A∩Z={1;-10/5}
A∩Q={1;-2,(6);-3/4;√(64/121);-10/5}
A∩(R\Q)={√3;√0.9;√1008}
A\Q=A∩(R\Q), pt ca toate sunt incluse in R
(R\Q)∩A= A∩(R\Q) intersectia e comutativa
Obs am tinut cont ca√0,9=√(9/10)=√9/√10=3/√10∈R\Q
√(64/121)=√64/√121=8/11∈Q
-10/5=-2∈Z
2 aplicatie la multimide numere
trebuie stiuta relatiade baza
N⊂Z⊂Q⊂R
atunci
N∪R=R
N\Z=∅
Q∩Z=Z
R∩Q=Q
(R\Q)\Z=R\Q
Q\R=∅
par usoare, dar pana nu le "tocesti" bine sunt usor cam subtile
