Triple Reverse Derivations on Semiprime Rings

Full text

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TRI
P
LE
R
EVER
S
E
D
ERIVA
T
IO
N S
O
NS
E
M
IPRI
M
E
R
I
N
G
S
C. JAYA SUBBA REDDY, S. MALLIKARJUNA RAO
&
S. VASANTHA KUMAR
Department of Mathematics, S. V. University, Tirupati, Andhra Pradesh, India
ABSTRACT
I
n thi
s paper
,
w
e p
r
ove that i
f
d
i
s
a
J
o
r
dan
r
eve
rs
e de
r
ivation o
f
a
s
emip
r
ime
r
ing
R
, then d I
s
a qua
r
te
r r
eve
rs
e
de
r
ivation.
Us
ing thi
s
,
w
e
s
ho
w
that d i
s
a t
r
iple
r
eve
rs
e de
r
ivation.
KEYWORDS:
Semi Prime Ring, Jordan Reverse Derivation, Triple Reverse Derivation, Quarter Reverse Derivation,
Center
INTRODUCTION
H
e
rs
tein
[
1
] s
tudied
Jor
dan de
r
ivation
s
o
f
p
r
ime
r
ing
s
and p
r
oved that eve
r
y
J
o
r
dan de
r
ivation on a p
r
ime
r
i
n
g o
f
cha
r
.≠2 i
s
a de
r
ivation and al
s
o
s
tudied
r
eve
rs
e de
r
ivation
s
o
f
p
r
ime
r
ing
s
.
K
.
S
uva
r
na and
D
.
S
.
Irf
ana
[
3
] s
tudied
s
ome
p
r
ope
r
tie
s
o
f J
o
r
dan de
r
ivation
s
on
s
emi p
r
ime
r
ing
s
.
W
e kno
w
that an additive mapping
d
:
R
→
R
i
s
called a
J
o
r
dan
r
eve
rs
e De
r
ivation i
f
d
(
x
2)
=
d
(
x
)
x
+
xd
(
x
)
f
o
r
all x in R.
A
n additive map d
fr
om a
r
ing R to R i
s
a t
r
iple
r
eve
rs
e de
r
ivation i
f
d
(
xyx
)
=
d
(
x
)
yx
+
xd
(
y
)
x
+
yxd
(
x
)
hold,
f
o
r
all x
,
y
∈
R
and
d
i
s
a qua
r
te
r r
eve
rs
e de
r
ivation i
f
d
(
xyzx
)
=
d
(
x
)
xyz
+
xzd
(
y
)
x
+
d
(
z
)
xy
+
xzyd
(
x
)
hold,
f
o
r
all
x
,
y
∈
R
.
Th
r
oughout thi
s Paper
,
R
w
ill de not ea
s
emi p
r
ime
r
ing and
z
it
s
cente
r
.
Le
mm
a 1:
Let
R
be any
r
ing and let
T
(
a
)
=
{
r
∈
R
/
r
(
ax
−
xa
)
=
0,
for al lx
∈
R
}
,
f
o
r
all
a
∈
R
. Then
T
(
a
)i
s
a
t
w
o
-s
ided ideal o
f
R
.
Proo
f:
Clea
r
ly
T
(
a
) i
s
a le
f
t ideal o
f
R
.
I
t
r
emain
s
to
s
ho
w
that i
f
u
∈
T
(
a
)
,
x
∈
R
, then
ux
∈
T
(
a
)
. But then,
f
o
r
all
r
∈
R
,
u
(
axr
−
xra
)
=
0
.
Thu
s
u
{(
ax
−
xa
)
r
+
x
(
ar
−
ra
)}
=
0
.
S
ince
u
∈
T
(
a
),
u
(
ax
−
xa
)
=
0
, and
s
o
w
e have that
ux
(
ar
−
ra
)
=
0
,
f
o
r
all
r
∈
R
.
Then
ux
∈
T
(
a
)
.
H
ence the Lemma i
s
p
r
oved.
■
Le
mm
a 2:
If
R
i
s
a p
r
ime
r
ing and i
f
a
∈
R
i
s
not in
Z
, the cente
r
o
f
R
, then
T
(
a
)
=
(0)
.
Proo
f:
S
ince
a
∉
Z
,
f
o
r s
ome
b
∈
R
,
ab
−
ba
≠
0
.
If
T
(
a
)
≠
(0)
, then
T
(
a
)(
ab
−
ba
)
=
(0)
.
S
o, the
r
ight ideal
T
(
a
) i
s
annihilated by an on
-
ze
r
oelement. By the de
f
inition o
f
a p
r
ime
r
ing, eithe
r
T
(
a
)
=
(0)
o
r
ab
−
ba
=
0
.
S
ince
a
∉
Z
,
ab
−
ba
≠
0
.
S
o,
T
(
a
)
=
(0)
.
■
N
o
w
,
w
e p
r
ove the
f
ollo
w
ing
r
e
s
ult
s
:
T
h
eore
m
1:
If
d
i
s
a
J
o
r
dan
r
eve
rs
ed e
r
ivation o
f
a
s
emi p
r
ime
r
ing
R
, then
d
i
s
a qua
r
te
r r
eve
rs
e de
r
ivation, that
i
s
,
d
(
abca
)
=
d
(
a
)
abc
+
acd
(
b
)
a
+
ad
(
c
)
ab
+
acbd
(
a
)
,
f
o
r
all
a
,
b
,
c
∈
R
.
International Journal of Applied Mathematics &
Statistical Sciences (IJAMSS)
ISSN(P): 2319-3972; ISSN(E): 2319-3980
Vol. 4, Issue 4, Jun - Jul 2015, 25-32
© IASET

26
C. Jaya Subba Reddy, S. Mallikarjuna Rao
&
S. Vasantha Kumar
Impact Factor (JCC): 2.0346 NAAS Rating: 3.19
Proo
f:
We have
d
(
a
2)
=
d
(
a
)
a
+
ad
(
a
)
We
r
eplace
a
by
a
+
bc
. Then
w
eget,
⇒
d
(
abc
+
bca
)
=
d
(
abc
)
+
d
(
bca
)
⇒
d
(
abc
+
bca
)
=
d
(
c
)
ba
+
cd
(
b
)
a
+
bcd
(
a
)
+
d
(
a
)
cb
+
ad
(
c
)
b
+
cad
(
b
),
(
1
)
f
o
r
all
a
,
b
,
c
∈
R
.
S
imila
r
ly,
⇒
d
(
abc
+
bca
)
=
d
(
c
)
ab
+
cd
(
b
)
a
+
cbd
(
a
)
+
d
(
a
)
bc
+
ad
(
c
)
b
+
acd
(
b
),…..
(
2
)
f
o
r
all
a
,
b
,
c
∈
R
By e
q
uating equ.
’s(
1
)
and
(
2
)
,
w
eget
⇒
d
(
c
)
ba
+
bcd
(
a
)
+
d
(
a
)
cb
+
cad
(
b
)
=
d
(
c
)
ab
+
cbd
(
a
)
+
d
(
a
)
bc
+
acd
(
b
),…..
(
3
)
f
o
r
all
a
,
b
,
c
∈
R
Con
s
ide
r
W
=
d
((
abc
+
bca
)
a
+
a
(
abc
+
bca
))
=
d
(
a
)(
abc
+
bca
)
+
ad
(
abc
+
bca
)
+
d
(
abc
+
bca
)
a
+
(
abc
+
bca
)
d
(
a
)
Fr
om equ.
(
2
)w
ehave,
=
d
(
a
)
abc
+
d
(
a
)
bca
+
a
[
d
(
c
)
ab
+
cd
(
b
)
a
+
cbd
(
a
)
+
d
(
a
)
bc
+
ad
(
c
)
b
+
acd
(
b
)]
+
[
d
(
c
)
ab
+
cd
(
b
)
a
+
cbd
(
a
)
+
d
(
a
)
bc
+
ad
(
c
)
b
+
acd
(
b
)]
a
+
abcd
(
a
)
+
bcad
(
a
)
⇒
W
=
d
(
a
)
abc
+
2
d
(
a
)
bca
+
ad
(
c
)
ab
+
2
acd
(
b
)
a
+
acbd
(
a
)
+
ad
(
a
)
bc
+
a
2
d
(
c
)
b
+
a
2
cd
(
b
)
+
d
(
c
)
aba
+
cd
(
b
)
a
2
+
..
(
4
)
cbd
(
a
)
a
+
ad
(
c
)
ba
+
abcd
(
a
)
+
bcad
(
a
)
O
n the othe
r
hand,
⇒
W
=
d
((
abc
+
bca
)
a
+
a
(
abc
+
bca
))
⇒
W
=
d
(2
abca
+
bca
2
+
a
2
bc
)
=
2
d
(
abca
)
+
d
(
a
2)
bc
+
a
2
d
(
bc
)
+
d
(
c
)
a
2
b
+
cd
(
a
2
b
)
⇒
W
=
2
d
(
abca
)
+
d
(
a
)
abc
+
ad
(
a
)
bc
+
a
2
d
(
c
)
b
+
a
2
cd
(
b
)
+
d
(
c
)
a
2
b
+
cd
(
b
)
a
2
+
cbd
(
a
)
a
+
cbad
(
a
) (5)

Tri
p
le
R
ever
s
E
d
eriva
t
io
n S
o
ns
E
m
ipri
m
e
R
i
n
g
s
27
www.iaset.us editor@iaset.us
By c
o
mpa
r
ing equ.
’s (
4
)
an
d (
5
)
,
f
o
r
W
,
w
e obtain,
⇒
2
d
(
a
)
bca
+
ad
(
c
)
ab
+
2
acd
(
b
)
a
+
acbd
(
a
)
+
d
(
c
)
aba
+
ad
(
c
)
ba
+
abcd
(
a
)
+
bcad
(
a
)
=
2
d
(
abca
)
+
d
(
c
)
a
2
b
+
c
bad
(
a
)
Fr
om the equality given in equ.
(
3
)
,
w
e have,
⇒
2
d
(
a
)
abc
+
2
ad
(
c
)
ab
+
2
acd
(
b
)
a
+
2
acbd
(
a
)
=
2
d
(
abca
)
⇒
d
(
abca
)
=
d
(
a
)
abc
+
acd
(
b
)
a
+
ad
(
c
)
ab
+
acbd
(
a
)
.
T
h
eore
m
2:
If
d
i
s
a
J
o
r
dan
r
eve
rs
e de
r
ivation o
f
a
s
emi p
r
ime
r
ing
R
, then
d
i
s
a t
r
iple
r
eve
rs
e de
r
ivation, that
i
s
,
d
(
aba
)
=
d
(
a
)
ba
+
ad
(
b
)
a
+
bad
(
a
)
,
f
o
r
all
a
,
b
,
c
∈
R
.
Proo
f:
We have
d
(
a
2)
=
d
(
a
)
a
+
ad
(
a
)
We
r
eplace
a
by
a
+
b
, then
w
eget,
⇒
d
(
ab
+
ba
)
=
d
(
ab
)
+
d
(
ba
)
=
d
(
b
)
a
+
bd
(
a
)
+
d
(
a
)
b
+
ad
(
b
)
,
f
o
r
all
a
,
b
,
c
∈
R
. …..
(
6
)
Con
s
ide
r
W
=
d
((
ab
+
ba
)
a
+
a
(
ab
+
ba
))
.
=
d
(
a
)(
ab
+
ba
)
+
ad
(
ab
+
ba
)
+
d
(
ab
+
ba
)
a
+
(
ab
+
ba
)
d
(
a
)
=
d
(
a
)
ab
+
d
(
a
)
ba
+
a
[
d
(
b
)
a
+
bd
(
a
)
+
d
(
a
)
b
+
ad
(
b
)]
+
[
d
(
b
)
a
+
bd
(
a
)
+
d
(
a
)
b
+
ad
(
b
)]
a
+
abd
(
a
)
+
bad
(
a
)
=
d
(
a
)
ab
+
d
(
a
)
ba
+
ad
(
b
)
a
+
abd
(
a
)
+
ad
(
a
)
b
+
a
2
d
(
b
)
+
d
(
b
)
a
2
+
bd
(
a
)
a
+
d
(
a
)
ba
+
ad
(
b
)
a
+
abd
(
a
)
+
bad
(
a
⇒
W
=
d
(
a
)
ab
+
2
d
(
a
)
ba
+
2
ad
(
b
)
a
+
2
abd
(
a
)
+
ad
(
a
)
b
+
a
2
d
(
b
)
+
d
(
b
)
a
2
+
bd
(
a
)
a
+
bad
(
a
)
….. .
(
7
)
O
n the othe
r
hand,
w
e get,
⇒
W
=
d
((
ab
+
ba
)
a
+
a
(
ab
+
ba
))
=
d
(
ba
2
+
a
2
b
+
2
aba
)
=
d
(
a
2)
b
+
a
2
d
(
b
)
+
d
(
b
)
a
2
+
bd
(
a
2)
+
2
d
(
aba
)
⇒
W
=
d
(
a
)
ab
+
ad
(
a
)
b
+
a
2
d
(
b
)
+
d
(
b
)
a
2
+
bd
(
a
)
a
+
bad
(
a
)
+
2
d
(
aba
)…..
(
8
)
By c
o
mpa
r
ing equ.
’s (
7
)
an
d (
8
) f
o
r
W
,
w
e obtain,
⇒
d
(
a
)
ab
+
2
d
(
a
)
ba
+
2
ad
(
b
)
a
+
2
abd
(
a
)
+
ad
(
a
)
b
+
a
2
d
(
b
)
+
d
(
b
)
a
2
+
bd
(
a
)
a
+
bad
(
a
)
=
d
(
a
)
ab
+
ad
(
a
)
b
+
a
2
d
(
b
)
+

28
C. Jaya Subba Reddy, S. Mallikarjuna Rao
&
S. Vasantha Kumar
Impact Factor (JCC): 2.0346 NAAS Rating: 3.19
d
(
b
)
a
2
+
bd
(
a
)
a
+
bad
(
a
)
+
2
d
(
aba
)
⇒
2
d
(
a
)
ba
+
2
ad
(
b
)
a
+
2
bad
(
a
)
=
2
d
(
aba
)
⇒
2[
d
(
a
)
ba
+
ad
(
b
)
a
+
bad
(
a
)]
=
2[
d
(
aba
)]
⇒
d
(
a
)
ba
+
ad
(
b
)
a
+
bad
(
a
)
=
d
(
aba
)
■
We linea
r
ize the
r
e
s
ult
s
o
f
Theo
r
em: 2 by
r
eplacing
a
by
a
+
c
,
w
e a
rr
ive at
T
h
eore
m
3:
F
o
r
all
a
,
b
,
c
∈
R
,
d
(
abc
+
cba
)
=
d
(
a
)
bc
+
d
(
c
)
ba
+
ad
(
b
)
c
+
cd
(
b
)
a
+
bad
(
c
)
+
bcd
(
a
)
T
h
eore
m
4:
F
o
r
all
a
,
b
∈
R
,
(
d
(
a
)
b
+
ad
(
b
))(
ba
−
ab
)
−
(
ba
−
ab
)
d
(
ab
)
=
0
.
Proo
f:
Con
s
ide
r
W
=
d
(
ab
(
ab
)
+
(
ab
)
ba
)
Byu
s
ing Theo
r
em:3,
w
ith
c
=
ab
,
w
e obtain,
⇒
W
=
d
(
ab
)
ab
+
(
ab
)
d
(
ab
)
+
d
(
ba
)(
ab
)
+
bad
(
ab
)
=
d
(
ab
)
ab
+
(
ab
)
d
(
b
)
a
+
(
ab
)
bd
(
a
)
+
d
(
a
)
b
(
ab
)
+
ad
(
b
)(
ab
)
+
bad
(
ab
)
(
9
)
H
o
w
eve
r
,
W
=
d
(
ab
(
ab
)
+
ab
2
a
)
=
d
(
ab
(
ab
))
+
d
(
ab
2
a
)
=
d
(
ab
)
ab
+
(
ab
)
d
(
ab
)
+
d
(
b
2
a
)
a
+
b
2
ad
(
a
)
⇒
W
=
d
(
ab
)
ab
+
(
ab
)
d
(
ab
)
+
d
(
a
)
b
2
a
+
ad
(
b
)
ba
+
abd
(
b
)
a
+
b
2
ad
(
a
).
(1
0
)
W
can al
s
o be
wr
itten a
s
,
⇒
W
=
d
(
ab
(
ab
))
+
d
(
ab
2
a
)
=
d
(
ab
)
ab
+
(
ab
)
d
(
ab
)
+
d
(
a
)
ab
2
+
ad
(
ab
2)
⇒
W
=
d
(
ab
)
ab
+
(
ab
)
d
(
ab
)
+
d
(
a
)
ab
2
+
ad
(
b
)
ba
+
abd
(
b
)
a
+
ab
2
d
(
a
)….
(1
1
)
By e
q
uating equ.
’s(
10
)
and
(
11
)
,
w
eget,
⇒
d
(
a
)
b
2
a
+
b
2
ad
(
a
)
=
d
(
a
)
ab
2
+
ab
2
d
(
a
)
⇒
d
(
a
)
b
2
a
=
d
(
a
)
ab
2
a
n
d al
s
o,
b
2
ad
(
a
)
=
ab
2
d
(
a
)
…. .
(1
2
)
By c
o
mpa
r
ing equ.
’s(
9
)
an
d (
10
) f
o
r
W
,
w
e obtain,

Tri
p
le
R
ever
s
E
d
eriva
t
io
n S
o
ns
E
m
ipri
m
e
R
i
n
g
s
29
www.iaset.us editor@iaset.us
⇒
d
(
ab
)
ab
+
(
ab
)
d
(
b
)
a
+
(
ab
)
bd
(
a
)
+
d
(
a
)
b
(
ab
)
+
ad
(
b
)(
ab
)
+
bad
(
ab
)
=
d
(
ab
)
ab
+
(
ab
)
d
(
ab
)
+
d
(
a
)
b
2
a
+
ad
(
b
)
ba
+
abd
(
b
)
a
+
b
2
ad
(
a
)
Bye
q
u.
(
12
)
,
w
ehave,
⇒
(
ab
)
bd
(
a
)
+
d
(
a
)
b
(
ab
)
+
ad
(
b
)(
ab
)
+
bad
(
ab
)
=
(
ab
)
d
(
ab
)
+
d
(
a
)
b
2
a
+
ad
(
b
)
ba
+
ab
2
d
(
a
)
⇒
d
(
a
)
b
(
ab
)
+
ad
(
b
)(
ab
)
+
bad
(
ab
)
=
(
ab
)
d
(
ab
)
+
d
(
a
)
b
2
a
+
ad
(
b
)
ba
By t
r
an
s
po
s
ing and collecting te
r
m
s
, it
f
ollo
ws
that
⇒
(
d
(
a
)
b
+
ad
(
b
))(
ba
−
ab
)
−
(
ba
−
ab
)
d
(
ab
)
=
0
.
■
T
h
eore
m
5:
If
R
i
s
a
s
emi p
r
ime
r
ing o
f
cha
r
. ≠2, then any
J
o
r
dan
r
eve
rs
e de
r
ivation o
f
R
i
s
an o
r
dina
r
y
r
eve
rs
e
de
r
ivation o
f
R
.
■
T
h
eore
m
6: Let
R
bea
s
emi p
r
ime
r
ing o
f
cha
r
.≠2and
d
i
s
a
J
o
r
dan
r
eve
rs
ede
r
ivationo
f
R
.
If
d
act
s
a
s
an
anti
-
homomo
r
phi
s
m o
f
R
, then d i
s
a cent
r
al de
r
ivation.
Proo
f:
S
ince d act
s
a
s
ananti
-
homom
or
phi
s
m,
w
e have,
d
(
xy
)
=
d
(
y
)
d
(
x
)
…..
(1
3
)
Fr
om Theo
r
em: 5,
w
e have,
d
(
xy
)
=
d
(
y
)
x
+
yd
(
x
)
,
f
o
r
allx,y
∈
R. Then
⇒
d
(
y
)
d
(
x
)
=
d
(
y
)
x
+
yd
(
x
)
…..
(1
4
)
By
s
ub
s
tituting
zy
f
o
r
y
i n equ.
(
14
)
,
w
e obtain,
d
(
zy
)
d
(
x
)
=
d
(
zy
)
x
+
zyd
(
x
)
⇒
d
(
zy
)
d
(
x
)
=
d
(
y
)
d
(
z
)
x
+
zyd
(
x
)
f
o
r
all
x
,
y
,
z
∈
R
…..
(1
5
)
S
ince d i
s
an anti
-
homomo
r
phi
s
m.
O
n the othe
r
hand,
w
e have,
d
(
zy
)
d
(
x
)
=
d
(
y
)
d
(
z
)
d
(
x
)
=
d
(
y
)
d
(
xz
)
=
d
(
y
)[
d
(
z
)
x
+
zd
(
x
)]
⇒
d
(
zy
)
d
(
x
)
=
d
(
y
)
d
(
z
)
x
+
d
(
y
)
zd
(
x
)
…..
(1
6
)
Fr
om equation
s (
15
)
and
(
16
)
,
w
e have,
d
(
y
)
d
(
z
)
x
+
zyd
(
x
)
=
d
(
y
)
d
(
z
)
x
+
d
(
y
)
zd
(
x
)
⇒
zyd
(
x
)
=
d
(
y
)
zd
(
x
)
…..
(1
7
)

30
C. Jaya Subba Reddy, S. Mallikarjuna Rao
&
S. Vasantha Kumar
Impact Factor (JCC): 2.0346 NAAS Rating: 3.19
P
ut
y
=
xy
in equ.
(
17
)
and u
s
ing equ.
(
17
)
zxyd
(
x
)
=
d
(
xy
)
zd
(
x
)
=
d
(
y
)
xzd
(
x
)
+
yd
(
x
)
zd
(
x
)
⇒
zxyd
(
x
)
=
zxyd
(
x
)
+
yd
(
x
)
zd
(
x
)
⇒
yd
(
x
)
zd
(
x
)
=
0
If w
e
r
e place
z
by
ry
, then
w
e
g
et,
yd
(
x
)
ryd
(
x
)
=
0
,
f
o
r
all
r
∈
R
⇒
yd
(
x
)
Ryd
(
x
)
=
0
By
s
emi p
r
imene
ss
o
f
R
,
y d
(
x
)
=
0
.
d
(
x
)
=
0
P
ut
x
=
y
implie
s
d
(
y
)
=
0
.
If w
e multiply
w
ith
[
x
,
y
]
fr
om the le
f
t hand
s
ide, then
w
eget,
[
x
,
y
]
d
(
y
)
=
0
,
f
o
r
all
x
,
y
∈
R
…..
(1
8
)
We
r
eplace
x
by
xz
in equ.
(1
8
)
and u
s
ing equ.
(
18
)
, then
[
xz
,
y
]
d
(
y
)
=
0
,
f
o
r
all
x
,
y
,
z
∈
R
⇒
[
x
,
y
]
zd
(
y
)
=
0
,
f
o
r
all
x
,
y
,
z
∈
R
…..
(1
9
)
O
n the othe
r
hand,a linea
r
ization o
f
equ.
(
18
)
lead
s
to
[
x
,
y
+
u
]
d
(
y
+
u
)
=
0
⇒
[
x
,
y
]
d
(
y
)
+
[
x
,
y
]
d
(
u
)
+
[
x
,
u
]
d
(
y
)
+
[
x
,
u
]
d
(
u
)
=
0
⇒
[
x
,
y
]
d
(
u
)
+
[
x
,
u
]
d
(
y
)
=
0
⇒
[
x
,
u
]
d
(
y
)
= −
[
x
,
y
]
d
(
u
)
=
[
y
,
x
]
d
(
u
)
…..
(2
0
)
Ifw
e
r
eplace
z
by
d
(
u
)
z
[
x
,
u
]
in equ.
(
19
)
and u
s
ingequ.
(
20
)
, then
w
e get,
⇒
[
x
,
y
]
d
(
u
)
z
[
x
,
u
]
d
(
y
)
=
0
⇒−
[
x
,
y
]
d
(
u
)
z
[
x
,
y
]
d
(
u
)
=
0
⇒
[
x
,
y
]
d
(
u
)
z
[
x
,
y
]
d
(
u
)
=
0
…..
(2
1
)
S
ince
R
i
s s
emip
r
ime, by equ.
(
21
) w
eget,
[
x
,
y
]
d
(
u
)
=
0
,
f
o
r
all
x
,
y
,
u
in
R
.
By
[
2
] (
i.e., Ring
s w
ithin volution–
I
.
N
.
H
e
rs
tein
)
,
d
(
u
)
∈
Z
,
f
o
r
all
u
∈
R
. Thi
s s
ho
ws
that
d
i
s
a
J
o
r
dan
r
eve
rs
ed
e
r
ivation on
R
w
hich map
s
R
into it
s
cente
r
.
■

Tri
p
le
R
ever
s
E
d
eriva
t
io
n S
o
ns
E
m
ipri
m
e
R
i
n
g
s
31
www.iaset.us editor@iaset.us
REFERENCES
1. Herstein. I. N. Jordan derivations of prime rings, Proc. Amer. Math. Soc. 8 (1957), 1104-1110.
2. Herstein. I. N. Rings with involution, Univ. of Chicago press, Chicago, 1976.
3. Suvarna. K. and Irfana. D. S. Triple derivations of semi prime rings, International Journal of Mathematical
Sciences and Engineering Application, Vol.5, May (2011), 117-121.