Theorem: -1 ´ a = -a.
Proof:
| 0 = 0 ´ a | Previously proved | ||
| 0 = (1 + -1) ´ a | Additive inverses (law 4) | 0 = (1 ´ a) + (-1 ´ a) | Distributive law (law 8) |
| 0 = a + (-1 ´ a) | Multiplicative identity (law 7) | ||
| -a = -1 ´ a | Additive inverses (law 4) |
Theorem: -1 ´ -1 = 1
Proof: By the previous theorem -1 ´ -1 = -(-1) = 1.
Theorem: -a ´ -a = a ´ a
Proof: -a ´ -a = (-1 ´ a) ´ (-1 ´ a) = (-1 ´ -1) ´ (a ´ a) = a ´ a.
So we see that in any Zn, the square roots come in plus/minus pairs. If a is a square root of b, then so is -a. For a field, if square roots exist there is only a single pair, which we now prove.
Theorem: If a ´ a = b ´ b, then a = ±b.
Proof:
| a ´ a = b ´ b | Given |
| (a ´ a) - (b ´ b) = 0 | Subtract b ´ b from both sides (law 4) |
| (a + b) ´ (a - b) = 0 | Factoring |
| a + b = 0 OR a - b = 0 | Previously proved (used law 8) |
| a = -b OR a = b | Add or subtract b from both sides (law 4) |
This time the proof depends on the existence of multiplicative inverses, so it only holds for fields. This says that if a square root exists, it is exactly a single plus/minus pair. If we look at a table of squares for a field, we can see the patterns guaranteed by the previous results.
| n | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| n2 | 0 | 1 | 4 | 2 | 2 | 4 | 1 |
The bottom row is the row of squares, and the top row the row of square roots. Note that the squares after the 0 form a palindrome, that is they read the same forward and backward, 1 4 2 2 4 1. This is a consequence of the rule that square roots come in plus/minus pairs. No terms repeat in the first half of the palindrome (1 4 2) so that exactly half the non-zero elements have square roots (1, 2, and 4 have square roots out of 1, 2, 3, 4, 5, 6). This follows from the fact that each square has two square roots (we are using the fact that our collection of elements is finite here). If we aren’t working in a field, then we will still have a palindrome because square roots still come in plus/minus pairs. But we might have some repetitions in the palindrome so we have fewer than half the numbers have square roots. On the other hand, we just proved things have to work out in a field. We didn’t prove things wouldn’t work out if we didn’t have a field, and indeed sometimes they do work out. The pattern of squares for non-zero elements of Z10 is 1 4 9 6 5 6 9 4 1, which has no repetitions in the first half, even though Z10 is not a field. On the other hand, the pattern of squares for non-zero elements of Z15 is 1 4 9 1 10 6 4 4 6 10 1 9 4 1, which does have repetitions. As usual, life is complicated if you aren’t in a field.
The reason we’ve been concerned about all these square roots is the quadratic formula. We can now state and prove our basic result for quadratic equations in a field.
Theorem: In any field, the quadratic equation ax2 + bx + c = 0, a ¹ 0, has roots if and only if the discriminant b2 - 4ac has a square root.
Proof: The key to the proof for a general field is the same as the usual proof of secondary algebra, completing the square. We start by assuming that x is a root of the quadratic. Then
| ax2 + bx + c = 0 | Given |
| 4a2x2 + 4abx + 4ac = 0 | Multiply by 4a on both sides |
| 4a2x2 + 4abx + b2 + 4ac = b2 | Add b2 to both sides |
| (2ax + b)2 + 4ac = b2 | Factoring |
| (2ax + b)2 = b2 - 4ac | Subtract 4ac from both sides |
| 2ax + b = Ö (b2 - 4ac) | Definition of square root |
So if the quadratic equation has a root, then the discriminant has a
square root. Note that we never used law 8 in this proof. This part of
the result is valid for any commutative ring, in particular it is true
for any Zn. Next, we assume the discriminant has a square
root, d2 = b2 - 4ac and
show the quadratic equation has a root x =
(d - b) ´
(2a)-1. Then x2 =
(d2 - 2db + b2)
´
(2a)-2 so
ax2 + bx + c = a´(d2 - 2db +
b2)´(2a)-2 + b´(d - b)´(2a)-1 + c
= a´(b2 - 4ac - 2db +
b2)´(2a)-2 + (bd -
b2)´(2a)-1 + c
= a´(2b2 - 4ac - 2db)´(2a)-2 +
(bd - b2)´(2a)-1 + c
= (2a)´(b2
- 2ac - db)´(2a)-2 +
(bd - b2)´(2a)-1 + c
= (b2 -
2ac - bd)´(2a)-1 + (bd -
b2)´(2a)-1 + c
= (b2 - 2ac - bd + bd -
b2)´(2a)-1 + c
= -2ac´(2a)-1 +
c
= -c + c
= 0
So the equation has a root if the discriminant has a square root, which completes the proof of the theorem. We do need to know that 2a has an inverse for this half of the proof, so we are using law 8 and also the assumption that a ¹ 0. Of course, if a = 0 we don’t actually have a quadratic equation, but rather only a linear equation, so this assumption is quite reasonable.