Question #262489

A-   For all integers a and b, if a + b is odd, then a is odd or b is odd.

B-  For any integer n the number (n3 - n) is even.

C-   Proof of De-Morgan’s Law


Expert's answer

A. For all integers aa  and bb let a+ba+b is odd. If aa and bb are both even than ∃p,l∈Za=2p,b=2l  ⟹  a+b=2(p+l)\exist p, l \in Z a=2p, b=2l\implies a+b=2(p+l) is even too. If aa and bb are both odd than ∃p,l∈Za=2p+1,b=2l+1  ⟹  a+b=2(p+l+1)\exist p, l \in Z a=2p+1, b=2l+1 \implies a+b=2(p+l+1) is even too. So we have only one of  aa  and bb is odd and ∃p,l∈Za=2p,b=2l+1  ⟹  a+b=2(p+l)+1\exist p, l \in Z a=2p, b=2l+1 \implies a+b=2(p+l)+1 .

B. For any integer nn  the number (n3−n)(n^3-n) equals to n(n+1)(n−1)n(n+1)(n-1) that is the product of three three consecutive natural numbers, one of which is even. In case (n−1)=0(n-1)=0 we have (n3−n)=0(n^3-n)=0 and it is even too.

C. Let's prove that for alll sets AA and BB the equality ¬(A∧B)=¬A∨¬B\lnot(A\land B)=\lnot A\lor \lnot B .

¬(A∧B)=(¬A∧B)∨(A∧¬B)∨(¬A∧¬B)=\lnot (A\land B)=(\lnot A\land B)\lor ( A\land \lnot B) \lor ( \lnot A\land \lnot B) =

((¬A∧B)∨(¬A∧¬B))∨((A∧¬B)∨(¬A∧¬B))=((\lnot A\land B) \lor ( \lnot A\land \lnot B))\lor (( A\land \lnot B) \lor ( \lnot A\land \lnot B)) =

(¬A∧(B∨¬B))∨(¬B∧(B∨¬B))=¬A∨¬B( \lnot A \land ( B\lor \lnot B))\lor (\lnot B \land ( B\lor \lnot B)) =\lnot A\lor \lnot B


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