# Rationalizing Denominators Using Gröbner Bases

Authors: Dongmei Li; Man Wu; Jinwang Liu; Yiman Gao
Journal: Complexity
 (2022)

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## Abstract

The problem of rationalizing denominators for two types of fractions is discussed in the paper. By using the theory and algorithms of Gröbner bases, we first introduce a method to rationalize the denominators of fractions with square root and cube root, and then, for the denominators with higher radical of the general form, the problem of rationalizing denominators is converted into the related problem of finding the minimal polynomials. Some interesting results and an executable algorithm for rationalizing the denominator of these type fractions are presented. Furthermore, an example is also established to illustrate the effectiveness of the algorithm.


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## Body

## 1. Introduction

A typical topic in algebra is rationalizing denominators [1]. Rationalizing the denominator avoids the problem known as “subtrative cancellation,” deals with the problem of recognizing equivalent expressions, and is commonly used in many of computer algebra systems. Also, rationalizing expression has applications in calculus [2]. More importantly, by rationalizing the denominator, mathematical operations and practical problems can be approximated more accurately [3]. Thus, there are circumstances in which it is advantageous to rationalize an expression.The general method of denominator rationalization is to seek the rationalized factor of the denominator first and then multiply the numerator and denominator by this factor at the same time [4]. Using this method, the denominator rationalization of quadratic radical fraction has been solved. However, it is difficult to find the rationalized factor for an expression which contains a root higher than a square root [5]. Therefore, a lot of literatures tend to deal with some special case of this problem. In 1929, Paradiso [6] showed that theoretically, in all cases, and practically, in many cases, a rationalizing factor may be found by the method of undetermined coefficients. In 1970, Fateman presented an algorithm named RADCAN that is implemented in MACSYMA for the simplification of expressions containing radicals [7]. Zhou rationalized the denominator for a class of fractions 1/gu by theory of minimal polynomials in 1986, where gu is a polynomial whose coefficient are rational and u is a complex root of a nonzero rational polynomial [8]. In 1989, Ma showed the possibility of denominator’s rationalization of the irrational expressions: A/a1b1n1+a2b2n2+⋯+ambmnm+c, where A is the combination of radical and rational addition, subtraction, and multiplication and a1b1n1+a2b2n2+⋯+ambmnm+c is an algebraic element in the field of rational numbers [9]. In 2000, Liu rationalized the denominator for a class of algebraic fractions as follows 1/fcn+1=1/a0+a1cn+1+⋯+ancnn+1 (where fx=a0+a1x+⋯+anxn,c,a1,a2,…,an∈Q,c>0) using the knowledge of determinants [10]. In 2002, using polynomials, Tang also discussed the algebraic fractions whose denominator is the same with the above fraction [11]. And in 2015, Berele and Catoiu produced an exact formula for rationalizing any fraction whose denominator is a linear combination with rational coefficients of square roots of rational numbers [1].Existing results mentioned above mainly deal with several kinds of denominators rationalization of radical fractions by using related theory of polynomials. And many other kinds of denominators of irrational fraction are unsolved, such as1/x+587+694. This kind of fraction is a very common form in mathematical calculation, so it is of high application value to study a general method to rationalize the denominator. In this paper, we will investigate this by the theory of Gröbner bases as it is an important tool to solve many problems in polynomial ideal [12, 13]. And it has been implemented in many computational softwares including Singular, Maple, CoCoA, Mathematica, Macaulay 2, etc. Many fundamental problems in commutative algebra, computational algebraic number theory, algebraic geometry, graph theory, image processing, cryptography and encoding, and science and engineering can be solved by it algorithmically [13–23]. And the minimal polynomial can be obtained by the reduced Gröbner basis algorithm easily. Based on this and results mentioned above, we consider using the theory of Gröbner bases to explore the relationship between minimal polynomials and denominator rationalization and discuss mainly denominator rationalization of the fraction with the form as(1)1x+c1⋅am+c2⋅bn,where m,n∈N+,m,n≥2,a,b,c1,c2∈Q,a,b>0. We hope to establish a simplified method for rationalizing the denominator of this type fraction.The rest of the paper is organized as follows. We present some preliminary knowledge, basic concepts, and a special method to rationalize the denominators with square root and cube root in Section2. Main results on rationalizing the denominator of a type fraction are shown in Section 3. A simplified algorithm and an example established to illustrate the algorithm are given in Section 4. Section 5 concludes the paper.


## 2. Denominators with Square Root and Cube Root

In what follows,Kx1,x2,…,xn will denote the polynomial ring in n variables x1,x2,…,xn with coefficients in a field K, L/K will be an algebraic extension of the field K, and L:K will denote the extension times of the extension field L/K. Q will denote the rational number field, N will be the set of integers, and lcmm,n will denote the least common multiple of m and n. For a nonzero polynomial fx∈Kx1,x2,…,xn, we use ltf, lcf, and lpf to denote the leading term, the leading coefficient, and the leading monomial of fx, respectively. For a set F⊂Kx1,x2,…,xn, we denote ltF=ltf|f∈F. Then, we introduce several related definitions and algorithm.Definition 1.
LetI be an ideal in Kx1,x2,…,xn. A finite subset G of I is called a Gröbner basis of I if(2)ltG=ltI.Definition 2 (see [13]).
Letf,g∈Kx1,x2,…,xn,f,g≠0, and L=lcmlpf,lpg, and the S-polynomial of f and g is defined as (Algorithm 1)(3)Sf,g=Lltff−Lltgg.
This section focuses on the problem of rationalizing the denominator with square root and cube root. For the sake of convenience in researching the problem, we put the coefficient in the radical sign, and then, the fraction can be reduced to one of the following forms:(4)11x+a2+b3a>0,anda,b∈Q,21x−a2+b3a>0,anda,b∈Q.
We first consider the denominators rationalization of form (1).Algorithm 1: Buchberger algorithm [14].
Input:F=f1,f2,…,fs.Output:G=g1,2,…,gs, G is a Gröbner basis of f1,f2,…,fs.Initialization:G≔F, Ω≔fi,fj|fi≠fj∈G.WhenΩ≠∅selectf,g∈ΩΩ≔Ω/f,gSf,g⟶G+h, h is reductive with respect to G.Ifh≠0, thenΩ≔Ω∪u,h|∀u∈GG≔G∪hLemma 1 (see [13]).
LetG=g1,g2,…,gs be a subset of an ideal I∈Kx1,x2,…,xn, and then, G is a Gröbner bases for I if and only if Sgi,gj⟶G0, for all i≠j,1≤i,j≤s.Theorem 1.
LetI be an ideal in Kx1,x2,…,xn, and then, fg−1∈I⟺1∈I,f.Proof.
Iffg−1∈I, then there exists h∈I, such that fg−1=h, that is, 1=fg−h∈I,f.
In turn, if1∈I,f, then there exists g∈Kx1,x2,…,xn and h∈I, such that 1=fg+h, that is, fg−1=−h∈I.
Next, we introduce how to rationalize the denominator of fraction as form (1). First, consider the following ideal:(5)I=y12−a,y23−b⊆Qxy1,y2,where Qxy1,y2 denotes the polynomial ring in variables y1 and y2 on Qx and Qx is an extension of the field Q with x as a variable. Let(6)f1y1,y2=y12−a,f2y1,y2=y23−b.
In the following, multivariate polynomialshx,y are denoted as h for convenience. And the term order of the multivariate polynomial ring is the lexicographical ordering, which is defined by y1>y2.
Note that(7)Sf1,f2=by12−ay23⟶f2by12+ab⟶f10.
By Lemma1, we have that f1,f2 is a Gröbner basis of I. Let(8)f3y1,y2=x+y1+y2,y1=a,y2=b3, and then, f3a,b3=x+a+b3, which is the denominator we want to deal with. Replace f by f3 in Theorem 1, and then, the condition f3⋅g−1∈I is equivalent to f3a,b3⋅ga,b3=1, where ga,b3∈Qa,b3 is the polynomial we need.
So, the key issue is whether1∈I,f3, that is, whether 1∈f1,f2,f3. If true, we can rationalize the denominator of 1/x+a2+b3 by using this method.
Now, we use Buchberger algorithm to obtain the Gröbner basis ofJ=f1,f2,f3 and then determine whether 1∈f1,f2,f3.
Initializef1=y12−a,f2=y23−b,andf3=x+y1+y2, and using the algorithm, we get the Gröbner basis of J is(9)f1,f2,f3,f4,f5,f6,where(10)f4=y22−xy1+xy2−a,f5=a+3x2y2+2x3−2ax−b,(11)f6=−2x3−2ax−b33x2+a3−b.
Observef6, and it is easy to find that f6 is independent of y1 and y2, and thus, f6∈Qx. Note that Qx is a field, and thus, f6−1=1/f6∈Qx. Consequently, 1=f6⋅1/f6∈J. Therefore, we can calculate the rationalizing result of 1/x+a+b3.
Then, expressf6 as the combination of f1,f2,f3,f4,f5:(12)f6=f2−y223x2+a+−2x3+2ax+b3x2+a2+2x3−2ax−b23x2+a3⋅f5,that is,(13)f2−y22/3x2+a+−2x3+2ax+b/3x2+a2+2x3−2ax−b2/3x2+a3⋅f5f6=1.
Hence, divide both sides of equation (13) by x+a+b3, and substitute the value of f2,f5 and y1=a,y2=b3 into the equation, and then, we obtain(14)1x+a+b3=2x3−2ax−b2⋅2x2+bx⋅3x2+a2+b2x3+2ax+b⋅3x2+a2x3−2ax−b3+b3x2+a3+b3x2+a2−2x⋅2x3−2ax−b22x3−2ax−b3+b3x2+a3⋅a+−b3x2+a2+2x2⋅−2x3+2ax+b⋅3x2+a+x2x3−2ax−b22x3−2ax−b3+b3x2+a3⋅b3+2x2⋅3x2+a2+x⋅3x2+a⋅−2x3+2ax+b−2x3−2ax−b22x3−2ax−b3+b3x2+a3⋅b23+2x⋅2x3−2ax−b⋅3x2+a+2x3−2ax−b22x3−2ax−b3+b3x2+a3⋅a⋅b3+−2x⋅3x2+a2+−2x3+2ax+b⋅3x2+a2x3−2ax−b3+b3x2+a3⋅a⋅b23.
Using Matlab to simplify the equation above, we obtain(15)1x+a+b3=1x6−3ax4+2bx3+3a2x2+6abx+b2−a3x5−2ax3+bx2+a2x+ab+−x4+2ax2+2bx−a2⋅a+−x4−bx+a2⋅b3+x3+3ax+b⋅b23+2x3−2ax−b⋅a⋅b3+−3x2−a⋅a⋅b23.
For form (2), we just need to change f3y1,y2=x+y1+y2 to f3y1,y2=x−y1+y2, and the other steps are the same.


## 3. Denominators with Higher Radical

In this section, we focus on rationalizing the denominator of general form as(16)1x+c1⋅am+c2⋅bn,where m,n∈N+,m,n≥2,a,b,c1,c2∈Q,a,b>0.By the method in Section2, we first construct the ideal J=<f1,f2,f3>∈Qxy1,y2, where f1=y1m−a,f2=y2n−n,andf3=x+c1y1+c2y2. Then, compute the Gröbner basis G and the reduced Gröbner basis G0 of J. If G0=1, then calculate fs=G∩Qx. Finally, express fs as a combination of G/fs.However, there are two important uncertainties, one is whetherG0=1 and the other is whether fs is a combination of G/fs. So, this method can only solve the case when c1,c2,m,n are specific values. Because of this limitation, we hope to find another method for the general form.First, we introduce several related definitions and lemmas.Definition 3 (see [24]).
LetL/K be an algebraic extension of the field K, L:K=n. Suppose σi:L⟶C1≤i≤n is n K-insertion of L, where C is the complex field. For α∈L, define(17)TL/Kα=∑i=1nσiα,as the trace of α to the expansion L/K.Definition 4 (see [24]).
LetL/K be an algebraic extension of the field K. Suppose σ1,…,σn is the K-insertion of L and α1,…,αn∈L, and we define(18)dL/Kα1,…,αn=det2σiαj,1≤i≤n,1≤j≤n,as the discriminant of α1,…,αn for the expansion L/K.Definition 5.
LetK be a field, u∈K, and fx∈Kx be a polynomial. We call fx is the minimal polynomial of u in K if fx satisfies the following:(1)
fx is monic and fu=0(2)
Ifgx∈Kx and gu=0, then fx|gxLemma 2 (see [24]).
LetL/K be the expansion of number field and L:K=n. Suppose α∈L,fx=xm−c1xm−1+⋯+−1mcm is the minimal polynomial of α in K, where m=Kα:K, and then, TL/Kα=nc1/m.Lemma 3 (see [23]).
LetdL/Kα1,…,αn be the discriminant of element α1,…,αn for the expansion L/K, and then, dL/Kα1,…,αn≠0 if and only if α1,…,αn is K-linearly independent.Lemma 4 (see [25], Eisenstein criterion).
Letfx=anxn+an−1xn−1+⋯+a1x+a0 be an integral coefficient univariate polynomial. If there is a prime number p such that(1)
p∤an(2)
p|an−1,an−2,…,a0(3)
p∤a02Then,fx is irreducible over the rational number field Q.Using the lemmas above, we can prove the following results.Theorem 2.
Ifp is prime number, then fx=xm−p is the minimal polynomial for pm in rational number filed Q, where m∈N+ and m≥2.Proof.
It is straightforward thatpm is a root of fx=xm−p=0. In the following, we prove that fx is irreducible.
Letgx=fx+p=xm+mpxn−1+⋯+Cmipixm−i+⋯+pm−p, where Cmi=m!/i!m−i!. It is obvious that p∤1, p|Cmipi1≤i≤m−1, and p|pm−p. In fact, we can prove that p2∤pm−p. Suppose p2|pm−p, and then, p|pm−1−1. Combining p|pm−1m∈N+,m≥2, we have that p|pm−1−pm−1−1, that is, p|1, and this is a contradiction. So, p2∤pm−p. By the Eisenstein criterion, gx is irreducible, so fx is irreducible, and it is the minimal polynomial for pm in Q.Theorem 3.
Ifp1,p2,…,ps are different prime numbers, then(19)p1t1m1⋅p2t2m2,⋯pstsms,ti<mi,ti,mi∈N+,i=1,2,…,s,is an irrational number.Proof.
It is straightforward thatp1t1m1⋅p2t2m2⋯pstsms is a real number. In the following, we prove that it is an irrational number. Suppose it is a rational number, and set fx=xm1m2⋯ms−p1t1m2⋯msp2m1t2⋯ms⋯psm1m2⋯ts. Then, p1t1m1⋅p2t2m2⋯pstsms is a positive rational root of fx=0. Note that fx is an integral coefficient polynomial and lcf=1, and then, the positive rational root of fx must be a factor of its constant term. So, it has the form as p1n1p2n2⋯psns, where 0≤ni≤m1⋯mi−1timi+1⋯ms,ni∈N, i=1,2,…,s. Then, we have that(20)p1t1m1⋅p2t2m2⋯pstsms=p1n1p2n2⋯psns.
From the equation above, we see that some ofni must be 0. Without loss of generality, we denote the elements whose power exponents are nonzero as p1,p2,…,pq, and equation (20) turns into the following:(21)p1t1m1⋅p2t2m2⋯pstsms=p1n1p2n2⋯pqnq,that is,(22)p1n1−t1/m1p2n2−t2/m2⋯pqnq−tq/mq=pq+1tq+1/mq+1pq+2tq+2/mq+2⋯psts/ms.
Multiply both sides of equation (22) by the m1m2⋯ms power, and then,(23)p1l1p2l2⋯psls=pq+1lq+1pq+2lq+2⋯psls,where li∈N+,i=1,2,…,s. Hence, p1∣pq+1lq+1pq+2lq+2⋯psls, and this contradicts p1,p2,…,ps are different prime numbers. So, the conclusion is correct.Theorem 4.
Ifp1,…,ps are different prime numbers and m1,…,ms are positive integers and no less than 2, then the finite extension times of Qp1m1,…,psms/Q is no more than m1m2…ms.Proof.
LetFi=Qp1m1,p2m2,…,pimi, 1≤i≤s and i∈N+, and then, Fi+1=Fipi+1mi+1, that is, Fi+1 can be viewed as the single extension of Fi. So, Fi+1:Fi=∂gi, where ∂gi denotes the degree of the polynomial gix, and gix is the minimal polynomial of pimi in the field Fi. Obviously, pimi is a root for fix=xmi−pi, 1≤i≤s. By the definition of minimal polynomial, we see that ∂gi≤∂fi, 2≤i≤s,i∈N+. From Theorem 3, we see that f1x is an irreducible polynomial in Q. By the property of domain extension, we have F1:Q=m1. Hence,(24)Qp1m1,p2m2,…,psms:Q=Fs:Fs−1Fs−1:Fs−2⋯F2:F1F1:Q=∂gs⋅∂gs−1⋯∂g2⋅m1≤m1m2⋯ms.Theorem 5.
Ifp1,p2,…,ps are different prime numbers and m1,m2,…,ms are positive integers and no less than 2, then xms−ps is the minimal polynomial of psms in the field Qp1m1,p2m2,…,ps−1ms−1.Proof.
Letbi=pimi,1≤i≤s,M=Qp1m1,p2m2,…,psms, and in the following, we prove that all the m1⋅m2⋯ms elements in set(25)A=b1n1b2n2⋯bsns|0≤ni≤mi−1,ni∈N,i=1,2,…,s⊆M,are linearly independent. First, sort the elements in A. It is easy to observe that each element in A corresponds to such an array n1,n2,…,ns. So, we can turn the problem into sorting the exponents n1,n2,…,ns, where the term order is lexicographic order. We denote the element that corresponds to the largest exponential as c1, the element that corresponds to the second largest exponential as c2,…, and the element that corresponds to the smallest exponential as cm1m2⋯ms, and then, we have sorted out all the elements in A.
Now, we calculate detTM/Qcicj. Suppose ci=b1ni1b2ni2⋯bsnis and cj=b1nj1b2nj2⋯bsnjs. If one of the following two situations is true,(1)
∀k=1,2,…,s,nik+njk=mk(2)
∀k=1,2,…,s,nik=njk=0
We obtain thatcicj∈Q by Theorem 3. Hence, TM/Qcicj=ncicj≠0, where n=M:Q. Otherwise, there exists k∈1,2,…,s does not satisfy either of the above two cases, and we can calculate TM/Qcicj by Lemma 2. We first want to obtain the minimal polynomial of cicj in rational number field Q. Set(26)l=lcmm1gcdm1,ni1+nj1,m2gcdm2,ni2+nj2,…,msgcdms,nis+njs.
Obviously,l>1. Let q=cicjl and B=m∈N+|cicjm∈Q, and by Theorem 3, we have that mk|nik+njkm, for any m∈B, where k=1,2,…,s. Hence, p is the smallest element in B.
In the following, we prove thatfx=xl−q is the minimal polynomial of cicj in the rational number field Q.
It is straightforward thatfcicj=cicjl−q=0. We factorize fx in the complex field C as(27)fx=x−qlx−εql⋯x−εl−1ql,where ε=e2πi/l=cos2π/n+isin2π/n.
Supposefx is reducible in Q, and then, some constant terms of the linear factor in the decomposition above are rational numbers, that is, there is a t∈N+ and k∈N such that −1tεkqt/l∈Q⊆R, i.e., e2kπi/l=cos2kπ/l+isin2kπ/l∈R. So, sin2kπ/l=0, that is, k=ln/2,n∈Z. Hence, εk=cos2kπ/l=cosnπ=±1,n∈Z, and there is an r∈Q such that qt/l=r. Note that q=cicjl, and then, cicjt=r∈Qt<l, and this contradicts the selection of l. Thus, fx=xl−q is irreducible in Q, and then, fx is the minimal polynomial of cicj in rational number field Q. Combined with Lemma 2, we see that TM/Qcicj=0.
Based on the discussion above, we obtain thatdM/Qc1,c2,…,cm1m2⋯ms=detTM/Qcicj≠0. Then, all elements in set A are linearly independent in Q by Lemma 3. Combined with Theorem 4, we see that(28)Qp1m1,p2m2,…,psms:Q=m1m2⋯ms.
So,xms−ps is the minimal polynomial of psms in the field Qp1m1,…,ps−1ms−1.
Based on the results above, we can present the steps for rationalizing denominators of the following form:(29)1x+c1am+c2bn,where mandn are integers and a,b∈Q+ and c1,c2∈Q.Step 1: writeaandb as fractions in the lowest term, a=d1/e1andb=d2/e2. Then, decompose d1,e1,d2,e2 into power product of different prime factors, respectively. We use p1,p2,…,pt to denote the common prime factors of d1 and d2, q1,q2,…,qk to denote the different prime factors in d1,e1, and v1,v2,…,vs to denote the different prime factors in d2,e2.Step 2: letl=lcmm,n, and(30)α1=p1l,α2=p2l,…,αt=plt,αt+1=q1m,αt+2=q2m,…,αt+w=qwm,αt+w+1=r1n,αt+w+2=r2n,…,αt+w+s=rsn.
We construct the rational extension field:(31)M=Qα1,…,αt,αt+1,…,αt+w,…,αt+w+s.
Letα=c1am+c2bn, and there are fx1,…,xt+w+s,gx1,…,xt+w+s∈Qx1,…,xt+w+s such that(32)α=fα1,…,αt+w+s/gα1,…,αt+w+s.
According to Theorem2, we have that h1=x1l−p1∈Qx1 is the minimal polynomial of α1 in the rational number field Q. It is also known, by Theorem 5, hi=xl−pi∈Qα1,…,αi−1xi is the minimal polynomial of αi in Qα1,…,αi−1 for 2≤i≤t, hi=xm−qi−t∈Qα1,…,αi−1xi is the minimal polynomial of αi in Qα1,…,αi−1 for t+1≤i≤t+s, and hi=xn−vi−t−w∈Qα1,…,αi−1xi is the minimal polynomial of αi in Qα1,…,αi−1 for t+w+1≤i≤t+w+s.
Next, we construct the following homomorphic maps:(33)φi:Qx1,x2,…,xi⟶Qα1,α2,…,αi−1xi,2≤i≤t+w+s.
For anyf=∑βλβx1β1x2β2…xiβi∈Qx1,x2,…,xi−1, where λβ∈Q,β=β1,β2,…,βi∈Ni. Define(34)φif=∑βλβα1β1α2β2⋯αi−1βi−1xiβi.
Under this map, the corresponding preimage ofhi∈Qα1,α2,…,αi−1xi is itself, where 2≤i≤t+w+s.Step 3: we find the minimal polynomial ofα in Q according to [26].
The term order≺ in Qx1,…,xt,xt+1,…,xt+w+s,y is the lexicographical ordering defined by y≺xt+w+s≺⋯≺xt+1≺xt≺⋯≺x1. Under this term order, we calculate the reduced Gröbner basis G of the ideal(35)I=h1,h2,…,ht+w+s,g⋅y−f,where(36)hi=xl−pi,1≤i≤t,xm−qi−t,t+1≤i≤t+w,xn−ri−t−w,t+w+1≤i≤t+w+s.
Then, we computehy=G∩Qy and take(37)my=hyλ,λ=lchy.
Hence,my is the minimal polynomial of α in Q. And then we can rationalize the denominator by applying the minimal polynomial.
Setny=x+y (x is a parameter) and divide my by ny such that(38)my=qyny+ry,∂ry<∂ny.
Then,∂ry=0 and ry≠0, otherwise, it contradicts that ny contains x. Correspondingly, equation (38) can be rewritten as(39)my=qyny+rx,rx∈Qx.
Substitutey=α into the equation above, and then, 0=x+αqα+rx, that is,(40)1x+α=−qαrx,which is the desired result of rationalizing denominators.


## 4. Algorithm and Example

According to the theorems and discussion in Section3, we obtain an algorithm for rationalizing the denominators of fractions with the form as 1/x+c1⋅am+c2⋅bn. We describe this algorithm in more detail in Figure 1.Figure 1
Denominator rationalized algorithm.In the following, we construct an example to show the effectiveness of the algorithm.Example 1.
Rationalizing the denominator of(41)1x+6+43.Step 1: we knowa = 6 and b = 4, and the command “format rat” can be omitted here. Carry out the prime factorization of 6 and 4 by using the function “factorization,” and we obtain(42)6=2×3,4=2×2.
It is easy to see thatp1=2 is a common prime factor of 6 and 4. In addition, 6 also contains the factor q1=3.Step 2: note thatm=2andn=3, and then, using the function “min_GBS,” we obtain(43)l=lcmm,n=6.
Let(44)α1=26,α2=3.
We construct the extension fieldQα1,α2 of the rational number field Q. Set α=6+43, and then, α=fα1,α2=α13α2+α14.Step 3: hence,h1=x16−2∈Qx1 is a minimal polynomial of α1 on Q and h2=x22−3∈Qα1x2 is a minimal polynomial of α2 on Qα1.
Set(45)I=x16−2,x22−3,y−x13x2−x14,take the lexicographical ordering y≺x2≺x1 as the term order in Qx1,x2,y, and calculate the reduced Gröbner basis G of ideal I in the software singular using the built-in function “groebner.” Then, the calculation results are as follows:(46)G=f1,f2,f3,where(47)f1=y6−18y4+108y2−144y−200,f2=x22−3,f3=7128x1+18x23y4−144x23y3−294x23y2+4756x23y+64x23−3x2y5−30x2y2−14904x2y+4536x2.
Therefore,G∩Qy=y6−18y4−8y3+108y2−144y−200. Hence, the minimal polynomial of α in the rational number field Q is my=y6−18y4−8y3+108y2−144y−200.
Next, dividemy by x+y in the software Matlab using the built-in function “polynomialReduce.” Then,(48)my=qy⋅x+y+rx,where(49)qy=y5−xy4+y3x2−18+y2−x3+18x−8+yx4−18x2+8x+108+−x5+18x3−8x2−108x−144,rx=x6−18x4+8x3+108x2+144x−200.
Substituteα=6+43 into(50)1x+α=−qαrx,then(51)1x+6+43=x5−12x3+4x2+36x+24rx+−x4+12x2+8x−36rx6+−x4−4x+36rx43+2x3+36x+8rx23+2x3−12x−4rx643+−6x2−12rx236.Remark 1.
Note thatm=2andn=3, and we can use the method in Section 2 to do this example. Replace a=6andb=4 and do all the steps as in Section 2, and we obtain(52)1x+6+43=1x6−18x4+8x3+108x2+144x−200x5−12x3+4x2+36x+24+−x4+12x2+8x−366+−x4−4x+36⋅43+2x3+36x+8⋅23+2x3−12x−4⋅6⋅43+−6x2−12⋅6⋅23.
Through the two examples above, it is easy to find that the results obtained by both methods are the same, which proves that the algorithm we designed is correct and the application scope is wider.Remark 2.
Using the method in Section2 or the method of undetermined coefficients, the problem of denominator rationalization of fractions with the form as 1/x+c1⋅am+c2⋅bn may not be solved or can be solved but will take a long time. However, from Example 1, it is straightforward that this problem can be solved efficiently and simply according to the three steps of Denominator rationalized algorithm by using the software Matlab and Singular.


## 5. Conclusion

In this paper, using theory of Gröbner bases, we have achieved in rationalizing denominators for two types of fractions, especially rationalizing the denominator of the fraction with the form as(53)1x+c1⋅am+c2⋅bn,where m,n∈N+,m,n≥2,a,b,c1,c2∈Q,a,b>0. We have presented some interesting results and an executable algorithm on rationalizing the denominators for this type fractions. Furthermore, we have established an example to illustrate the effectiveness of the algorithm.The method that we proposed on rationalizing denominators can be realized in computer system such as Maple and Singular, which makes the related computation more quick. Furthermore, it can improve the performance of the related algorithms. For example, we find that, for some new swarm intelligence algorithms proposed in recent years, such as monarch butterfly optimization (MBO) [27], earthworm optimization algorithm (EWA) [28], elephant herding optimization (EHO) [29], etc. the fraction of the form 1/x+c1⋅am+c2⋅bn occurs during the implementation of these algorithms. The denominator rationalized algorithm we proposed can rationalize the denominator of this type fraction, which can improve the accuracy of these algorithms on calculation.Future work will investigate in rationalizing denominators of more general forms of radical fraction. The type of fractions discussed in the paper is unnested radical expressions, and the problem of the nested radical fractions is not involved here, which is also what we will do.



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*Source: 1288357-2022-01-06.xml*