Here \(\displaystyle{G}={R}×{Z}\) and the binary operation defined as follows

\((a,m)\times(b,n)=(ab,m+n)\)

where \((a,m),(b,n)\in R\times Z\).. Since \((1,0)\in R\times Z\), so G is non empty.

Associative: Let \((a,m),(b,n),(c,p)\in R\times Z\),

\((a,m)\times((b,n)\times(c,p))=(a,m)\times(bc,n+p)=(abc,m+n+p)=(ab,m+n)\times(c,p)=((a,m)\times(b,n))\times(c,p)\).

\((a,m)\times((b,n)\times(c,p)) =(a,m)\times(bc,n+p) =(abc,m+n+p) =(ab,m+n)\times(c,p) =((a,m)\times(b,n))\times(c,p)\).

Identity: Claim (1,0) is the identity of G, since for any \((a,m)\in G(a,m)\in G\) we have

\((a,m)\times(1,0)=(a,m)=(1,0)\times(a,m)\).

Inverse:

For any \((a,m)\in\)

\((a,m)\times(1.a,−m)=(a⋅1.a,m−m)=(1,0)=(1.a,−m)\times(a,m)\). Hence, G is a group