HWSet-Chapter3 - Math 326 Fall 2010 Homework Set Chapter 3...

Math 326 Fall 2010
Homework Set Chapter 3
Solutions
1

3.3-3 Does 5 | 0?
Yes. 0 = 0 · 5.
3.3-4 Is (3 k + 1)(3 k + 2)(3 k + 3) divisible by 3?
Yes, because (3 k + 1)(3 k + 2)(3 k + 3) = 3(3 k + 1)(3 k + 2)( k + 1) by factoring a 3 from the
last term.
3.3-5 Is 6 m (2 m + 10) divisible by 4?
Yes, because 6 m (2 m + 10) = 2 · 3 m · 2( m + 5) = 4 · 3 m ( m + 5)
3.3-13 If n = 4 k + 3, does 8 divide n
2
- 1? Yes, because
n
2
- 1 = (4 k + 3)
2
- 1
= 16 k
2
+ 24 k + 9 - 1
= 16 k
2
+ 24 k + 8
= 8(2 k
2
+ 3 k + 1)
3.3-16 Prove the following directly from the definition of divisibility: For all integers a , b , and c , if
a | b and a | c , then a | ( b - c ).
Since a | b then ∃ k ∈ Z such that b = ka (def of divisibility)
Since a | c then ∃ m ∈ Z such that c = ma (def of divisibility)
Hence
b - c = ka - ma (substitution)
= ( k - m ) a (distributive)
Let q = k - m . Then by closure of Z under subtraction, q ∈ Z . Hence b - c = qa . Hence
∃ q ∈ Z such that b - c = qa . Hence a | ( b - c ).
3.3-17 True (prove) or false (counterexample): The sum of any three consecutive integers is divisible
by 3.
Proof: Let k ∈ Z and let S = k + ( k + 1) + ( k + 2) = 3 k + 3 = 3( k + 1)
Let p = k + 1 ∈ Z because Z is closed under addition. Hence S = 3 p . Hence 3 | S . Hence the
sum of any 3 consecutive integers is divisible by 3.
3.3-18 True (prove) or false (counterexample): The product of any two even integers is a multiple
of 4.
Proof: Let p, q be any even integers. Then by the definition of even, ∃ m, n ∈ Z such that
p = 2 m and q = 2 n . Hence pq = 2 m · 2 n = 4 mn . Since pq ∈ Z by closure of Z under
multiplication, 4 | pq .
3.3-20 True (prove) or false (counterexample): A sufficient condition for an integer to be divisible
by 8 is that it be divisible by 16.
Another way to write this is ( ∀ a ∈ Z )(16 | a = ⇒ 8 | a ).
Proof: Suppose a ∈ Z and 16 | a . Then ∃ k ∈ Z such that a = 16 k = 2 · 8 · k . Then let p = 2 k .
Since k ∈ Z , by closure of Z , p ∈ Z . By substitution, a = 8 p . Hence 8 | a .
2

3.3-26 True (prove) or false (counterexample): For all integers a and b , if a | b then a
2
| b
2
.
Proof: Let a, b ∈ Z such that a | b . By definition of divisibility, ∃ k ∈ Z such that b = ak .
Squaring both sides of this equation gives b
2
= ( ak )
2
= a
2
k
2
. But k
2
is an integer (being a
product of the integer k times itself). Hence by definition of divisibility, a
2
| b
2
.
3.3-28 True (prove) or false (counterexample): For all integers a and b , if a | 10 b then a | 10 or a | b .
Counterexample : Let a = 25 and b = 5. Then 10 b = 50. Since 25 | 50, a | 10 b . But neither
a | 10, which is 25 | 10 nor a | b , which is 25 | 5, are true.
3.3-34 Use the unique factorization theorem to write the following integers in standard factored
form.
(a) 1176 = 2
3
· 3 · 7
2
(b) 5733 = 3
2
· 7
2
· 13
(c) 3675 = 3 · 5
2
· 7
2
3.3-35 Suppose that in standard factored form a = p
e
1
1
p
e
2
2
· · · p
e
k
k
where k is a positive integer;
p
1
, p
2
, . . . are prime numbers; and e
1
, e
2
, . . . are positive integers.
(a) What is the standard factored form for a
2
?
a
2
= p
2 e
1
1
p
2 e
2
2
· · · p
2 e
k
k
(b) Find the least positive integer n such that 2
5
· 3 · 5
2
· 7
3
· n is a perfect square.