Daca \( M \) este o multime finita vom nota cu \( n(M) \) numarul elementelor sale . Fie \( A,B,C \) trei multimi . Dovediti ca :
\( n(A\cup B\cup C)=n(A)+n(B)+n(C)-[n(A\cap B)+n(A\cap C)+n(B\cap C)]+n(A\cap B\cap C) \)
Trei multimi
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Trei multimi
. A snake that slithers on the ground can only dream of flying through the air.
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Laurian Filip
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Desenand cele 3 multimi ca niste diagramele van euler, afirmatia este evidenta.
In desen am notat cu \( M1,M2,M3,M4 \) multimile de puncte situate in cea mai mica suprafata inchisa care le contine.
\( n(A\cup B \cup C)=n(A)+n(B)+n(C)-n(M1)-n(M2)-n(M3)-2n(M4)=n(A)+n(B)+n(C)-[(n(M1)+n(M4)) + (n(M2)+n(M4))+(n(M3)+n(M4))]+n(M4)=n(A)+n(B)+n(C)-[n(A\cap B)+n(A\cap C)+n(B\cap C)]+n(A\cap B\cap C) \)
In desen am notat cu \( M1,M2,M3,M4 \) multimile de puncte situate in cea mai mica suprafata inchisa care le contine.
\( n(A\cup B \cup C)=n(A)+n(B)+n(C)-n(M1)-n(M2)-n(M3)-2n(M4)=n(A)+n(B)+n(C)-[(n(M1)+n(M4)) + (n(M2)+n(M4))+(n(M3)+n(M4))]+n(M4)=n(A)+n(B)+n(C)-[n(A\cap B)+n(A\cap C)+n(B\cap C)]+n(A\cap B\cap C) \)
Eu cred ca merge si astfel :
Folosind proprietatea \( n(A\cup B)=n(A)+n(B)-n(A\cap B) \)
Deci\( n[(A\cup B)\cup C]=n(A\cup B)+n(C)-n[(A\cup B)\cap C]=n(A)+n(B)-n(A\cap B)+n(C)-n[(A\cap C)\cup (A\cap B)]=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(A\cap B)+n(A\cap B\cap C) \) , ceea ce trebuia demonstrat .
Folosind proprietatea \( n(A\cup B)=n(A)+n(B)-n(A\cap B) \)
Deci\( n[(A\cup B)\cup C]=n(A\cup B)+n(C)-n[(A\cup B)\cap C]=n(A)+n(B)-n(A\cap B)+n(C)-n[(A\cap C)\cup (A\cap B)]=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(A\cap B)+n(A\cap B\cap C) \) , ceea ce trebuia demonstrat .
. A snake that slithers on the ground can only dream of flying through the air.