After drawing a diagram of this statement, I believe it to be false. However, I'm having trouble approaching how to disprove this. Do I try to prove the negation? Or what else can I do?
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$x\in (A'\cap B')\to x\in A'$ and $x\in B'$. If $x\in A'$ then $x\notin A$ and if $x\in B'$ then $x\notin B$. In both cases, $x\notin A\cup B$. This means that $x\in (A\cup B)'$. The other direction is analogous. You can do it. |
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$x\in(A\cup B)'$ $\Rightarrow$ $(x\notin A\text{ and } x\notin B)$ $\Rightarrow$ $(x\in A'\text{ and } x\in B')$ $\Rightarrow$ $x\in A'\cap B'$ $\Rightarrow (A\cup B)'\subseteq A'\cap B'$ $x\in A'\cap B'$ $\Rightarrow$ $(x\in A'\text{ and } x\in B')$ $\Rightarrow$ $(x\notin A\text{ and }x\notin B)$ $\Rightarrow$ $x\in(A\cup B)'$ $\Rightarrow A'\cap B'\subseteq (A\cup B)'$ And $(A\cup B)'\subseteq A'\cap B'$ and $A'\cap B'\subseteq (A\cup B)'$ $\Leftrightarrow$ $(A \cup B)' = A' \cap B'$ |
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