Apply a real example to divide-and-conquer algorithm

Question

This question comes from Levitin, A. Introduction to the Design and Analysis of Algorithms. Pearson. Exercise 5.1 Q1.

Write pseudocode for a divide-and-conquer algorithm for finding the position of the largest element in an array of n numbers.

The solution given is as below:

function MaxIndex(A, l, r)
if l = r return l
else temp1 <- MaxIndex(A, l, (l+r)/2 )
     temp2 <- MaxIndex(A, (l+r)/2+1, r)
     if A[temp1] >= A[temp2]
         return temp1
     else return temp2

However, I meet some problem when I use a random example to understand this algorithm

For example, I have an array A[20, 10, 30, 15,6] at position 0 to 4

function MaxIndex (A[.], l=0, r=4)
if l=r return l  // pass
else temp1 <- MaxIndex(A, l=0, (l+r)/2=(0+4)/2=2 ) // So MaxIndex (A, 0, 2)
     temp2 <- MaxIndex(A, ((l+r)/2)+1=((0+4)/2)+1 =3, r=4) // So MaxIndex(A,3,4)
     if A[temp1] >= A[temp2]
         return temp1
     else return temp2

then

function MaxIndex (A[.], l=0, r=2)
if l=r return l  // pass
else temp1_1 <- MaxIndex(A, l=0, (l+r)/2=(0+2)/2=1) // So MaxIndex (A, 0, 1)
     temp1_2 <- MaxIndex(A, ((l+r)/2)+1=((0+2)/2)+1 =2, r=2) // So MaxIndex(A,2,2)
     if A[temp1_1] >= A[temp1_2]
         return temp1_1
     else return temp1_2

then

function MaxIndex (A[.], l=0, r=1)
if l=r return l  // pass
else temp1_1_1 <- MaxIndex(A, l=0, (l+r)/2=(0+1)/2=0.5) // So MaxIndex (A, 0, 0.5)?????????
     temp1_1_2 <- MaxIndex(A, ((l+r)/2)+1=((0+1)/2)+1 =1.5, r=1) // So MaxIndex(A,1.5,1)??????????
     if A[temp1_1] >= A[temp1_2]
         return temp1_1
     else return temp1_2

I feel like I cannot move any further since then. Maybe my understanding about apply this algorithm is wrong. Should I think by default, when either j or i is not an integer, round down to the nearest integer? And when ((1+r)/2)+1 > r, I will keep r value instead?