Question 1007930
<pre>      
Divide each member of the set A by 3, using
ordinary division, to find the remainder
when dividing by 3:

         <u> 0</u> = quotient 
0/3     3)0  
          <u>0</u>
          0 = remainder

         <u> 0</u> = quotient 
1/3     3)1  
          <u>0</u>
          1 = remainder

         <u> 1</u> = quotient 
3/3     3)3  
          <u>3</u>
          0 = remainder

         <u> 1</u> = quotient 
4/3     3)4  
          <u>3</u>
          1 = remainder

         <u> 2</u> = quotient 
6/3     3)6  
          <u>6</u>
          0 = remainder

         <u> 2</u> = quotient 
8/3     3)8  
          <u>6</u>
          2 = remainder

D = {(x, y) | x/3 and y/3 have the same remainder}

0/3 has remainder 0
1/3 has remainder 1
3/3 has remainder 0
4/3 has remainder 1
6/3 has remainder 0
8/3 has remainder 2

0/3 and 0/3 have the same remainder 0, so (0,0) is an element of D.
0/3 and 3/3 have the same remainder 0, so (0,3) is an element of D.
0/3 and 6/3 have the same remainder 0, so (0,6) is an element of D.
1/3 and 1/3 have the same remainder 1, so (1,1) is an element of D.
1/3 and 4/3 have the same remainder 1, so (1,4) is an element of D.
3/3 and 0/3 have the same remainder 0, so (3,0) is an element of D.
3/3 and 3/3 have the same remainder 0, so (3,3) is an element of D.
3/3 and 6/3 have the same remainder 0, so (3,6) is an element of D.
4/3 and 1/3 have the same remainder 1, so (4,1) is an element of D.
4/3 and 4/3 have the same remainder 1, so (4,4) is an element of D.
6/3 and 0/3 have the same remainder 0, so (6,0) is an element of D.
6/3 and 3/3 have the same remainder 0, so (6,3) is an element of D.
6/3 and 6/3 have the same remainder 0, so (6,6) is an element of D.
8/3 and 8/3 have the same remainder 2, so (8,8) is an element of D.

So<font size=1>

D = { (0,0),(0,3),(0,6),(1,1),(1,4),(3,0),(3,3),(3,6),(4,1),(4,4),(6,0),(6,3),(6,6),(8,8) }</font> 
Edwin</pre>