Consider the set Q$[5]=\{$a$+$b$5$:a$,$b in Q$\}$Q$[5]=\{$a$+$b$5$:a$,$b in Q$\}.$ The fact that for all a$,$b in Qa$,$b in Q with $($a$,$b$)(0,0)($a$,$b$)(0,0)$ we have $1($a$+$b$5)=($aa$25$b$2)+($ba$25$b$2)51($a$+$b$5)=($aa$25$b$2)+($ba$25$b$2)5$ justifies that Q$[5]$Q$[5]$ satisfies the additive closureadditive associativityadditive commutativityadditive identityadditive inversemultiplicative closuremultiplicative associativitymultiplicative commutativitymultiplicative identitymultiplicative inversedistributivity additive closureadditive associativityadditive commutativityadditive identityadditive inversemultiplicative closuremultiplicative associativitymultiplicative commutativitymultiplicative identitymultiplicative inversedistributivity law.