SET® Mathematics
Mathematical Proof of the Magic Square
by Llewellyn Falco

One day, while sitting by myself with a deck of SET® cards, I began to wonder whether or not I could construct a 3x3 square which made a set regardless of which direction you looked. I sorted the deck into single colors, and then started constructing a square. To my surprise it worked. I tried to make another one. It worked. As a test, I made a 3x3 square with all three colors, and sets involving no similarities, and other sets with only one difference. When it ended up working out, I was convinced that no matter which cards you started with, you could always construct a 3x3 square that made a set in every direction.   Being an educated man, and a lover of mathematics, I decided that I should be able to prove this theory. So I set out to work; this is the fruit of my labor...   First, we need a convention in which to label the cards. Thus, if we look at each characteristic on each card separately, and denote all variation to 1, 2, or 3...

 Number[X1]  Color[X2]  Symbol[X3]  Shading[X4] {1,2,3,}

So the vector x=[p,q,r,s] completely describes the card.

For example: the card with one, red, empty, oval might be Number[X1] = 1, Color[X2] =1, Symbol[X3] = 1, Shading[X4] = 1, or x = [1,1,1,1,].

For shorthand, I use the notation Cx to represent the card.
Where Cx = Number[X1] , Color[X2] , Symbol[X3], Shading[X4], and x = [p,q,r,s].  

 

If I wanted to make the third card which makes a set from two cards Ca and Cb, I would have the card C(ab) where
ab = [a1b1, a2b2, a3b3, a4b4]

and the rule for the operator is:
If an = bn, then bn = xn and an = xn If anbn, then bnxn and anxn

For Example: 1*1=1, 1*2=3, 1*3=2
[1,1,2,2,] [1,2,2,3] = [1,3,2,1]  

 

This holds consistent with the rules of SET®. If the first two cards are red the third must also be red; if the first one is a squiggle, and the second a diamond, the third must be an oval.  

Here are some basic theorems in this group, linked to their proofs:

 an * bn = bn * an  (1.1)
 (an * bn) cnan * (bn * cn)  (1.2)
 (an * cn) (an * bn) = an(cn * bn)  (1.3)
 a(a * b) = b  (1.4)
 
 
The Square
So let us begin by choosing any three cards:
a, b, and c, and placing them in positions 7, 5, 9.

     1

    2

    3

    4

    Cc

    6

    Ca

    8

    Cb

Now we need to fill in the blanks for the remaining cards. Starting with card 8; it needs to complete the set with the cards Ca and Cb. We now look at the multiplier. The new card will be the product of Ca operating on Cb which is Cab. Likewise, filling in slots 1 and 3 leaves us with the square below.

 Cbc

2

Cac

4

Cc

6

Ca

Cab

Cb

We now know that we have three sets on the board (7,8,9; 7,5,3; 1,5,9). However, how should we go about filling slot two? I chose to combine 5 and 8 and give slot 2 the card Cc(ab). Now I have a set going down 2,5,8, but the theory states that 1,2,3 should form a set as well. This means if I choose to combine Cbc with Cac, it would equal C(bc)(ac), which must be the smae card as Cc(ab). Therefore we must show that:

(bc)(c(bc)) = (ac)  
c(b(ab)) = ac by 1.3 and 1.1
c(a) = ac by 1.4 and 1.1
ac = ac by 1.1

Now we fill in the two remaining slots of 4 and 6 by combining down to end up with 4 equaling Ca(bc) and 6 equaling Cb(ac). So now we have the following square:

 Cbc

Cc(ab)

Cac

Ca(bc)

Cc

Cb(ac)

Ca

Cab

Cb

Now that all other rows, columns, and diagonals have been accounted for, we only have to prove that 4,5,6 is a set. This means

(a(bc))*c = b(ac)  
b(bc) = c by 1.4
(a(bc))(b(bc)) = b(ac) substitution of b(bc) for c
(bc)(ab) = b(ac) by 1.3
(bc)(ba) = b(ac) by 1.1
b(ca) = b(ac) by 1.3
b(ac) = b(ac) by 1.1

This completes the proof of the square. Four sets still remain unaccounted for (1,6,8; 3,4,8; 7,2,6; 9,2,4). We note that if we prove one of these all must be true since now we can reconstruct this square by placing any card from 1,3,7, or 9 in the beginning slot and still get the same square. Follow this link to see proof...