Another consideration that supports understanding numerical negation as logical negation, is the consideration of how integers can be constructed from pairs of naturals. Recall that integers can be identified as equivalence classes of
natural number pairs, e.g
an instance of '2' can be any of (2,0), (3,1), (4,2) , ...
Here, the numbers in a pair (a,b) can be thought of as denoting the scores of two players A and B.
Negation switches the scores the other way around
-2 := any of { (0,2) (1,3), (2,4) ,.. }
Zero represents tied results where A and B's scores are identical, and these results lie on a 45 diagonal line (call it the 'zero line') running through the centre of the positive quadrant of euclidean space, dividing the quadrant into two non-overlapping 'victory zones', one for each player.
The magnitude m of a general score (a,b) is it's distance from the zero line, and measures by how much the winning player won by. Hence we can view this as the score of an adversarial zero-sum game of tug-of-war between A and B, with rope length m, along the axis perpendicular to the zero-line.
Compare to the case of 'Complex Number Games'. In contrast,
i) A game with scores (a,b) is written a + j*b, where j is the imaginary unit.
ii) Either or both of a and b can be positive or negative, which means A and B face a common opponent C.
iii) B's score is perpendicular to A's due to multiplication by j, which means that A and B might play cooperatively.
iv) The magnitude n of the score (a,b) is the Euclidean length, i.e. sqrt( a^2 + b ^2). This represents the total reward with respect to an n-square-sum three player game.
v) The phase angle of the result determines how the reward is distributed among A, B and C.
vi) The imaginary unit j serves as negation for three-player games, dividing the 2D Euclidean space of real-valued score outcomes into the following quadrants (where a quadrant is taken to include it's clockwise-next axis and excludes zero):
{A doesn't lose and B wins, A loses and B doesn't lose, A doesn't win and B loses, A wins and B doesn't win}
Multiplying any of these quadrants by j yields the next quadrant to the right (using circular repetition).