Tweak code example in binary-search approach

The switch statement in the loop-with-switch approach had
two explicit cases and a default.

The three cases that exist are:
- equal
- less than
- greater than

If we want to have two explicit cases and a default, then it would
feel more symmetrical to have the less-than and greater-than cases
be defined explicitly with the equal case being the default.

However, it's hard to argue that any of the three are actually a
default here.

Instead I've changed it to have three explicit cases.

exercises/practice/binary-search/.approaches/loop-with-switch/content.md

@@ -8,13 +8,13 @@ package binarysearch
 func SearchInts(list []int, key int) int {
 	for left, right := 0, len(list); left != right; {
 		mid := (left + right) / 2
-    
+
 		switch {
 		case list[mid] == key:
 			return mid
 		case key < list[mid]:
 			right = mid
-		default:
+		case key > list[mid]:
 			left = mid + 1
 		}
 	}
@@ -37,7 +37,7 @@ A [`switch` with no condition][switch-no-condition] is used to check the value o
 - If the value being searched for is less than the element at the index of the middle value, then `right` is set to the middle value
 so that the next iteration will look at lower numbers.
 
-- Otherwise, the value being searched for must be greater than the element at the index of the middle value, so `left` is set to the middle value
+- Otherwise, the value being searched is greater than the element at the index of the middle value, so `left` is set to the middle value
 plus one so that the next iteration will look for higher numbers.
 
 If `left` and `right` are changed during the iterations so that they equal other, then the value being searched for is not in the slice of `int`s.

exercises/practice/binary-search/.approaches/loop-with-switch/snippet.txt

@@ -3,6 +3,6 @@ case list[mid] == key:
   return mid
 case key < list[mid]:
   right = mid
-default:
+case key > list[mid]:
   left = mid + 1
 }