Understanding geometric improvement approach for obtaining polynomial-time algorithms - algorithm

Question

I'm reading Network Flows - Theory, Algorithms, and Applications and I'm stuck on the proof of the following theorem (Ch. 3, Page 67):
Theorem. Suppose that 𝑧^𝑘 is the objective function value of some solution of a minimization problem at the 𝑘𝑡ℎ iteration of an algorithm and 𝑧^∗ is the minimum objective function value. Furthermore, suppose that the algorithm guarantees that for every iteration 𝑘,
(1) (𝑧^𝑘 − 𝑧^(𝑘+1)) ≥ 𝛼(𝑧^𝑘 − 𝑧^∗)
(i.e., the improvement at iteration 𝑘 + 1 is at least 𝛼 times the total possible improvement)for some constant 𝛼 with 0 < 𝛼 < 1 (which is independent of the problem data). Then the algorithm terminates in 𝑂((𝑙𝑜𝑔𝐻)/𝛼) iterations, where 𝐻 is the difference between the maximum and minimum objective function values.
Proof. The quantity (𝑧^𝑘 − 𝑧^∗) represents the total possible improvement in the objective function value after the 𝑘𝑡ℎ iteration. Consider a consecutive sequence of 2/𝑎 iterations starting from iteration 𝑘. If each iteration of the algorithm improves the objective function value by at least 𝑎(𝑧^𝑘 − 𝑧^∗)/2 units, the algorithm would determine an optimal solution within these 2/𝑎 iterations. Suppose, instead, that at some iteration 𝑞 + 1, the algorithm improves the objective function value by less than 𝑎(𝑧^𝑘 − 𝑧^∗)/2 units. In other words,
(2) 𝑧^𝑞 − 𝑧^(𝑞+1) ≤ 𝑎(𝑧^𝑘 − 𝑧^∗)/2.
The inequality (1) implies that
(3) 𝛼(𝑧^𝑞 − 𝑧^∗) ≤ 𝑧^𝑞 − 𝑧^(𝑞+1)
The inequalities (2) and (3) imply that
(𝑧^𝑞 − 𝑧^∗) ≤ (𝑧^𝑘 − 𝑧^∗)/2,
so the algorithm has reduced the total possible improvement (𝑧^𝑘 − 𝑧^∗) by a factor at least 2. We have thus shown that within 2/𝑎 consecutive iterations, the algorithm either obtains an optimal solution or reduces the total possible improvement by a factor of at least 2. Since 𝐻 is the maximum possible improvement and every objective function value is an integer, the algorithm must terminate within 𝑂((𝑙𝑜𝑔𝐻)/𝛼) iterations.
Why does the author focus on 2/a iterations?

Answer 1

I think it's written this way for computer scientists to understand more easily that we're halving an integer quantity every 2/α rounds, hence the number of 2/α-round super-rounds will be O(log). No particular reason we can't do the math more directly:
T(n) is the gap between the nth solution and an optimal solution.
T(n) ≤ (1 - α) T(n-1) [assumption]
n
T(n) ≤ T(0) (1 - α) [by induction]
-α n x
< T(0) (e ) [by 1 + x < e for x ≠ 0]
-α (ln T(0))/α
T((ln T(0))/α) < T(0) e
= 1,
so (ln T(0))/α rounds suffice.