M403 Homework 13
Enrique Areyan
December 5, 2012
(2.52)
(i) True. We need to check three properties: 1) e H. This is true because H is a subgroup of K and K is a
subgroup of G and hence, K inherits the identity of G and H inherits the same identity from K

M403 Homework 11
Enrique Areyan
November 14, 2012
(2.21)
(i) False. Let n = 3. Then |S3 | = 3! = 6 > 3 = n.
(ii) True. We can write as a product of cycles. Then n = lcm of the lengths of all cycles.
(iii) True. This is the standard notation of composition

M403 Homework 9
Enrique Areyan
October 31, 2012
(2.1)
(i) True. By hypothesis if x S then x T . Also, if x T then x X. Hence, if x S then x X S X
(ii) False. Since f g is not a well-defined function.
(iii) True. Since g f : X Z is a well-defined function.

M403 Homework 8
Enrique Areyan
October 24, 2012
(1.77) (iii) False. Let a = 2. Then a6 = 64 = 10 6 + 4 64 4 (mod 6).
(iv) False. Let a = 2. Then, a4 = 16 = 4 4 + 0 16 0 (mod 4).
(vii) True. On the one hand, n 1 (mod 100) n = 100p + 1, where p Z.
On the ot

M403 Homework 2
Enrique Areyan
September 5, 2012
(1) Let b and h be the base and height of
2b+2h
2b2h
2
40
4bh
2
20
2
bh
bh
10
102
bh
a rectangle respectively. Suppose the perimeter is 2b + 2h = 40:
Inequality of the means
By hypothesis the perimeter

M403 Homework 5
Enrique Areyan
October 3, 2012
(1.54)
(i) Proof by contradiction:
suppose that n is square free and is also a rational number. Then, we can write it
in lowest terms: n = ab , where a.b Z, b 6= 0, gcd(a, b) = 1. Now,
n=
a
a2
n = 2 nb2 = a2

M403 Homework 7
Enrique Areyan
October 17, 2012
(1.77)
(i) True. This is a restatement of Corollary 1.59
(ii) False. Let a = 1, b = 5 and m = 4. Then (1+5)4 = 64 = 1296 0 mod 4, and 14 +54 = 1+625 = 626 2 mod 4
(v) False. Using the fact that if a b mod m

M403 Homework 4
Enrique Areyan
September 21, 2012
(1.46)
(i) False. Suppose for a contradiction that 6|2. Then, 2 = 6 q for some q Z. But, solving for q we get that
/ Z, a contradiction. Hence 2 - 6.
q = 13
(ii) True. 6 = 2 3 and 3 Z. Hence, 2|6.
(iii) T

M403 Homework 3
Enrique Areyan
September 12, 2012
(1.28)
(i) True.
(ii) False.
(iii) True.
(vi) True.
7
1
7
4 =75
10
10!
For n = 10 and r = 2 we have that: 2 = 8!2!
= 90
2 = 45 is not a multiple of n =
10
10
10
There are 4 quartets of dogs and 6 sextets o

M403 Homework 1
Enrique Areyan
August 29, 2012
(1.1)
(i) True. Let C be an arbitrary nonempty set of negative integers. Define a new set D as follow:
D = cfw_d : d = c for some c C
By definition, D contains the additive inverses of C, hence D N and D 6= .