MSU MATHEMATICS |
MTH 310.002 |
Spring 2012 |
+ | 0 | 1 | 2 | 3 | 4 |
0 | 0 | 1 | 2 | 3 | 4 |
1 | 1 | 2 | 3 | 4 | 0 |
2 | 2 | 3 | 4 | 0 | 1 |
3 | 3 | 4 | 0 | 1 | 2 |
4 | 4 | 0 | 1 | 2 | 3 |
• | 0 | 1 | 2 | 3 | 4 |
0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 2 | 3 | 4 |
2 | 0 | 2 | 4 | 1 | 3 |
3 | 0 | 3 | 1 | 4 | 2 |
4 | 0 | 4 | 3 | 2 | 1 |
1b. (a+b)2 = a2 + 2ab + b2 ;
(a+b)3 = a3 + 3a2b + 3ab2 + b3 ;
(a+b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
= a4 − a3b + a2b2 − ab3 + b4 ;
(a+b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
= a5 + b5 (!)
1c. (a−b)(a+b) = a(a+b) − b(a+b) = a.a + a.b − b.a − b.b
= a2 + a.b − a.b − b2 = a2 − b2 .
1d. The mulitplication table has a single 1 in each row (except the row for mult by 0).
This means precisely that each a ≠ 0 has a unique a−1.
1e. (2x - 1) / (3x + 2) = 2 ⇔  2x − 1 = 2 (3x + 2) ⇔ 4x = −5 = 0 ⇔  x = 4−10 = 0 (unique solution).
Check: x = 0 plugs into the equation: (2(0) − 1)/(3(0) + 2) = (−1)/(2) = (4)(2−1) = (4)(3) = 2
1f. Taking 02,...,42, we find only 0, 1, and 4 = −1 have square roots.
If x2 = a, then (−x)2 = (−1)2x2 = 1(x2) = a, so −x is also a square root.
1g. x = ( −b±√(b2−4ac) ) / 2a = ( −1 ± √(1−4(3)(1)) ) / 2(3) = = ( −1 ± √(4) ) / 1 = 1 or −3 = 2.
1h. There is no 0 in any row of the mulitplication table, except in the row and column of multiplication by 0.
Thus ab = 0 only for a = 0 or b = 0.
2. Z6 has no reciprocals except 1−1 = 1 and −1−1 = −1. Also (2)(3) = 6 = 0. Similarly for Z9.
A I = | a1 | a2 | • | 1 | 0 | = | a1 1 + a2 0 | a1 0 + a2 1 | = | a1 | a2 | = A |
a3 | a4 | 0 | 1 | a3 1 + a4 0 | a3 0 + a4 1 | a3 | a4 |
a | b |
−b | a |
Poly factored in: | Q[x] | R[x] | C[x] |
x2 − 2 | x2 − 2 | (x − √2)(x + √2) | (x − √2)(x + √2) |
x2 + x − 6 | (x − 2)(x + 3) | (x + 3)(x − 2) | (x + 3)(x − 2) |
x2 − 2x − 5 | x2 − 2x − 5 | (x − 1 − √6)(x − 1+ √6) | (x − 1− √6)(x − 1 + √6) |
x2 − 2x + 5 | x2 − 2x + 5 | x2 − 2x + 5 | (x − 1 − 2i) (x − 1 + 2i) |
3x3 − 2x2 − 4x + 1 | 3(x − 1/3)(x2 + x − 1) | 3(x − 1/3)(x + ½(1+√5))(x + ½(1 − √5)) | 3(x − 1/3)(x + ½(1+√5))(x + ½(1−√5)) |
x3 − 2 | x3 − 2 | (x − 21/3)(x2 + 21/3x + 22/3) | (x − 21/3)(x2 − 2−2/3(1+i√3)) (x2 − 2−2/3(1−i√3)) |
x4 − 2 | x4 − 2 | (x2 + √2)(x − 4√2)(x + 4√2) | (x − i 4√2)(x + i 4√2)(x − 4√2)(x + 4√2) |
0 | x |
0 | 0 |
a | −b |
b | a |
Remember, your assignment is to do all of the steps above not for f(z), but for g(z) = z3 + iz + 1.
The moral: Consider any polynomial f(z) = a0 + a1z + ... + zn with a0 ≠ 0 and an = 1. For small values of z, we have w = f(z) ≈ a0 + a1z, so a very small circle around z = 0 is mapped to a small circle around w = a0, staying far from w = 0. As we enlarge the z-circle, the corresponding w-path acquires n loops, and eventually each loop enlarges to enclose w = 0. Each time this happens, we have a root f(z0) = 0. It must happen somewhere, because for a very large z, we have f(z) ≈ zn, which makes n loops enclosing w = 0.
[f(x)] | = { f(x) + (x2+1)q(x) for all q(x) ∈ R[x] } |
= { g(x) ∈ R[x] s.t. g(x) ≡ f(x) mod (x2+1) } | |
= { g(x) ∈ R[x] s.t. (x2+1) | (g(x) − f(x)) } |
× | 0 | 1 | α | α+1 | α2 | α2+1 | α2+α | α2+ α+1 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | α | α+1 | α2 | α2+1 | α2+α | α2+α+1 |
α | 0 | α | α2 | α2+α | α+1 | 1 | α2+α+1 | α2+1 |
α+1 | 0 | α+1 | α2+α | α2+1 | α2+α+1 | α2 | 1 | α |
α2 | 0 | α2 | α+1 | α2+α+1 | α2+α | α | α2+1 | 1 |
α2+1 | 0 | α2+1 | 1 | α2 | α | α2+α+1 | α+1 | α2+ α |
α2+α | 0 | α2+α | α2+α+1 | 1 | α2+α | α+1 | α | α2 |
α2+α+1 | 0 | α2+α+1 | α2+1 | α | 1 | α2+α | α2 | α+1 |
Let us illustrate why this is so. Let R = Z and I = (4) = {4q for all q ∈ Z}, all multiples of 4, and R/I = Z/(4) = Z4 = {[0], [1], [2], [3]}.
1 | 2 | 3 | 4 |
2 | 3 | 4 | 1 |
2 | 3 | 4 | 1 |
1 | 2 | 3 | 4 |
1 | 2 | 3 | 4 |
4 | 1 | 2 | 3 |