Binary search in C ++ with an array

Question

Here is an array with exactly 15 elements:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 

Suppose we do a binary search for an element. Indicate any elements that will be found by examining two or less numbers from the array.

What I have: since we are doing a binary search, so the number found by only one comparison will be the 7th element = 7. For two comparisons, this leads to the second division of the array. That is, the found number can be either 3 or 11.

Am I right or not?

Answer 1

You are almost right, the first number is not seven, but eight.

The remaining 2 will be equal to 4 and 12.

The correct answer will be 4, 8, 12

Answer 2

ʻI found answer 8, which is the 7th element, the rest of the elements found were the 3.5th and 10.5th elements of the sorted array. So, the following two numbers: 4 and 11.

An explanation of how I received the answers.
this array is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

 head=1 tail=15 middle= 0+14/2=7th element **0 is the index value of 1 and 14 is of 15** middle value turns to be 8 as it is the 7th element. solving value for first half head=1 tail=8 middle= 0+7/2=3.5 or 3rd element **0 is the index value of 1 and 7 is of 8** middle value now turns to be 4 as it is the 3rd element. solving value for second half head=8 tail=15 middle= 7+14/2=10.5 or 10th element **7 is the index value of 8 and 14 is of 15** middle value now turns to be 11 as it is the 10th element of the array` 

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