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u
mb
e
r
o
na
n
u
mb
e
r
l
i
n
e
.
We
u
s
e
s
y
mb
o
l
s
t
oh
e
l
pu
s
e
f
f
i
c
i
e
n
t
l
y
c
o
mmu
n
i
c
a
t
e
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e
l
a
t
i
o
n
s
h
i
p
s
b
e
t
w
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e
nn
u
mb
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r
s
o
nt
h
e
n
u
mb
e
r
l
i
n
e
.
T
h
e
s
y
mb
o
l
s
u
s
e
dt
od
e
s
c
r
i
b
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a
n
 
e
q
u
a
l
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t
y
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e
l
a
t
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o
n
s
h
i
p
 
b
e
t
w
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n
n
u
mb
e
r
s
f
o
l
l
o
w
:
=
i
s
 
e
q
u
a
l
 
t
o≠
i
s
 
n
o
t
 
e
q
u
a
l
 
t
o≈
i
s
 
a
p
p
r
o
x
i
ma
t
e
l
y
 
e
q
u
a
l
 
t
o
T
h
e
s
e
s
y
mb
o
l
s
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r
e
u
s
e
da
n
di
n
t
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p
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n
t
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f
o
l
l
o
w
i
n
gma
n
n
e
r
:
5
=
5
0
≠
5
π
≈
3
.
1
4
5
 
i
s
 
e
q
u
a
l
 
t
o
 
5
0
 
i
s
 
n
o
t
 
e
q
u
a
l
 
t
o
 
5
p
i
 
i
s
 
a
p
p
r
o
x
i
ma
t
e
l
y
 
e
q
u
a
l
 
t
o
 
3
.
1
4
We
n
e
x
t
d
e
f
i
n
e
s
y
mb
o
l
s
t
h
a
t
d
e
n
o
t
e
a
n
 
o
r
d
e
r
 
r
e
l
a
t
i
o
n
s
h
i
p
 
b
e
t
w
e
e
n
 
r
e
a
l
n
u
mb
e
r
s
.
<
L
e
s
s
 
t
h
a
n
>
G
r
e
a
t
e
r
 
t
h
a
n
≤
L
e
s
s
 
t
h
a
n
 
o
r
 
e
q
u
a
l
 
t
o
≥
G
r
e
a
t
e
r
 
t
h
a
n
 
o
r
 
e
q
u
a
l
 
t
o
T
h
e
s
e
s
y
mb
o
l
s
a
l
l
o
wu
s
t
oc
o
mp
a
r
e
t
w
on
u
mb
e
r
s
.
F
o
r
e
x
a
mp
l
e
,
−
1
2
0
<
−
1
0
N
e
g
a
t
i
v
e
 
1
2
0
 
i
s
 
l
e
s
s
 
t
h
a
n
 
n
e
g
a
t
i
v
e
 
1
0
S
i
n
c
e
t
h
e
 
g
r
a
p
h
 
o
f
 
−
1
2
0
 
i
s
t
ot
h
e
l
e
f
t
o
f
t
h
e
 
g
r
a
p
h
 
o
f
 
–
1
0
 
o
nt
h
e
n
u
mb
e
r
l
i
n
e
,
t
h
a
t
n
u
mb
e
r
i
s
l
e
s
s
t
h
a
n
 
−
1
0
.
We
c
o
u
l
dw
r
i
t
e
a
n
e
q
u
i
v
a
l
e
n
t
s
t
a
t
e
me
n
t
a
s
f
o
l
l
o
w
s
:
−
1
0
>
−
1
2
0
N
e
g
a
t
i
v
e
 
1
0
 
i
s
 
g
r
e
a
t
e
r
 
t
h
a
n
 
n
e
g
a
t
i
v
e
 
1
2
0
S
i
mi
l
a
r
l
y
,
s
i
n
c
e
t
h
e
 
g
r
a
p
h
 
o
f
z
e
r
oi
s
t
ot
h
e
r
i
g
h
t
o
f
t
h
e
 
g
r
a
p
h
 
o
f
a
n
y
n
e
g
a
t
i
v
e
n
u
mb
e
r
o
n
t
h
e
n
u
mb
e
r
l
i
n
e
,
z
e
r
oi
s
g
r
e
a
t
e
r
t
h
a
na
n
y
n
e
g
a
t
i
v
e
n
u
mb
e
r
.
0
>
−
5
0
Z
e
r
o
 
i
s
 
g
r
e
a
t
e
r
 
t
h
a
n
 
n
e
g
a
t
i
v
e
 
5
0
T
h
e
s
y
mb
o
l
s
 
<
 
a
n
d
 
>
 
a
r
e
u
s
e
dt
od
e
n
o
t
e
 
s
t
r
i
c
t
i
n
e
q
u
a
l
i
t
i
e
s
,
a
n
dt
h
e
s
y
mb
o
l
s
a
n
da
r
e
u
s
e
dt
od
e
n
o
t
e
 
i
n
c
l
u
s
i
v
e
i
n
e
q
u
a
l
i
t
i
e
s
.
I
ns
o
me
s
i
t
u
a
t
i
o
n
s
,
mo
r
e
t
h
a
n
o
n
e
s
y
mb
o
l
c
a
n
b
e
c
o
r
r
e
c
t
l
y
a
p
p
l
i
e
d
.
F
o
r
e
x
a
mp
l
e
,
t
h
e
f
o
l
l
o
w
i
n
gt
w
os
t
a
t
e
me
n
t
s
a
r
e
b
o
t
ht
r
u
e
:
−
1
0
<
0
 
a
n
d
 
−
1
0
≤
0
I
na
d
d
i
t
i
o
n
,
t
h
e
“
o
r
e
q
u
a
l
t
o
”
c
o
mp
o
n
e
n
t
o
f
a
ni
n
c
l
u
s
i
v
e
i
n
e
q
u
a
l
i
t
y
a
l
l
o
w
s
u
s
t
o
c
o
r
r
e
c
t
l
y
w
r
i
t
e
t
h
e
f
o
l
l
o
w
i
n
g
:
−
1
0
≤
−
1
0
T
h
e
l
o
g
i
c
a
l
u
s
e
o
f
t
h
e
w
o
r
d“
o
r
”
r
e
q
u
i
r
e
s
t
h
a
t
o
n
l
y
o
n
e
o
f
t
h
e
c
o
n
d
i
t
i
o
n
s
n
e
e
db
e
t
r
u
e
:
t
h
e
“
l
e
s
s
t
h
a
n
”
o
r
t
h
e
“
e
q
u
a
l
t
o
.
”
E
x
a
mp
l
e
 
2
F
i
l
l
i
n
t
h
e
b
l
a
n
k
w
i
t
h
 
<
,
=
,
o
r
 
>
: −
2
 
_
_
_
_
 
−
1
2
.
S
o
l
u
t
i
o
n
U
s
e
>
b
e
c
a
u
s
e
t
h
e
 
g
r
a
p
h
 
o
f
 
−
2
 
i
s
t
ot
h
e
r
i
g
h
t
o
f
t
h
e
 
g
r
a
p
h
 
o
f
 
−
1
2
 
o
n
a
n
u
mb
e
r
l
i
n
e
.
T
h
e
r
e
f
o
r
e
,
 
−
2
>
−
1
2
,
w
h
i
c
hr
e
a
d
s
“
n
e
g
a
t
i
v
e
t
w
oi
s
g
r
e
a
t
e
r
t
h
a
n
n
e
g
a
t
i
v
e
t
w
e
l
v
e
.
”
A
n
s
w
e
r
:
−
2
>
−
1
2
A
b
s
o
l
u
t
e
V
a
l
u
e
T
h
e
 
a
b
s
o
l
u
t
e
v
a
l
u
e
 
o
f
a
r
e
a
l
n
u
mb
e
r
 
a
,
d
e
n
o
t
e
d
 
|
a
|
,
i
s
d
e
f
i
n
e
da
s
t
h
e
d
i
s
t
a
n
c
e
b
e
t
w
e
e
n
z
e
r
o(
t
h
e
 
o
r
i
g
i
n
)
a
n
dt
h
e
 
g
r
a
p
h
 
o
f
t
h
a
t
r
e
a
l
n
u
mb
e
r
o
nt
h
e
n
u
mb
e
r
l
i
n
e
.
S
i
n
c
e
i
t
i
s
a
d
i
s
t
a
n
c
e
,
i
t
i
s
a
l
w
a
y
s
p
o
s
i
t
i
v
e
.
F
o
r
e
x
a
mp
l
e
,
|
−
4
|
=
4a
n
d|
4
|
=
4
B
o
t
h
 
4
 
a
n
d
 
−
4
 
a
r
e
f
o
u
r
u
n
i
t
s
f
r
o
mt
h
e
o
r
i
g
i
n
,
a
s
i
l
l
u
s
t
r
a
t
e
db
e
l
o
w
:
A
l
s
o
,
i
t
i
s
w
o
r
t
hn
o
t
i
n
gt
h
a
t
|
0
|
=
0
T
h
e
 
a
b
s
o
l
u
t
e
v
a
l
u
e
 
c
a
nb
e
e
x
p
r
e
s
s
e
dt
e
x
t
u
a
l
l
y
u
s
i
n
gt
h
e
n
o
t
a
t
i
o
n
a
b
s(
a
)
.
We
o
f
t
e
n
e
n
c
o
u
n
t
e
r
n
e
g
a
t
i
v
e
a
b
s
o
l
u
t
e
v
a
l
u
e
s
,
s
u
c
ha
s
 
−
|
3
|
 
o
r
 
–
 
a
b
s
(
3
)
.
N
o
t
i
c
e
t
h
a
t
t
h
e
n
e
g
a
t
i
v
e
s
i
g
n
i
s
i
n
f
r
o
n
t
o
f
t
h
e
 
a
b
s
o
l
u
t
e
v
a
l
u
e
 
s
y
mb
o
l
.
I
nt
h
i
s
c
a
s
e
,
w
o
r
k
t
h
e
 
a
b
s
o
l
u
t
e
v
a
l
u
e
 
f
i
r
s
t
a
n
dt
h
e
n
f
i
n
dt
h
e
 
o
p
p
o
s
i
t
e
 
o
f
t
h
e
r
e
s
u
l
t
.
−
|
3
| −
|
−
3
|
=
−
3 =
−
3
T
r
y
n
o
t
t
oc
o
n
f
u
s
e
t
h
i
s
w
i
t
ht
h
e
 
d
o
u
b
l
e
-
n
e
g
a
t
i
v
e
p
r
o
p
e
r
t
y
,
w
h
i
c
h
s
t
a
t
e
s
t
h
a
t
 
−
(
−
7
)
=
+
7
.
A
t
t
h
i
s
p
o
i
n
t
,
w
e
c
a
n
d
e
t
e
r
mi
n
e
w
h
a
t
 
r
e
a
l
n
u
mb
e
r
s
 
h
a
v
e
a
p
a
r
t
i
c
u
l
a
r
 
a
b
s
o
l
u
t
e
v
a
l
u
e
.
F
o
r
e
x
a
mp
l
e
,
|
?
|
=
5
T
h
i
n
k
o
f
a
r
e
a
l
n
u
mb
e
r
w
h
o
s
e
d
i
s
t
a
n
c
e
t
ot
h
e
 
o
r
i
g
i
n
 
i
s
 
5
 
u
n
i
t
s
.
T
h
e
r
e
a
r
e
t
w
os
o
l
u
t
i
o
n
s
:
t
h
e
d
i
s
t
a
n
c
e
t
ot
h
e
r
i
g
h
t
o
f
t
h
e
 
o
r
i
g
i
n
 
a
n
dt
h
e
d
i
s
t
a
n
c
e
t
ot
h
e
l
e
f
t
o
f
t
h
e
 
o
r
i
g
i
n
,
n
a
me
l
y
,
 
{
±
5
}
.
T
h
e
s
y
mb
o
l
 
(
±
)
 
i
s
r
e
a
d“
p
l
u
s
o
r
mi
n
u
s
”
a
n
di
n
d
i
c
a
t
e
s
t
h
a
t
t
h
e
r
e
a
r
e
t
w
oa
n
s
w
e
r
s
,
o
n
e
p
o
s
i
t
i
v
e
a
n
do
n
e
n
e
g
a
t
i
v
e
.
|
−
5
|
=
5
 
a
n
d|
5
|
=
5
N
o
wc
o
n
s
i
d
e
r
t
h
e
f
o
l
l
o
w
i
n
g
:
|
?
|
=
−
5
H
e
r
e
w
e
w
i
s
ht
of
i
n
da
v
a
l
u
e
f
o
r
w
h
i
c
ht
h
e
d
i
s
t
a
n
c
e
t
ot
h
e
 
o
r
i
g
i
n
 
i
s
n
e
g
a
t
i
v
e
.
S
i
n
c
e
n
e
g
a
t
i
v
e
d
i
s
t
a
n
c
e
i
s
n
o
t
d
e
f
i
n
e
d
,
t
h
i
s
e
q
u
a
t
i
o
n
h
a
s
n
o
 
s
o
l
u
t
i
o
n
.
I
f
a
ne
q
u
a
t
i
o
nh
a
s
Before absolute modes there's not negative sign which
means only a value within mode should be in such a
way that the result will be -5 which is not possible.
n
o
 
s
o
l
u
t
i
o
n
,
w
e
s
a
y
t
h
e
s
o
l
u
t
i
o
n
 
i
s
t
h
e
e
mp
t
y
s
e
t
:
 
∅.
R
e
a
l
,
I
ma
g
i
n
a
r
y
a
n
dC
o
mp
l
e
x
N
u
mb
e
r
s
R
e
a
l
n
u
mb
e
r
s
a
r
e
t
h
e
u
s
u
a
l
p
o
s
i
t
i
v
e
a
n
dn
e
g
a
t
i
v
e
n
u
mb
e
r
s
.
I
f
w
e
mu
l
t
i
p
l
y
a
r
e
a
l
n
u
mb
e
r
b
y
i
,
w
e
c
a
l
l
t
h
e
r
e
s
u
l
t
a
n
i
ma
g
i
n
a
r
y
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z
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(
b+d
)
z
1-
z
2=
a
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b-
c
-
i
d=
a
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c
+i
(
b-
d
)
E
x
a
mp
l
e
(
a
)
(
1+i
)
+(
3+i
)
=1+3+i
(
1+1
)
=4+2
i
(
b
)
(
2+5
i
)
-
(
1-
4
i
)
=2+5
i
-
1+4
i
=
1+9
i

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