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

This Question is Closed

LolWolfBest ResponseYou've already chosen the best response.2
Since \(x\) 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.
 one year ago

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

LolWolfBest ResponseYou've already chosen the best response.2
In any case: \[ A=\{x1\leq x\leq 5\}=\{x1\leq x \wedge x\leq 5\} \]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
i can't believe i didn't think of that first reason....
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.0
Unless your x's can quantum tunnel . . .
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
uhoh....math conversation.....time to leave..
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.0
(Sorry, I've been hanging out in the Physics section too long)
 one year ago

LolWolfBest ResponseYou've already chosen the best response.2
Go go, quantum entanglement and random transmission of data!
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.0
Strange things can happen in this here multiverse.
 one year ago
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