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 Sep 20 comment Why is it that $\left|b_n - b \right| < \frac{\left|b \right|}{2} \Rightarrow \left| b_n \right| > \frac{\left|b \right|}{2}$? Thanks to all of you for your help! Part of what I overlooked is that $|b_n - b| = |b - b_n|$... I think I was trying to do this, but I kept getting $\frac{3|b|}{2}$ or something... Anyway, I think it makes sense to me finally :) Sep 7 comment What is the explanation for the elements of this set? by the way I hope that $f(A)$ is the correct notation for the range of the function. Sep 7 comment What is the explanation for the elements of this set? thank you, that definitely helps a lot. I still have to ask: in order to "note that $B$ is not in the range of the function" we then need to consider the range of the function to be $\{ \{ a\}, \{ a,c\} \{ a,b,c\}\}$ as I wrote in the question? And that's why $B = \{b\} \notin f(A)$? Is that correct? In other words $B$ is a set, but the important observation is that it is not an element of the range of the function? Sep 7 comment What is the explanation for the elements of this set? Sorry, those weren't actually my solutions, I've updated the question to hopefully make that more clear... Aug 20 comment What are the explanations for certain steps in these proofs for the irrationality/rationality of certain numbers? @kahen: sorry, but I was having a little difficulty understanding the right to left direction of the statement you wrote. The way I was interpreting it, I would have thouhgt that if $p = 4$, $a=4$, $b=2$ so $ab = 8$, then $p$ divides $ab$ as well as $a$... But is part of the requirement that this is true for every $ab$? So if $p=4$ it divides $ab=12$ yet $4$ divides neither $a=6$ nor $b=2$ and therefore $p=4$ is not prime? Is that correct? Aug 20 comment What are the explanations for certain steps in these proofs for the irrationality/rationality of certain numbers? thank you rbojohn Aug 20 comment What are the explanations for certain steps in these proofs for the irrationality/rationality of certain numbers? thank you for this Aug 20 comment What are the explanations for certain steps in these proofs for the irrationality/rationality of certain numbers? thank you very much Aug 19 comment What are the explanations for certain steps in these proofs for the irrationality/rationality of certain numbers? @Bill: for context, this is me trying to go through the book Understanding Analysis by Stephen Abbott. Unfortunately I do not know any number theory or abstract algebra... Aug 19 comment What are the explanations for certain steps in these proofs for the irrationality/rationality of certain numbers? @kahen and John, thank you for your comments! Jul 6 comment For which $k \in \mathbb{R}$ will this be a euclidean vector space? thank you for the explanation! Jul 4 comment Help understanding this example of a linear operator which rotates each vector $v$ about the z-axis by an angle $\theta$ [at]Jason: thank you! Jul 3 comment Help understanding this example of a linear operator which rotates each vector $v$ about the z-axis by an angle $\theta$ @Mark: thank you for the comment, unfortunately I do not know the formulae for rotations in any dimension at the moment... Jul 1 comment Understanding this partial derivative problem thank you again! Jul 1 comment Understanding this partial derivative problem sorry, i had read your answer and thought it was a done deal, but i seem to still be stuck: taking the step you suggest, i get $\frac{\partial \theta}{\partial x} = - \frac{-3 \sin \theta}{e^{2r}} \frac{1}{3\sin^{2}\theta + 2\cos^{2} \theta}$. is there some way that i can change $3\sin^{2}\theta + 2\cos^{2}\theta$ into the expression $2+\sin^{2}\theta$ as in the solution above? Jul 1 comment Understanding this partial derivative problem thank you for helping me with this arturo Jul 1 comment Understanding this partial derivative problem this definitely does help, thank you Jun 4 comment Does this equality always hold? thank you mixedmath Jun 4 comment Does this equality always hold? excellent, thank you Jun 4 comment Does this equality always hold? @Fabian: thank you