Maurice Tutor

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Teaching Since: May 2017
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  • MCS,PHD
    Argosy University/ Phoniex University/
    Nov-2005 - Oct-2011

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  • Professor
    Phoniex University
    Oct-2001 - Nov-2016

Category > Management Posted 13 Jan 2018 My Price 4.00

arbitrary member

What is wrong with the following “proof”?

(Non)Theorem: If binary relation R is symmetric and transitive, then R is reflexive.

(Non)Proof: Let x be some member of the domain of R. Pick y such that xRy. By symmetry, yRx. By transitivity, xRy and yRx imply xRx. Since x is an arbitrary member of R’s domain, we have shown thatxRx for every element in the domain of R, which “proves” that R is reflexive.

 

Answers

Maurice Tutor
(5)
Status NEW Posted 13 Jan 2018 10:01 PM My Price 4.00

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