I have the domain $A = \{1, 2, \ldots , 1000\}$. I need to figure out how many non zero entries are in each relation:
a. $R_1 = \{\;(a, b) \;|\; a \le b\;\}$
b. $R_2 = \{\;(a, b) \;|\; a + b = 1000\;\}$
c. $R_3 = \{\;(a, b) \;|\; a + b \le 1001\;\}$
I did these entries for $A = \{1,2,\ldots,10\}$ and got for (a.) $55$ because you get a $10\times 10$ matrix and you get $10$ 1's for the first column, then $9$ in the next and so on.
For (b.) I did the same. I shrunk the relation down so I could do it on paper so used $A = \{1,2,\ldots,10\}$, and used this relation $R_2 = \{\;(a, b)\;|\; a + b = 10\}$. I got $9$ 1 entries because in each row there was only one pair of numbers that would equal $10$, so I got $91=100-9$ zero entries.
I did the same thing for (c.) and got $55$ again. I just need help on expanding these relations, so I can get the value of the entries above.
You are generally on the right track. The easiest set to count is the relation $R_2$ of part (b.). How many $a\in A$ allow for $b\in A$ such that $a+b=1000$? Since zero is not in $A$, only $999$ of the elements $a\in A$ allow this, and there is clearly exactly one $b = 1000-a$ for each such $a$.
Similar to what you found for the simple case with ten elements in $A$, the counts for $R_1$ and $R_3$ will be equal. This is because $a \le b$ in $A$ is equivalent to:
$$ a + (1001 - b) \le 1001 $$
That is, $(a,b) \in R_1$ if and only if $(a,1001-b) \in R_3$. If you verify that $b\in A$ if and only if $1001-b \in A$, then you should be convinced the relation $R_3$ has exactly the same number of ordered pairs as $R_1$ does.