Answer
**(a)** \(C \subseteq D\) is **true**.
**(b)** \(D \subseteq C\) is **false**, because elements of \(D\) with odd \(s\) do not lie in \(C\).
Work Step by Step
First, let us restate the sets in a clearer form:
\(C = \{\,6r - 5 : r \in \mathbb{Z}\}\).
\(D = \{\,3s + 1 : s \in \mathbb{Z}\}\).
We want to check:
1. Is \(C \subseteq D\)?
2. Is \(D \subseteq C\)?
---
## (a) **Prove or disprove \(C \subseteq D\).**
To show \(C \subseteq D\), we must show that **every** element of \(C\) is also in \(D\).
Take an arbitrary element \(n \in C\). By definition, \(n = 6r - 5\) for some integer \(r\). We want to see if \(n\) can be expressed in the form \(3s + 1\).
Starting with \(n = 6r - 5\), notice:
\[
6r - 5
\;=\; 3\cdot (2r) - 5
\;=\; 3\bigl(2r - 2\bigr) + 1
\quad\text{(since }-5 = 3(-2) + 1\text{)}.
\]
Hence
\[
6r - 5
\;=\; 3\bigl(2r - 2\bigr) + 1.
\]
Let \(s = 2r - 2\). Because \(r\) is an integer, \(2r - 2\) is also an integer. Thus
\[
n = 6r - 5
\;=\; 3s + 1,
\]
where \(s\in \mathbb{Z}\). That precisely matches the form of elements in \(D\).
Therefore **every** \(n\in C\) is also in \(D\). We conclude
\[
\boxed{C \subseteq D \text{ is true.}}
\]
---
## (b) **Prove or disprove \(D \subseteq C\).**
To show \(D \subseteq C\), we would need **every** element of \(D\) to lie in \(C\).
Take an arbitrary element \(m \in D\). Then \(m = 3s + 1\) for some integer \(s\). We want to see if \(m\) can always be written in the form \(6r - 5\) for some integer \(r\).
Set up the equation:
\[
3s + 1 \;=\; 6r - 5.
\]
Rearrange to solve for \(r\):
\[
6r \;=\; 3s + 6
\;\;\Longrightarrow\;\;
r \;=\; \frac{3s + 6}{6}
\;=\; \frac{3s}{6} + 1
\;=\; \frac{s}{2} + 1.
\]
For \(r\) to be an integer, \(\tfrac{s}{2}\) must be an integer, i.e. \(s\) must be even.
- **If \(s\) is even**, then indeed \(r\) is an integer, so \(3s+1 \in C\).
- **If \(s\) is odd**, then \(\tfrac{s}{2}\) is not an integer, and we cannot solve \(3s+1 = 6r-5\) with \(r\in\mathbb{Z}\). In that case, \(3s+1 \notin C\).
### **Counterexample**
For instance, take \(s=1\) (which is odd). Then
\[
3s + 1 = 3\cdot1 + 1 = 4.
\]
Check if \(4\) lies in \(C\). That would require
\[
4 = 6r - 5 \;\;\Longrightarrow\;\; 6r = 9 \;\;\Longrightarrow\;\; r=\tfrac{9}{6}=1.5,
\]
which is not an integer. Hence \(4 \in D\) but \(4 \notin C\).
This shows there exists at least one element of \(D\) (actually infinitely many) not in \(C\). Therefore
\[
\boxed{D \subseteq C \text{ is false.}}
\]
