Negation of "either $x = 0$ or $y = 0$" Why is the negation of "either $x = 0$ or $y = 0$" both $x \neq 0$ and $y \neq 0$?
Or is inclusive here, I suppose?
 A: It is inclusive, unless specified otherwise.
Let $x=0$ be $A$ and $y=0$ be $B$.
By De Morgan's Law
$$
\neg (A \lor B) = (\neg A \land \neg B)
$$
However, it does say either. This might indicate it is exclusive. In that case, that negation is not correct.
A: Yes it is inclusive if it is not precised.  
In general the negation of "$A$ or $B$" is "not $A$ and not $B$". To convince yourself that it is true, you can compare the truth tables. 
A: statement A: either x=0 or y=0.
negation of statement A: Its not true that either x=0 or y=0.
       as negation means opposite of statement A , so we are saying that statement-A is not true.
      now you read negation statement once it says that both are false it is similar to I wont eat either pasta or french fries which is equivalent to say that I wont eat pasta and I wont eat french fries. So if you equate you will get as
Its not true that x=0 and its not true that y=0.
to write in short:
~(p or q) = ~p And ~q
~(p and q) = ~p or ~q
 if you know about truth tables you can even compare them with truth tables.
if anyone needs detailed proof using truth tables do ask :-)
