split fibonacci quaternions

Adv. Appl. Cliff ord Algebras 23 (2013), 535–545© 2013 Springer Basel0188-7009/030535-11published online June 13, 2013DOI 10.1007/s00006-013-0401-9

Split Fibonacci Quaternions

Mahmut Akyigit∗, Hidayet Huda Kosal and Murat Tosun

Abstract. Starting from ideas given by Horadam in [5], in this paper,we will define the split Fibonacci quaternion, the split Lucas quaternionand the split generalized Fibonacci quaternion. We used the well-knownidentities related to the Fibonacci and Lucas numbers to obtain the rela-tions between the split Fibonacci, split Lucas and the split generalizedFibonacci quaternions. Moreover, we give Binet formulas and Cassiniidentities for these quaternions.

Keywords. Split Fibonacci Quaternion, Split Lucas Quaternion.

1. Introduction

The split quaternions or coquaternions are the elements of a 4-dimensionalassociative algebra introduced by James Cockle. The quaternions introducedby Hamilton form a 4-dimensional real vector space equipped with a mul-tiplicative operation. Unlike the quaternion algebra, the split quaternionscontain zero divisors, nilpotent elements and nontrivial idempotent.A split quaternion q is an expression of the form

q = a + bi + cj + dk

where a, b, c and d are real numbers and i, j, k are split quaternionic unitswhich satisfy the non-commutative multiplication rules

i2 = −1, j2 = k2 = 1,ij = −ji = k, jk = −kj = −i, ki = −ik = j.

(1.1)

The scalar and the vector part of a split quaternion q are denoted by Sq = a

and−→Vq = bi + cj + dk, respectively. Thus, a split quaternion q is given by

q = Sq +−→Vq. Define qn = an + bni + cnj + dnk (n = 0, 1). Addition and

subtraction of the split quaternions is defined by

q0 ∓ q1 = (a0 + b0i + c0j + d0k)∓ (a1 + b1i + c1j + d1k)= (a0 ∓ a1) + (b0 ∓ b1) i + (c0 ∓ c1) j + (d0 ∓ d1) k.

∗Corresponding author.This work was completed with the support of our TEX-pert.

Advances inApplied Cliff ord Algebras

536 M. Akyigit, H. Huda Kosal and M. Tosun Adv. Appl. Cliff ord Algebras

Multiplication of the split quaternions is defined by

q0.q1 = (a0 + b0i + c0j + d0k) . (a1 + b1i + c1j + d1k)

= (a0a1 − b0b1 + c0c1 + d0d1) + (a0b1 + b0a1 − c0d1 + d0c1) i

+ (a0c1 − b0d1 + c0a1 + d0b1) j + (a0d1 + b0c1 − c0b1 + d0a1) k

= Sq0Sq1 + g(−→Vq0 ,

−→Vq1

)+ Sq0

−→Vq1 + Sq1

−→Vq0 +

−→Vq0 ∧

−→Vq1

whereg(−→Vq0 ,

−→Vq1

)= −a0a1 + b0b1 + c0c1 + d0d1

and

−→Vq0 ∧

−→Vq1 =

∣∣∣∣∣∣−i j kb0 c0 d0b1 c1 d1

∣∣∣∣∣∣.

The conjugate of split quaternion q is denoted by q and it is

q = a− bi− cj − dk.

The norm of q is defined as

Nq = qq = a2 + b2 − c2 − d2.

The Fibonacci and Lucas sequence are defined as

F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2 (1.2)

andL0 = 2, L1 = 1, Ln = Ln−1 + Ln−2 (1.3)

for n ≥ 2. Here, Fn is the nth Fibonacci number and Ln is the nth Lucasnumber.The numerous relations connecting Fibonacci and Lucas numbers have beengiven by Dunlap [2] and Vajda [12].In addition to these, by using Fibonacci numbers, the generalized Fibonaccisequences were defined by Horadam (in the relation(7) of [5]) as follows

H1 = x , H2 = x + y , Hn = Hn−1 + Hn−2 (n ≥ 3) (1.4)

where x and y are arbitrary integers and it follows that

Hn = (x− y)Fn + yFn+1. (1.5)

In equation (1.5), Hn is a Fibonacci sequence if x = 1 and y = 0, Hn is aLucas sequence if x = 1 and y = 2, [5].Horadam [6] defined the nth Fibonacci and nth Lucas quaternions

Qn = Fn + iFn+1 + jFn+2 + kFn+3 (1.6)

andKn = Ln + iLn+1 + jLn+2 + kLn+3 (1.7)

respectively, where

i2 = j2 = k2 = −1,ij = −ji = k, jk = −kj = i, ki = −ik = j.

Vol. 23 (2013) Split Fibonacci Quaternions 537

In [11], M.N. Swamy gave a few interesting relations for the nth Fibonacciquaternions which are defined by Horadam. In [7, 8], M.R. Iyer derived re-lations connecting the Fibonacci quaternions and Lucas quaternions. In [3],C. Flaut and V. Shpakivskyi investigated some properties of generalized Fi-bonacci quaternions and Fibonacci-Narayana quaternions. There have beennumerous studies in the literature on the split quaternions [10], except thatfor the split Fibonacci quaternions.In the present paper our main objective is to derive the relations connectingthe split Fibonacci quaternion, split Lucas quaternion and split generalizedFibonacci quaternion, by using a similar terminology with the Fibonacci andLucas numbers [2, 12, 9] as well as the interrelations between them.

2. Split Fibonacci Quaternions

The split Fibonacci quaternion, the split Lucas quaternion and the split gen-eralized Fibonacci quaternion are given by the following formulae,

Qn = Fn + iFn+1 + jFn+2 + kFn+3 (2.1)

Tn = Ln + iLn+1 + jLn+2 + kLn+3 (2.2)and

Pn = Hn + iHn+1 + jHn+2 + kHn+3 (2.3)where Fn is the nth Fibonacci number, Ln is the nth Lucas number and i, j, kare split quaternionic units which satisfy the non-commutative multiplicationrules:


(2.4)

The scalar and the vector part of a split Fibonacci quaternion Qn are denotedby SQn = Fn and

−−→VQn = iFn+1 + jFn+2 + kFn+3, respectively. If SQn = 0,

then Qn is called as a pure split Fibonacci quaternion.Let Qn and Kn be two split Fibonacci quaternions such that

Qn = Fn+ iFn+1+jFn+2+kFn+3 and Kn = Xn+ iXn+1+jXn+2+kXn+3.

The addition, subtraction and multiplication of these quaternions are definedby

Qn ∓Kn = (Fn ∓Xn) + i (Fn+1 ∓Xn+1)+j (Fn+2 ∓Xn+2) + k (Fn+3 ∓Xn+3)

(2.5)

andQnKn = (Fn + iFn+1 + jFn+2 + kFn+3) (Xn + iXn+1 + jXn+2 + kXn+3)

= SQnSKn + g(−−→VQn ,

−−→VKn

)+ SQn

−−→VKn + SKn

−−→VQn +

−−→VQn ∧

−−→VKn (2.6)

respectively. The conjugates of Qn, Tn and Pn are defined by

Qn = Fn − iFn+1 − jFn+2 − kFn+3

Tn = Ln − iLn+1 − jLn+2 − kLn+3

Pn = Hn − iHn+1 − jHn+2 − kHn+3.

(2.7)


The norm of the split Fibonacci quaternion Qn is defined by

NQq= QqQq = Fn

2 + Fn+12 − Fn+2

2 − Fn+32. (2.8)

Similar notations hold for split Lucas quaternions Tn and split generalizedFibonacci quaternions Pn. Therefore, considering the equations above relatedto the addition, the subtraction and the multiplication of the split quaternionswe can give the following theorems.

In the following theorem, the first identical is analogous to the recursive rela-tion defining the Fibonacci and Lucas numbers and that fourth is analogousto FnFm + Fn+1Fm+1 = Fn+m+1 , (see Dunlap-10, [2] or Vajda-8, [12]).

Theorem 2.1. Let Fn and Qn be the Fibonacci number and the split Fibonacciquaternions, respectively. In this case, for n ≥ 1 we can give the followingrelations:

Qn + Qn+1 = Qn+2,Qn − iQn+1 − jQn+2 − kQn+3 = −5Fn+3,QnQm + Qn+1Qm+1 = 5Fn+m+4 + 2Qn+m+1.

Proof. From equations (2.1) and (2.5), we get

Qn +Qn+1 = (Fn + iFn+1 + jFn+2 + kFn+3) + (Fn+1 + iFn+2 + jFn+3 + kFn+4)

= Fn+2 + iFn+3 + jFn+4 + kFn+5.

It results that

Qn + Qn+1 = Qn+2. (2.9)

Using equations (1.2), (2.1) and (2.4), we obtain that

Qn − iQn+1 − jQn+2 − kQn+3 = (Fn + iFn+1 + jFn+2 + kFn+3)

−i (Fn+1 + iFn+2 + jFn+3 + kFn+4)

−j (Fn+2 + iFn+3 + jFn+4 + kFn+5)

−k (Fn+3 + iFn+4 + jFn+5 + kFn+6)

= Fn + Fn+2 − Fn+4 − Fn+6 = −5Fn+3.

(2.10)

Lastly, considering equations (2.4), (2.5) and (2.6), we have


QnQm+Qn+1Qm+1 = (Fn + iFn+1 + jFn+2 + kFn+3]).

(Fm + iFm+1 + jFm+2 + kFm+3)

+ (Fn+1 + iFn+2 + jFn+3 + kFn+4) (Fm+1 + iFm+2 + jFm+3 + kFm+4)

= FnFm + 2Fn+3Fm+3 + Fn+4Fm+4

+i

⎛⎝ [FnFm+1 + Fn+1Fm+2] + [Fn+1Fm + Fn+2Fm+1]

− [Fn+2Fm+3 + Fn+3Fm+4] + [Fn+3Fm+2 + Fn+4Fm+3]

⎞⎠

+j

⎛⎝ [Fn+2Fm + Fn+3Fm+1] + [Fn+3Fm+1 + Fn+4Fm+2]

+ [FnFm+2 − Fn+2Fm+4] + [Fn+1Fm+3 − Fn+1Fm+3]

⎞⎠

+k

⎛⎝ [FnFm+3 + Fn+1Fm+4] + [Fn+1Fm+2 + Fn+2Fm+3]

+ [Fn+3Fm + Fn+4Fm+1]− [Fn+2Fm+1 + Fn+3Fm+2]

⎞⎠ .

Thus, putting the identity of Fibonacci numbers Fn+m+1=FnFm+Fn+1Fm+1

(see Dunlap-10, [2] or Vajda-8, [12]) into the last equation and making basicsimplifications simplifying we get the following equation

QnQm + Qn+1Qm+1 = 5Fn+m+4 + 2Qn+m+1. (2.11)

�

The identities in following theorem are analogous to Ln = Fn−1+Fn+1 , (seeVajda-6, [12]) and Ln = Fn+2 − Fn−2 , (see Vajda-7a, [12]).

Theorem 2.2. Let Qn be a split Fibonacci and Tn be a split Lucas quaternion.The following relations are satisfied

Qn−1 + Qn+1 = Tn,Qn+2 −Qn−2 = Tn.

Proof. From equations (2.1) and (2.5), it follows thatQn−1 + Qn+1 = (Fn−1 + iFn + jFn+1 + kFn+2) + (Fn+1 + iFn+2 + jFn+3 + kFn+4)

= (Fn−1 + Fn+1) + i (Fn + Fn+2) + j (Fn+1 + Fn+3) + k (Fn+2 + Fn+4) .

Using the identity of Fibonacci numbers Ln = Fn−1 + Fn+1 (see Vajda-6,[12]), the last equation becomes

Qn−1 + Qn+1 = Tn , n ≥ 1. (2.12)

In the same manner, we can see thatQn+2 −Qn−2 = (Fn+2 + iFn+3 + jFn+4 + kFn+5)− (Fn−2 + iFn−1 + jFn + kFn+1)

= (Fn+2 − Fn−2) + i (Fn+3 − Fn−1) + j (Fn+4 − jFn) + k (Fn+5 − Fn+1)

= Ln + iLn+1 + jLn+2 + kLn+3.

Combining the identity Ln = Fn+2 − Fn−2 (see Vajda-7a, [12]) with the lastequation yields

Qn+2 −Qn−2 = Tn. (2.13)�


The first identity in the following theorem is analogous to Horadam’s identityfor ordinary Fibonacci quaternions QnQn = 3F2n+3; the second one is anal-ogous to Horadam’s 2Fn = Qn +Qn; the third one is the same as Horadam’sQ2

n = 2QnFn −QnQn, [6].

Theorem 2.3. Let Qn be split Fibonacci quaternions and Qn be conjugate ofQn. Then, we can give the following relations between these quaternions

QnQn = −L2n+3,Qn + Qn = 2Fn,Q2

n = 2QnFn + L2n+3.

Proof. By equations (2.1), (2.4) and (2.7), it is obvious that

QnQn = F 2n + F 2

n+1 − F 2n+2 − F 2

n+3. (2.14)

Using the identities of Fibonacci numbers F 2n +F 2

n+1 = F2n+1 (see Dunlap-7,[2] or Vajda-11, [12]) and Fn−1 +Fn+1 = Ln (see Vajda-6, [12]) into the lastequation, it may be concluded that

QnQn = −L2n+3 (2.15)

andQn + Qn = 2Fn. (2.16)

From equation (2.16), it follows that

Q2n = QnQn = Qn(2Fn −Qn) = 2QnFn −QnQn.

Finally, since (2.15) is satisfied, the last equation becomes

Q2n = 2QnFn + L2n+3. (2.17)

�The first, third and fourth identities in the following theorem are analogousto Corollary 2 of [4].

Theorem 2.4. Let Qn be split Fibonacci quaternion. Then,n∑

s=1Qs = Qn+2 −Q2,

p∑s=0

Qn+s + Qn+1 = Qn+p+2,

n∑s=1

Q2s−1 = Q2n −Q0,

n∑s=1

Q2s = Q2n+1 −Q1.

Proof. Since∑n

t=a Ft = Fn+2 − Fa+1 (see Hoggatt-I1, [13]) we haven∑

s=1

Qs =n∑

s=1

Fs + i

n∑s=1

Fs+1 + j

n∑s=1

Fs+2 + k

n∑s=1

Fs+3

= (Fn+2 − F2) + i (Fn+3 − F3) + j (Fn+4 − F4) + k (Fn+5 − F5)= Qn+2 −Q2.


Hence we can writep∑

s=0Qn+s + Qn+1 =

n+p∑r=1

Qr −n−1∑r=1

Qr + Qn+1,

it is obvious that from the last equation,p∑

s=0

Qn+s + Qn+1 = Qn+p+2. (2.18)

Using (2.9), we obtainn∑

s=1Q2s−1 =

n∑s=1

(Q2s −Q2(s−1)) = Q2n −Q0.

We haven∑

s=1

Q2s =n∑

s=1

(Q2s+1 −Q2s−1) = Q2n+1 −Q1. �

Theorem 2.5. Let Qn and Pn be the conjugate of split Fibonacci quaternionsQn = Fn+iFn+1+jFn+2+kFn+3 and split generalized Fibonacci quaternionsPn = Hn + iHn+1 + jHn+2 + kHn+3, respectively. Then,

PnQn − PnQn = 2 [FnPn −HnQn] ,Pn Qn + Pn Qn = 2 [PnFn + HnQn − PnQn] ,PnQn − Pn Qn = 2 [HnQn + PnFn −HnFn] .

Proof. Using equations (2.5) and (2.6), after simple computations one caneasily reach that

PnQn − PnQn = 2Hn [−iFn+1 − jFn+2 − kFn+3] + 2Fn [iHn+1 + jHn+2 + kHn+3]

= 2 [Hn (Fn −Qn) + Fn (Pn −Hn)]

= 2 [FnPn −HnQn] .

Considering equations (2.5) and (2.6) and by direct computation we getPn Qn + Pn Qn = (Hn + iHn+1 + jHn+2 + kHn+3) (Fn − iFn+1 − jFn+2 − kFn+3)

+ (Hn − iHn+1 − jHn+2 − kHn+3) (Fn + iFn+1 + jFn+2 + kFn+3)

= 2 [HnFn − (iHn+1 + jHn+2 + kHn+3) (iFn+1 + jFn+2 + kFn+3)]

= 2 [HnFn − (Pn −Hn) (Qn − Fn)] .

Lastly, in the same manner, we can see that

PnQn − PnQn = 2 [HnQn + PnFn −HnFn] . (2.19)

�

The characteristic equation of the relation (2.9) is as follows

x2 − x− 1 = 0. (2.20)

So, the roots of equation (2.20) are α = 1+√5

2 and β = 1−√52 . For the split

Fibonacci quaternions Qn and Qn−1, we getαQn + Qn−1 = (αFn + Fn−1) + i (αFn+1 + Fn) + j (αFn+2 + Fn+1) + k (αFn+3 + Fn+2)

where n ≥ 0.


Putting the equation αn = αFn + Fn−1 (see [9]) into the last equation, weget

αQn + Qn−1 = αn + iαn+1 + jαn+2 + kαn+3

=(1 + iα1 + jα2 + kα3

)αn

= ααn(2.21)

where α = 1 + iα + jα2 + kα3.In addition to that, by considering the equation βn = βFn + Fn−1 (see [9]),the following equation may be indicated in much the same way as equation(2.21)

βQn + Qn−1 = β βn (2.22)

where β = 1 + iβ + jβ2 + kβ3.The first and second formulas in the following theorem are analogous to theTheorem 3.1 in [4].

Theorem 2.6. Let Qn and Tn be the the split Fibonacci and the split Lucasquaternions. For n ≥ 0 the Binet formulas for these quaternions are as follows

Qn =ααn − β βn

α− β

andTn = ααn + β βn

Proof. Let Qn be a the split Fibonacci quaternion. Subtracting equations(2.21) and (2.22) side by side, we obtain

ααn − β βn = (αQn + Qn−1)− (βQn + Qn−1) = αQn − βQn.

Therefore, we get

Qn =ααn − β βn

α− β. (2.23)

Moreover, for the split Lucas quaternion Tn, adding equations (2.21) and(2.22) side by side yields

ααn + β βn = (αQn + Qn−1) + (βQn + Qn−1)= (α + β)Qn + Qn−1 + Qn−1.

Since α+β = 1, taking into consideration equations (2.9) and (2.12) togetherwith the last equation it follows

Tn = ααn + β βn. (2.24)

�

Theorem 2.7. Let Qn and Tn be the split Fibonacci and the split Lucas quater-nions. In this case, for n ≥ 1, the Cassini Identities for Qn and Tn are asfollows

Qn−1Qn+1 −Q2n = (−1)n (2Q1 − 2i− 3k)

andTn−1Tn+1 − T 2

n = 5(−1)n+1 (2Q1 − 2i− 3k)respectively.


Proof. By definition of the split Fibonacci quaternion given in equation (2.1),we have

Qn−1Qn+1 −Q2n = (Fn−1 + iFn + jFn+1 + kFn+2) (Fn+1 + iFn+2 + jFn+3 + kFn+4)

− (Fn + iFn+1 + jFn+2 + kFn+3) (Fn + iFn+1 + jFn+2 + kFn+3)

for n ≥ 1. Considering equations (2.5) and (2.6) and using the identities ofFibonacci numbers Fn−1 Fn+1 − F 2

n = (−1)n (see Dunlap-9, [2]) andFn+2 = Fn+1 + Fn (see Dunlap-1, [2]), we obtain

Qn−1Qn+1 −Q2n = (−1)n (2Q1 − 2i− 3k) .

For the split Lucas quaternions Tn = Ln + iLn+1 + jLn+2 + kLn+3,similar argument applies and using the identities of Lucas numbers we reachLn−1Ln+1 − L2

n = 5(−1)n+1 (see Benjamin-60, [1]) and Ln+2 = Ln+1 + Ln

(see Dunlap-2, [2]) leads to

Tn−1Tn+1 − T 2n = 5(−1)n+1 (2Q1 − 2i− 3k) . �

We will give an example in which we check in a particular case the Cassiniequality for split Fibonacci quaternions and for split Lucas quaternions.

Example 2.8. Let Q1, Q2 and Q3 be the split Fibonacci quaternions suchthat Q1 = 1+ i+2j+3k, Q2 = 1+2i+3j+5k and Q3 = 2+3i+5j+8k.In this case,

Q1Q3 −Q22 = (1 + i + 2j + 3k) (2 + 3i + 5j + 8k)− (1 + 2i + 3j + 5k)2.

Considering equations (2.4) and (2.6) together with the last equation we have

Q1Q3 −Q22 = (−1)2 = 2Q1 − 2i− 3k.

Suppose that T1, T2 and T3 are the split Lucas quaternions such that T1 =1 + 3i + 4j + 7k, T2 = 3 + 4i + 7j + 11k and T3 = 4 + 7i + 11j + 18k. Then,using the same argument as above we get

T1T3 − T 22 = 5(−1)3 (2Q1 − 2i− 3k) = −5 (2Q1 − 2i− 3k)

where Q1 is a split Fibonacci quaternion and Q1 = 1 + i + 2j + 3k.

3. Conclusion

The difference between the Fibonacci and the split Fibonacci quaternionsoriginated from the quaternionic units, i.e., the quaternionic units for a Fi-bonacci quaternion are

i2 = j2 = k2 = −1,ij = −ji = k, jk = −kj = i, ki = −ik = j.

Whereas for a split Fibonacci quaternions they are



Generally, taking the quaternionic units i, j and k as

i2 = −α , j2 = −β, k2 = −αβ ,ij = −ji = k, jk = −kj = βi, ki = −ik = αj

we may define Fibonacci generalized quaternions and Lucas generalized qua-ternions. In addition to that one may give the Binet formula and Cassiniidentities for these quaternions.

Acknowledgements

The authors would like to thank the anonymous referees for their helpfulsuggestions and comments which improved significantly the presentation ofthe paper.

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Mahmut Akyigit, Hidayet Huda Kosal, Murat TosunSakarya UniversityFaculty of Arts and SciencesDepartment of MathematicsSakaryaTurkeye-mail: [email protected]

[email protected]

[email protected]

Received: October 23, 2012.

Accepted: May 16, 2013.

split fibonacci quaternions

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