PairSum problem 
Input: Array of numbers A of length n, number x.
Output: true if for some indices i,j (not equal) we have x = A_i + A_j, otherwise false is returned.
Exercise 2.3-7 is to find an algorithm to solve the PairSum problem in O(nlg(n)) time.

My solution:
PairSum(A,n,x) 
{
1. Sort A.
2. return PairSumSorted(A,n,x);
} 

Correctness:
PairSumSorted has the same output requirement as PairSum, while assuming A is sorted.  Thus PairSum is correct if PairSumSorted is correct.  

I will show below that PairSumSorted is correct and runs in linear time (Theta(n)).

Complexity:
Step 1, sorting, can be done in Theta(nlg(n)), which dominates step 2. Thus PairSum runs in Theta(nlog(n)) as required for the problem.
QED.

PairSumSorted(A,n,x) 
Input: Array of numbers A of length n sorted in increasing order, number x.
Output: true if for some indices i,j (not equal) we have x = A_i + A_j, otherwise false is returned.
{  
1. if (n < 2) then return false;
2. y := A_1 + A_n;
3. if (y == x) then return true;
4. if (y < x) then return PairSumSorted(A+1,n-1,x);
5. if (y > x) then return PairSumSorted(A,n-1,x);
}

Correctness:
The result returned in step 1 or step 3 is clearly correct.
Suppose the sum y is less than x. Every other sum involving A_1 is smaller (at least no larger), so we may remove A_1 from consideration.  This is precisely what line 4 does.  We similarly justify removing A_n from consideration in line 5.

Runtime:
Let T(n) be the cost for array of size n.  
We have, for some constant C, the recurrence
T(1) <= C, and
T(n) <= T(n-1) + C, for n > 1.
This is because only one recursive call is made (in line 4 or 5) and the rest of the work is of constant amount.

It is easy to see by induction that T(n) <= Cn: 
a) it is true for the base case n = 1.
b) Using the inductive hypothesis, T(n-1) + C <= C(n-1) + C = Cn.
Thus T(n) <= T(n-1) + C <= Cn.