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μ m , with 1
μ m x ,
if x
=
0
1
or
μ n ÷ μ m = μ n
(
x
) =
0 ,
and
μ m
0
,
if x
=
1
μ m m =
1.
Let's compute the example
μ 3 ÷ μ 3 , with
μ 3 as in Sect. 6.2.2 . It will be:
4 ,
3
1
1
4
y ,
if y
3
3 ,
2
1
μ
1
1
(
y
) =
y
2
,
if y
,
.
0
otherwise
1
μ
Hence, for
μ 3 ÷ μ 3 = μ 3
3 , it will be
3 ÷ μ 3 )(
1
) =
1, and
1
y
= ʱ + 2
4
4 t
2
L ,
ʱ =
4
=
x
2
t
=
x
·
y
ʱ ʱ =
t
+
1
1
4
ʱ
ʱ +
4
2 t
R ,
ʱ =
4
x
=
y
2
t
=
x
·
y
=
ʱ =
1 .
2
t
+
That is
4 t
2
1
t + 1 ,
if t
∈[
2 ,
1
]
4
2 t
3 ÷ μ 3 )(
) =
= μ 1 (
)
t
1 ,
if t
∈[
1
,
2
]
t
t
+
0
,
otherwise
Graphically,
Remark 6.2.3 Obviously, like in the product's case, the result is not a linear-
triangular fuzzy number, and the basic interval
[
a
,
b
]
is not symmetrical respect
n
to the point
m .
6.2.3 Note
] R , and
As it is easy to see, if
μ, ˃ ∈[
0
,
1
∗∈{+ , , · , :}
, it follows the relation
ʱ
between the corresponding
-cuts:
 
 
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