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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.

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  • $\begingroup$ This seems to be basically a duplicate of this question. (And I would not be surprised if you find several other posts about the same question.) $\endgroup$ – Martin Sleziak Jun 20 '16 at 8:41
  • $\begingroup$ Your answer is correct. $\endgroup$ – Ashwin Ganesan Jun 20 '16 at 16:08
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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 $B$.

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