Exercise 3.  Show that, if  ,  then    has a unique value.

Solution 3.

See text and/or instructor's solution manual.

Solution.   All the values for    are  .  

Recall that    which collapses to the single element zero, i.e.   .  

Therefore, for  ,   we have   .

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell