FRIZELL: FOUNDATIONS OF ARITHMETIC. 395 



a = bx =^ b {x o v) = bx o bv = aobv for every a. That 

 is hv is also a modulus. But there can not be more than 

 one modulus (///). Therefore hv = v for every h so 

 that when h' is not = h we have h'v = v = hv contrary 

 to definition. 



56. Corollary. It is riot possible to define a set of 

 symbols constituting a semigroup with modulus as re- 

 gards both rules of § 55. 



57. If Prop. XX is applied to the set of natural 

 numbers and the rule of addition, we lose the semi- 

 group on multiplication and can not recover it. This 

 semigroup, in fact, is destroyed if we only annex a 

 modulus for addition. Consequently we give the pref- 

 erence to the semigroup on multiplication and proceed 

 with the group G obtained by applying XX to it. 



58. Proposition XXIII. Given a higher rule dis- 

 tributive over a lower, and a set of symbols forming an 

 abelian semigroup with respect to each rule, if we build 

 the group G of § 52 with reference to the higher rule, 

 then a necessary and suflftcient condition of having a 

 lower rule over which the higher shall be distributive 

 and with regard to which G shall constitute an abelian 

 semigroup is the formula of definition (m, q) o{n, r) 

 = {rrir onq, qr). 



Proof. The necessity of this relation is obvious. To 

 show that it is sufficient we first let {m' , q') = (m, q) 

 SLYid {n^, r') = (n, r) .'. (m'q = mq') and n'r = nr'. 

 Therefore m'r'qr = m'qrr' = mrq'r' and n'q'qr = 

 nqq'r'. Hence (m'r' on' q)qr ^ (mr onq) q'r' by 

 distributive law and .'. by definition {m' r' on' q' q'r') 

 = {mr o nq, qr) or equals with equals give equals and 

 we have a C-rule. Similarly equals with unequals give 

 unequals. For the associative law {k, l)o\{m, q)o 

 {n, r) \ = {k, I) o {mr o nq, qr) = {kqr o Imr o Inq, l(/r) 

 = (kqo Im, lq)o{n, r) = {k, l)o(m, q) o{n, r). 



