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Fie multimile
A= {2^0 ; 2^0+2^1 ; 2^0+2^1+2^2 ; 2^0+2^1+2^2+2^3'....} si
B={3^0 ' 3^0+3^1 ; 3^0+3^1+3^2 ' 3^0+3^1+3^2+3^3'...}
Determinati multimea A intersectat cu B
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Răspuns :

Bunicaluiandreiid

pentru ca {A}intersectat {B }≠ Ф   trebuie ca cele doua multimi sa aiba macar un termen comun      Ta= Tb

Ta=2^o +2^1 +2^2 +..............+2^a          Tb = 3^o +3^1 + 3^2 +...............+3^b                  Ta  = 1+2^1 + 2^2 +...............+2^a          Tb=1 +3^1 +3^2 +......................+3^b                 Ta +!=(2 +2^1) +2^2 +........+ 2^a           3Tb=3+3^2+ 3^3 (......................+3^(b+1)          Ta +1=(2^2 +2^2) +2^3+......+2^a             3Tb-Tb=2Tb=3^(b+1) -1                                    Ta +1=(2^3 +2^3)+...............+2^a              Tb=[3^(b+1) -1]/2                                              Ta = ..........................2^a +2^a -1               Ta =2^(a+1)

2^(a+!) -1 = [3^(b+1) -1]/2   si inmultind  relatia cu 2 ⇒   2^(a+2)  - 2 = 3^(b+1) - 1     adica,  2^(a+2) =3^(b+1) + 1      ptr. b=0    2^(a+2) =4 = 2^2    a+2 =2 ⇒ a=0 ⇒Ta=2^0=1    Tb =3^0 =1   ⇒⇒{A}int.{B}={1}    se dau si alte valori  lui  b  si se verifica daca  Ta= Tb        

                                                                                                   

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