Exercise 4.2a(1), p. 92

Base Case: e = 0. Then 0 mults are done, which is at least e.

Induction Hypothesis (I.H.): Assume that (mod-expt b k m) does at least k multiplications, for k < e.

Induction Step: Consider (mod-expt b e m).

• Case 1: e is even. Then the number of multiplications done is:

                1   +  e/2  +  e/2
                |	       \    /
        	      |		\  /
             by mod*   by (mod-expt b e/2 m)
  		     and the I.H.
  
              = 1 + e,  which is at least e
  

• Case 2: e is odd. Then the number of multiplications done is:

                1   +  e-1
                |	      |
       	      |	      |
             by mod*   by (mod-expt b e-1 m)
  		     and the I.H.
  
              = e, which is at least e