# HW4

## Set 6.1

7.

Given:
A = {x $\in$ Z \| x = 6a+4 for some integer a}
B = {y $\in$ Z \| y = 18b-2 for some integer b}
C = {z $\in$ Z \| z = 18c+16 for some integer c}

(a) $A \subseteq B$ False.

Let x = 4. Then x $\in$ A because $x = 6 \cdot 0 + 4$, but x $\notin$ B because there is no integer b such that 4 = 18b-2. For if there were such an integer, then

18b-2 = 4
18b = 6 by algebra
b = 6/18 = 1/3

but 1/3 is not an integer. Thus 4 $\in$ A but 4 $\notin$ B. Hence $A \not\subseteq B$

(b) $B \subseteq A$ True. Suppose y in a particular but arbitrary element of B. We must show that y $\in$ A.

We know from the definition of B that there is some value of y which can be written as y = 18b-2, for some integer b. We must show that y $\in$ A. This means we must show that y = 6(some integer) + 4.

Given that y = 18b - 2, can y also be expressed as 6(some integer) + 4?

I.e., is there an integer, say a, such that 6a+4=18b-2?

We solve for a and get a = (18b-2-4)/6 = (18b-6)/6 = 6(3b-1)/6 = 3b-1. Yes, a is an integer because difference of products of integers is an integer.

Check to see if y = 6a+4 works. We now have, a = 3b-1. so, 6(3b-1)+4 = 18b-6+4 = 18b-2 = y. Thus, by definition of A, y is an element of A as was to be shown.

\(c\) $B = C$ True. By the definition of set equality, $B = C \leftrightarrow (B \subseteq C) \cap (C \subseteq B)$

First consider to show $B \subseteq C$:

Suppose y is a particular but arbitrary element of B. We must show that y $\in$ C. We know from the definition of B that there is some value of y which may be written as y=18b-2, for some integer b.

We must show that y $\in$ C. This means we must show that y = 18(some integer) + 16

Given that y=18b-2, can y also be expressed as 18(some integer)+16?
I.e., is there an integer, say c, such that 18c+16=18b-2?
We solve for c and get c=(18b-2-16)/18 =(18b-18)/18 = 18(b-1)/18 = b-1. Yes, c is an integer because the difference of integers is an integer.

Check to see if y = 18c+16 works. We now have, c = b-1. So, 18(b-1)+16=18b-18+16=18b-2=y. Thus, by definition of C, y is an element of C which is what to be shown. Thus, $B \subseteq C$ is true.

Now consider to show $C \subseteq B$:

Suppose z is a particular but arbitrary element of C. We must show that z $\in$ B. We know from the definition of C that there is some value of z which may be written as z=18c+16, for some integer c.

We must show that z $\in$ B. This means we must show that z = 18(some integer) - 2

Given that z=18c+16, can z also be expressed as 18(some integer)-2?
I.e., is there an integer, say b, such that 18b-2=18c+16?
We solve for b and get b=(18c+16+2)/18=(18c+18)/18=18(c+1)/18=c+1. Yes, b is an integer because the sum of integers is an integer.

Check to see if z=18b-2 works. We now have, b=c+1. So, 18(c+1)-2=18c+18-2=18c+16=z. Thus, by definition of B, z is an element of B which is what to be shown. Thus, $C \subseteq B$ is true.

Therefore, B = C is true.

26.

$S_1 = (1, 1+\frac{1}{1}) = (1, 2)$
$S_2 = (1, 1+\frac{1}{2}) = (1, \frac{3}{2})$
$S_3 = (1, 1+\frac{1}{3}) = (1, \frac{4}{3})$
$S_4 = (1, 1+\frac{1}{4}) = (1, \frac{5}{4})$

(a)

$\cup_{i=1}^4 S_i = (1, 2) \cup (1, \frac{3}{2}) \cup (1, \frac{4}{3} \cup (1, \frac{5}{4}) = (1,2)$

(b)

$\cap_{i=1}^4 S_i = (1, 2) \cap (1, \frac{3}{2}) \cap (1, \frac{4}{3} \cap (1, \frac{5}{4}) = (1, \frac{5}{4})$

\(c\)

$S_1$, $S_2$, $S_3$, ... are not mutually disjoint. In fact, each $S_k \subseteq S_{k-1}$

29.

Yes. Every real number is eigher positive, negative, or 0, and no real number is both postive and negative, positive and 0, or negative and 0.

30.

33.
(b)

\(c\)

34.
(b)

35.
(d)