 - b_3$ and the right bound is $a_3 - b_1$!

§5.2 Multiplication, Division, Reciprocals and Extreme Bounds

1. Multiplication of Fuzzy Numbers ($A \cdot B$)

By the Extension Principle:

$$\mu_{A \cdot B}(z) = \sup_{z = x \cdot y} \min(\mu_A(x), \mu_B(y))$$

In terms of $\alpha$-cuts, for positive fuzzy numbers ($A_\alpha, B_\alpha > 0$):

$$[A \cdot B]_\alpha = [a_1(\alpha) \cdot b_1(\alpha), \; a_2(\alpha) \cdot b_2(\alpha)]$$
The Quadratic Shape Distortion:

For two TFNs $A = (a_1, a_2, a_3)$ and $B = (b_1, b_2, b_3)$ with positive vertices:

$$a_1(\alpha) \cdot b_1(\alpha) = (a_1 + \alpha(a_2 - a_1))(b_1 + \alpha(b_2 - b_1))$$

This expression contains a quadratic term in $\alpha$ ($\alpha^2$)! Consequently:

Crucial Observation: The product of two Triangular Fuzzy Numbers is NOT a Triangular Fuzzy Number! Its left and right membership branches become non-linear (parabolic curves). However, for practical approximations, engineers often approximate the product as a TFN: $(a_1 b_1, a_2 b_2, a_3 b_3)$.


2. Reciprocal and Division ($A / B$)

For a strictly positive fuzzy number $B$ ($b_1 > 0$):

$$[B^{-1}]_\alpha = \left[ \frac{1}{b_2(\alpha)}, \; \frac{1}{b_1(\alpha)} \right] = \left[ \frac{1}{b_3 - \alpha(b_3 - b_2)}, \; \frac{1}{b_1 + \alpha(b_2 - b_1)} \right]$$

Then the division $A / B = A \cdot B^{-1}$ on $\alpha$-cuts is:

$$[A / B]_\alpha = \left[ \frac{a_1(\alpha)}{b_2(\alpha)}, \; \frac{a_2(\alpha)}{b_1(\alpha)} \right] = \left[ \frac{a_1 + \alpha(a_2 - a_1)}{b_3 - \alpha(b_3 - b_2)}, \; \frac{a_3 - \alpha(a_3 - a_2)}{b_1 + \alpha(b_2 - b_1)} \right]$$

The branches of $A / B$ are rational functions of $\alpha$, exhibiting hyperbolic curvature.

§5.3 Min and Max Operations on Fuzzy Numbers

1. Extended Min and Max

Beyond standard arithmetic, one can apply the Extension Principle to the crisp binary functions $\min(x, y)$ and $\max(x, y)$.

Let $A$ and $B$ be fuzzy numbers. We define:

$$\text{MIN}(A, B)(z) = \sup_{z = \min(x, y)} \min(\mu_A(x), \mu_B(y))$$
$$\text{MAX}(A, B)(z) = \sup_{z = \max(x, y)} \min(\mu_A(x), \mu_B(y))$$

2. $\alpha$-Cut Characterization

For any intervals $I = [a_1, a_2]$ and $J = [b_1, b_2]$:

$$\min([a_1, a_2], [b_1, b_2]) = [\min(a_1, b_1), \; \min(a_2, b_2)]$$
$$\max([a_1, a_2], [b_1, b_2]) = [\max(a_1, b_1), \; \max(a_2, b_2)]$$

Therefore:

$$[\text{MIN}(A, B)]_\alpha = [\min(a_1(\alpha), b_1(\alpha)), \; \min(a_2(\alpha), b_2(\alpha))]$$
$$[\text{MAX}(A, B)]_\alpha = [\max(a_1(\alpha), b_1(\alpha)), \; \max(a_2(\alpha), b_2(\alpha))]$$

Both $\text{MIN}(A, B)$ and $\text{MAX}(A, B)$ are valid fuzzy numbers. Together with the fuzzy number ordering $A \le B \iff a_1(\alpha) \le b_1(\alpha) \text{ and } a_2(\alpha) \le b_2(\alpha)$, the set of fuzzy numbers forms a distributive lattice.

§5.4 Linear Fuzzy Equations: Solving $A + X = B$ and Non-Invertibility of Fuzzy Subtraction

1. The Fundamental Problem of Fuzzy Equations

Consider the simple linear algebraic equation where $A$ and $B$ are known fuzzy numbers, and $X$ is an unknown fuzzy number to be determined:

$$A + X = B$$

In classical algebra, one simply subtracts $A$ from both sides: $X = B - A$. In fuzzy mathematics, setting $X = B - A$ generally FAILS to solve $A + X = B$!

Why $B - A$ Fails:

Compute $A + (B - A)$:

$$[A + (B - A)]_\alpha = A_\alpha + (B_\alpha - A_\alpha) = [a_1, a_2] + [b_1 - a_2, b_2 - a_1] = [b_1 - (a_2 - a_1), \; b_2 + (a_2 - a_1)]$$

The width of $A + (B - A)$ is $(b_2 - b_1) + 2(a_2 - a_1)$, which is strictly wider than $B_\alpha$! Thus $A + (B - A) \ne B$ whenever $A$ is non-crisp.


2. Exact Solvability Criterion for $A + X = B$

Theorem 5.3 (Solvability of Linear Fuzzy Equation $A + X = B$): Let $A$ and $B$ be fuzzy numbers with $\alpha$-cuts $A_\alpha = [a_1(\alpha), a_2(\alpha)]$ and $B_\alpha = [b_1(\alpha), b_2(\alpha)]$. The equation $A + X = B$ has an exact fuzzy number solution $X$ if and only if:

  1. For all $\alpha \in (0, 1]$, $x_1(\alpha) \equiv b_1(\alpha) - a_1(\alpha)$ is non-decreasing in $\alpha$.
  2. For all $\alpha \in (0, 1]$, $x_2(\alpha) \equiv b_2(\alpha) - a_2(\alpha)$ is non-increasing in $\alpha$.
  3. $x_1(1) \le x_2(1)$.

In terms of spreads (uncertainty widths $\Delta A_\alpha = a_2(\alpha) - a_1(\alpha)$ and $\Delta B_\alpha = b_2(\alpha) - b_1(\alpha)$): An exact solution exists if and only if the uncertainty of $B$ is greater than or equal to the uncertainty of $A$ at every level:

$$\Delta B_\alpha \ge \Delta A_\alpha \quad (\forall \alpha \in [0, 1])$$

When this condition holds, the unique solution $X$ has $\alpha$-cuts:

$$X_\alpha = [b_1(\alpha) - a_1(\alpha), \; b_2(\alpha) - a_2(\alpha)]$$

For Triangular Fuzzy Numbers $A = (a_1, a_2, a_3)$ and $B = (b_1, b_2, b_3)$: The candidate solution is $X = (b_1 - a_1, \; b_2 - a_2, \; b_3 - a_3)$. It is a valid TFN if and only if:

$$b_1 - a_1 \le b_2 - a_2 \le b_3 - a_3 \iff b_2 - b_1 \ge a_2 - a_1 \text{ and } b_3 - b_2 \ge a_3 - a_2$$

§5.5 Solving $A \cdot X = B$ and Non-Linear Fuzzy Algebraic Systems

1. Multiplicative Fuzzy Equations ($A \cdot X = B$)

Consider the multiplicative equation for strictly positive fuzzy numbers $A, B > 0$:

$$A \cdot X = B$$

In terms of $\alpha$-cuts: $[A \cdot X]_\alpha = [a_1(\alpha) x_1(\alpha), \; a_2(\alpha) x_2(\alpha)] = [b_1(\alpha), \; b_2(\alpha)]$

Theorem 5.4 (Solvability of Multiplicative Equation $A \cdot X = B$): For strictly positive fuzzy numbers $A, B$, the equation $A \cdot X = B$ possesses an exact fuzzy number solution $X$ if and only if:

  1. $x_1(\alpha) = \frac{b_1(\alpha)}{a_1(\alpha)}$ is non-decreasing with $\alpha \in (0, 1]$.
  2. $x_2(\alpha) = \frac{b_2(\alpha)}{a_2(\alpha)}$ is non-increasing with $\alpha \in (0, 1]$.
  3. $x_1(1) \le x_2(1)$.

When these conditions hold, the unique solution is given by:

$$X_\alpha = \left[ \frac{b_1(\alpha)}{a_1(\alpha)}, \; \frac{b_2(\alpha)}{a_2(\alpha)} \right]$$

2. Fuzzy Polynomial Equations

For general equations such as $A X^2 + B X = C$, solutions are obtained by solving the coupled non-linear interval equations at each $\alpha$-cut level:

$$a_1(\alpha) [x_1(\alpha)]^2 + b_1(\alpha) x_1(\alpha) = c_1(\alpha)$$
$$a_2(\alpha) [x_2(\alpha)]^2 + b_2(\alpha) x_2(\alpha) = c_2(\alpha)$$

and verifying monotonicity to ensure the resulting family of intervals $[x_1(\alpha), x_2(\alpha)]$ defines a legitimate fuzzy number.

Rigorous Tiered Solved Examination Problems

Step-by-step unskipped derivations, complete proofs, and verification across Foundational, Advanced, and Honors tiers.

Fundamentals Example 5.1: Problem 5.1: Addition and Subtraction of Triangular Fuzzy Numbers

Let $A = (2, 5, 8)$ and $B = (3, 6, 10)$ be two Triangular Fuzzy Numbers.

  1. Compute the sum $C = A + B$. Determine its parameters and verify its modal core and support.
  2. Compute the difference $D = A - B$. Determine its parameters, modal core, and support.
  3. Compute $E = A - A$. Explain why $E \ne 0$ and state its modal value and support.

1. Fuzzy Addition $C = A + B$

$$A = (2, 5, 8), \qquad B = (3, 6, 10)$$

Using Theorem 5.1:

$$C = (2 + 3, \; 5 + 6, \; 8 + 10) = (5, 11, 18) \qquad \blacksquare$$
  • Core: $\{11\}$
  • Support: $(5, 18)$
  • Spread: $18 - 5 = 13$ (which equals $(8 - 2) + (10 - 3) = 6 + 7 = 13$).

2. Fuzzy Subtraction $D = A - B$

Using Theorem 5.2:

$$D = (a_1 - b_3, \; a_2 - b_2, \; a_3 - b_1) = (2 - 10, \; 5 - 6, \; 8 - 3) = (-8, -1, 5) \qquad \blacksquare$$
  • Core: $\{-1\}$
  • Support: $(-8, 5)$
  • Spread: $5 - (-8) = 13$.

3. Self-Difference $E = A - A$

$$E = (a_1 - a_3, \; a_2 - a_2, \; a_3 - a_1) = (2 - 8, \; 5 - 5, \; 8 - 2) = (-6, 0, 6) \qquad \blacksquare$$
  • Modal value (Core): $\{0\}$.
  • Support: $(-6, 6)$.
  • Explanation: Although the peak is centered at 0, the support $(-6, 6)$ has non-zero width because subtracting independent uncertainties doubles the span. This proves $A - A \ne 0$. $\blacksquare$
Computational Triad Example 5.2: Problem 5.2: Solvability of Linear Fuzzy Equations $A + X = B$

Consider the linear fuzzy equation $A + X = B$. Let $A = (1, 3, 5)$ be a Triangular Fuzzy Number.

  1. Suppose $B_1 = (4, 8, 14)$.
  • Check the solvability criteria for $A + X = B_1$.
  • If solvable, find the exact solution $X$.
  • Verify that $A + X = B_1$.
  1. Suppose $B_2 = (4, 8, 9)$.
  • Check the solvability criteria for $A + X = B_2$.
  • What happens to the candidate vertices of $X$? Explain why no exact solution exists in $\mathcal{F}_N(\mathbb{R})$.
  1. For $B_2$, evaluate the naive subtraction candidate $\tilde{X} = B_2 - A$ and compute $A + \tilde{X}$ to show explicitly that it fails to solve the equation.

1. Case 1: $B_1 = (4, 8, 14)$

$A = (1, 3, 5) \implies a_1 = 1, a_2 = 3, a_3 = 5$. $B_1 = (4, 8, 14) \implies b_1 = 4, b_2 = 8, b_3 = 14$.

Solvability Check:
  • Left spread of $A$: $a_2 - a_1 = 3 - 1 = 2$.

Left spread of $B_1$: $b_2 - b_1 = 8 - 4 = 4 \ge 2$. (Satisfied).

  • Right spread of $A$: $a_3 - a_2 = 5 - 3 = 2$.

Right spread of $B_1$: $b_3 - b_2 = 14 - 8 = 6 \ge 2$. (Satisfied).

Candidate solution:

$$X = (b_1 - a_1, \; b_2 - a_2, \; b_3 - a_3) = (4 - 1, \; 8 - 3, \; 14 - 5) = (3, 5, 9)$$

Since $3 \le 5 \le 9$, $X = (3, 5, 9)$ is a valid TFN.

Verification:
$$A + X = (1, 3, 5) + (3, 5, 9) = (1 + 3, \; 3 + 5, \; 5 + 9) = (4, 8, 14) = B_1 \qquad \blacksquare$$

2. Case 2: $B_2 = (4, 8, 9)$

Here $b_3 - b_2 = 9 - 8 = 1$, whereas $a_3 - a_2 = 5 - 3 = 2$. Since $1 < 2$, the right spread condition FAILS! Candidate vertices:

$$x_1 = 4 - 1 = 3, \quad x_2 = 8 - 3 = 5, \quad x_3 = 9 - 5 = 4$$

Notice that $x_2 = 5 > x_3 = 4$. The tuple $(3, 5, 4)$ is NOT an ordered sequence of real numbers; its membership function would fold backwards, which violates fuzzy convexity! Therefore, NO exact solution exists in the space of fuzzy numbers. $\blacksquare$


3. Failure of Naive Subtraction $\tilde{X} = B_2 - A$

$$\tilde{X} = B_2 - A = (4, 8, 9) - (1, 3, 5) = (4 - 5, \; 8 - 3, \; 9 - 1) = (-1, 5, 8)$$

Now substitute $\tilde{X}$ back into the left-hand side:

$$A + \tilde{X} = (1, 3, 5) + (-1, 5, 8) = (1 + (-1), \; 3 + 5, \; 5 + 8) = (0, 8, 13)$$

Comparing:

$$A + \tilde{X} = (0, 8, 13) \ne (4, 8, 9) = B_2$$

The resulting fuzzy number is far wider and has a completely different support $[0, 13]$ instead of $[4, 9]$! This demonstrates conclusively that $B - A$ does NOT solve $A + X = B$. $\blacksquare$

Honors / Proof Challenge Example 5.3: Problem 5.3: Exact Non-Linear Membership Function of Product of Two TFNs

Consider two symmetric Triangular Fuzzy Numbers:

$$A = (1, 2, 3), \qquad B = (1, 2, 3)$$
  1. Using interval arithmetic on $\alpha$-cuts, determine the exact $\alpha$-cut representation $[A \cdot B]_\alpha$ of their product.
  2. Invert the $\alpha$-cut boundaries to derive the exact analytical membership function $\mu_{A \cdot B}(z)$ for all $z \in [1, 9]$.
  3. Prove that the graph of $\mu_{A \cdot B}(z)$ consists of two parabolic arcs and calculate the curvature at the peak $z = 4$.
  4. Compare the exact product with the standard linear heuristic approximation $C_{\text{approx}} = (1, 4, 9)$ at $z = 2.25$.

1. $\alpha$-Cut Representation of $A \cdot B$

Since $A = B = (1, 2, 3)$:

$$A_\alpha = B_\alpha = [1 + \alpha, \; 3 - \alpha], \quad \alpha \in [0, 1]$$

Since all values are positive, the product of the intervals is:

$$[A \cdot B]_\alpha = [ (1 + \alpha)^2, \; (3 - \alpha)^2 ] \qquad \blacksquare$$
  • For $\alpha = 0$: $[A \cdot B]_0 = [1^2, 3^2] = [1, 9]$.
  • For $\alpha = 1$: $[A \cdot B]_1 = [2^2, 2^2] = [4, 4] = \{4\}$.

2. Inversion to Membership Function $\mu_{A \cdot B}(z)$

Left Branch ($1 \le z \le 4$):

On the left boundary:

$$z = (1 + \alpha)^2 \implies \sqrt{z} = 1 + \alpha \implies \alpha = \sqrt{z} - 1$$
Right Branch ($4 \le z \le 9$):

On the right boundary:

$$z = (3 - \alpha)^2 \implies \sqrt{z} = 3 - \alpha \implies \alpha = 3 - \sqrt{z}$$
Exact Piecewise Membership Function:
$$\mu_{A \cdot B}(z) = \begin{cases} 0, & z < 1 \\ \sqrt{z} - 1, & 1 \le z \le 4 \\ 3 - \sqrt{z}, & 4 \le z \le 9 \\ 0, & z > 9 \end{cases} \qquad \blacksquare$$

3. Curvature Analysis

  • On $[1, 4]$: $\frac{d\mu}{dz} = \frac{1}{2\sqrt{z}}$, and $\frac{d^2\mu}{dz^2} = -\frac{1}{4 z^{3/2}} < 0$.

The curve is strictly concave (parabolic arc), NOT linear!

  • On $[4, 9]$: $\frac{d\mu}{dz} = -\frac{1}{2\sqrt{z}}$, and $\frac{d^2\mu}{dz^2} = \frac{1}{4 z^{3/2}} > 0$.

The curve is strictly convex.

At the peak $z = 4$:

  • Left derivative: $\left.\frac{d\mu}{dz}\right|_{4^-} = \frac{1}{2\sqrt{4}} = \frac{1}{4} = 0.25$.
  • Right derivative: $\left.\frac{d\mu}{dz}\right|_{4^+} = -\frac{1}{2\sqrt{4}} = -\frac{1}{4} = -0.25$.

The peak is a sharp corner with derivative jump $\Delta = -0.5$. $\blacksquare$


4. Comparison with Heuristic Approximation

The heuristic linear approximation is $C_{\text{approx}} = (1, 4, 9)$:

$$\mu_{\text{approx}}(z) = \frac{z - 1}{4 - 1} = \frac{z - 1}{3}, \quad z \in [1, 4]$$

Evaluate at $z = 2.25$:

  • Exact membership:
$$\mu_{\text{exact}}(2.25) = \sqrt{2.25} - 1 = 1.5 - 1 = 0.5000$$
  • Linear approximation:
$$\mu_{\text{approx}}(2.25) = \frac{2.25 - 1}{3} = \frac{1.25}{3} \approx 0.4167$$

The linear approximation underestimates the true membership grade by over 16.7%! This highlights the importance of using exact $\alpha$-cuts in critical engineering applications. $\blacksquare$