Transcript fractals

Chapter three:


Far rings and some results
In This chapter , we introduce some
results about far rings and period profiles,
and we prove some properties of far rings
which are known for the usual rings and
some other results.
Section one:



Far rings and period profiles
In this section we introduce the meaning
of the period profile and we explain the
relation between them . This relation is
important in the subject of faster way to
get fractal images.
First we give the definition of period
profile.
Definition 3.1:


Let (ZN,+, ๏) be a far ring; and α1,…, αk
be in ZN. The period profile of α1,…, αk is
a list in descending order of the period
lengths of the k-sequences which are
obtained by recurrence relation over the
far ring:
Sn= α1๏Sn-1, α2๏ Sn-2+…+ αk๏Sn-k
n≥k


Whatever the initial values S0S1…Sk-1
which are elements of ZN.
To illustrate the concept of period profile,
we take the simple case k=2 and we
consider the far ring of order 4.


From examples 2.17 in chapter2 , there
are four possible far rings of order 4. They
are FRA, FRB,FRC, and FRD.
Let α1, α2 be in Z4. Recurrence relations of
length k=2 over the far ring (Z4,+,๏) are
given by Sn= α1๏Sn-1+ α2๏Sn-2
n ≥ 2;


Where any two elements S0S1from ZN can
be taken as initial values to obtain each of
these recurrence relation .
The period of the sequence S0S1 … cannot
exceed 42=16 , which is the maximum
number of distinct pairs SiSi+1 which can
be encountered before repetition.



Further, any such pair may be taken as
initial so is found in some sequence Thus
the period lengths sum to 16.
The period profile of (α1, α2) is a list of
period lengths in decending order, and the
set of lists for all
(α1, α2) in any order , is the profile of the
multiplication table of the far ring itself .


For example when the coefficients α1, α2
=0,1 are taken in FRA, then we get the
sequence 120322…, 102300 ,133…,and
1…, which are periodic of period 6,6,3,and
1 respectively.
Therefore the period profile for 0,1 is
6631.
 The
following tables give the
period profile for all coefficients αi,
αj in Z4and for the four far rings
(of order four) FRA,FRB,FRC, and
FRD.
Table 3.1
coefficients
Sequences over far ring FRA (period
length)
Period
profile
11
12
13
10
21
22
23
20
31
32
33
30
01
02
03
00
112310(6);130332(6);220(3);0(1)
113120032330102(15);2(1)
132230020(9);1103(4);12(2);3(1)
122100(6);232030(6);133(3);1(1)
113233020121003(15);2(1)
1220300231332(13);110(3)
123200102213033(15)1(1)
131032230(9);11202(5);3(1);0(1)
110301312232002(15);3(1)
123033100(9);1322(4);20(2);1(1)
11203210(8);133023(6);2(1);0(1)
113003(6);10220(5);233(3);12(2)
120322(6);102300(6);133(3);1(1)
13022320(8);112103(6);3(1);0(1)
1131003012202(13);233(3)
110(3);123(3);132(3);200(3);330(3);2(1)
6631
15,1
9421
6631
15,1
13,3
15,1
9511
15,1
9421
8611
6532
6631
8611
13,3
333331
Table 3.2
coefficients
Sequences over far ring FRB (period
length)
Period
profiles
11
12
13
10
21
22
23
20
31
32
33
30
01
02
03
00
112310(6);130332(6);220(3);0(1)
120102(6);223003(6);113(3);3(1)
110302331320012(15);2(1)
122100203(9);1330(4);23(2);1(1)
122320(6);100320(6);113(3);3(1)
110(3);123(3);132(3);200(3);330(3);2(1)
12031022(8);133230(6);1(1);0(1)
1300103220233(13);112(3)
110203230013312(15);2(1)
122103130(9);23320(5);1(1);0(1)
1123220100213(13);330(3)
113212003022310(15);3(1)
120023301031322(15);1(1)
133100(6);22030(5);112(3);23(2)
113003210(9);1223(4);20(2);3(1)
12303320(8);110213(6);2(1);0(1)
6631
6631
15,1
9421
6631
333331
8611
13,3
15,1
9511
13,3
15,1
15,1
6532
9421
8611
Table 3.3
coefficients
Sequences over far ring FRC (period
length)
Period
profile
11
12
13
10
21
22
23
20
31
32
33
30
01
02
03
00
112310(6);130332(6);220(3);0(1)
120300(6);102322(6);113(3);3(1)
110312(6);132330(6);200(3);2(1)
122302(6);100320(6);133(3);1(1)
120300(6);102322(6);113(3);3(1)
110312(6);132330(6);200(3);2(1)
122302(6);100320(6);133(3);1(1)
112310(6);130332(6);220(3);0(1)
110312(6);132330(6);200(3);2(1)
122302(6);100320(6);133(3);1(1)
112310(6);130332(6);220(3);0(1)
120300(6);102322(6);113(3);3(1)
122302(6);100320(6);133(3);1(1)
112310(6);130332(6);220(3);0(1)
120300(6);102322(6);113(3);3(1)
110312(6);132330;200(3);2(1)
6631
6631
6631
6631
6631
6631
6631
6631
6631
6631
6631
6631
6631
6631
6631
6631
Table 3.4
coefficients
Sequences over far ring FRD(period
length)
Period
profile
11
12
13
10
21
22
23
20
31
32
33
30
01
02
03
00
112310(6);130332(6);220(3);0(1)
113(3);120102(6);223003(6);3(1)
110312(6);132330(6);200(3);2(1)
122100(6);232030(6);133(3);1(1)
122320(6);100302(6);113(3);3(1)
110(3);123(3);132(3);200(3);330(3);2(1)
120322(6);102300(6);133(3);1(1)
112(3);130(3);103(3);220(3);233(3);0(1)
110312(6);132330(6);200(3);2(1)
122100(6);232030(6);133(3);1(1)
112310(6);130332(6);220(3);0(1)
120102(6);223003(6);113(3);3(1)
120322(6);102300(6);133(3);1(1)
112(3);130(3);103(3);220(3);233(3);0(1)
122320(6);100302(6);113(3);3(1)
110(3);123(3);132(3);200(3);330(3);2(1)
6631
6631
6631
6631
6631
333331
6631
333331
6631
6631
6631
6631
6631
333331
6631
333331
Definition 3.2:



Let each of (G1,๏) and (G2,*) be a
quasigroup. An isomorphism Ψ from G1
into G2 is a bijection or a permutation such
that for all x,y in G1
Ψ(x ๏ y)=Ψ(x) *Ψ(y).
Then (G1,๏)and(G2, *) are called
isomorphic.


Our aim is to obtain far rings (of the same
order) which have the same period
profiles.
First the following example shows that
the isomorphism of quasigroups does not
suffice for equal profiles.
Example 3.3

Consider (Z4,๏)and (Z4, *) such that ๏ and
* are two binary operations defined by the
following tables
๏
1230
*
1230
1
2
3
0
1230
2013
3102
0321
1
2
3
0
1230
2301
3012
0123




Define Ψ from the quasigroup (Z4,๏) into
the quasigroup (Z4, *)by
Ψ(0)=3, Ψ(1)=1,Ψ(2)=2, and Ψ(3)=0.
It is easy to show that Ψ is an
isomorphism.
But (Z4,๏)=FRA and (Z4, *)=FRC have
different period profiles ;


see table 3.1 and table 3.3.
We shall see that the following
definition of isomorphism of far rings is
sufficient to make isomorphic far rings of
the same period profiles.
Definition 3.4:



An isomorphism of far rings
(ZN,+,๏)→(ZN,+, *) is a pair of
permutations Ψ , ø of ZN such that
Ψ(1)=1 and for each α,β,a,b in ZN,
ø (α ๏ a+ β๏b)= Ψ(α ) * ø(a) + Ψ(β) *
ø(b) .
Then (ZN,+,๏) and (ZN,+, *) are called
isomorphic far rings .

The following theorem proves that any
two isomorphic far rings have the same
period profiles.
Theorem 3.5

Isomorphic far rings have the same period
profiles .In particular for k=2 If(Ψ,ø) is an
isomorphism from the far ring( ZN,+, ๏ )
into the far ring (ZN,+, *) then the
coefficient pairs α,β and Ψ(α), Ψ(β)in ZN
have the same period profile, and similarly
for k-tuples with k>2.
Proof



First let k=2 and suppose α,β give a
sequence {an} of period t.That is , with
addition as usual mod N, we have
an=α๏an-1+ β๏an-2.
Using ø(α๏a+β๏b) =Ψ(α )*ø(a)+Ψ
(β)+ø(b)for any α,β, a, b in ZN ….…(1)



Then the sequence
{an}={a0,a1,…}satisfies that at=a0,at+1=a1,
and so on.
Thus , the coefficients Ψ(α),Ψ(β)
give the sequence
{ø(an)}={ø(a0),ø(a1),…,ø(an),…}

Now, we need to prove that {ø(an)}is
periodic of period t; we must prove that
Ø(at)=ø(a0),ø(at-1)=ø(a1) and there is no
positive integer r such that r<t and
Ø(ar)=ø(a0),ø(ar+1)=ø(a1).

It is clear that ø(at)=ø(a0)and
ø(at+1)=ø(a1)…,since {an} is periodic of
period t.If there is r<t such that
ø(ar)=ø(a0), ø(ar+1)=ø(a1)and since ø is
bijection, then we get that there is r<t
such that ar=a0,ar+1=a1,


but this is a contradiction to the fact that
the period of {an} is t and r≠t; and t is the
least positive integer making {an} periodic
.Therefore we get that the
sequence{ø(an)} is periodic of period t;
Which generated by Ψ(α),Ψ(β) with * as
multiplication.



Finally , we may infer that these two far
rings have the same period peofile ;since
the bijection Ψ defines a bejiction of pairs
(α,β )→(Ψ(α),Ψ(β)).
This complete the proof for the case k=2 .



The proof for the case k>2 is by
induction .
It suffices to illustrate the proof for the
case k=3. Let α,β,γ,a,b,c be in ZN and
x= α๏a+β๏b+γ๏c,
then





Ø(x) =ø(α๏a+β๏b+γ๏c)
= ø(α๏a+1๏(β๏b+γ๏c))
Since 1 acts as identify for ๏,
=Ψ(α) *ø(a)+Ψ(1) *ø(β๏b+γ๏c) ; since
Ψ(1)=1 is an identity for *.
=Ψ(α) *ø(α)+Ψ(β) *ø(b)+Ψ(γ) *ø(c);
since +is associative and by using (1).

By using the same notions in case k=2; it
is easy to show that the coefficients
Ψ(α),Ψ(β),Ψ(γ)generate the sequence
{ø(an)}of period t,if the coefficients
α,β,and γ generate the sequence {an}of
period t.

We get the same result for k>3 by the
same method. This means that if the
coefficients α1,α2,…,αk generate a
sequence {an} of period t.

under the multiplication๏, then the
coefficients Ψ(α1),Ψ(α2),…,Ψ(αk) generate
the sequence {ø(an)} of period t under
multiplication *.
Consequently ,isomorphic far rings have
the same period profiles for all k≥2.
Note:


The important of theorem 3.5 is the result
that if any one of these isomorphic far
rings gives an M-sequence, then the
others do the same.
Now we search for what is required to
prove the converse of theorem 3.5 .We
give the following preparatory theorem for
k=2.
Theorem3.6

let each of (ZN,+,๏) and (ZN,+, *) be
a far ring which are denoted by FRX,FRY
respectively . Suppose that Ψ,ø are
permutations of ZN,Ψ(1)=1 and ø sends
sequences over FRX generated by
coefficients α,β to sequences over FRY
generated by coefficients Ψ(α), Ψ(β) for
any α,β in ZN .
Then (Ψ,ø) is an isomorphism from FRX
into FRY.
Proof:




In order to prove this theorem, we must
prove the following :
For any α,β,a,b in ZN;
Ø(α๏a+β๏b)=Ψ(α) *ø(a)+Ψ(β) *ø(b).

Now consider α,β,a,b in ZN and take k to
be 2 to obtain a k-sequences over FRX
particularly when k=2. Then a,b must
appear as successive members of some
sequence with coefficents α,β (we can
take a,b as initial members if necessary).



Let c=α๏a+β๏b; c will be in ZN. Then ø
sends this sequence with coefficients α,β
to one over FRY with coefficients
Ψ(α),Ψ(β) which results in
ø ( c) = ø(α๏a+β๏b)
= Ψ(α) *ø(a)+Ψ(β) *ø(b); As
required.


This is true for any four elements of ZN .
And since Ψ(1)=1, we get (Ψ,ø) is An
isomorphism and the proof is complete .
Remark 3.7: let Ψ and ø be permutations
of ZNand each of FRX,FRY is a far ring of
order N.

The statement that (Ψ,ø) preserves kprofiles from FRX to FRY means that ø
sends a sequence defined over FRX by a
recurrence with coefficients αi(1≤i≤k) to
the sequence over FRY defined by a
recurrence with coefficients ø(αi) (1≤i≤k)

The following theorem gives some
conditions to get isomorphic far rings
Theorem 3.8:

Let Ψ,ø be permutations of ZN with
Ψ(1)=1 and ø(0)=0, and let FRX, FRY be
far rings of order N. Suppose that for
some k>2,(Ψ,ø) preserves k-profiles from
FRX to FRY then (Ψ,ø) is a far ring
isomorphism from FRX to FRY.
Proof:

Let (Ψ,ø) preserves k-profiles from FRX=
(ZN,+,๏) to FRY= (ZN,+, *) for k=2 . Then
, from Remark 3.7 , that means ø Sends
sequences over FRX generated by
coefficients α,β to sequences over FRY
generated by coefficients Ψ(α),Ψ(β); for
any α,β in ZN.

And since Ψ(1)=1,then we get that (Ψ,ø)
is an isomorphism. It is sufficient to show
that if for some k>2 ,(Ψ,ø) preserves kprofiles from FRX to FRY then (Ψ,ø)
preserves 2- profiles from FRX to FRY ;
and then the theorem is proved by
theorem 3.6





Let (Ψ,ø) preserves 3-profiles from FRX to
FRY. If c=α๏a+β๏b
α,β,a,b€ZN
Then we can put that
C=α๏a+β๏b+1๏0
And since (Ψ,ø) preserves 3-profiles from
FRX to FRY, then we get
Ø(c) =Ψ(α) * ø(a)+Ψ(β) * ø(b)+Ψ(1) *
ø(0).




Since Ψ(1)=1 the identity and ø(0)=0,
then
Ø(c) =Ψ(α) * ø(a)+Ψ(β) * ø(b)
This shows that (Ψ,ø) preserves 2profiles from FRX to FRY.
In a similar method if for any k>3,(Ψ,ø)
preserves k-profiles from FRX to FRY,

then , we can show that (Ψ,ø)
preserves 2-profiles from FRX to FRY
.Thus the theorem is proved .
Section two



Properties of far rings
In this section ; we shall prove some
results for the far rings which are knows
for the usual rings .
First we give the following theorem.
Theorem 3.9:

The composition of far ring isomorphisms
(Ψ,ø): FRX→FRY and (Ψ’,ø’):FRY→FRZ is
an isomorphism
Proof

Let each of (Ψ,ø) and (Ψ’,ø’) be an
isomorphism from FRX into FRY and from
FRY into FRZ respectively .the composition
of (Ψ,ø) and (Ψ’,ø’) consists of two
permutations from FRX into FRZ will be
denoted by (Ψ’Ψ,ø’ø).


It is clear that each of Ψ’Ψ and ø’ø is a
permutation from FRX into FRZ such that
Ψ’Ψ is a composition map of Ψ and Ψ’;
and ø’ø is a composition map of ø and ø’.
Since Ψ,Ψ’,ø and ø’ are bijections , then
we get that Ψ’Ψ and ø’ø are bijections.
We have also that




Ψ’Ψ (1)= Ψ’(Ψ(1))= Ψ’(1)=1, since Ψ(1)=
Ψ’(1)=1
Now we consider that FRX=(ZN,+,°),
FRY=(ZN,+, *), and FRZ=(ZN,+,๏).
To prove that (Ψ’Ψ,ø’ø) is an
isomorphism from FRX into FRZ, it is
sufficient to prove that for any α,β,a,b in
ZN,
ø’ø(α°a+β°b)= Ψ’Ψ(α) ๏ ø’ø (a)+ Ψ’Ψ(β) ๏
ø’ø(b).
=ø’(ø(α° a+ β°b))
= ø’(Ψ(α) * ø(a)+Ψ(β) * ø(b)),
Since (Ψ,ø) is an isomorphism from FRX into
FRY.
=Ψ’(Ψ(α))๏ ø’ø(a))+Ψ’(Ψ(β))๏
ø’(ø(b))
Since (Ψ’,ø’) is an isomorphism from
FRY into FRZ
=Ψ’Ψ(α) ๏ ø’ø(a)+Ψ’Ψ(β) ๏ ø’ø(b)


Therefore we get that composition of any
two far ring isomorphisms is an
isomorphism and the proof is complete.
By induction ,one can easily prove
the following theorem
Theorem 3.10:



The composition of any finite number of
far ring isomorphisms is an isomorphism .
If (Ψ,ø)is an isomorphism from FRX into
–1
FRY then the inverse of(Ψ,ø) is(Ψ,ø) .
We can
nominate
the
inverse
of
(Ψ,ø)
to
-1 -1
be (Ψ ,ø )which consists of two

-1
-1
Permutations ø and Ψ from FRY
into FRX . Now we give the following
theorem about the inverse of an
isomorphism of far rings
Theorem 3.11

The inverse of an isomorphism (Ψ,ø) from
FRX into FRY is an isomorphism from FRY
into FRX.
Proof


Consider (ZN,+,๏) to be the far ring FRX
and (ZN,+, *)to be the far ring FRY.
Suppose that (Ψ,ø)is an isomorphism from
-1 -1
FRX into FRY .To prove that (Ψ ,ø )is an
isomorphism from FRY into FRX.
We have Ψ and ø which are two bijections
from ZN into ZN, Ψ(1)=1, and for any
α,β,a,b in ZN



Ø(α๏a+β๏b)=Ψ(α) *ø(a)+Ψ(β) *ø(b)
Since Ψ and ø are bijections from ZN into
-1
-1
ZN; then Ψ and ø are also bijections
from ZN into ZN ;and
-1
-1
-1
-1
ΨΨ = IZN = Ψ Ψ, øø = IZN =ø ø, where
IZN is the identity map over ZN.



-1
Since Ψ(1)=1, we get Ψ (1)=1
-1 -1
To prove that (Ψ ,ø ) is an isomorphism
from FRY into FRX , it is sufficient to prove
that for any α’,β’,a’,b’ in ZN,
-1
-1
-1
-1
Ø (α’*a’+β’*b’)=Ψ (α’)๏ø (a’)+Ψ (β’)๏ø
1
(b’)………… 1
Since,
-1
-1
-1
-1
Ø(Ψ (α’)๏ø (a’)+Ψ (β’)๏ø (b))
-1
-1
-1
= Ψ(Ψ (α’)) *ø(ø (a’))+ Ψ(Ψ (β’))
-1
*ø(ø (b’))
-1
-1
-1
-1
= ΨΨ (α’) *øø (a’)+ ΨΨ (β’) *øø
(b’)
= IZN(α’) * IZN(a’)+ IZN(β’) * IZN(b’)
= α’ *a’+β’ *b’
Thus 1 is satisfied and this completes the
-1 -1
-1
proof that the inverse (Ψ ,ø )=(Ψ,ø)
of far ring isomorphism (Ψ,ø) is also far ring
isomorphism

Note:


The far ring isomorphism (Ψ,ø) from FRX
into itself is called an automorphism of
FRX
Now we prove the following property of
the set of automorphisms of far ring
Theorem 3.12
The set of automorphisms of the far ring
FRX forms a group, under composition
0,denoted by Aut(FRX).
 Proof :
consider the far ring FRX=(ZN,+,๏).
To prove that (Aut(FRX), o) is a group
where o is the composition of maps.


From theorem 3.9 , we get that o is closed
binary operation on the set of maps
Aut(FRX).It is clear that o is associative If
we take ø=Ψ = IZN the identity over ZN,
then it is clear that (IZN, IZN) is an
automorphism of FRX .It is easy to show
that ,for any (Ψ,ø) in aut(FRX), the
composition of (IZN, IZN) and (Ψ,ø) is the
automorphism (Ψ,ø);

that means there is (IZN, IZN) the identity
element of (Aut(FRX),o).From theorem
3.11 the inverse exists for any element in
Aut(FRX) . Finally we can say that o is
closed binary operation and associative on
Aut(FRX);and the identity element exists
in Aut(FRX) and each element in Aut(FRX)
has an inverse in Aut(FRX)with respect to
o.

Therefore (Aut(FRX), o) forms a
group and the proof is finished.
Note

We denote the number of elements in the
set Aut(FRX) by |Aut(FRX)| .And we are
going to prove the following theorem
Theorem 3.13:


If FRX and FRY are isomorphic far rings,
then the number of isomorphisms
between them equals both |Aut(FRX)| and
by |Aut(FRY)|
Proof: Consider FRX and FRY to be two
isomorphic far rings. Then there is an
isomorphism (Ψ,ø) denotes by f fom FRX
into FRY.
If the isomorphism f is the only from FRX
into FRY, then the only automorphism of
FRX is the identity ; since if there is an
automorphism of FRX denoted by
(Ψ’,ø’)=k which is different from the
identity , then by the above theorems
-1
fok is an isomorphism from FRX into FRY

-1
where k is the inverse of k.
-1
 Then f ok must be f.

f

→FRY

FRX

↓↑k

-1
k
FRX


Therefore k must be the identity .
Thus we get that the identity is the only
automorphism of FRX . Similarly if f is the
only isomorphism from FRX into FRY ,
then the only automorphism of FRY is the
identity .


Now if there exists more than one
isomorphism from FRX into FRY. Let f be one
isomorphism from FRX into FRY and the other is
g=(Ψ″,ø″) which is different fromf. Let the set
of isomorphisms from FRX into FRY be A. Define
-1
a map Ө from A into Aut(FRX) by Ө(g)=g f for
-1
any g in A where g is the inverse map of g.
It is clear that Ө is well defined map.
To prove that Ө is bijection (1-1 and onto)
 First to prove that Ө is 1-1
Let Ө(g1)=Ө(g2), for any g1,g2 in A
-1
-1
-1
-1
-1
g
g
g
g
Then 1 of=2 of.this yields to 1 ofof =2
-1
๏f๏f
-1
-1
g
g
 Therefore 1
=2 and then we get that
g
g
1 =2 This proves that Ө is 1-1.Now to
prove that Ө is onto.



Let h be in Aut(FRX),and h is not the
identity over FRX .Then there is g in A
-1
-1
,g≠f such that g =hf and
-1
-1
-1
-1
Ө(g)=g f=hf f=h;(if g=f,then g =f and
-1
-1
-1
g =f =hf , therefore h is the identity )
this proves that Ө is onto .thus Ө is 1-1
and onto well defined mapping. This
proves that the number of 1 to 1

Isomorphisms from FRX into FRY equals to
the number of automorphisms of FRX .one
can similarly prove that the number of
isomorphisms from FRX into FRY equals to
the number of automorphisms of FRY by
defining the map Ө from A into Aut(FRY)
-1
by Ө(g)=fg for any g in A Thus the
theorem is proved.
Section three



Unsolved problems there are several
unsolved problems concerning our topics:
The first problem is to find a test for
isomorphic far rings based on inspecting
Latin squares .
To give something about this for
N=1,2,3,4.


First : Let N=1, then the only Latin aquare
is 0 which gives only one far ring of order
1 if N=2, then also there is only one Latin
square 1 0, which gives only one far ring
of order 2
0 1 If N=3, then also there is only one
Latin Square





1 2 0
2 0 1
0 1 2
which gives only one far ring of order 3
In case N=1 or N=2 or N=3 , we have
only one far ring which is isomorphic to
itself by the identify isomorphism .If N=4,
in this case we have four Latin squares
A,B,C, and D as show before in
Examples: 2.17

and therefore have four far ring FRA ,
FRB,FRC, and FRD which are of order
4.FRA and FRB are isomorphic by the
isomorphism (Ψ,ø) which is defined by
Ψ=(1)(234) and ø=(0)(2)(13).But FRC
and FRD are not isomorphic ; since if they
were , then they have the same period
profiles by theorem 3.5


But they have different period profiles (see
tables 3.3 and similarly, we obtain that
FRA,FRC are not isomorphic ,FRA,FRD are
also not isomorphic and FRB ,FRC are not
isomorphic
A second problem is to reduce to
minimum the number of 4-tuples we must
check to verify
Ø(α๏a+β๏b)=Ψ(α) *ø(a)+Ψ(β)
*ø(b)………….1
 Where α,β,a,b are in ZN and (ZN,+,๏)
, (ZN,+, *) are two far rings .


We have two far rings of ordered N.(ZN,+,
๏) and (ZN,+, *). And we have each of Ψ
and ø is a permutation of ZN. To prove
one , we need to prove that (Ψ,ø) is an
isomorphism from (ZN,+,๏) into (ZN,+, *).
In general the number f 4-tuples we must
check to verify 1 is N2.N2=N4.

But if the multiplication of far ring is
commutative as for example in case
N=1,2,3,4, then the number of 4-tuples
that we must check will reduce the
number N2 will reduce toN+(N2-N)/2 or
less than this number . To show in an
explicit manner for cases N≤4.



In case N=1, it is trivial case we need only
one check which is happened by taking
α=β=a=b=0
αβ
ab
00
00


In case N=2, we have a far ring of order
two with elements of Z2={1,0}
The following table gives the required
cases that we need to prove 1






αβ
ab
ab
11
11
10
10
10
00
00
00
N2.N2=22.22=24=16
16 reduces to 7
ab
01
ab
00

in case N=3, we have a far ring of order 3
wih elements of Z3={1,2,0} the following
table gives the cases that we need to
prove 1







αβ
11
12
10
22
20
00
ab
11
12
10
22
23
00
ab
12
10
20
20
00
ab
13
22
01
00
ab
22
20
00
ab
20
02
ab
00
00
The number N4=34=81 reduces to
22
 In case N=4, we have a far ring of
order 4 with elements of
Z4={1,2,3,0}
 The following tale gives the cases
that we need to prove 1


αβ
11
12
13
10
22
23
20
33
30
00

The number N2.N2=44=16x16=256reduce to 55










ab
11
12
13
10
22
23
20
33
30
00
ab
12
13
10
20
23
20
30
30
00
ab
13
10
23
33
20
31
00
00
ab
10
22
20
30
33
33
ab
22
23
33
00
30
30
ab
23
20
20
ab
20
31
03
ab ab ab ab
33 30 00
32 33 30 00
00
03 00
00