542 



Mr DE MORGAN, ON SOME POINTS 



It matters nothing which set of symbols is made progressive in the formation : thus 



(ABC^ is A p {BC V) + A q (BC rp) + A r (BC m , 



If necessary, the two modes of progression may be distinguished as in {(ABC Mr) and (ABC , 

 or as in (ABC^ and {pqr ABC). 



In (ABCD ( we have constructed the 24 ways of distributing p, q, r, s, among A, B, 

 C, D, in such manner that A p B q C r D 3 , suggested by the order in the symbol, shall be positive, 

 and that any two terms which are formed from each other by one interchange of contiguous 

 symbols shall have different signs. Thus, B p C q A r D a and B p A q C r D s have different signs. When 

 this is always true, it is clear that one contiguous interchange, such as is seen in passing from 

 (ABCDp^^ to (ACBDj^^ changes the sign of the whole : whence* any odd number changes 

 the sign, while any even number restores it. Hence it is easily proved that if the property be 

 true for the n th eliminant, it is true for the (n + l) tb . Suppose, for instance, that it is found 

 true for every fifth eliminant. Then, since (ABCDEF^^ = A p (BCDEF vstu) - ... it is 

 clear that the theorem is true for all interchanges which take place in fifth eliminants without 

 affecting the multipliers. When the interchange affects a multiplier, take 



(ABCDEF^ = A p (BCDEF V!tu) - B p (CDEFA^ + ... 



For {CDEFA ttu) write its equal (ACDEF^^ formed by four contiguous interchanges, 

 and then resolve the fifth eliminants one step, which gives 



A p B q (CDEF rttu) t ... - B p A q (CDEF Tatu) - ... 

 whence the theorem is obvious as to terms beginning with A p B q and B p A q . 



In passing from an even eliminant to the following odd one, the change of sign arises from 

 the process which brings A from the end to the beginning, requiring an odd number of 

 interchanges. This incomplete outline of a proof is merely given to exhibit the facility with 

 which the notation applies. 



It immediately follows that when two symbols of either set are the same, the eliminant 

 vanishes. Let there be n homogeneous linear equations, A p a + B p (5 + ...= 0, A q a + B q fi +... > 

 between n + 1 quantities a, (3, ... : for instance, five equations between six quantities. Mul- 

 tiply the equations by (CDEF qrat) , (CDEF rttp} , &c. and add, as the terms of fifth eliminants 

 are all additive. 



* In order to ascertain, by a mechanical process, and in any 

 case however complicated, whether the number of contiguous 

 interchanges which convert one arrangement into another be 

 even or odd, proceed as follows. Write down one arrangement 

 under the other, and beginning at one letter in one of the lines, 

 mark the companion letter in the other line, pass on to that 

 companion letter in the first line, mark its companion, and so 

 on, until we arrive at a letter already marked. Call this 

 sequence a chain, each mark being one link. Having formed 

 one chain, begin at a letter not yet used and form another 

 chain ; and so on, until every letter has been used. Then, 

 according as the number of chains with even numbers of links 



is odd or even, the number of displacements required is odd or 

 even. The demonstration is very easy. As an example take 



ABCDEFGHIJKLMNOPQ 

 HMOGQBKLJPFC INADE 

 12123221222124123 



Under A is H, under H is L, under L is C, under C is O, 

 under O is A, already taken : the first chain has five links. 

 The second is found to have nine, the third two, the fourth 

 one. The number of chains having even links is one, an odd 

 number; hence the number of contiguous interchanges by 

 which either arrangement is converted into the other, is odd. 



