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Let $A = \left\{x \in \mathbb{Z} \mid \exists a\in\mathbb{Z}: x = 6a + 4\right\}$

and $B = \left\{y \in \mathbb{Z} \mid \exists b\in\mathbb{Z}: y = 18b - 2\right\}$

and $C = \left\{z \in\mathbb{Z} \mid \exists c\in\mathbb{Z}:z = 18c + 16\right\}$

Prove or disprove each of the following statements:

a. $A\subseteq B$

b. $B\subseteq A$

c. $A = B$

Did I Prove these correctly?

a. A ⊆ B. Suppose x is a particular, but arbitrarily chosen element of A.
By definition of A, there is an integer a, such that 6a+4.
Suppose that a = 1 then 6(1)+4=10=x. Suppose y is a particular, but arbitrarily chosen element of B. By definition of B, there is an integer b, such that 18b-2. If x, or 10 in this case is an element of A, then it must be an element of B for A⊆B. However if 18b-2=10 then 18b=12 b=12/18 b=⅔ Therefore, b is not an integer thus 10 is an element of A, but not of B, therefore A ⊈ B. [This is what needed to be shown.]

b. B ⊆ A. Suppose x is an arbitrarily chosen element of A. By definition of A, there is an integer a, such that 6a+4. Suppose y is a particular, but arbitrarily chosen element of B. By definition of B, there is an integer b, such that 18b-2. Suppose that y =1 then y= 18(1)-2 = 16. If y, or 16 in this case is an element of B then it must be an element of A for B ⊆ A. If 6a+4=16 then, 6a=12 a=12/6 a=2. Therefore, 16 is an element of both B and A, thus B ⊆ A. [This is what needed to be shown.]

c. B=C Suppose y is an element of B and z is an element of C. By definition of B, there is an integer b, Such that 18b-2. By definition of C, there is an integer c, such that 18c+16. Suppose that b = 1 Then 18(1) -2 = 16. If y, or in this case 16 ∈ B, then it must be an element of C for B ⊆ C. If 18c+ 16=16 c=0. 0 is an integer, thus y is an element of both B and C, thus B ⊆ C. For B=C, C must also be a subset of B. We know that 0∈ C thus, 18b-2=0 b=1/9 B is not an integer, therefore C⊈B, thus B≠C.

Michael Hardy
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antman1p
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1 Answers1

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In order to show that $A \nsubseteq B$ it is enough to find a single element $x \in A$ which is not an element of $B$. You don't need to write "If x, or 10 in this case is an element of A, then it must be an element of B for A⊆B." Although it's fine to show $a \subseteq B$ results in a contradiction, it's simpler to just show $10 \in A$ but $10 \notin B$

In order to prove that one set is a subset of another, e.g., $B \subseteq A$, you show that any element of $B$ must be an element of $A$. You have successfully shown that 16 is an element of both sets, but not that every $b \in B$ is also in $A$.

In step c you're making an error when you write that "we know that $0 \in C$." $0$ is not in $C$, it is the integer in $\mathbb{Z}$ which we use as $c$ in the definition of $C$ to see that $16 \in C$ as $16 = 18*0 + 16$ and $0 \in \mathbb{Z}$

Try parts b and c again with this information. And remember that your sets, $A, B, C$ contain those elements $x, y, z$ which satisfy the predicates in their definitions, not the $a$,$b$,$c \in \mathbb{Z}$

n1000
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