therealcain
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- Mar 13, 2021
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Hey guys ( I'm sorry in advance if this isn't the correct place to ask discrete math questions ),
I need to prove that [MATH]If\ A \cup B^c \neq U\ and\ B \cup A^c \neq U\ then\ |A\Delta B| \geq 2[/MATH].
Here is my attempt to prove it:
[MATH] Suppose\ that\ A \neq B\ ( since\ A = B \rightarrow A \cup B^c = B \cup A^c = U )\ and\ suppose\ A \wedge B \neq \emptyset. \\ Since\ A \wedge B \neq \emptyset\ then\ |A| \land |B| \geq 1,\ and\ since\ A \neq B\ then\ \exists x(x \in A \wedge x \notin B)\ or\ \exists (x \notin A \wedge x \in B). \\ Therefore\ ( |A| \Delta |B| = | \{ x : x \in A \oplus x \in B \}| ) \geq 2. [/MATH]I'm quite new to proofs and this was my attempt to prove it, but i feel like something is off...
Just a note on the syntax:
|A| is the cardinality of A.
Delta symbol is symmetric difference.
I need to prove that [MATH]If\ A \cup B^c \neq U\ and\ B \cup A^c \neq U\ then\ |A\Delta B| \geq 2[/MATH].
Here is my attempt to prove it:
[MATH] Suppose\ that\ A \neq B\ ( since\ A = B \rightarrow A \cup B^c = B \cup A^c = U )\ and\ suppose\ A \wedge B \neq \emptyset. \\ Since\ A \wedge B \neq \emptyset\ then\ |A| \land |B| \geq 1,\ and\ since\ A \neq B\ then\ \exists x(x \in A \wedge x \notin B)\ or\ \exists (x \notin A \wedge x \in B). \\ Therefore\ ( |A| \Delta |B| = | \{ x : x \in A \oplus x \in B \}| ) \geq 2. [/MATH]I'm quite new to proofs and this was my attempt to prove it, but i feel like something is off...
Just a note on the syntax:
|A| is the cardinality of A.
Delta symbol is symmetric difference.