UDK 54
Original scientific article/Izvirni znanstveni članek
ISSN 1580-2949 MTAEC9, 44(4)193(2010)
A NEW COMPLEX VECTOR METHOD FOR BALANCING CHEMICAL EQUATIONS
NOVA KOMPLEKSNA METODA ZA URAVNOTEŽENJE KEMIČNIH
ENAČB
Ice B. Risteski
2 Milepost Place # 606, Toronto, Ontario, Canada M4H 1C7 ice@scientist.com
»Just as tensor analysis is the proper language of kinematics, so linear algebra is the proper language of stoichiometry.« Aris, R.; Mah, R. H. S.; Ind. Eng. Chem. Fundam. 1963, 2, 90.
Prejem rokopisa - received: 2009-10-22; sprejem za objavo - accepted for publication: 2010-01-21
In this article, the author discovers a paradox of balancing chemical equations. The many counterexamples illustrate that the considered procedure of balancing chemical equations given in the paper1 is inconsistent. A new complex vector method for paradox resolution is given too.
Keywords: paradox, balancing, chemical equations, vector analysis.
V članku avtor opisuje paradoks pri uravnoteženju kemijskih reakcij. Več primerov dokazuje, da je procedura uravnoteženja kemijskih reakcij v viru 1 inkonsistentna. Predstavljena je nova kompleksna vektorska metoda za rešitev paradoksa.
Ključne besede: paradoks, uravnoteženje, kemijske reakcije, vektorska analiza
1 INTRODUCTION
Balancing chemical equations is a basic matter of chemistry, if not one of its most important issues, and it plays a main role in its foundation. Indeed, it is a subtle question which deserves considerable attention.
Also, this topic was a magnet for a number of researchers around the world, because the general problem of balancing chemical equations was considered as one of the hardest problems in chemistry, as well as in mathematics.
In chemical literature there are a great number of papers which consider the problem of balancing chemical equations in different chemical ways, but all of them offer only some particular procedures for balancing simple chemical equations. Very often, these chemical procedures generate absurd results, because most of them are founded on presumed chemical principles, but not on genuine exact principles. And these presumed chemical principles generate paradoxes! The balancing chemical equations does not depend on chemical principles, it is a mathematical operation which is founded on true mathematical principles.
There is a number of paradoxes in chemistry about balancing chemical equations, and these will be systematized and studied in a special article of higher level by the same author which will appear in the near future. In this article only one paradox in balancing chemical equations, discovered in1 is considered.
The author would like to emphasize very clearly, that balancing chemical equations is not chemistry; it is just
linear algebra. Although it is true, that it is not chemistry, it is very important for chemistry! In this particular case the following question comes up: If balancing chemical equations is not chemistry, then why is it considered in chemistry? Or maybe more important is this question: If it is linear algebra, then it should be studied in mathematics, so why bother chemists?
The author will address the above questions assuming that: the problem of balancing chemical equations was a multidisciplinary subject and for its solution both mathematicians and chemists are needed. The job of chemists is to perform reactions, while balancing their equations is a job for mathematicians. Mathematics, as a servant to other sciences, is a problem solver, but chemistry is a result user.
The skepticism about balancing chemical equations by chemical principles appeared a long time ago. Let's quote the opinions of three chemists.
In 1926 the American chemist Simons2 wrote: The balancing of an equation is a mathematical process and independent of chemistry. The order of steps in the process is as essential as the order of steps in long division and the process is much simpler than is ordinary assumed. Seventy-one years later, the Dutch chemist Ten Hoor3, made a similar statement: Balancing the equation of the reaction is a matter of mathematics only.
Now is the right place to paraphrase the criticism of the American chemist Herndon4: The major changes that have taken place over seven decades are substitutions of the terms 'change in oxidation number' and 'algebraic
method' for the terms 'valence change' and 'method of undetermined coefficients', respectively. This assertion shows very clearly the picture of chemists' contributions to balancing chemical equations by chemical principles. Probably expressing his satisfaction with the chemists' contributions to balancing chemical equations, Herndon, the editor of the Journal of Chemical Education4, decided that further discussion of equation balancing will not appear in the Journal unless it adds something substan-tively new to what has already appeared.
In view of the above assertions this question arises: Are there in chemistry 'chemical principles' capable of offering solutions of the general problem of balancing chemical equations? Maybe more interesting is the question: What are 'chemical principles' - a rhetorical sophism of chemists or their hopes?
More interesting for us is to give an answer to both questions from a scientific view point. Perhaps, the more appropriate short answer to the first question is: 'Chemical principles' are not defined entities in chemistry, and so this term does not have any meaning. They are not capable to provide solutions of the general problem of balancing chemical equations, because they are founded on an intuitive basis and they represent only a main generator for paradoxes. However, the necessary and sufficient conditions for a complete solution of the general problem of balancing chemical equations lie outside of chemistry, and we must look for them in an amalgamated theory of n-dimensional vector spaces, linear algebra, abstract algebra and topology. It is a very hard problem of the highest level in chemistry and mathematics, which must be considered only on a scientific basis. Like this should look the answer to the first question.
The answer to the second question will be described in this way: Actually, 'chemicalprinciples' are a remnant of an old traditional approach in chemistry, when chemists were busy with the verification of their results obtained in chemical experiments. It is true, that until second half of 20'h century there was no mathematical method for balancing chemical equations in chemistry, other than Bottomley's algebraic method5. Chemists balanced simple particular chemical equations using only Johnson's change in oxidation number procedure6, Simons - Waldbauer - Thrun's partial reactions procedure2,7 and other slightly different modifications derived from them. The 'chemical principles' were an assumption of traditional chemists, who thought before the appearance of Jones' problem8, that the solution of the general problem of balancing chemical equations is possible by use of chemical procedures. But, practice showed that the solution of the century old problem is possible only by using contemporary sophisticated mathematical methods.
These questions require a deeper elaboration than given here, and it will be a main concern of the author in his future research.
2 PRELIMINARIES
The exact statements about balancing chemical equations agree with the following well-known results.
Theorem 1. Every chemical reaction can be reduced in a matrix equation Ax = 0, where A is a reaction matrix, x is a column-vector of the unknown coefficients and 0 is a null column-vector.
Proof. The proof of this theorem immediately follows from9.
Remark 2. The coefficients satisfy three basic principles (corresponding to a closed input-output static model10'11)
•	the law of conservation of atoms,
•	the law of conservation of mass, and
•	the time-independence of the reaction.
Theorem 3. Every chemical reaction can be presented as a matrix Diophantine equation Ax = By, where A and B are matrices of reactants and products, respectively, and x and y are column-vectors of unknown coefficients. The proof of this theorem is given in.1213
This theorem is actually the Jones'8 problem founded by virtue of Crocker's procedure for balancing chemical equations14, and from this problem the formalization of chemistry began. This problem is a milestone in chemistry and mathematics as well, and just it opened the door in chemistry to enter a new mathematical freshness. Jones8 with his problem transferred the general problem of balancing chemical equations from the field of chemistry into the field of mathematics and opened way to solve this problem with mathematical methods founded on principles of linear algebra.
Before the appearance of this problem, the approach of balancing chemical equations was intuitive and useful only for some elementary chemical equations. Now, some well-known results from the theory of complex n-dimensional vector spaces1516, for resolution of balancing chemical equations will be introduced. Here, by C is denoted the set of complex numbers and by C" is denoted the Euclidian n-dimensional vector space with complex entries.
Definition 4. A vector space over the field C consists of a set V of objects called vectors for which the axioms for vector addition hold
(A1) If u, v G V then (u + v) G V;
(A2) u + v = v + u, Vu, v G V
(A3) u + (v + w) = (u + v) + w, Vu, v, w G V,
(A4) u + 0 = u = 0 + u, Vu G V,
(A5) - u + u = 0 = u + (- u), Vu G V; and the axioms for scalar multiplication
(51)	If u G V, then au G V, Va G C,
(52)	a(u + v) = au + av, Vu, v G V A Va G C,
(53)	(a + b)u = au + bu, Vu G V A Va,b G C,
(54)	a(bu) = (abu), Vu G V A Va,b G C,
(55)	1u = u, Vu G ^
Remark 5. The content of axioms (A1) and (S1) is described with thes assertion that V is closed under vector addition and scalar multiplication. The element 0 in axiom A4 is called the zero vector.
Definition 6. IfV is a vector space over the field C, a subset U of V is called a subspace of V if U itself is a vector space over C, where U uses the vector addition and scalar multiplication of V.
Definition 7. Let V be a vector space over the field C, and let Vi G V (1 < i < n). Any vector in V of the form v = aiVi + a2V2 + ^ + anVn, where at G C, (1 < i < n) is called a linear combination of vi, (1 < i < n).
Definition 8. The vectors Vi, V2, Vn are said to span or generate V or are said to form a spanning set of VifV = span{Vi, V2, Vn}. Alternatively, Vi G V (1 < i < n) span V, if for every vector v G V there exist scalars ai G C (1 < i < n) such that
V = aiVi + a2V2 + ^ + anVn,
i. e., v is a linear combination of
aiVi + a2V2 + _ + anVn.
Remark 9. If V = span{Vi, V2, Vn}, then each vector V G V can be written as a linear combination of the vectors vi, V2, Vn. Spanning sets have the property that each vector in V has exactly one representation as a linear combinations of these vectors.
Definition 10. Let Vbe a vector space over afield C. The vectors Vi G V (1 < i < n) are said to be linearly independent over C, or simply independent, if it satisfies the following condition: if
siVi + s2V2 + _ + snVn = 0,
then
s1 = s2 = ^ = sn = 0.
Otherwise, the vectors that are not linearly independent, are said to be linearly dependent, or simply dependent.
Remark 11. The trivial linear combination of the vectors vi, (1 < i < n) is the one with every coefficient zero
0v1 + 0v2 + _ + 0v„.
Definition 12. A set of vectors {e1, e2, en} is called a basis of V if it satisfies the following two conditions
1° e1, e2, en are linearly independent,
2° V = span{e1, e2, e„}.
Definition 13. A vector space V is said to be of finite dimension n or to be n-dimensional, written dim V = n, if V contains a basis with n elements.
Definition 14. The vector space {0} is defined to have dimension 0.
3 PARADOX APPEARANCE
Chemistry as other natural sciences is not immune of paradoxes. Unlike other natural sciences, in chemistry
paradoxes appeared some time later, and it has only two, while in other sciences many such contradictions are met. These paradoxes are well-known Levinthal's paradox17: The length of time in which a protein chain finds its folded state is many orders of magnitude shorter than it would be if it freely searched all possible configurations, and Structure-Activity Relationship (SAR) paradox18: Exceptions to the principle that a small change in a molecule causes a small change in its chemical behavior are frequently profound.
However, these paradoxes are not alone and there are more, but now another will be mentioned, which appears in the balancing of chemical equations.
In the paper1, the so-called formal balance numbers (FBN) are introduced, like this: Formal balance numbers are an aid that may grossly facilitate the problem of balancing complex redox equations. They may be chosen as being equal to the traditional values of oxidation numbers, but not necessarily. An inspection of the redox equation may suggest the optimal values that are to be assigned to formal balance numbers. In most cases, these optimal values ensure that only two elements will 'change their state' (i. e. the values of the formal balance numbers), allowing the use of the oxidation number technique for balancing equations, in its simplest form. Just like for oxidation numbers, the algebraic sum of the formal balance numbers in a molecule/neutral unit is 0, while in an ion it is equal to its charge (the sum rule).
Promptly, it was detected that the procedure given in1 boils down to using of well-known unconventional oxidation numbers, which previously were advocated by Toth19 and Ludwig20.
Consider this sentence from previous definition: They may be chosen as being equal to the traditional values of oxidation numbers, but not necessarily. It is a paradox! If the formal balance numbers can be the same as oxidation numbers or not, then the whole definition is illogical. This definition represents only a contradictory premise, which does not have any correlation with balancing chemical equations. The above definition does not speak anything about balancing chemical reactions in a chemical sense of the word, or their solution in a mathematical sense. In order for a chemical equation to be balanced the first necessary and sufficient condition is its solvability, but the above definition is far from it.
The so-called formal balance numbers, which actually are the same as the well-known oxidation numbers, are not any criterion for balancing chemical equations.
If the chemical equation is not formalized, then it generates only paradoxes. To support this assertion some ordinary counterexamples will be given.
1° Balancing chemical equations by using of change in oxidation number procedure has a limited usage! It holds only for some simple equations. Is it possible to determine valence of elements in some complex organic molecules as they are C2952H4664N8i2O832S8Fe4, C738H1166N812O203S2Fe, and so on?
Not yet! For instance, one may show the power-lessness of that procedure by the following two counterexamples.
Example 1.
2993 C2952H4664N816O832S8Fe4 + 9300568 CsaO3 + 408724 Li3PO4 ^ 11972 Li4Fe(CN)6 + 8763504CO2 + 2370456CsNO3 + 1178284 LiHa2 + 408724 H3PO4 + 6944000 Cs0.998Cl + 23944 H2S + 5753504 H2O.
What is the valence of C, N, S and Fe?
Example 2.
249000 C738H1166N812O203S2Fe + 1098133767 Pb(NO3)2 + 46812000 HSiCls + 7716178 P2O5 + 23148534 Coa2 ^ 46812000 Sia3.989 + 7716178 Co3(PO4)2 + 124500 H2SO4 + 549616500 Pb1.998O3 + 124500 Fe2(SO4)3 + 15313500 C12H22O11 + 2398455534 NO2.
The same previous question holds for this counterexample too. This means, the solution of the above equations is based on very sophisticated matrix methods21-23, but in no case on the change in oxidation number procedure!
As consequence of that, same holds for so-called formal balance numbers. That procedure is useless to balance complex chemical equations as it does not give effects for balancing chemical equations. Still, there are many causes for arguing why that elementary procedure is useless, but the above mentioned two counterexamples are enough to show that it is ineffective.
2° Consider this simple reaction
Example 3.
X1 NaNO2 + X2 FeSO4 + X3 H2SO4 ^ X4 NaHSO4 + X5 Fe2(SO4)3 + X6 NO.
In the above chemical reaction N and Fe change the valence, but chemical equation is impossible! Thus, is the procedure of formal balance numbers useful for balancing all chemical equations? Obviously, not! Since it does not have the capability to detect if some chemical equation is possible or not.
3° Now, one more counterexample will be given, where that particular procedure of formal balance numbers is impossible.
For instance, consider this chemical reaction
Example 4.
128147987406 FeSO4 + 27736706439 PrTlTe3 + 286573109604 H3PO4 ^ 128404797000 Fe0.998(H2PO4)2-H2O + 13945051000 Tl1.989(SO3)3 + 13889187000 Pr1.997(SO4)3 + 83210119317 TeO2 + 14881757802 P2O3 + 44645273406 H2S.
Is it possible to balance the above chemical equation by the procedure of formal balance numbers?
Not yet! Since it is not possible to determine the valence of Fe, Tl and Pr. Then, on what basis the author of the paper1 states, that his procedure (there called method) is proposed for fast and easy balancing of complex redox equations? On top of all, he states that the procedure (there called method) is probably the fastest of all possible methods! Really, a very modest statement is offered by the author, which is wrong, not only because of today's current balancing methods view point, but also from an earlier view point, when that procedure was published.
Is it a method when somebody can find on every step counterexamples? Obviously, the answer is negative! It is merely a picture of the old chemical traditionalism. Or, maybe it is a lonely case lost in the newest progressive and contemporary mathematical generalism!
4 PARADOX RESOLUTION BY A NEW COMPLEX VECTOR METHOD
With the purpose of solution of the paradox, in this section a new complex vector method of balancing chemical equations will be developed. This method is founded on the theory of n-dimensional complex vector spaces.
Theorem 15. Suppose that chemical equation
x1V1 + x2V2 + ^ + xnvn = 0,
(1)
where vi (1 < i < n) are the molecules and xi (1 < i < n) are unknown coefficients is a vector space V over the field C spanned by the vectors of the molecules vi (1 < i < n). If any set of m vectors of the molecules in V is linearly independent, then m < n.
Proof. Let be V = span{v1, V2, v„}. We must show that every set {«1, «2, Um} of vectors in V with m > n fails to be linearly independent. This is accomplished by showing that numbers x1, x2, xm can be found, not all zero, such that
x,U, + x2U2+ ^ + xmUm = 0.
Since V is spanned by the vectors V1, V2, v„, each vector Uj can be expressed as a linear combination of vi
«j = a1jv1 + a2jv2 + ^ + a„jv„.
Substituting these expressions into the preceding equation gives
m / n
0 = X xj X
j=1
'aiiv j
X aijxj
v
This is certainly the case if each coefficient of vi is zero, i. e., if
m
X a^x. = 0, (1 < i < n).
j=1
This is a system of n equations with m variables x1, x2, xm, and because m > n, it has a nontrivial solution.
This is what we wanted. Now we shall prove the following results.
Theorem 16. Let U be a subset of a vector space V of the chemical equation (1) over the field C. Then U is a subspace of V if and only if it satisfies the following conditions
0 G U, where 0 is the zero vector of V (2) If Mj, «2 G U, then (u, + u^) G U,	(3)
If u G U, then au G U, Va G C.	(4)
Proof. If U is a subspace of V of the chemical equation (1), it is clear by axioms (Aj) and (Sj) that the sum of two vectors in U is again in U and that any scalar multiple of a vector in U is again in U. In other words, U is closed under the vector addition and scalar multiplication of V. The converse is also true, i. e., if U is closed under these operations, then all the other axioms are automatically satisfied. For instance, axiom (A2) asserts that holds uj + u2 = u2 + uj, Vuj, u2 G U. This is clear because the equation is already true in V, and U uses the same addition as V. Similarly, axioms (A3), (S2), (S3), (S4) and (S5) hold automatically in U, because they are true in V. All that remains is to verify axioms (A4) and (A5).
If (2), (3) and (4) hold, then axiom (A4) follows from
(2)	and axiom (A5) follows from (4), because - u = (-1)u lies in U, Vu G U. Hence, U is a subspace by the above discussion. Conversely, if U is a subspace, it is closed under addition and scalar multiplication and this gives
(3)	and (4). If z denotes the zero vector of U, then z = Gz in U. But, Gz = 0 in V; so 0 = z lies in U. This proves (2).
Remark 17. If U is a subspace of V of the chemical equation (j) over the field, then the above proof shows that U and V share the same zero vector. Also, if u G U, then - u = (-1)u G U, i. e., the negative of a vector in U is the same as its negative in V.
Proposition 18. If V is any vector space of the chemical equation (j) over the field C, then {0} and V are subspaces of V.
Proof. U = V clearly satisfies the conditions of the Theorem jč. As to U = {0}, it satisfies the conditions because 0 + 0 = 0 and a0 = 0, Va G C.
Remark 19. The vector space {0} is called the zero subspace of V of the chemical equation (j) over the field C. Since all zero subspaces look alike, we speak of the zero vector space and denote it by G. It is the unique vector space containing just one vector.
Proposition 20. If v is a vector of some molecule in a vector space V of the chemical equation (j) over the field C, then the set Cv = {av, Va G C} of all scalar multiplies of v is a subspace ofV.
Proof. Since 0 = Gv, it is clear that 0 lies in Cv. Given two vectors av and bv in Cv, their sum av + bv = (a + b)v is also a scalar multiple of v and so lies in Cv. Therefore Cv is closed under addition. Finally, given av, r(av) = (ra)v lies in Cv, so Cv is closed under scalar multipli-
cation. Now, if we take into account the Theorem jč, immediately follows the statement of the proposition.
Theorem 21. Let U = span{vl, v2, vn} in a vector space V of the chemical equation (j) over the field C. Then,
U is a subspace of V containing each of vi (j < i <
n),(5)
U is the smallest subspace containing these vectors in the sense that any subspace of V that contains each of vi (j < i < n), must contain U.(6)
Proof. First we shall proof (5). Clearly
0 = Gvj + Gv2 + _ + Gv„
belongs to U. If
and
w = bjvj + b2v2+ ^ + b„v„ are two members of U and a G U, then v + w
= (aj + bl)Vl + (a2 + b2)v2 + _ + (a„ + b„)v„, av = (aaj)vj + (aa2)v2 + _ + (aan)vn,
so both v + w and av lie in U. Hence, U is a subspace of V. It contains each of vi (j < i < n). For instance,
v2 = Gvj + jv2 + Gv3 + _ + Gv„.
This proves (5).
Now, we shall prove (6). Let W be subspace of V that contains each of vi, (j < i < n). Since W is closed under scalar multiplication, each of aivi (j < i < n) lies in W for any choice of ai (j < i < n) in C. But, then aivi (j < i < n) lies in W, because W is closed under addition. This means that W contains every member of U, which proves (6).
Theorem 22. The intersection of any number of subspaces of a vector space V of the chemical equation (j) over the field C is a subspace of V.
Proof. Let {Wi: i G I} be a collection of subspaces of V and let W = n (Wi: i G I). Since each Wi is a subspace, then 0 G W, Vi G I. Thus 0 G W. Assume u, v G W. Then, u, v G Wi, Vi G I. Since each Wi is a subspace, then (au + bv) G Wi, Vi G I. Therefore (au + bv) G W. Thus W is a subspace of V of the chemical equation (j).
Theorem 23. The union Wj U W2 of subspaces of a vector space V of the chemical equation (j) over the field C need not be a subspace of V
Proof^. Let V = C2 and let Wj = {(a, G): a G C} and W2 = {(G, b): b G C}. That is, Wj is the x-axis and W2 is the y-axis in C2. Then Wj and W2 are subspaces of V of the chemical equation (j). Let u = (j, G) and v = (G, j). Then the vectors u and v both belong to the union Wj U W2, but u + v = (j, j) does not belong to Wj U W2. Thus Wj U W2 is not a subspace of V.
Theorem 24. The homogeneous system of linear equations obtained from the chemical equation (j), in n
v = alvl + a2v2+ ^ + anvn
unknowns x1, X2, Xn over the field C has a solution set W which is a subspace of the vector space Cn.
Proof. The system is equivalent to the matrix equation Ax = 0. Since AO = 0, the zero vector 0 G W. Assume u and v are vectors in W, i. e., u and v are solutions of the matrix equation Ax = 0. Then Au = 0 and Av = 0. Therefore, Va, b G C, we have A(au + bv) = aAu + bAv = a0 + b0 = 0 + 0 = 0. Hence au + bv is a solution of the matrix equation Ax = 0, i. e., au + bv G W. Thus W is a subspace of Cn.
Theorem 25. If S is a subset of the vector space V of the chemical equation (1) over the field C, then
1° the set span{S} is a subspace of V of the chemical equation (1) over the field C which contains S.
2° span{S} C W, if W is any subspace of V of the chemical equation (1) over the field C containing S.
Proof. 1°. If S = 0, then span{S} = {0}, which is a subspace of V containing the empty set 0. Now assume S ^ 0. If v G S, then 1v = v G span {S}, therefore S is a subset of span{S}. Also, span{S} ^ 0 because S ^0. Now assume v, w G span {S}; say
and
v = a1v1 + _ + amvm
v = b1W1 + _ + bnWn
where v;, Wj G S and ai, bj are scalars.
Then
v + W = a1v1 + _ + amvm + b1W1 + _ + bnWn and for any scalar k,
kv = k(a1v1 + _ + amvm) = ka1v1 + _ + kamvm belong to span{S} because each is a linear combination of vectors in S. Thus span{S} is a subspace of V of the chemical equation (1) over the field C which contains S.
2°. If S = 0, then any subspace W contains S, and span{S} = {0} is contained in W. Now assume S ^0 and assume v; G S C W (1 < i < m). Then all multiples aivi G W (1 < i < m) where ai G C, and therefore the sum (a1v1 + _ + amvm) G W. That is, W contains all linear combinations of elements of S. Consequently, span{S} C W, as claimed.
Proposition 26. If W is a subspace of V of the chemical equation (1) over the field C, then span{W} = W.
Proof. Since W is a subspace of V of the chemical equation (1) over the field C, W is closed under linear combinations. Hence, span{W} C W. But W C span{W}. Both inclusions yield span{W} = W.
Proposition 27. If S is a subspace of V of the chemical equation (1) over the field C, then span{span{S}} = span{S}.
Proof. Since span{S} is a subspace of V; the above Proposition 26 implies that span {span {S}} = span{S}.
Proposition 28. If S and T are subsets of a vector space V of the chemical equation (1) over the field C, such that S C T, then span {S} C span {T}.
Proof. Assume v G span {S}. Then
v = ajuj + _ + aru„
where ai G C, (1 < i < r) and u; G S (1 < i < r). But S C T, therefore every ui G T (1 < i < r). Thus v G span {T}. Accordingly, span {S} C span {T}.
Proposition 29. The span {S} is the intersection of all the subspaces of a vector space V of the chemical equation (1) over the field C which contains S.
Proof. Let {W;} be the collection of all subspaces of a vector space V of the chemical equation (1) containing S, and let W = n W;. Since each W; is a subspace of V; the set W is a subspace of V. Also, since each Wi contains S, the intersection W contains S. Hence span{S} C W. On the other hand, span{S} is a subspace of V containing S. So span{S} = W^ for some k. Then W C Wk = span{S}. Both inclusions give span{S} = W.
Proposition 30. If span{S} = span{S U {0}}, then one may delete the zero vector from any spanning set.
Proof. By Proposition 28, span{S} C span {S U {0}}. Assume v G span {S U {0}}, say
v = ajuj + _ + anun + b-0
where ai, b G C (1 < i < n) and u; G S (1 < i < n).
Then v = a1u1 + _ + anun, and so v G span {S}. Thus span {S U {0}} C span {S}. Both inclusions give span {S} = span {S U {0}}.
Proposition 31. If the vectors vt G V (1 < i < n) span a vector space V of the chemical equation (1) over the field C, then for any vector W G V, the vectors W, vi (1 < i < n) span V.
Proof. Let v G V. Since the v; (1 < i < n) span V; there exist scalars a; (1 < i < n) such that v = a1v1 + _ + anvn + 0w. Thus w, v; (1 < i < n) span V.
Proposition 32. If vi (1 < i < n) span a vector space V of the chemical equation (1) over the field C, and for k > 1, the vector vk is a linear combination of the preceding vectors v; (1 < i < k - 1), then v; without vk span V, i. e., span{v1, v2, vk-1, vk+1, vn} = V.
Proof. Let v G V. Since the v; (1 < i < n) span V; there exist scalars ai (1 < i < n) such that
v = a1v1 + _ + anvn.
Since vk is a linear combination of v; (1 < i < k - 1), there exist scalars bi (1 < i < k - 1) such that
Thus
vk = b1v1 + ^ + ak-1vk-1.
v = a,v, + _ + atvt + ^ + a„v„
= ajvj + _ + ak(b1v1 + _ + bk-jvk-j) + _ + anvn = (a1 + akb1)v1 + _ + (ak-1 + akbk-i)vk-i + ak+ivk+i + _ + anvn.
Therefore,
span{v1, v2, vk_1, vk+1, vn } = ^
Proposition 33. IfWi, (1 < i < k) are subspaces of a vector space V of the chemical equation (1) over the field C, for which C W2 C _ C Wkand
W = Wj U W2 U _ U Wk,
then W is a subspace of V
Proof. The zero vector 0 G Wj, hence 0 G W. Assume u, v G W. Then, there exist iJ and j2 such that u G Wjj and v G W;2. Let j = max(/j, j2). Then Wjj C Wj and W/2 C W, and so u, v G Wj. But Wj is a subspace. Therefore (u + v) G Wj and for any scalar s the multiple su G Wj. Since Wj C W, we have (u + v), su G W. Thus W is a subspace of ^
Proposition 34. IfW, (1 < i < k) are subspaces of a vector space V of the chemical equation (1) over the field C and Si (1 < i < k) span Wi (1 < i < k), then S = 81 U S2 U _ U Sk spans W
Proof. Let v G W. Then there exists j such that v G Wj. Then v G span {Sj} C span {S}. Therefore W C span {S}. But S C W and W is a subspace. Hence span {S} C W. Both inclusions give span {S} = W, i. e., S spans W.
Theorem 35. Let {vJ, v2, vn} be a linearly independent set of vectors in a vector space V of the chemical equation (1) over the field C, then the following conditions
1° {v, VJ, v2, vn} is a linearly independent set, 2° v does not lie in {vJ, v2, v„}, are equivalent for a vector v in V.
Proof. Assume 1° is true and assume, if possible, that v lies in span{vJ, v2, v„}, say,
v = a,v, +
Then
v -aJvJ + _ + a„v„ = 0
is a nontrivial linear combination, contrary to 1°. So 1° implies 2°. Conversely, assume that 2° holds and assume that
av + aJvJ + _ + anvn = 0. If a ^ 0, then
v = (-aJ/a)vJ + _ + (-a„/a)v„, contrary to 2°. So a = 0 and
aJvJ + _ + a„v„ = 0.
This implies that
aJ = _ = an = 0,
because the set {vJ, v2, vn} is linearly independent. By this is proved that 2° implies 1°.
Proposition 36. Let {vJ, v2, vn} be linearly independent in a vector space V ofthe chemical equation (1) over the field C, then {aJVJ, a2v2, anvn}, such that the numbers at (1 < i < n) are all nonzero, is also linearly independent.
Proof. Suppose a linear combination of the new set vanishes
SJ(aJVJ) + s2(a2v2) + _ + s„(a„v„) = 0,
where si (1 < i < n) lie in C.
Then
SJaJ = s2a2 = ^ = snan = 0
by the linear independence of {vJ, v2, vn}. The fact that each ai ^ 0 (1 < i < n) now implies that SJ = s2 = ^ = sn = 0.
Proposition 37. No linearly independent set of vectors of molecules can contain the zero vector.
Proof. The set {0, VJ, v2, vn} cannot be linearly independent, because
10 + 0vJ + _ + 0v„ = 0,
is a nontrivial linear combination that vanishes.
Theorem 38. A set {vJ, v2, vn} of vectors ofmole-cules in a vector space V of the chemical equation (1) over the field C is linearly dependent if and only if some vi is a linear combination of the others.
Proof. Assume that {vJ, v2, vn} is linearly dependent. Then, some nontrivial linear combination vanishes, i. e.,
aJVJ + a2v2+ _ + anvn = 0,
where some coefficient is not zero. Suppose aJ ^ 0.
Then
VJ = (-a2/aJ)v2 + _ + (-a„/aJ)v„,
gives VJ as a linear combination of the others. In general, if ai ^ 0, then a similar argument expresses Vi as linear combination of the others.
Conversely, suppose one of the vectors is a linear combination of the others, i. e.,
vJ = a2v2 + _ + a„v„.
Then, the nontrivial linear combination 1v1 - a2v2 -^ - anVn equals zero, so the set {vJ, V2, Vn} is not linearly independent, i. e., it is linearly dependent. A similar argument works if any v i (1 < i < n) is a linear combinations of the others.
Theorem 39. Let V ^ 0 be a vector space of the chemical equation (1) over the field C, then
1° each set of linearly independent vectors is a part of a basis of V,
2° each spanning set V contains a basis of V,
3° V has a basis and dim V < n.
Proof. 1° Really, if V is a vector space that is spanned by a finite number of vectors, we claim that any linearly independent subset S = {vJ, V2, Vk} of V is contained in a basis of V. This is certainly true if V = span{S} because then S is itself a basis of V. Otherwise, choose Vk+J outside span{S}. Then SJ = {vJ, V2, Vk, Vk+J} is linearly independent by Theorem 35. If V = span{SJ}, then S1 is the desired basis containing S. If not, choose Vk+2 outside span{vJ, V2, Vk, Vk+J} so that S2 = {vJ, V2,
+ anVn
Vk, Vk+1, Vk+2} is linearly independent. Continue this process. Either a basis is reached at some stage or, if not, arbitrary large independent sets are found in V. But this later possibility cannot occur by the Theorem 15 because V is spanned by a finite number of vectors.
2° Let V = span{vi, V2, v^}, where (as V ^ 0) we may assume that each Vi ^ 0. If {vi, V2, v m} is linearly independent, it is itself a basis and we are finished. If not, then according to the Theorem 38, one of these vectors lies in the span of the others. Relabeling if necessary, it is assumed that Vi lies in span{v2, v m} so that V = span{v2, Vm}. Now repeat the argument. If the set {v2, v m} is linearly independent, we are finished. If not, we have (after possible relabeling) V = span{v3, Vm}. Continue this process and if a basis is encountered at some stage, we are finished. If not, we ultimately reach V = span {v m}. But then {vm } is a basis because Vm ^ 0 (V ^ 0).
3° V has a spanning set of n vectors, one of which is nonzero because V ^ 0. Hence 3° follows from 2°.
Corollary 40. A nonzero vector space V of the chemical equation (1) over the field C is finite dimensional only if it can be spanned by finitely many vectors.
Theorem 41. Let V be a vector space of the chemical equation (1) over the field C and dim V = n >0, then
1° no set of more than n vectors in V can be linearly independent,
2° no set of fewer than n vectors can span V. Proof. V can be spanned by n vectors (any basis) so 1° restates the Theorem 15. But the n basis vectors are also linearly independent, so no spanning set can have fewer than n vectors, again by Theorem 15. This gives 2°.
Theorem 42. Let V be a vector space of the chemical equation (1) over the field C and dim V = n >0, then
1° any set of n linearly independent vectors in V is a basis (that is, it necessarily spans V),
2° any spanning set of n nonzero vectors in V is a basis (that is, they are necessarily linearly independent).
Proof. 1° If the n independent vectors do not span V; they are a part of a basis of more than n vectors by property 1° of the Theorem 39. This contradicts Theorem 41.
2° If the n vectors in a spanning set are not linearly independent, they contain a basis of fewer than n vectors by property 2° of Theorem 39, contradicting Theorem 41.
Theorem 43. Let V be a vector space of dimension n of the chemical equation (1) over the field C and let U and W denote subspaces of V, then
1° U is finite dimensional and dim U < n, 2° any basis of U is a part of a basis for V, 3° ifU C ^ and dim U = dimW; then U = W: Proof. 1° If U = 0, dimU = 0 by Definition 14. So assume U ^ 0 and choose U1 ^ 0 in U. If U = span{H1}, then dimU = 1. If U ^ span{u1}, choose u2 in U outside
span{u1}. Then {u1, U2}, is linearly independent by Theorem 35. If U = span{u1, u2}, then dimU = 2. If not, repeat the process to find u3 in U such that {u1, u2, u3} is linearly independent and continue in this way. The process must terminate because the space V (having dimension n) cannot contain more than n independent vectors. Therefore U has a basis of at most n vectors, proving 1°.
2° This follows from 1° and Theorem 39.
3° Let dimU = dimW = m. Then any basis {u1, u2, um} of U is an independent set of m vectors in W and so is a basis of Wby Theorem 42. In particular, {u1, u2, um} spans W so, because it also spans U, W = span{u1, u2, um} = U. By this is proved 3°.
Proposition 44. If U and W are subspaces of a vector space V of the chemical equation (1) over the field C, then U + W is a subspace of V.
Proof. Since U and W are subspaces, 0 G U and 0 G W. Hence 0 = 0 + 0 G U + W. Assume v, v' G U + W. Then there exist u, u' G U and v, v' G W such that v = u + w and v' = u' + w'. Since U and W are subspaces, u + u' G U and w + w' G Wand for any scalar k, ku G U and kw G W. Accordingly, v + v' = (u + w) + (u' + w') = (u + u') + (w + w') G U + W and for any scalar k, kv = k(u + w) = ku + kw G U + W. Thus U + W is a subspace of V.
Proposition 45. IfU and Ware subspaces of a vector space V of the chemical equation (1) over the field C, then U and W are contained in U + W.
Proof. Let u G U. By hypothesis W is a subspace of
V	and so 0 G W. Hence u = u + 0 G U + W. Accordingly, U is contained in U + W. Similarly, W is contained in U + W. By this the proof is finished.
Proposition 46. If U and W are subspaces of a vector space V of the chemical equation (1) over the field C, then U + W is the smallest subspace of V containing U and V, i. e., U + W = span{U, W}.
Proof. Since U + W is a subspace of V containing both U and W, it must also contain the linear span of U and W, i. e., span{U, W} C U + W.
On the other hand, if v G U + W, then v = u + w = 1u + 1w, where u G U and w G W. Hence, v is a linear combination of elements in U U W and so belongs to span{U, W}. Therefore, U + W C span{U, W}. Both inclusions give us the required result.
Proposition 47. If W is a subspace of a vector space
V	of the chemical equation (4. 2) over the field C, then W + W = W.
Proof. Since W is a subspace of V; we have that W is closed under vector addition. Therefore, W + W C W. By Proposition 45, W C W + W. Thus, W + W = W. By this the proof is finished.
Proposition 48. If U and W are subspaces of a vector space V of the chemical equation (1) over the field C, such that U = span{5} and W = span{I}, then U + W = span{S U T}.
Proof^. Since S C U C U + W and T C W C U + W, we have S U T C U + W. Hence span{S U T} C U + W.
Now assume v G U + W. Then v = u + W, where u G U and w G W. Since U = span{5} and W = span{I}, u = aiui + _ + arUr and w = biwi + _ + bsWs, where at, bj G C, uj G S, and Wt G T. Then v = u + w = aiui + _ + arur + biwi + _ + bsws. Thus, U + W C span {S U T}. Both inclusions yield U + W = span {S U T}.
Proposition 49. IfU and W are subspaces of a vector space V of the chemical equation (1) over the field C, then V = U + W tf every v G V can be wrttten tn the form v = u + w, where u G U and w G W.
Proof. Assume, for any v G V, we have v = u + w where u G U and w G W. Then v G U + W and so V C U + W. Since U and V are subspaces of V, we have U + W C V. Both inclusions imply V = U + W.
By this, in whole is given the skeleton of the complex vector method.
5 AN APPLICATION OF THE MAIN RESULTS
In this section the above complex vector method will be applied on some chemical equations for their balancing. All chemical equations balanced here appear for the first time in professional literature and they are chosen with an intention to avoid all well-known to date chemical equations which were repeated many times in the chemical journals for explanation of certain particular techniques for balancing of some chemical equations using only atoms with integer oxidation numbers.
1° First, we shall consider the case when the chemical reaction is infeasible.
Example 5. Consider this chemical reaction
ZlK4Fe(CN)6 + Z2K2S2O3 ^ Z3 CO2 + Z4 K2SO4 + Z5 NO2 + Zg FeS, Zi G C, (1 < i < 6).
From the following scheme
	6 N) (C Fe(4 =	O32 S22 =	2 C =	04 5 2 =	C) N =	S e F =
	v1	v2	3 v	v4	v5	6 v
K	4	2	0	2	0	0
Fe	1	0	0	0	0	1
C	6	0	1	0	0	0
N	6	0	0	0	1	0
S	0	2	0	1	0	1
O	0	3	2	4	2	0
follows this vector equation
v1 + Z2v2 = Z3v3 + Z4v4 + Z5v5 + Z6v
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i. e.,
Zi (4, 1, 6, 6, 0, 0)T + Z2 (2, 0, 0, 0, 2, 3)T = Z3 (0, 0, 1, 0, 0, 2)T + Z4 (2, 0, 0, 0, 1, 4)T + Z5 (0, 0, 0, 1, 0, 2)T + Z6 (0, 1, 0, 0, 1, 0)T,
or
(4z1 + 2z2, Z1, 6Z1, 6Z1, 2z2, 3z2)t
= (2z4, Z6, Z3, Z5, Z4 + Z6, 2z3 + 4z4 + 2z5) .
From the system of linear equations
4Z1 + 2z2 = 2z4, Z1 = Z6,
6Z1 = Z3, 6Z1 = Z5,
2Z2 = Z4 + Z6, 3z2 = 2z3 + 4z4 + 2z5,
one obtains the contradictions Z2 = 3 Z1 and Z2 = 44z1/3, that means that the system is inconsistent. According to Definition 8, the vectors v1, v2, v3, v4, v5 and v6 of the molecules of the chemical reaction (5. 1) do not generate a vector space V; and according to the Definition 10 they are linearly independent, i. e., we have only a trivial solution Zi = 0, (1 < i < 6), that means that the chemical reaction is infeasible.
2° Next, we shall consider the case when the chemical reaction is feasible and it has a unique solution.
This type of chemical equations really is the most appropriate for study the process of balancing chemical equations, because it gives an excellent opportunity for application of the group theory.
At once, we would like to emphasize here, that by application of groups theory one may determine Sylow subgroups, conjugacy classes of maximal subgroups, proper normal subgroups, and so one. The main reason why we confined ourselves to the next group of calculations is limitation of the size of the work.
Example 6. Consider this chemical reaction
Z1 Fe2(SO4)3 + Z2 PrTlTe3 + Z3 H3PO4 ^ Z4 Fe0,996(H2PO4)2-H2O + Z5 Tl1,98v(SO3)3
+ Z6 Pr1,998(SO4)3 + Zv Te2O3 + Z8 P2O5 + Z9 H2S, Zt G C, (1 < i < 9).
According to the scheme given below
O2 H2
				O	)3 )3	)3			
	O)34 (S =	leT3 PrTl =	PO43 =	P 2 (H2 (6 9 c^ CD =	O3 (S =	)O4 (S (8 9 =	O32 =	5 O52 oP =	S2 ti! =
	v1	v2	v3	v4	v5	v6		v8	v9
Fe	2	0	0	0.996	0	0	0	0	0
S	3	0	0	0	3	3	0	0	1
O	12	0	4	9	9	12	3	5	0
Pr	0	1	0	0	0	1.998	0	0	0
Tl	0	1	0	0	1.987	0	0	0	0
Te	0	3	0	0	0	0	2	0	0
H	0	0	3	6	0	0	0	0	2
P	0	0	1	2	0	0	0	2	0
one obtains the following vector equation
Z1V1 + Z2V2 + Z3V3 = Z4V4 + Z5V5 + Z6V6 + Z7V7 + Z8V8 + Z9V9,
i. e.,
Z1 (2,3,12,0,0,0,0,0)T + Z2 (0,0,0,1,1,3,0,0)T + Z3 (0,0,4,0,0,0,3,1)T = Z4 (0.996,0,9,0,0,0,6,2)T + Z5 (0,3,9,0,1.987,0,0,0)T + Z6 (0,3,12,1.998,0,0,0,0)T + Z7 (0,0,3,0,0,2,0,0)T + Z8 (0,0,5,0,0,0,0,2)T
+ Z9 (0,1,0,0,0,0,2,0)T, from where follows this system of linear equations 2z1 = 0.996Z4, 3z1 = 3z5 + 3z6 + Z9, 12Z1 + 4z3 = 9z4 + 9z5 + 12z6 + 3Z7 + 5z8, Z2 = 1.998Z6, Z2 = 1.987Z5, 3z2 = 2z7,
3z3 = 6z4 +
Z9,
Z3 = 2z4 + 2z8, From the last system one obtains the required solution. This show that the vectors Vi (1 < i < 9) generate a vector space V and they are linearly dependent. The balanced reaction has this form
17839883133 Fe2(SO4)3 + 12843034110 PrTlTe3 + 81542933266 H3PO4 ^ 35823058500 Feo.996(H2PO4)2-H2O + 6463530000 Tl1.987(SO3)3 + 6427945000 Pr1.998(SO4)3 + 19264551165 Te2O3 + 4948408133 P2O5 + 14845224399 H2S. 3° Next, the case when the chemical reaction is non-unique will be considered, i. e., when it has infinite number of solutions.
Example 7. Consider double reduction of titanium dioxide with carbon and chlorine given by the reaction
Xi TiO2 + x2 C + x3 a2 ^ + CO + x6 CO,
X4 Tia4
This reaction plays an important role in theory of metallurgical processes, especially in processes of direct reduction of metal oxides. Sure, that this reaction is not unique, and there are many other reactions of that kind, which may to be used for analysis of this particular case.
From the following scheme
	c:)			l4 iCl	O	2
	=	C =	=	iT =	C =	C =
		2 V	V3	V4		V6
Ti	1	0	0	1	0	0
O	2	0	0	0	1	2
C	0	1	0	0	1	1
Cl	0	0	2	4	0	0
follows this vector equation
Z1V1 + Z2V2 + Z3V3 = Z4V4 + Z5V5 + Z6V6,
i. e.,
(Z1, 2Z1, 0, 0)T + (0, 0, Z2, 0)T
+ (0, 0, 0, 2z3)t = (Z4, 0, 0, 4z4)t + (0, Z5, Z5, 0)T + (0, 2z6, z6, 0)T,
or
(z1, 2Z1, Z2, 2z3)T = (z4, Z5 + 2Z6, Z5 + Z6, 4z4) ,
i. e., immediately follows this system of linear equations
Z1 = Z4,
2Z1 = Z5 + 2z6, Z2 = Z5 + Z6, 2z3 = 4z4,
which general solution is Z3 = 2z1, Z4 = Z1, Z5 = -2z1 + 2z2, z6 = 2z1 - Z2, where Z1 and Z2 are arbitrary complex numbers.
Now, balanced general chemical reaction has this form
Z1 TiO2 + Z2 C + 2Z1 Cl2 ^ Z1 TiCl4
+ (- 2Z1 + 2Z2) CO + (2Z1 - Z2) CO2, (VZ1, Z2 G C).
According to the Definition 8, the vectors Vi, v2, v3, V4, V5 and V6 of the molecules of the chemical reaction generate infinite number of vector spaces V„, and according to the Definition 10 they are linearly dependent, i. e., we have an infinite number of solutions (Z1, Z2, 2z1, Z1, - 2z1 + 2z2, 2z1 - Z2), (Vz1, Z2 G C), that means that the this chemical reaction is non-unique.
6	DISCUSSION
In his previous work22, the author announced a decrease of barren intuitionism from of chemistry and its substitution by an elegant formalism from one side, and substitution of the old chemical traditionalism by a new mathematical generalism, from other side. This announcement is realized in this work that gives a new contribution to the theory as well as practice of balancing chemical equations.
The complex vector method of balancing chemical equations augmented the research field in chemistry and made obsolete the old traditional approach, and gave reliable results for paradox resolution.
By this work will begin consideration of paradoxes in chemistry as a serious object, and it will increase researchers' carefulness to avoid appearance of paradoxes.
7	CONCLUSION
The new complex mathematical method of balancing chemical equations, which was used for the solution of a paradox is farewell to the chemical tradition, which still
respects the composers of general chemistry textbooks, that affirm that chemistry is a science which studies the structure of substances, how they react when combined or in contact, and how they behave under different conditions. These subjects include a great part of the matter to which chemistry was applied.
In this work the foundation of chemistry is enriched by one more new topic, and a contribution to a new formalization of chemistry founded by virtue of a new complex vector method of balancing chemical equations is offered. This work opens doors for the next research in chemistry for diagnostic of paradoxes and their resolution. It will accelerate the newest mathematical research in chemistry and it will surmount the barriers hampering the development of chemistry.
This work is a critical survey that requires changes of chemical thinking. Hence, it must be distinguished from the uncritical penetration, in which chemistry itself is developed sometimes.
ACKNOWLEDGEMENT
Author would like to thanks to Prof. Dr Valery C. Covachev, from Bulgarian Academy of Sciences and to the Dutch chemist Marten J. Ten Hoor for reading the article and for their remarks.
In addition, author would like to thanks to Prof. Dr Franc Vodopivec from University of Ljubljana for his helpful suggestions.
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