Question 916124
<pre>No.  Not (b) since absolute values are never negative!
Let's go through them:
  
(a) is onto because for any k &#8712; Z, f(k,n) = k, for any n &#8712; Z

(b) is not onto because for any k &#8712; Z where k < 0, 
    f(m,n) = |n| &#8800; k for all m,n &#8712; Z

(c) is onto because for any k &#8712; Z, f(k+1,1) = (k+1)-1 = k

Edwin</pre>