A number system with five numbers
Tonight's idea
On a 5-hour clock the only numbers are 0, 1, 2, 3, 4. Add or multiply as usual, then keep the remainder on dividing by 5. This is arithmetic mod 5, and \(\equiv\) means "equal on the clock". So \(3 + 4 \equiv 2\) and \(2 - 4 \equiv 3\).
Division works too. Every number except 0 has an inverse, a partner that multiplies with it to give 1:
\(1 \times 1 \equiv 1, \qquad 2 \times 3 \equiv 1, \qquad 4 \times 4 \equiv 1 \pmod 5\).
To divide by 2, multiply by 3 (see Dividing in clock arithmetic). A system where you can add, subtract, multiply, and divide by anything but 0 is a field.
In a field, if \(ab = 0\) then \(a = 0\) or \(b = 0\): when \(a \ne 0\), multiply both sides by its inverse and \(b = 0\). So a quadratic \(a(x - r)(x - s)\) is zero only at \(r\) and \(s\). Granting that the factor theorem still holds mod 5 (a root \(r\) gives a factor \(x - r\)), every quadratic has at most two roots, exactly as in school algebra.
Geometry comes along. A point is a pair \((x, y)\), so there are 25. A line is the points with \(y = mx + c\); a parabola, those with \(y = x^2\).
Watch one: where \(y = x + 2\) meets \(y = x^2\)
Set them equal: \(x^2 \equiv x + 2\), so \(x^2 - x - 2 \equiv 0\). Factorise as usual:
\[(x - 2)(x + 1) \equiv 0, \quad\text{so } x \equiv 2 \text{ or } x \equiv -1 \equiv 4.\]
The points are \((2, 4)\) and \((4, 1)\). Check the second: \(4^2 = 16 \equiv 1\) and \(4 + 2 = 6 \equiv 1\). The line wraps round the grid, but it still meets the parabola twice and no more.
Your turn
- Work out mod 5: \(4 + 4\), \(1 - 3\), \(3 \times 3\), and \(3 \div 2\).
- List the five points on the line \(y = 2x + 1\). Then find the one point where it meets the line \(y = 3x + 3\).
- For each \(c\) from 0 to 4, how many points does the horizontal line \(y = c\) share with the parabola \(y = x^2\)? Which numbers have square roots mod 5?
- Now use a 6-hour clock. Test all six numbers to find every solution of \(x^2 \equiv x \pmod 6\). A quadratic with more than two roots: which step of the argument above fails mod 6? Give two numbers that show it.
Stretch: the quadratic formula on a clock
Solve \(2x^2 + 3x + 1 \equiv 0 \pmod 5\) with the quadratic formula, where dividing by \(2a\) means multiplying by its inverse. Check your roots. Then use \(b^2 - 4ac\) to explain why \(x^2 + x + 1\) has no roots mod 5.
Answers
- \(4 + 4 = 8 \equiv 3\). \(1 - 3 = -2 \equiv 3\). \(3 \times 3 = 9 \equiv 4\). \(3 \div 2 \equiv 3 \times 3 \equiv 4\) (check: \(4 \times 2 = 8 \equiv 3\)).
- \((0, 1), (1, 3), (2, 0), (3, 2), (4, 4)\). They meet where \(2x + 1 \equiv 3x + 3\), so \(x \equiv -2 \equiv 3\) and \(y \equiv 2\): the point \((3, 2)\). Check: \(3 \times 3 + 3 = 12 \equiv 2\). Two lines with different slopes still meet exactly once.
- The squares are \(0^2 \equiv 0\), \(1^2 \equiv 1\), \(2^2 \equiv 4\), \(3^2 \equiv 4\), \(4^2 \equiv 1\). So \(y = 0\) shares 1 point, \(y = 1\) shares 2, \(y = 2\) and \(y = 3\) share none, and \(y = 4\) shares 2. Only 0, 1 and 4 have square roots: \(\sqrt 4\) is 2 or 3, and 2 and 3 have none.
- \(0, 1, 3\) and \(4\): for instance \(3^2 = 9 \equiv 3\) and \(4^2 = 16 \equiv 4\), while \(2^2 \equiv 4\) and \(5^2 \equiv 1\) fail. The step that fails is "if \(ab = 0\) then \(a = 0\) or \(b = 0\)". Mod 6, \(2 \times 3 \equiv 0\) with neither factor zero. So \(x(x - 1) \equiv 0\) can hold with neither factor zero, as at \(x = 3\): \(3 \times 2\). The numbers 2 and 3 have no inverses mod 6, so this is not a field. Five is prime; six is not.
Stretch. \(b^2 - 4ac = 9 - 8 = 1\), with square roots \(\pm 1\). Dividing by \(2a = 4\) means multiplying by 4. So \(x \equiv (-3 + 1) \times 4 = -8 \equiv 2\) or \(x \equiv (-3 - 1) \times 4 = -16 \equiv 4\). Check: \(2(4) + 6 + 1 = 15 \equiv 0\) and \(2(16) + 12 + 1 = 45 \equiv 0\). For \(x^2 + x + 1\), \(b^2 - 4ac = -3 \equiv 2\), and 2 has no square root mod 5, so there are no roots.