Lizwi Posted July 26, 2019 Share Posted July 26, 2019 (edited) Please help check my answers, it is not cut and paste, I wrote it through latex and I used a snipping tool to cut and paste it Edited July 26, 2019 by Lizwi Link to comment Share on other sites More sharing options...

ydoaPs Posted July 30, 2019 Share Posted July 30, 2019 Reflexivity and transitivity look good, but "obviously" won't cut it for symmetry. You should cite that set intersection is commutative for (b). Link to comment Share on other sites More sharing options...

Lizwi Posted July 30, 2019 Author Share Posted July 30, 2019 Thanks very much, I see the mistake in (b). Because the relation means an empty set. Link to comment Share on other sites More sharing options...

wtf Posted July 30, 2019 Share Posted July 30, 2019 (edited) This site supports Mathjax with the "math" and "\math" tags in square brackets surrounding your markup. In (1) your proof is fine but I'd reword it for clarity. You wrote, "I assumed [math]A = \{2,3,4,5]\}[/math] is the subset of the power set ..." I would reword that as, "Suppose [math]A = \{2, 3, 4, 5\}[/math]. Then [math]A \cap A = A \neq \emptyset[/math]. Or you could just say, For any nonempty set [math]A[/math], we have [math]A \cap A = A \neq \emptyset[/math]. That would be sufficient for me. Also note that you meant that [math]A[/math] is an element of the power set. It's a subset of the integers, but an element of the power set. Edited July 30, 2019 by wtf 2 Link to comment Share on other sites More sharing options...

ydoaPs Posted July 30, 2019 Share Posted July 30, 2019 10 hours ago, wtf said: Also note that you meant that A is an element of the power set. It's a subset of the integers, but an element of the power set. That's a good point. This is an important distinction and one that technically makes the proof incorrect. A is an element of P(Z), not a subset. A is a subset of Z. The elements of P(Z) are the subsets of Z. 1 Link to comment Share on other sites More sharing options...

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