# Math 037 Homework 4 Solutions

## Section 3.1

3.a. Yes. Let k = 2rs. Since r and s are integers, so it k. Now 4rs = 2(2rs)
= 2k. Thus, by the

definition of even, 4rs is even.

b. Yes. Let k = 3r+2s^2+1. Since r and s are integers, so is k. Now 6r+4s^2+3
= 2(3r+2s^2+1)+1 =

2k + 1. Thus, by the definition of odd, 6r + 4s^2 + 3 is odd.

c. Yes. We know r^2 +2rs+s^2 = (r +s)(r +s). Since r and s are both positive
integers, r +s must

be a positive integer greater than 1. Thus, we have written r^2 + 2rs + s^2 as
the product of two

positive integers greater than 1. Thus, r^2 + 2rs + s^2 is composite by the
definition of composite.

6. Let a = b = 0. Then a and b are real numbers such that

12. Let n = 9. Then n is odd, but (n - 1)/2 = 4 is even.

16. This property is sometimes true and sometimes not true. For example, it
is true for the pair 7

and 11. It is not true for the pair 7 and 9.

28. Thm: is n is odd, then n^2 is odd.

**Proof:** Let n be any odd integer. Then, from the definition of odd, there is
an integer k such that

n = 2k +1.Let k_{1} = 2k^2 +2k. Since k is an integer, so is k_{1}. Now, n^2 = (2k
+1)^2 = 4k^2 +4k +1 =

2(2k^2 + 2k) + 1 = 2k_{1} + 1 for some integer k_{1}. Thus, by the definition of odd,
n^2 is odd.

46. Thm: if n - m is even, then n^3 - m^3
is even.

Proof: Let n and m be integers such that n - m is even. Thus, by the definition
of even, there

exists an integer k such that n-m = 2k. Let k_{1} = k(n^2 +nm+m^2).
Since k, n, and m are integers,

so is k_{1}. Now, n^3 - m^3 = (n - m)(n^2 + nm + m^2) = 2k(n^2 +
nm + m^2) = 2 k_{1}, for some integer k_{1}.

Thus, by the definition of even, n^3 = m^3 is even.

48. This is false. Let m = 3. Note m is an integer but m^2 - 4 = 5 is prime.

53. This is false. Let m = 8 and n = 2. Then m and n are both positive
integers and mn = 16

which is a perfect square, but neither m nor n is a perfect square.

54. Thm: The difference of the squares of any two consecutive integers is
odd.

Proof: Let n and n + 1 be any two consecutive integers. Then the difference of
their squares can

be written aswhere n is an integer. Thus,
from the

definition of odd, the difference of the squares of any two consecutive numbers is
odd.

## Section 3.2

7. 5246197/99990

8. a. if xy = 0, then x = 0 or y = 0.

c. There exist a pair of nonzero real numbers x and y such that their product is zero.

17. Thm:

Proof: Let r and s be rational numbers. Then, from the definition of rational
numbers, there exist

integers a, b, c, d such that b ≠ 0, d ≠
0, r = a/b and s = c/d. Let A = ad + bc and B = 2bd. Since

a, b, c, d are integers, so are A and B. Since neither b nor d are zero, neither
is B.

Now,where A and B are integers and B is
not

zero. Thus, by the definition of a rational number, (r + s)/2 is rational.

19. Thm:

Proof: Let r and s be any two rational numbers. From #17, the number t = (r +
s)/2 is also a

rational number. From #18, r < t < s. Thus, between any two distinct rational
numbers, there

exists a rational number strictly between them.

## Section 3.3

3. yes. Pick k = 0, Now, 3 * k = 3 * 0 = 0.

9. yes. We know that 2a * 34b = 4 * 17ab. Since a and b are integers, so is
17ab. So 4|(2a * 34b) by

the definition of divisibility. So 4 is a factor of 2a * 34b.

16. Thm: For all integers a, b, and c, if a|b and a|c, then a|(b
- c).

Proof: Let a, b, and c be integers such that a|b and a|c. Thus, by the
definition of divisibility, there

exist integers k_{1} and k_{2} such that ak_{1} = b and ak_{2} = c. Let k_{3} = k_{1} - k_{2}.
Note that since k_{1} and k_{2}

are integers, so is k_{3}. Now, b - c = ak_{1} - ak_{2} = a(k_{1} -
k_{2}) = ak_{3} where k_{3} is an
integer. Thus, by

the definition of divisibility, a |(b - c).

20. Thm: If an integer n is divisible by 16, then n is divisible by 8.

Proof: Let n be an integer that is divisible by 16. Then,
by the definition of divisibility, there

exists an integer k such that n = 16k. Let k_{1} = 2k.Since k is an
integer, so is k_{1}. Now,

n = 16k = 8(2k) = 8k_{1} where k_{1} is an integer. Thus, by the
definition of divisibility, n is divisibly by 8.

25. This is not true. Let a = 6, b = 2, and c = 3. Then