user23699: In Silman's Complete Endgame Course, the author gives this rule in order to gain the Opposition when there isn't a direct connection in between the kings:

In this type of situation, the rule is to make a square or rectangle in which each color is the same color with the other guy to move.

But later in the book, this practice problem is given which seems to contradict this:
[FEN "2k5/8/8/8/1PK5/8/8/8 b - - 0 1"]

1...Kb8 2.Kd4 Kc7?? 3.Kc5


2.Kd4 Kb7 with a draw (and not 2.Kc7?? 3.Kc5 when White gets the opposition and wins).

Black moved his king to a square of the same color as White's king (forming a rectangle with corners c7-d7-d4-c4), but White still got the Opposition and won. Doesn't this invalidate the rule given above?

user4418: There is a crucial part missing from the definition. To have the opposition you must have an odd number of squares between the kings when it is your opponent's turn to move.
This is obvious when you think about it. Suppose you move your king to be diagonally opposite your opponent's king on a square of side 4. The two kings are on the same diagonal and so on the same colour. Then there are two squares between the kings and your opponent can simply move the king one square diagonally closer to your king giving your opponent the opposition.
According to Silman's quoted rule each player had the opposition on their turn. This is nonsense. When you moved your king onto the same coloured square diagonally opposite your opponent's king but two squares away you lost the opposition.
There is a good YouTube video by Hanging Pawns on the opposition here.

user27187: 
the rule is to make a square or rectangle in which each color is the
same color with the other guy to move.

I'm not quite sure how this sentence makes sense other than there is a typo, and it actually says "...rectangle in which each corner is the same color with the other guy to move". If this is indeed the case, then obviously b4-d4-d8-b8 works and c4-d4-d7-c7 doesn't (two white and two black corners).
An example to this would be if it's white to move, they play 1. Kd4 and black plays either Kd8 (direct long opposition) or Kb8 (indirect, rectangular opposition).

23699: <p>In <em>Silman's Complete Endgame Course</em>, the author gives this rule in order to gain the Opposition when there isn't a direct connection in between the kings:</p>
<blockquote>
<p>In this type of situation, the rule is to make a square or rectangle in which each color is the same color <em>with the other guy to move.</em></p>
</blockquote>
<p>But later in the book, this practice problem is given which seems to contradict this:</p>
<pre><code>[FEN &quot;2k5/8/8/8/1PK5/8/8/8 b - - 0 1&quot;]

1...Kb8 2.Kd4 Kc7?? 3.Kc5
</code></pre>
<blockquote>
<p>2.Kd4 Kb7 with a draw (and not 2.Kc7?? 3.Kc5 when White gets the opposition and wins).</p>
</blockquote>
<p>Black moved his king to a square of the same color as White's king (forming a rectangle with corners c7-d7-d4-c4), but White still got the Opposition and won. Doesn't this invalidate the rule given above?</p>
 4418: <p>There is a crucial part missing from the definition. <em>To have the opposition you must have an odd number of squares between the kings when it is your opponent's turn to move.</em></p>
<p>This is obvious when you think about it. Suppose you move your king to be diagonally opposite your opponent's king on a square of side 4. The two kings are on the same diagonal and so on the same colour. Then there are two squares between the kings and your opponent can simply move the king one square diagonally closer to your king giving your opponent the opposition.</p>
<p>According to Silman's quoted rule each player had the opposition on their turn. This is nonsense. When you moved your king onto the same coloured square diagonally opposite your opponent's king but two squares away you lost the opposition.</p>
<p>There is a good YouTube video by Hanging Pawns on the opposition <a href="https://youtu.be/SoZPZdnYYk8?t=474" rel="nofollow noreferrer">here</a>.</p>
 27187: <blockquote>
<p>the rule is to make a square or rectangle in which each color is the
same color with the other guy to move.</p>
</blockquote>
<p>I'm not quite sure how this sentence makes sense other than there is a typo, and it actually says &quot;...rectangle in which each <strong>corner</strong> is the same color with the other guy to move&quot;. If this is indeed the case, then obviously b4-d4-d8-b8 works and c4-d4-d7-c7 doesn't (two white and two black corners).</p>
<p>An example to this would be if it's white to move, they play 1. Kd4 and black plays either Kd8 (direct long opposition) or Kb8 (indirect, rectangular opposition).</p>
