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$x^a - 1$ divides $x^b - 1$ if and only if $a$ divides $b$ Let $x &gt; 1$ and $a$, $b$ be positive integers. I know that $a$ divides $b$ implies that $x^a - 1$ divides $x^b - 1$. If $b = qa$, then  $$x^b - 1 = (x^a)^q - 1^q = (x^a - 1)((x^a)^{q-1} + \ldots + x^a + 1).$$  I'm interested in the converse of the statement. If $x^a - 1$ divides $x^b - 1$, does this imply that $a$ divides $b$? Came across the following exercise in Bartle's Elements of Real Analysis. This is the solution I came up with. Would be grateful if someone could verify it for me and maybe suggest better/alternate solutions.  I also looked up these related questions - , ,  - but was not happy with proofs given there. I seem to need some help understanding these. Any such help is appreciated.      Show that every sequence in $\Bbb R$ either has a monotone increasing sub-sequence or a monotone decreasing sub-sequence.   Let $(x_n)$ be a sequence in $\Bbb R$. Suppose $(x_n)$ is not bounded. Without loss of generality we may assume that $(x_n)$ is not bounded above. Therefore given any real number there is a member of the sequence which is greater. Let $x_{n_1}$ be any member of the sequence.   There is $x_{n_2} \gt \sup\{x_1, x_2, ..., x_{n_1} \}$. For $i \gt 1$ let $x_{n_i} = \{x_1, x_2, ..., x_{n_{i - 1}}\}$ then $(x_{n_k})$ forms a monotone subsequence of $(x_n)$.   Now suppose instead that $(x_n)$ is bounded. By the Bolzano-Weierstrass Theorem there is a subsequence $(y_n)$ of $(x_n)$ which converges to a limit $y$. Without loss of generality there are infinitely many distinct values in $(y_n)$ that are unequal to $y$. Let $y_{k1}$ be the first such element. Let $y_{k2}$ be any element in $\{ y' \in (y_n) \ \ | \ \ |y' - y | \lt |y - y_{k1}|  \}$.   For $i \gt 1$ let $y_{ki} \in \{ y' \in (y_n)  \ \ | \ \ |y' - y | \lt |y - y_{k \ i - 1}|  \}$. Such $y_{ki}$ exists for every $i \in  \Bbb N$ since $ \lim (y_n) = y $. Now let $(y_{kn})$ be the sub-sequence of $(y_n)$ thus formed. At least one of the two following sets must contain infinitely many elements.    $\{ y \in (y_{kn}) \ \ | \ \ y \gt x\}$ $\{ y \in (y_{kn}) \ \ | \ \ y \lt x\}$   The one which does forms a monotone subsequence. Show that holomorphic $f_1, . . . , f_n $ are constant if $\sum_{k=1}^n \left| f_k(z) \right|$ is constant. While studying for an exam in complex analysis, I came across this problem. Unfortunately I was not able to solve it. Any help would be greatly appreciated.  Let $U ⊂ \mathbb{C}$ be a domain and $f_1, . . . , f_n : U \rightarrow \mathbb{C}$ holomorphic functions, such that $\sum_{k=1}^n \left| f_k(z) \right|$ is constant on $U$. Show that $f_1, . . . , f_n$ are constant. Manually clear saved question drafts I've seen lately that new question editing is stored somewhere for later use if we wander off the new question without posting it and then return back to the page. Which is a good functionality addition in it's own right.  What I'd like to also have is the Discard this draft functionality that would clear form data and forget it. Somehow we find an answer in the meantime and the question isn't valid any more.  So having a Discard saved draft or something similar would be a nice contra-functionality to auto-save to remove stored data. The way it's implemented now seems too hidden hidden and unpredictable to users.  Additional edit  When adding hidden auto functionality usually means creating the opposite as well by either giving the possibility to preference of not having automatism at all or giving the possibility to clear what's been auto saved.  Let's see how GMail doesn't work (and I think we all agree that from UX perspective it does the job as expected:   User clicks Compose message User starts writing a new email message User decides to not send the message and checks their inbox User clicks Compose message again A clean form is presented to user (old one being part of drafts)   I'm not suggesting we should have something like Saved drafts but we surely should have the possibility to clear previous question draft.  Since this new functionality was introduced I should point out that I would use the discard functionality more frequently than the saved one. I do welcome saving messages, but discarding of saved drafts is anything but a rare situation one might want to do. It wouldn't be a problem if this was a 1% against 99%. To me it's more like 50/50.  Best possible solution  When providing saved drafts there should (from the UX perspective) be the opposite functionality as well. Namely discarding of it. But first thing we should be asking ourselves is:    do users more frequently need pre-filled saved question draft or do users more frequently need empty form with the possibility to populate it with a saved draft   User testing should provide the usage frequency between the two and the one that's more likely to happen (my guess is that there would be a big difference between the two) should be implemented. So one of these:   Pre-populate question fields with saved draft on question form load, while also providing a button to clear the draft and start writing anew without the need to manually clear all the fields (which would be the optimal solution in case of a single field form) Question form should always load empty while also having a button that would pre-populate it with a saved draft.   I think this would be the best consensus related to saved question drafts and their usage. But as we've seen this issue has been swept under the rug for an indefinite time. I'd like to see the full How-To on how to use manual partitioning during Ubuntu installation. The existing guides (at least those I found here) cover only automatic part and leave untouched the manual part (or extremely short and contain no pictures).  I'd like to cover such situations:     If your disk contains other systems: Can you create a strong blockcipher with small blocksize, given a strong blockcipher of conventional blocksize? Suppose I want a strong 20-bit blockcipher.  In other words, I want a function that takes a key (suppose the key is 128 bits), and implements a permutation from 20 bits to 20 bits.  The set of permutations should be close to a randomly-chosen subset of size $2^{128}$ of all $2^{20}!$ permutations on 20 bits.  I don't want you to build a new blockcipher from scratch.  Instead, assume you have a "strong" blockcipher like AES (128-bit blocksize, 128-bit keysize) and use this to build your 20-bit cipher.   Of course, your construction should be practical (i.e., you should be able to run it in both directions with reasonable amounts of time and memory). Enter from password field doesn't submit SE login form On  page for a site where I don't yet have an account, if I select the "Log in using Stack Exchange" box, when I hit enter from the password form, nothing happens.  Minor, but... What is the purpose of wrapping whole Javascript files in anonymous functions like “(function(){ … })()”? $$I(\alpha)=\int_0^1 \frac{x^\alpha-1}{\ln x}dx.$$  As the title says, if someone could solve this without using the differentiation under the integral sign technique, I would be very grateful. Why does PGFplots not recognize my expression? How prove this two symmetric matrices $AB=0$ Let $A,B$ be real symmetric matrices, and for any $n\in \Bbb N^{+}$, and for all $x,y\in \Bbb R$, we have $$tr(xA+yB)^n=x^ntr(A^n)+y^ntr(B^n).$$  Show that $AB=0$.  My try:since $$(xA+yB)^n=x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n$$ so $$tr\left(x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n\right)=x^ntr(A^n)+y^ntr(B^n)$$  then  $$tr\left(\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}\right)=0$$ so $$\binom{n}{1}x^{n-1}ytr(A^{n-1}B)+\cdots+\binom{n}{n-1}xy^{n-1}tr(AB^{n-1})=0$$ then I can't  I found this is a nice result, maybe this is an odd problem, and I can't solve it. Thank you. "Install Spotify" can't be opened because Apple cannot check it for malicious software I'm running macOS 10.15 Catalina Beta.  When I try to install Spotify I see the following error:    Is this a Spotify issue or a Catalina issue?  Thanks. EDIT: Summary  I think I've found the best definition of what I want thanks to the discussion with @M. Tibbits:  When I open or refresh the SO, I get a notification of new activities that are of interest to me. Today these are (hard-coded): answers to my questions and comments addressed to me. I want a bit more control over it, that's all.     Yes, I've read  and , but I think I add new (good :) ) reasons for re-asking it:  The first question was answered before this so handy "inbox" notification feature was implemented. So now using favorites for "watching" a question is wrong, IMHO, for 2 reasons:   it splits essentially the same functionality (of getting notified when there is an activity you are interested in) between 2 very different user interfaces it's cumbersome - I need to: 1) add question to favorites; 2) click on my profile dropdown; 3) click on the "favorites" line; 4) see all my favorites with recent activity; 5) select the one I watch; 6) repeat this process occasionally to get updates; and 7) possibly remove the question from favorites at the end. All this as opposed to this little red button when there are news.   Favorites are different in nature that watching. I add a question to Favorites when I find the answers useful to me in my work, so I may want to have it handy in the future. By watching an interesting question I will be able to decide whether I want it in faviorites, after I see the answers.  EDIT: I see that I was not clear enough.  I want to add a question to Favorites when:   it has some good answers the question and answers are relevant in my work I want to keep them handy as my long-term (persistent) knowledge base   I want to watch a question for short while (matter of days/weeks) when:   it is interesting / relevant I want to follow it to see what are the answers. Then I can decide whether I want it in my favorites   As you see, the main differentiators are good answers and long/short term. For short-term, automatic notification is better.  Bottom line, all I ask is "watch this question" checkbox (which may expire after say a week if there is a need to free resources) Sorry if I keep asking for proof checks. I'll try to keep it to a minimum after this. I know this has a well-known proof. I understand that proof as well but I thought I'd do a proof that made sense to me and seemed, in some ways, simpler. Trouble is I'm not sure if it's totally correct. It's quite short though. I was just hoping someone could look it over and see if it is a valid proof. Thank you!  Lemma: Every Cauchy sequence is bounded.  Proof: Let $(a_{n})$ be Cauchy. We choose $ 0&lt;\epsilon_{0}$. So $ \forall \; n&gt;m\geq N_{0}$ we have that $\vert a_{n}-a_{m} \vert &lt; \epsilon_{0}$. Therefore $(a_{n})$ is bounded for all $ m \geq N_{0} $ by $ \epsilon_{0} $. Since $ \mathbb{N}_{N_{0}}$ is finite, it is bounded. So, for all $ m&lt;N_{0} $, $ (a_{n})$ is bounded. Therefore $(a_{n})$ is bounded.  I realize I haven't said what the bounds are but I think that's sort of irrelevant. So long as we know it's bounded. Any help is much appreciated! On a quantum scale the smallest unit is the , which is a discrete measure.  There several question that come to mind:   Does that mean that particles can only live in a discrete grid-like structure,  i.e. have to "magically" jump from one pocket to the next? But where are they in between? Does that even give rise to the old paradox that movement as such is impossible (e.g. )? Does the same hold true  for time (i.e. that it is discrete) - with all the ensuing paradoxes? Mathematically does it mean that you have to use difference equations instead of differential equations? (And sums instead of integrals?) From the point of view of the space metric do you have to use a discrete metric (e.g. the ) instead of good old Pythagoras?   Thank you for giving me some answers and/or references where I can turn to.  Update: I just saw this call for papers - it seems to be quite a topic after all: Is Reality Digital or Analog? FQXi Essay Contest, 2011. , , . One can find some pretty amazing papers over there. I've never really understood why math induction is supposed to work.  You have these 3 steps:   Prove true for base case (n=0 or 1 or whatever) Assume true for n=k.  Call this the induction hypothesis. Prove true for n=k+1, somewhere using the induction hypothesis in your proof.   In my experience the proof is usually algebraic, and you just manipulate the problem until you get the induction hypothesis to appear.  If you can do that and it works out, then you say the proof holds.  Here's one I just worked out,  Show $\displaystyle\lim_{x\to\infty} \frac{(\ln x)^n}{x} = 0$  So you go:   Use L'Hospital's rule. $\displaystyle\lim_{x\to\infty} \frac{\ln x}{x} = 0$. Since that's $\displaystyle\lim_{x\to\infty} \frac{1}{x} = 0$. Assume true for $n=k$. $\displaystyle\lim_{x\to\infty} \frac{(\ln x)^k}{x} = 0$. Prove true for $n=k+1$.  You get $\displaystyle\lim_{x\to\infty} \frac{(\ln x)^{k+1}}{x} = 0.$   Use L'Hospital again: $\displaystyle\lim_{x\to\infty} \frac{(k+1)(\ln x)^{k}}{x} = 0$.  Then you see the induction hypothesis appear, and you can say this is equal to $0$.  What I'm not comfortable with is this idea that you can just assume something to be true ($n=k$), then based on that assumption, form a proof for $n=k+1$ case.  I don't see how you can use something you've assumed to be true to prove something else to be true. Can graduated Stack Exchange sites get closed too? Django - specify which model manager Django admin should use Images instead of radio buttons or dropdown list? How to align the camera in a solved motion tracked scene? I have been having an issue with my floor plane and object when using a solved camera after tracking some footage. I get a solve error of between 0.05 and 0.4 on my solves which is good and not the issue I've been having. However when I go to orient the camera I will snap it to an origin and adjust the scale. The plane and object are not aligned, but could be fixed if I could just rotate around them (much like in PF Track). I would think the pivot point for the camera, if I have it locked to view, would be the origin (where my plane and object are), but it seems to be just some arbitrary point out in space which makes manually fixing the alignment very difficult. I'm wondering if there is any way to fix this so the pivot point is in fact in the center, or if there are any other suggestions to go about this problem.