Is $\{x \in \mathbb{Z} \mid x = 6a + 4 \}$ a subset of $\{y \in \mathbb{Z} \mid y = 18b - 2\}$?

Question

Here is the question:

Let $A = \{x \in \mathbb{Z} \mid x = 6a + 4, \text{for some integer } a\}$

And let $B = \{y \in \mathbb{Z} \mid y = 18b - 2, \text{for some integer } b\}$

Prove or disprove the claim that $A \subseteq B$.

So far, I've managed to do the following:

$6a + 4 = 18b - 2$ by substitution

$6a + 6 = 18b$

$a+1 = 3b$

$b = \frac{a+1}{3}$ by algebra

The solution I read says $b$ is not an integer since $b = \frac{a+1}{3}$ is not an integer and therefore $A \not\subseteq B$. However, for $a = 2, b = 3$. So there is some integer $a$ and $b$ then and therefore $A \subseteq B$?

Answer 1

Before trying to prove the claim, it is helpful to try a few examples to see whether or not it makes sense. This is particularly important as you asked to prove or disprove it.

Let's pick a few simple examples of elements in $A$. If the claim is true and $A \subseteq B$, then these should also be elements in $B$.

If we choose $a=0, a=1, a=2$ (just as a starting point), then by substitution, we find that: $$4, 10, 16 \in A$$

We can now check whether or not it makes sense for these elements to also be in the set $B$ (which they must be if the claim is true).

$$\text{If }18b-2 = 4 \text{, then }b = \frac{3}{9}$$

Immediately, we have a problem. We have $4 \in A$ but we cannot have $4 \in B$ since this is only true when $b = \frac{3}{9}$ which is not an integer. So $A \not \subseteq B$ since there are elements of $A$ that are not elements of $B$.

Therefore the claim is False (using a proof by counterexample).

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Note: we didn't have to consider the element $4$, we could have done the same procedure for the other elements that we found above and this would have also shown the desired result

Answer 2

Alternate perspective:

$A = \{x \in \mathbb{Z} \mid x = 6a - 2, \text{for some integer } a\}.$
$B = \{y \in \mathbb{Z} \mid y = 18b - 2, \text{for some integer } b\}.$

Every number of the form $~18b~$ can be expressed as $~6a~$ where $~a = 3b.~$ However, the reverse is not true. That is, if $~a~$ does not happen to be a multiple of $~3,~$ then $~6a~$ can not be expressed as $~18b.~$

So, from this perspective, it is clear that $~B \subseteq A~$ and that an element $~x \in A \setminus B~$ can be constructed by setting $~a~$ to be any integer that is not a multiple of $~3.~$

For example, taking $~a = 1 \implies 4 = x \in A \setminus B.~$

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A somewhat different perspective is to examine the congruency classes $\pmod{18}$ which may be expressed as

$$\{0,1,2,\cdots,17\}.~$$

The only corresponding congruency class represented by $~B~$ is $~16,~$ while $~A~$ represents the congruency classes of $~4,~10,~16.$