For associative property, elements of set must satisfy following condition :
Zi+(Zj+Zk)=(Zi+Zj)+Zk , where Zi,Zj&Zk are the different elements of a given set.
let's check..
Z1+(Z2+Z3)=?(Z1+Z2)+Z3
LHS:
LHS:5i^−2j^+(3i^+3j^+4i^−j^)=12i^RHS:(5i^−2j^+3i^+3j^)+4i^−j^=12i^
Hence, Z2,Z2 and Z3 are associative under addition operation.
For commutative property, we must have ...
Zi+Zj=Zj+Zi for all i and j.
let's check...
Z1+Z2=?Z2+Z1
LHS:(5i^−2j^)+(3i^+3j^)=8i^+j^RHS:(3i^+3j^)+(5i^−2j^)=8i^+j^
similarly we can show that...
Z1+Z3=Z3+Z1=9i^−3j^Z3+Z2=Z2+Z3=7i^+2j^
Hence Z1,Z2 and Z3 are commutative under addition operation.
For distributive property, we must have...
Zi.(Zj+Zk)=Zi.Zj+Zi.Zk for all i,j and k.
let's check...
Z1.(Z2+Z3)=?Z1.Z2+Z1.Z3
LHS:(5i^−2j^).(7i^+2j^)=31RHS:(15−6)+(20+2)=31
Similarly we can show that...
Z2.(Z3+Z1)=Z2.Z3+Z2.Z1Z3.(Z1+Z2)=Z3.Z1+Z3.Z2
Hence Z1, Z2 and Z3 follow distributive property.
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