Let's analyze each statement based on the sets provided:

1. $F = \{2, 4, 6, 8\}$, which contains even numbers only.
2. $G = \{\text{odd numbers}\}$, meaning it contains all odd numbers.

Now, let's evaluate each statement:

Statement (a) $F \cup G = \emptyset$

• $F \cup G$ represents the union of sets $F$ and $G$, which combines all unique elements from both sets. Since $F$ contains even numbers and $G$ contains all odd numbers, the union $F \cup G$ would include all elements from both sets. Therefore, $F \cup G$ is not empty, as it includes both even and odd numbers.
• Conclusion: The statement $F \cup G = \emptyset$ is false.

Statement (b) $F \cap G = \emptyset$

• $F \cap G$ represents the intersection of sets $F$ and $G$, meaning it includes elements that are in both $F$ and $G$.
• Since $F$ only has even numbers and $G$ only has odd numbers, there are no elements that are both in $F$ and $G$. Therefore, $F \cap G$ is indeed empty.
• Conclusion: The statement $F \cap G = \emptyset$ is true.

Summary:

• (a) False
• (b) True

Would you like further explanation or have any questions?

Here are some related questions to deepen your understanding:

1. What is the difference between the union and intersection of two sets?
2. Can a union of two non-empty sets ever be empty?
3. What would $F \cup G$ look like if $G$ contained only odd numbers less than 10?
4. If $G = \{1, 3, 5, 7\}$, what would $F \cup G$ and $F \cap G$ be?
5. How do we represent the complement of a set, and what would it be in this context?

Tip: Remember, the union of two sets includes all elements in either set, while the intersection includes only those in both sets simultaneously.