Question

soution
(i) Commutativity For all ( m, n in N, ) we have ( operatorname{gcd}(m, n)=operatorname{gcd}(n, m) )
[
therefore m * n=n+m forall m, n in N
]
Hence, ( * ) is commutative on ( N ).
(ii) Associativity Let ( m, n, p in N . ) Then,
[
begin{aligned}
(m * n) * p &=[operatorname{gcd}(m, n)]^{*} p
&=operatorname{gcd}[g operatorname{cd}{(m, n), p}]
&=operatorname{gcd}[m, operatorname{gcd}(n, p)}]
end{aligned}
]
( [because text { gcd of three numbers }=operatorname{gcd}{(g c d text { of any two, th }) )
[
=operatorname{gcd}(m, n * p)=m *(n * p)
]
Hence, ( ^{*} ) is associative on ( N ).

# On the set N of all natural numbers, define the operation * on N by min=gcd (m, n) for all m,ne N. Show that is commutative as well as associative.

Solution