MAT 240H
Homework 1
1. Suppose a and b are nonzero elements of a field F. Using only the field axioms, prove
that a −1b −1 is a multiplicative inverse of ab. State which axioms are used in your proof.
Solution
∀a, b ≠ 0 ∈ F
(a −1b −1 )(ab) = (ab)(a −1b −1 )
−1 −1
= (ba )(a b )
−1
−1
= (baa )(b )
−1
= (b)(1)(b )
= bb
−1
=1
−1
(by property F1)
(by property F1)
(by property F 2)
(by property F 4)
(by property F 3)
(by property F 4)
−1
Since (ab) × (a b ) = 1 , therefore, by property F4(Existence of Multiplicative Inverse),
a −1b −1 is multiplicative inverse of ab.
2. Write the following complex numbers in the form a + ib , with a, b ∈ ℜ :
Solution
(a)
1 −2i
1(5 − i )
−2i (2i )
+
=
+
2i 5 − i 2i (5 − i ) (5 − i )(2i )
5−i + 4
=
10i + 2
9−i
=
10i + 2
(9 − i )(1 − 5i )
=
2(1 + 5i )(1 − 5i )
4 − 46i
=
2(26)
1 − 23
= +
i
13 26
Thus,
1 −2i
1 −23
i.
= +
+
2i 5 − i 13 26
(b)By using Binomial Theorem,
⎛ 5⎞ ⎛ 5⎞ ⎛ 5⎞
⎛ 5⎞
⎛5⎞
⎛ 5⎞
(1 + i ) 5 = ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟i + ⎜⎜ ⎟⎟i 2 + ⎜⎜ ⎟⎟i 3 + ⎜⎜ ⎟⎟i 4 + ⎜⎜ ⎟⎟i 5
0
1
2
3
4
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ 5⎠
= 1 + 5i + 10i 2 + 10i 3 + 5i 4 + i 5
= 1 + 5i − 10 − 10i + 5 + i
= −4 − 4i
= −4 + (−4)i
Thus, (1 + i )5 = −4 + (− 4)i .
Page 1 of 5
MAT 240H
Homework 1
3(a). Prove that the set F1 = {a + b 3 : a, b ∈ Q} (endowed with the addition and
multiplication inherited from ℜ ) is a field.
Solution
As a field, F1 must has elements 0, and 1.
Since a, b ∈ Q , if we let a = 0, b = 0 :
0 = 0 + 0 3 ∈ F1 .
And, if we let a = 1, b = 0 :
1 = 1 + 0 3 ∈ F1 .
Since F1 ⊆ ℜ and endowed with the addition and multiplication inherited from ℜ , it
automatically holds the Commutative Property(F1), Associative Property(F2),
Distributive Property(F5), Additive Identity, and Multiplicative Identity(F3).
Assume that the additive inverse of a + b 3 is c ∈ F1 .
a+b 3 +c = 0
− (a + b 3 ) + (a + b 3 ) + c = 0 − (a + b 3 )
(
c = − a+b 3
)
= −a − b 3
Since a, b ∈ Q , and Q is a field set, the additive inverse of a is -a and for b is -b.
This implies that c ∈ F1 if a, b ∈ Q . Therefore, additive inverse of a + b 3 exists(F4).
Assume that the multiplicative inverse of a + b 3 is d ∈ F1 .
(a + b
(a + b 3 )d = 1
3 ) (a + b 3 )d = 1(a + b 3 )
−1
−1
d=
1
(a + b 3 )
(a − b 3 )
=
(a + b 3 )(a − b 3 )
=
a
a 2 − 3b 2
−
b
a 2 − 3b 2
3
When a 2 − 3b 2 = 0 , a = ± 3b . Since ± 3 is an irrational number, there is no solution in Q
for a 2 − 3b 2 = 0 unless a = 0 and b = 0 . In fact, if a = 0 and b = 0 , then a 2 − 3b 2 = 0 , and it
will be a 0 divisor which is excepted in the definition of fields. Therefore,
a
a − 3b
2
2
∈ Q and
−b
a − 3b 2
2
∈ Q if
and only if a 2 − 3b 2 ≠ 0 . This implies that d ∈ F1 if a, b ∈ Q .
Therefore, multiplicative inverse of a + b 3 exists(F4).
Since F1 = {a + b 3 : a, b ∈ Q} holds all the field properties, it is a field. (proved)
Page 2 of 5
MAT 240H
Homework 1
3(b). Is the set F2 = {a + b 3 : a, b ∈ Z } (with the same addition and multiplication) also a
field?
Solution
As in 3(a), F2 has the special elements, 0 and 1, and holds the properties of F1, F2, F3,
and F5. Elements in F2 , a + b 3 also has additive inverse for a, b ∈ Z .
Assume that the inverse of a + b 3 is p ∈ F2 .
p=
a
a − 3b
2
2
−
b
a − 3b 2
2
3
If a ≠ 0 and b = 0 ,
p=
=
a
a − 3(0)
a
2
2
−
0
a − 3(0) 2
2
3
a2
1
= .
a
1
a
Consider that ∀a ∈ Z , p = ∈ Q , p is a rational number which is contradicted to
F2 = {a + b 3 : a, b ∈ Z } .
Since a + b 3 does not has multiplicative inverse when a ≠ 0 and
b = 0 , a ∈ Z , thus, F2 = {a + b 3 : a, b ∈ Z } is not a field.
Page 3 of 5
MAT 240H
Homework 1
4. Let F4 = {0, 1, a, b} be a field containing 4 elements. Assume that 1 + 1 = 0 . Prove that
b = a −1 = a 2 = a + 1 . (Hint: For example, for the first equality, show that a ⋅ b cannot equal 0,
a, or b.)
Solution
Statement 1: Before we start, we have to aware that neither a nor b can be equal to 0 or 1.
This is because of a field consists only distinct elements. Since F4 is a field, a and b have
to be distinct from each other as well as 0, and 1.
b = a −1 = a 2 = a + 1 has been broken down to three equalities:
(a) b = a −1 ;(b) b = a 2 ; (c) b = a + 1 .
(a) Proof of first equality:
(i) Let ab = 0 :
−1
(ii) Let ab = b :
ab = b
ba = b
−1
a ab = a 0
b=0
( By F 4)
−1
( By F1)
−1
b ba = b b
a =1
( By F 4)
This is contradicted to statement 1.
This is contradicted to statement 1.
(iii) Let ab = a :
−1
(iv) Let ab = 1 :
−1
a −1 ab = a −11
a ab = a a
b =1
b = a −1 ( By F 4)
( By F 4)
This is contradicted to statement 1.
There is no contradiction.
Thus, according to (iv), ab = 1 and b = a −1 (First equality proved).
(b) Proof of second equality:
(i) Let a 2 = 0 :
(ii) Let a 2 = a :
a −1 a 2 = a −1 0
a=0
a −1 a 2 = a −1 a
a =1
( By F 4)
( By F 4)
This is contradicted to statement 1.
This is contradicted to statement 1.
(iii) Let a 2 = 1 :
(iv) Let a 2 = b :
a −1 a 2 = a −11
a = a −1 = b
a −1 a 2 = a −1b
a = a −1b
( By F 4)
This is contradicted to statement 1.
There is no contradiction.
Thus, according to (iv), b = a 2 (Second equality proved).
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( By F 4)
MAT 240H
Homework 1
(c) Proof of third equality:
(i) Let a + 1 = 0 :
(ii) Let a + 1 = a :
a +1 = 0
a −1 (a + 1) = a −1 a
a +1 = 1+1
a =1
(1 + 1 = 0)
This is contradicted to statement 1.
1 + a −1 = 1
( By F 5 and F 4)
1+ b = 1
1+1+ b = 1+1
(b = a −1 )
b=0
(1 + 1 = 0)
This is contradicted to statement 1.
(iv) Let a + 1 = b :
(iii) Let a + 1 = 1 :
a +1 = b
1+ a = b
1+1+ a = b +1
a = b +1
a +1 = 1
1+ a = 1
1+1+ a = 1+1
(1 + 1 = 0)
a=0
( By F1)
(1 + 1 = 0)
There is no contradiction.
This is contradicted to statement 1.
Thus, according to (iv), b = a + 1 (Third equality proved).
By transitivity, b = a −1 = a 2 = a + 1 . (proved)
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