Page 4-1
Chapter 4 Fuzzy Arthmetic
4.1 Fuzzy Numbers ( or Fuzzy intervals )
--- Fuzzy sets defined on R
Membership function A:R[0,1]
e.g., i, close to a real number
ii, around a given interval of real numbers
◎ Applications : fuzzy control, decision making approximate reasoning,
optimization, statistics with imprecise probabilities.
◎ Required properties of a fuzzy number A
( i ) A : a normal fuzzy set
(ii)
(iii)

A
0
: a convex set,    0,1
A : bounded support
Page 4-2
◎ Cases of fuzzy numbers
Crisp
fuzzy
an
ordinary
member
an
interval
of
numbers
◎ Membership functions of fuzzy numbers need not be symmetric
large number
small number
Page 4-3
◎Membership function may
be piecewise-defined
Theorem 4.1 : Let A  F ( R)  A : a fuzzy number
fuzzy power
iff  a, b  
Normal
b.
Convex
c.
bounded
x   a, b
1

A( x)  l ( x)
r ( x)

Where a.
s.t.
a.
x   , a 
----- ( 4.1 )
x   b,  
b, Monotonic increasing

l :  , a   0,1 : c, Continuous from right
d , l ( x)  0, x  (, w )
1

 Monotonic decreasing

r : (b, )   0,1 :  Continuous from left
r ( x)  0, x  ( w )
2,

Proof:
I, hecessity =>
1.
A : normal
 1A  
Let 1 A   a, b
2. prove that the left part of A agree with l
(a )l :
( a , )  0 , 1

(b) Monotonic increasing

(c) Continuous from right
(d )l ( x)  0, x  (, w )

1
Page 4-4
a, 0  A( x)  1, x  (, a)
( A (a ) 1 )
with l : (, a)  0,1
 A : ( a, 
)  0 , agree
1
b. Let x  y  a,  A( y)  min  A( x), A(a)  A( x)
Theorem1.1
(1)
A(a)  1
(A:convex)
 A : increasing agree with l : monotonic increasing
c. assume A(x): not continuous from the right
Let x0  (, a) , be a discontinuous pt.
  a sequence
xn  , s.t.
xn  x0 , n
and lim xn  x0
n 
increasing
Let lim A( xn )    A( x0 )
n 

A : closed interval (
 x0   A
A : convex)
contradict
A( x0 )  

 A( x) : continuous from right
d.
A : fuzzy number
0
A : bounded
w1 , s.t.
A( x)  0, x   , w1 
Page 4-5
3, prove r :
monotonic decreasing
Continuous from the left
r : (b 
, 
)
[ 0 , 1]
r ( x) 0 x  (2w , )
II.
Sufficiency
i.
A defined by (4.1) is normal -----(a)
ii.
iii.
and its support 0 A : bounded----(b)
prove   (0,1],  A : a closed interval (i.e. A: convex)
proof : let
X   inf{x | l ( x)   , x  a} ----(c)
Y  s u p {x r| x( )
x, ---(d)
b }
Show that

A : closed (i.e.,

A  [ X  , Y ] )
(A) prove A  [ X  , Y ]
proof : x0   A , if x0  a  A( x0 )  l ( x0 )  
i.e., x0 {x | l ( x)   , x  a}
 x0 i n f {x l| x( )
x, a }---(i)
x
Page 4-6
if x0  b  A( x0 )  r ( x0 )  
i.e., x0 {x | r ( x)   , x  b}
 x0 s u p {x r| x( )
x, b } y
 x0 [ X , Y ]  A  [ X , Y ]
(B) prove [ X  , Y ]   A (idea [ x , y ]   A :)
proof : from (c) , i.e. X   inf{x | l ( x)   , x  a}
 a sequence {xn } in {x | l ( x)   , x  a}
s.t.
lim xn  x
n 
l : continuous from right
l ( x ) x ( l ixm
ll xi nm  ( )
n  )
n 
n 
 x   A similarly
 [ x , y ]
( A ) (B 
)

x[

y   A
A
y,  ] :Aa closed interval

(i)(ii)(iii)=> complete sufficiency
(I)(II) => complete theorem
※ Every fuzzy number can be represented in the form of (4.1)
Page 4-7
 Examples
(a)
w1  a  b  w2  1.3
x  (,1.3), l ( x)  0
x  (1.3, ), r ( x)  0
(b)
w1  a  1.25, w2  b  1.35
x  (,1.25), l ( x)  0
x  (1.35, ), r ( x)  0
(c)
a  b  1.3,
w1  1.2, w2  1.4
l ( x) 
r ( x) 
x  (,1.2)
0
1 0 x(  1 .3 )
1 x [ 1 . 2 , 1 . 3 )
1 0 ( 1 . 3x  )
1 x( 1 . 3 , 1 . 4 ]
x  (1.4, )
0
(d)
a  1.28, b  1.32
w1  1.2, w2  1.4
l ( x) 
r ( x) 
0
1 2 . 5x (
x  (,1.2)
1 .2 8 )
1 2 . 5 ( 1 . 2x 8
0
)
1x  [ 1 . 2 , 1 . 2 8 )
1x  ( 1 . 3 2 , 1 . 4 ]
x  ( 1 . 4,
)
Page 4-8
a = 90, b=100
w1  77.5, w2  100
x   , 77.5 
0
l ( x)  
0.08( x  90)  1 x   77.5,  
r ( x)  0, x  (100, )
◎ Fuzzy Cardinality A
A. a fuzzy number defined on N
A
Example


A
Page 4-9.
| D2 |
0.13 0.27 0.4 0.53 0.67 0.8 0.93 1.0







19
17
15
13
11
9
7
5
e.g.
x {30,50}, D2 ( x)  0.67
50  30
 1  11
2
B. Scalar Cardinality | A |
|  A |
| A |

A( x)
xsup( A)
Example :
| D2 
| 0 . 1 3 * 2
0 . 27 * 2
0 . 4 * 2  0 . 5 3 * 2
0 . 67 ** 22 10*. 85 * 21 2 .04. 69 3
Page 4-10
4.3 Arithmetic Operations on Fuzzy Numbers
Idea: Define operations in terms of arithmetic operations on α-cuts which are
a subject of interval analysis of classical mathematics.
※ Fuzzy number can uniquely be represented by theirα-cuts.
⊙Classical interval analysis
‧ 4 operations：
+,-,‧,/
Let ＊ denote any of the 4 operations→
a, b d , e  f  g | a  f
except a, b/d , e when
 b, d  g  e
0  [ d , e]
※ The result of an operations on closed intervals is a closed interval.
Page 4-11
●Definitions of operations
[a, b]  [d , e]  [a  b, b  e]
[a, b]  [d , e]  [a  b, b  e]
[a, b]  [d , e]  [min( ad , ae, bd , be), max( ad , ae, bd , be)]
[a, b] /[ d , e]  [a, b]  [1 / e,1 / d ]  [min(
where 0  [d , e]
●Examples：
1.[2,5]+[1,3]=[3,8]
2.[2,5]-[1,3]=[-1,4]
3.[3,4]‧[2,2]=[6,8]
4..[4,10]/[1,2]=[2,10]
a a b b
a a b b
, , , ), max( , , , )]
d e d e
d e d e
Page 4-12
●Properties of Operations
Let A  [a1 , a2 ] , B  [b1 , b2 ] , C  [C1 , C2 ] , 0  [0,0] , 1  [1,1]
A  B  B  A
1. Commutativity 
 A B  B  A
( A  B)  C  A  ( B  C )
2. Associativity 
 ( A  B)  C  A  ( B  C )
A  0  A  A  0
3. Identity 
 A  1 A  A 1
4. Subdistributivity A  ( B  C )  A  B  A  C
5. Distributivity
If b  B, c  C , b  c  0,  A  ( B  C )  A  B  A  C
6. 0  A  A,1  A / A,0  A
7. Inclusion monotonicity
If A  E , B  F
A  B 
A  B 


 A B 
 A / B 
EF
EF
EF
E/F
Page 4-13
● Prove Subdistrubutivity： ( A  ( B  C )  A  B  A  C )
Proof： A  ( B  C )  {a  (b  c) | a  A, b  B, c  C}
 {a  b  a  c) | a  A, b  B, c  C}
 Less Re striction (loose )  {a  b  a   c) | a, a   A, b  B, c  C}  A  B  A  C
 ( A  (B  C)  A  B  A  C)
● Example：
Let A=[0,1], B=[1,2], C=[-2,-1]
 A  B  [0,1]  [1,2]  [0,2]

  A  C  [0,1]  [2,1]  [2,0]
B  C  [1,2]  [2,1]  [1,1]

 A  ( B  C )  [0,1]  [1,1]  [1,1]

 A  B  A  C  [0,2]  [2,0]  [2,2]
 A  (B  C)  A  B  A  C
Page 4-14
˙ Prove Distributivity : ( f b  B, c  C , b c  0,
 A (B  C)  A B  A C )
Proof : Let A  [a1 , a2 ], B  [b1 , b2 ], C  [c1 , c2 ]
1. assume
B,C
i.e , b1 , c1  0
0
Case 1 : a1  0
 A ( B  C )  [a1 (b1  c1 ), a2 (b2  c2 )]
 [a1b1  a1c1 ), a2b2  a2c2 )]
 [a1b1 , a2b2 ]  [a1c1 , a2c2 )]
A
[
]
0
Case2 : a2  0
 ( A)  [a2 , a1 ]
( A) ( B  C ) )( A) B  ( A) C
A
 A (B  C)  A B  A C
[
]
0
Case3 : a1  0, a2  0
 A ( B  C )  [a1 (b2  c2 ), a2 (b2  c2 )]
 [a1 b2 , a2 b2
] [ a1 c2, a2 c]2
 A B A C
A
[
]
0
B,C
, i.e., b2 , c2  0
2. Similarity ,
0
Page 4-15
4.4 . Arithmetic Operations on Fuzzy Numbers
Fuzzy Arithmetic :
Interval arithmetic
Extension principle
⊙Interval arithmetic method

( A  B)   A   B
see theorem 4.2 (c)
*: arithmetic operation
From Theorem 2.5
A B 

( A  B)
[0,1]
˙Example :
Let A(x) =
x  1, x  3
0
1  x  1
1 x  3
(x+1)/2
(3-x)/2
B(x) =
x  1, x  5
0
1 x  3
(5-x)/2
3 x 5


 A  [2  1,3  2], B  [2  1,5  2]
now,
(x-1)/2

( A  B)  [4,8  4]
 ( 0 , 1 ]

( A  B)  [4  6, 2  4]
 ( 0 , 1 ]

( A B) 
=

( A / B) 
=
  0.5
Both  A,  B  0
[4 2  12  5, 4 2  16  15]
[4  1, 4  16  15]
[ ( 2  1 ) / ( 2 1 ) , ( 3 2 ) / ( 2
[ ( 2  1 ) / (5  2 ) , ( 3 2 ) / ( 2
2
2
( 0 , . 5 ]
 ( . 5 , 1 ]
1)(]0 , . 5 ]
1)(]. 5 , 1 ]
Page 4-16
From [4,8  4]
f (4 )  
g ( 8 4 ) 
f ( ) 

4
g ( ) 
8
4
f ( x) 
x
4
g ( x) 
8 x
4
 0   1 


 0  4  4 
 i.e., 0  x  4 


(A+B)(x)=
 0   1 


 0  4  4 
 0  4  4 


 8  8  4  4 
 i.e., 4  x  8 


0
x  0, x  8
x
4
0 x4
8 x
4
4 x 8
Page 4-17
 ( A  B)( x) 
0
x  0, x  8
x
4
0 x4
8 x
4
( A  B )( x) 
( A B )( x) 
4 x 8
0
x  6, x  2
x6
4
6  x  2
2 x
4
2  x  2
x  5, x  15
0
1
2
3  ( 4 x )
2
5  x  0
1
( 1 x 2)
2
0 x3
1
2
4  ( 1 x )
2
3 x  1 5
Page 4-18
0

[3  (4  X )1 / 2 ] / 2

Show ( A  B)( X )  
1/ 2
 (1  X ) / 2
[4  (1  X )1 / 2 ] / 2
X  5, X  15
5 X  0
0 X 3
3  x  15
[4 2  12  5,4 2  16  15],   (0,0.5]
From： ( A  B)  
[4 2  1,4 2  16  15],   (0.5,1]


A. Let x  4 2  12  5  4 2  12  (5  x)  0 by 公式：
x
 b  b 2  4ac
2a
 
3 4 x
   (0,0.5]
2
As
 00
0 
3 4 x
3 4 x
其中
(不能成立)( 3  4  x  0, 4  x  3, 不可能)
2
2
3 4 x
 x  5
2
As   0.5  0.5 
3 4 x
1 3 4 x
,0.5  
(不能成立)
2
2
2
1 3 4 x

x0
2
2
 5  x  0
 0.5 
B.
Let x  4 2  16  15  4 2  16  15  x  0
 
4  1 x
   (0,0.5]
2
As   0  0 
0 
4  1 x
4  1 x
,0 
(不能成立)
2
2
4  1 x
 x  15
2
As   0.5  0.5 
4  1 x
1 4  1 x
,0.5  
(不能成立)
2
2
2
1 4  1 x

 x8
2
2
8  x  15
 0.5 
Page 4-19
x  4 2  16  15  4 2  16  15  x  0
4  (1  x)
2
  (0.5,1] 
 
x  4 2  1  4 2  1  x  0
C. Let
4  (1  x)
2 4  (1  x)
,1  
2
2
2
1 4  (1  x)
 0.5  
 x8
2
2
4  (1  x)
5  8  x  15  x 
x8
2
 11
(1  x)
2
  (0.5,1]
 
As
  0.5  0.5 
(1  x)
1

2
2
x  0
0  x  3
D. Let x  4 2  16  15  4 2  16  15  x  0
4  (1  x)
2
  (0.5,1]
 
As   0.5 , 0.5 
 0.5 
4  (1  x)
(不能成立)
2
1 4  (1  x)

 x8
2
2
As   1  1 
4  (1  x)
2 4  (1  x)
,1  
(不可能成立)
2
2
2
2 4  (1  x)

 x3
2
2
3  x  8
1 
By (A),(B),(C),(D)
(A).when 5  x  0  x 
3  (4  x)
2
(B).when 8  x  15  x 
4  (1  x)
2
(C).when 0  x  3  x 
(D).when 3  x  8  x 
i.e
4  (1  x)
2
x  5, x  15
0
3
( A B) 
(1  x)
2
( 4 x )
2
( 1 x )
2
4
( 1 x )
2
5  x  0
0 x3
3 x  1 5
Page 4-20
(A/B)(x)=
0
Derivation : from
x
let x=
(5x+1)/(2x+2)
0
(3-x)/(2x+2)
1/ 3  x  3

( A / B) 
[



x

0

-1
2  1 3 - 2
,
] 
2 + 1 2 + 1
2  1 3 - 2
,
]
5  2 2 + 1

1/3

x

[0,0.5]
[0.5,1]
2  1
,  (2 +1)x=2  1  2x +x=2 -1
2 +1
 ( 2x  2)   1x  
as   0  0 
1 x
2  2x
, 
(0, 0.5]
1 x
 x  1 ( )
2  2x
as   0.5  0.5 
ii, Let

, x3
(x+1)/(2-2x)
[
i.

-1

1 x
 1 x  1 x  x  0 [ ]
2  2x
3  2
3 x
,  
2  1
2x  2
  (0.5,1]  1  x  3
x
f3 ( x)
2  1
5x 1
,  
5  2
2  2x
1
  (0.5,1]  0  x 
3
x
iii, Let
f 4 ( x)
3  2
3 x
,  
2  1
2x  2
1
  (0.5,1]   x  1
3
x
iv, Let
f 4 ( x)
1
 x3
3