FRIZELL: FOUNDATIONS OF ARITHMETIC. 395 



a = bx==h{xov) = bxobv = aohv tor every a. That 

 is l>v is also a modulus. But there can not be more than 

 one modulus (///). Therefore hv = v for every 6 so 

 that v^hen h' is not = b we have h'v = v = hv contrary 

 to definition. 



56. Corollary. It is not possible to define a set of 

 symbols constituting a semigroup v^ith 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, v^e 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 w^e give the pref- 

 erence to the semigroup on multiplication and proceed 

 with the group G obtained by applying XX to it. 



58. Preposition 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 sufl[icient 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) 

 = {mronq, qr). 



ProoJ. The necessity of this relation is obvious. To 

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

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

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

 nqq'r'. ^ence (m' r' on' q)qr = {mronq) q'r' by 

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

 = {mro 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 

 (Uf r) \ = (k, I) o (mr o nq, qr) = {kqro Imr o Inq, Iqr) 

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



