Cameron Buie
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 15h answered Prove that there is no largest irrational number 15h answered Prove that if A and B are sets such that $A \cup B \neq \emptyset$, then $A \neq \emptyset$ or $B \neq \emptyset$ 1d awarded proof-verification 1d revised Set of rational numbers bounded between two irrationals is a closed set? adjusted tags 1d comment Set of rational numbers bounded between two irrationals is a closed set? @MeesdeVries: It's close, but not quite there. 1d answered Set of rational numbers bounded between two irrationals is a closed set? Apr 25 awarded Guru Apr 25 comment How many partially ordered sets(poset) does a set have on 4 elements? As a result, your answer is not correct, though it seems you've correctly determined all the possible Hasse diagrams (I haven't had the chance to sit down and work it all out), which would mean you've determined all partial orders on a four-element set up to isomorphism. Apr 25 comment How many partially ordered sets(poset) does a set have on 4 elements? A reflexive relation on a non-empty set cannot be trichotomous. I think you mean "total," instead. Totality is used to prove that every total order on a four element set is of "type P." As for why there are $24$ such: How many options are there for the least element? Once that's chosen, how many options are there for the second-least element? (etc.) Apr 25 comment How many partially ordered sets(poset) does a set have on 4 elements? The answer to the latter is: $4!=24$ distinct orderings, but only one such ordering up to isomorphism. The latter distinction is something you must take into account for partial orders, too, since (for example) all $24$ total orders are of "type P." Apr 23 revised How to evaluate $\arctan(1)$ edited body Apr 22 answered How to evaluate $\arctan(1)$ Apr 22 revised How can we write (2,5) in the countable family of disjoint open intervals? added 280 characters in body Apr 22 answered How can we write (2,5) in the countable family of disjoint open intervals? Apr 22 comment Find the Laurent expansion of f (z) = 1/( z(z − 1)(z − 2)), (in powers of z) for: a. 0 < |z| < 1 b. 1 < |z| < 2 c. |z| > 2. @JessyCat: I'm happy to help you, but I need to know what you want me to help you do. Apr 22 comment What would moving in the 4th dimension look like in 3d? Well, arguably, you and the cup could be thought of as moving through a 4th dimension (time) in tandem. Apr 21 comment Injective: In what circumstances would there be less than one pre-image of an image? @Will: For non-surjective functions, there will be elements of a codomain which have no preimages. There will never be elements of the range which have no preimages. Apr 21 comment Does it hold/can you prove that if $\frac{1}{x} + x$ is an integer, then $x = 1$? @lulu: I think you mean $x 2. @Jessy: Show you how to do what in the region$0<|z|<1\$? Apr 20 comment There are 56 teams in a knockout tournament, then how many matches has to be played to select the champion? @Kenny: You seem to be using "match" to mean the same thing as the OP's "round." You're right that there must be at least 6 rounds, but there may be as many as 55. See my answer.