Tables where every equation has an answer
Tonight's idea
In a finished Sudoku, every row and every column holds each symbol exactly once. A square grid like that (forget Sudoku's little boxes) is called a Latin square.
Now use a 4 by 4 Latin square as the "times table" of a made-up operation \(*\) on the numbers 0, 1, 2, 3. To find \(a * b\), look in row \(a\), column \(b\).
Why bother? Because equations get solved. To solve \(a * x = c\), search row \(a\) for \(c\). Every number is in that row exactly once, so there is exactly one answer. To solve \(y * b = c\), search column \(b\) the same way.
Ordinary multiplication also obeys the grouping rule: \((a \times b) \times c = a \times (b \times c)\), so you may do either multiplication first. A Latin square promises nothing of the kind. Tonight you catch one breaking that rule.
Watch one: the take-away clock
| \(a\) ↓ \(b\) → | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| 0 | 0 | 3 | 2 | 1 |
| 1 | 1 | 0 | 3 | 2 |
| 2 | 2 | 1 | 0 | 3 |
| 3 | 3 | 2 | 1 | 0 |
Picture a clock with only four hours, marked 0, 1, 2, 3 in a circle. Let \(a * b\) mean: start at \(a\) and step back \(b\) hours. So \(1 * 2\) starts at 1 and steps back to 0, then to 3: \(1 * 2 = 3\). Here is the whole table.
Is it Latin? The rows read 0 3 2 1, then 1 0 3 2, then 2 1 0 3, then 3 2 1 0. Each holds 0, 1, 2, 3 once. The columns read 0 1 2 3, then 3 0 1 2, then 2 3 0 1, then 1 2 3 0. The same. Yes, it is a Latin square.
Solve \(1 * x = 3\). Search row 1 for a 3. It sits in column 2 and nowhere else, so \(x = 2\) is the one answer.
Test the grouping rule on 3, 2, 1. Brackets first each time.
\[(3 * 2) * 1 = 1 * 1 = 0, \qquad 3 * (2 * 1) = 3 * 1 = 2.\]Two different answers. This table solves every equation of the two kinds above, and still the grouping rule fails.
Your turn
Here is a different table on 0, 1, 2, 3: multiply \(a\) by \(b\), divide by 4, and keep the remainder. Which rows hold each of 0, 1, 2, 3 exactly once? Is the table a Latin square? Then say how many answers each equation has in this table, and list them: \(2 * x = 2\), \(2 * x = 1\), \(3 * x = 2\).\(a\) ↓ \(b\) → 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 - Go back to the take-away clock table. Solve each equation by searching the right row or column: \(2 * x = 3\), \(0 * x = 1\), \(y * 3 = 1\), \(y * 1 = 2\).
- Still with the take-away clock, work out \((a * b) * c\) and \(a * (b * c)\) for these three choices of \(a, b, c\): 1, 1, 1; then 3, 1, 2; then 2, 0, 3. A friend tried only 3, 1, 2 and announced that the grouping rule holds for this table. What is wrong with that?
Now the adding clock: \(a * b\) means start at \(a\) and step forward \(b\) hours. Fill in the table. Is it a Latin square? Test the grouping rule on the same three choices as in question 3.\(a\) ↓ \(b\) → 0 1 2 3 0 1 2 3
Stretch: a number that does nothing
In ordinary multiplication, 1 does nothing: \(a \times 1 = a\) and \(1 \times a = a\). Which number does nothing on both sides in the adding clock table? In the take-away clock table, find a number \(e\) with \(a * e = a\) for every \(a\). Does it also do nothing on the other side, so that \(e * a = a\) for every \(a\)? Does any number? Last, explain why a Latin square can never have two different numbers \(e\) and \(f\) with \(a * e = a\) and \(a * f = a\) for every \(a\).
Answers
- Rows 1 and 3 hold each number once. Row 0 is all 0s and row 2 reads 0 2 0 2, so the table is not a Latin square. (Columns 0 and 2 fail in the same way.) \(2 * x = 2\) has two answers, \(x = 1\) and \(x = 3\), because 2 appears twice in row 2. \(2 * x = 1\) has no answer, because row 2 has no 1. \(3 * x = 2\) has exactly one answer, \(x = 2\), because row 3 happens to hold each number once. Without the Latin property an equation can have too many answers or none.
- \(2 * x = 3\): row 2 has its 3 in column 3, so \(x = 3\). \(0 * x = 1\): row 0 has its 1 in column 3, so \(x = 3\). \(y * 3 = 1\): column 3 has its 1 in row 0, so \(y = 0\). \(y * 1 = 2\): column 1 has its 2 in row 3, so \(y = 3\). Check one on the clock: from 3, one step back is 2.
- 1, 1, 1: \((1 * 1) * 1 = 0 * 1 = 3\) but \(1 * (1 * 1) = 1 * 0 = 1\). Different. 3, 1, 2: \((3 * 1) * 2 = 2 * 2 = 0\) and \(3 * (1 * 2) = 3 * 3 = 0\). Equal. 2, 0, 3: \((2 * 0) * 3 = 2 * 3 = 3\) but \(2 * (0 * 3) = 2 * 1 = 1\). Different. The friend found one choice that works, but a rule has to hold for every choice, and a single failure breaks it. In this table the two sides agree exactly when \(c\) is 0 or 2. That is 32 of the 64 possible choices.
- The rows are 0 1 2 3, then 1 2 3 0, then 2 3 0 1, then 3 0 1 2. Every row and every column holds each number once, so it is a Latin square. 1, 1, 1: \((1 * 1) * 1 = 2 * 1 = 3\) and \(1 * (1 * 1) = 1 * 2 = 3\). 3, 1, 2: \(0 * 2 = 2\) and \(3 * 3 = 2\). 2, 0, 3: \(2 * 3 = 1\) and \(2 * 3 = 1\). Equal every time. In fact the rule holds for all 64 choices here, because both sides mean: start at \(a\) and step forward \(b + c\) hours. A Latin square that also obeys the grouping rule is the table of a group. So the adding clock is a group, and the take-away clock is a quasigroup that is not a group.
Stretch. In the adding clock, 0 does nothing on both sides: stepping forward 0 hours leaves \(a\) alone, and starting at 0 and stepping forward \(a\) hours lands on \(a\). In the take-away clock, \(e = 0\) works on the right: column 0 reads 0 1 2 3, so \(a * 0 = a\). On the left it fails: \(0 * 1 = 3\), not 1. No number does nothing on the left, because that would need a row reading 0 1 2 3 and no row does. So the take-away clock has a do-nothing number on one side only. For the last part, suppose \(e\) and \(f\) both worked. Then \(0 * e = 0\) and \(0 * f = 0\), so the equation \(0 * x = 0\) would have two different answers. In a Latin square it has exactly one, so \(e\) and \(f\) cannot be different.