Sets - Complement of a Set-advanced

Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, $A = \{1, 3, 5, 7, 9\}$ and $B = \{2, 3, 5, 7, 8\}$. Find $(A \cup B)'$.

1. First, find $A \cup B$ by listing all elements present in either $A$ or $B$: $A \cup B = \{1, 2, 3, 5, 7, 8, 9\}$
2. Now, find the complement $(A \cup B)'$ by identifying elements in $U$ that are not in $A \cup B$: $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ $A \cup B = \{1, 2, 3, 5, 7, 8, 9\}$ $(A \cup B)' = \{4, 6, 10\}$

To find the complement of the union, we first determine the combined set of $A$ and $B$, then subtract those elements from the Universal set $U$.

Verify De Morgan's Law $(A \cap B)' = A' \cup B'$ if $U = \{x : x \in \mathbb{N}, x \leq 6\}$, $A = \{2, 3\}$, and $B = \{3, 4, 5\}$.

1. List elements of $U$: $U = \{1, 2, 3, 4, 5, 6\}$
2. Find LHS: $(A \cap B)'$ $A \cap B = \{3\}$ $(A \cap B)' = U - \{3\} = \{1, 2, 4, 5, 6\}$
3. Find RHS: $A' \cup B'$ $A' = U - \{2, 3\} = \{1, 4, 5, 6\}$ $B' = U - \{3, 4, 5\} = \{1, 2, 6\}$ $A' \cup B' = \{1, 4, 5, 6\} \cup \{1, 2, 6\} = \{1, 2, 4, 5, 6\}$
4. Since $LHS = RHS = \{1, 2, 4, 5, 6\}$, the law is verified.

We calculate the intersection's complement and compare it to the union of the individual complements. Both result in the same set, confirming the law.

If $U$ is the set of all real numbers and $A = \{x : x \in \mathbb{R}, x > 5\}$, what is $A'$?

The set $A$ consists of all real numbers strictly greater than $5$. The complement $A'$ consists of all elements in the Universal set $U$ (real numbers) that are NOT in $A$. Therefore, $A' = \{x : x \in \mathbb{R}, x \leq 5\}$.

The complement of 'greater than' ($>$) is 'less than or equal to' ($\leq$) within the domain of real numbers.