Here's how I did it:
Let A = {1,2,3} Let B = {1,2,3,4,5}
Since A ∪ B, everything in A would also be in B thus the simplified form would be B?
If it's wrong, please let me know how to go about this, thanks.
Here's how I did it:
Let A = {1,2,3} Let B = {1,2,3,4,5}
Since A ∪ B, everything in A would also be in B thus the simplified form would be B?
If it's wrong, please let me know how to go about this, thanks.
You are correct. Since all elements in $A$ are also in $B$, there is no need to write down $A \cup B$ (except when e.g. specifically needed for proving a theorem) as that would simply be $B$ again.
You do not even need the examples, you can do this in a fully general way. Since for every $a \in A$, we know that also $a \in B$, it follows that the set $A \cup B = \{x | x \in A \lor x \in B \}$ is equal to $\{x | x \in B \}$, which is just $B$.