1 Natural Numbers 7We define thesumofmandnto be
m+n=sm(n).It is not difficult to deduce from this definition and the axioms(N1)–(N3)the usual
rules foraddition:for all a,b,c∈N,
(A1)if a+c=b+c,then a=b; (cancellation law)
(A2)a+b=b+a; (commutative law)
(A3)(a+b)+c=a+(b+c). (associative law)
By way of example, we prove the cancellation law. LetMbe the set of allc∈N
such thata+c=b+conly ifa=b.Then1∈M,sincesa( 1 )=sb( 1 )implies
S(a)=S(b)and hencea=b. Supposec∈M.Ifa+S(c)=b+S(c),i.e.sa(S(c))=
sb(S(c)),thenSsa(c)=Ssb(c)and hence, by(N1),sa(c)=sb(c).Sincec∈M,this
impliesa=b. Thus alsoS(c)∈M. Hence, by(N3),M=N.
We now show that
m+n=n for allm,n∈N.For a givenm∈N,letMbe the set of alln∈Nsuch thatm+n=n.Then1∈M
since, by(N2),sm( 1 )=S(m)=1. Ifn∈M,thensm(n)=nand hence, by(N1),
sm(S(n))=Ssm(n)=S(n).Hence, by(N3),M=N.
By Proposition 1 again, for eachm∈Nthere exists a unique mappm:N→N
such that
pm( 1 )=m,
pm(S(n))=sm(pm(n)) for everyn∈N.We define theproductofmandnto be
m·n=pm(n).From this definition and the axioms(N1)–(N3)we may similarly deduce the usual
rules formultiplication:for all a,b,c∈N,
(M1)if a·c=b·c,then a=b; (cancellation law)
(M2)a·b=b·a; (commutative law)
(M3)(a·b)·c=a·(b·c); (associative law)
(M4)a· 1 =a. (identity element)
Furthermore, addition and multiplication are connected by
(AM1)a·(b+c)=(a·b)+(a·c). (distributive law)
As customary, we will often omit thedot when writing products and we will give
multiplication precedence over addition. With these conventions the distributive law
becomes simply
a(b+c)=ab+ac.