Download Homework 1: Page 19: 2.1, 2.2, 2.3(let n 8), 2.5, 2.6, 2.9 Extra points

Homework 1: Page 19: 2.1, 2.2, 2.3(let n8), 2.5, 2.6, 2.9
Extra points: 2.11,
2.1 Verify that  2 is a field.
 2  a  2 b; a, b ∈ Q
Check: identities for , : e   0  2 0, e   1  2 0
Check: inverse for  : I   −a  2 −b for a  2 b
I   2 a 2 − 2 2 b 2 for a  2 b
a − 2b
a − 2b
2.2 Solve the equation 3x  7  6 in ℤ 13 .
3x ≡ 13 − 1 ≡ 13 12. Now we need to find x in ℤ 13  0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 such that
3x ≡ 13 12.
Try out 3x for each element in ℤ 13 : 34 ≡ 13 12. x  4.
2.3 Prove that ℤ n is a field if and only if n is a prime. Consider n  8.
Z 8  0, 1, 2, 3, 4, 5, 6, 7 is not a field because 2 −1 does not exist.
2.4 If d is in  and is not a perfect square show that 
(1) 

d
a  d b; a, b are in 
is a subfield of .
d
is not empty.
(2) Let u  a  d b and v  x  d y. Both are in 
u  v  a  x  d b  y in 
d
uv  ax  dby  d xb  ya in 

d .
d
is closed under  and .
d
(3) 0  0  d 0 and 1  1  d 1 are in 
d .
d
contains identities: 0, 1.
(4) Let u  a  d b. Then u  −u  0 so v  −u is the additive inverse of u.
Let w  x  d y. If w is the multiplicative inverse of u then
wu  ax  dby  d xb  ya  1  d 0. We need to find x and y such that
a db
b
a
x
y

1
0
,
x
y

1
a − db 2
2
a
−db
1
−b
a
0
a
− d 2 b 2 in  d
a 2 − db 2
a − db
contains identities additive and multiplicative inverses of u  a  d b.


d
Therefore, 
d
is a subfield of .
2.5 Let w 3  1, w ≠ 1. Show then that 1  w  w 2  0.
w 3  1  w 3 − 1  0  w 3 − 1  w − 1w 2  w  1  0  w  1 or w 2  w  1  0.
Since w ≠ 1, w 2  w  1  0.
2.6 Let w 3  1, w ≠ 1 (so w 2  w  1  0). Let w 
w is a subfield of ℂ.
(1) w is not empty.
a  bw; a, b are in  . Show that
(2) Let u  a  w b and v  x  w y. Both are in w.
u  v  a  x  wb  y in w
uv  ax  w 2 by  wxb  ya 
ax  −1 − wby  wxb  ya  ax − by  wxb  ya − by in w
w is closed under  and .
(3) 0  0  w0 and 1  1  w1 are in w. w contains identities: 0, 1.
(4) Let u  a  w b. Then u  −u  0 so v  −u is the additive inverse of u.
Let w  x  wy. If w is the multiplicative inverse of u then
wu  ax − by  wxb  ya − by  1  w0. We need to find x and y such that
a
−b
b a−b
x
y

1
0
,
x

y
1
aa − b  b 2
a−b b
−b
a
1
0
a−b
 w 2 −b2
in  d
a 2  b 2 − ab
a  b − ab
w contains identities additive and multiplicative inverses of u  a  wb.
Therefore, w is a subfield of .

2.9 Let z  4  7i, w  6 − i. Compute the following.
(i) z  4 − 7i, w  6  i, |z|  4 2  7 2  65 , |w| 
(ii) z  w  10  6i, zw  24  7  i42 − 4  31  38i,
62  1 
37 .
6 i 1
 1 24 − 7  i42  4  1 17  i46
37
37
37
37
1/4 i 2
1/4 i 2 
7
z  65 e
, z  65 e
where   arctan 4   1.
z
−1
w  zw  4  7i
(iii) z  65 e i ,
051 650 21
(iv) z  65 cos  i sin, where   1. 051 650 21.
(v) z 5  65 5/2 cos5  i sin5, where   1. 051 650 21.
(vi) Solve the equation zZ  6w  1 − i.
4  7iZ  1 − i − 66 − i  −35, Z  − 35 4 − 7i
65
2.10 Prove Lemma 2.2.1.
Let z  a  ib  |z|e i z and w  x  iy  |w|e i w .
(1) |z|  a 2  b 2 ≥ 0. |z|  0 if and only if a 2  b 2  0 if and only if a  b  0, that is,
z  0.
(2) zw  |z| |w| e i z  w  , |zw|  |z| |w|
(3) |z  w| 2  a  x 2  b  y 2  a 2  2ax  x 2  b 2  2by  y 2
|z|  |w| 2  |z| 2  2|z||w|  |w| 2  a 2  b 2  2 a 2  b 2 x 2  y 2  x 2  y 2
Is ax  by ≤ a 2  b 2 x 2  y 2 or ax  by 2 ≤ a 2  b 2 x 2  y 2 
ax  by 2  a 2 x 2  2abxy  b 2 y 2 , a 2  b 2 x 2  y 2   a 2 x 2  b 2 x 2  a 2 y 2  b 2 y 2
Because ax − by 2  a 2 x 2 − 2axby  b 2 y 2 ≥ 0, a 2 x 2  b 2 y 2 ≥ 2axby.
2.11 Prove Lemma 2.2.2.
Let z  a  ib  |z|e i z and w  x  iy  |w|e i w .
(1) z  w  a  x  ib  y  a  x − ib  y  a − ib  x − iy  z  w .
(2) zw  a  ibx  iy  ax − by  ibx  ay  ax − by − ibx  ay
z w  a − ibx − iy  ax − by − iay  bx
So, zw  z w .
(3) z  x − iy, z  x  iy  z.
(4) z  |a − ib|  a 2  b 2  z
(5) z  z, if and only if y  −y, that is, 2y  0, y  0 so z  x real.