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lgbasallote
 4 years ago
Given:
\[A = \{x: 1 \le x \le 5\}\]
why is \[A' = \{x : x < 1 \vee x > 5\}\]
and not \[A' = \{x: x< 1 \wedge x > 5\}\]
lgbasallote
 4 years ago
Given: \[A = \{x: 1 \le x \le 5\}\] why is \[A' = \{x : x < 1 \vee x > 5\}\] and not \[A' = \{x: x< 1 \wedge x > 5\}\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since \(x\) cannot be both less than one and greater than five at one time (otherwise we have a logical contradiction). Also, De Morgan's laws are useful for this.

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0so does that mean \[A = \{x : x \ge 1 \wedge x \le 5\}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In any case: \[ A=\{x1\leq x\leq 5\}=\{x1\leq x \wedge x\leq 5\} \]

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0i can't believe i didn't think of that first reason....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Unless your x's can quantum tunnel . . .

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0uhoh....math conversation.....time to leave..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(Sorry, I've been hanging out in the Physics section too long)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Go go, quantum entanglement and random transmission of data!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Strange things can happen in this here multiverse.
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