import time  for i in range(100):      time.sleep(0.1)      print("sleeping is happening")   I think it will print the string every 0.1 seconds, but actually it prints the string 100 times on the output client after the code is done.  The answer is a little different (python version 3.5):  following is not useful   import sys sys.stdout.flush()    and   print("sleeping is happening", flush = True)   is corret.
How do I force Python's print function to output to the screen?
How to prove that the set of rational numbers are countable? Showing that $a^n - 1 \mid a^m - 1 \iff n \mid m$ Let $a\ge 2$ be an integer. Show that for positive integers $m,n$, we have $a^n - 1$ divides $a^m - 1$ if and only if $n$ divides $m$.   I am having trouble showing this. I've seen a similar problem on here but it only shows one direction for the proof. I would like to see both directions.  This is what I have so far: Let $m = nq + r$, then $a^m - 1 = a^{nq + r}- 1 = a^{nq}a^r - 1 = a^{nq}a^r - a^r + a^r - 1 = a^r(a^{nq} - 1) + a^r - 1$. Since $0\le r\lt n$, $r = 0$ then $a^m - 1$ = $a^{nq} - 1$ so $a^m$ = $a^{nq}$ . Thus $m = nq$ hence $n\mid m$. I've been trying to solve the following problem:  Suppose that $f$ and $f&#39;$ are continuous functions on $\mathbb{R}$, and that $\displaystyle\lim_{x\to\infty}f(x)$ and $\displaystyle\lim_{x\to\infty}f&#39;(x)$ exist. Show that $\displaystyle\lim_{x\to\infty}f&#39;(x) = 0$.  I'm not entirely sure what to do. Since there's not a lot of information given, I guess there isn't very much one can do. I tried using the definition of the derivative and showing that it went to $0$ as $x$ went to $\infty$ but that didn't really work out. Now I'm thinking I should assume $\displaystyle\lim_{x\to\infty}f&#39;(x) = L \neq 0$ and try to get a contradiction, but I'm not sure where the contradiction would come from.   Could somebody point me in the right direction (e.g. a certain theorem or property I have to use?) Thanks Have the guys from the show Impractical Jokers gotten into any major trouble because of the nature of their show?   Eg: have they been physically assaulted or has anyone thrown a law suit on them? The phrase "sleeping with someone" often means "having sex." What is the origin of this sexual connotation? Is there a non-sexual equivalent of this phrase to express sleeping with someone without sexual intercourse? What is the C# equivalent of NaN or IsNumeric? I found this heart while exploring the crimson biome - what is it? Solidify conundrum? I stumbled across a solidify issue, I cannot wrap my head around. I have a simple mesh, consisting of three faces.  When adding a solidify modifier, I expect to get an object with three walls, where every angle is square. However, whatever I try, tweaking the various solidify parameters, what I get is something like the following, with a skewed rim.  Did I hit a bug, or is it a limitation of the solidify modifier? Is there a workaround? How can I make a deepcopy of a function in Python? In Lieb and Loss's Analysis, I saw that they mentioned  $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$ dense in $ L^2(\mathbb{R}^n)$ (dense wrt the $L^2$ norm, I think). But I didn't find its proof in the book. So I wonder why it is?    Does this conclusion hold for $L^1(\Omega, \mathcal{F}, \mu) \cap L^2(\Omega, \mathcal{F}, \mu)$ for any measure space $(\Omega, \mathcal{F}, \mu)$?  Are there similar statements if replace $L^1$ with $L^p$, and  $ L^2$ with $L^q$ for $p \leq q \in (0, \infty ]$ or $\in [1, \infty]$?  Thanks and regards! How to move Ubuntu to an SSD Tag scores are not updated and the issue is getting worse I'm playing as a khajit in skyrim, going to assassinate Beitlid in Dawnstar. Now, the first time I did it on her porch and just killed her with the Blade of Woe. I got caught somehow, whatever, so I loaded a previous save because mid-jailbreak the guards decided to just attack instead of attempt an arrest.  Then I tried sniping her... and now my arrows have FED, Flight Erectile Dysfunction. They fly... like two feet before suddenly dropping off. It's weird. It doesn't matter how long I hold the trigger, it just droops. I am looking for QGIS function to extract (or detect) edge from a raster, which was referred in here,  as answer by @MappaGnosis.  The name could have changed since then, or it may have been integrated into other tool... I cannot find it in my QGIS Processing Toolbox (2.14.10 and 2.18.2).  There are similar tools such as GRASS "i.zc Zero-crossing" and SAGA "Gaussian filter", but I am hoping to find this QGIS geoalgorithm. I have realized that to pronounce the plural form of words ending in -th, we have to drop the letter "h". e.g. Months is pronounced /mʌnts/. But I have never seen this point mentioned in any context. I just want to make sure that this rule is correct and can be always applied. I was looking for a tree or graph data structure in C# but I guess there isn't one provided.  explains a bit about why.  Is there a convenient library which is commonly used to provide this functionality?  Perhaps through a strategy pattern to solve the issues presented in the article.  I feel a bit silly implementing my own tree, just as I would implementing my own ArrayList.     I just want a generic tree which can be unbalanced.  Think of a directory tree.  C5 looks nifty, but their tree structures seem to be implemented as balanced red-black trees better suited to search than representing a hierarchy of nodes. Windows 10 non-persistent installation (live ISO) We've got a Windows 10 installation running in an environment where it can occasionally lose power and restart. However, this sometimes leads to corruption of files.  I've had a look at Windows "To-Go" but this seems to be persistent. And I've also checked out virtualisation, but that would require us to run a non-persistent Windows 10, in another Windows 10 environment.  Is it possible to install Windows 10 as a "non-persistent live CD"? So that a user will have a seemingly fresh Windows 10, but doesn't require to re-install it at each startup? is it true if $f,\overline f$ are both analytic in a domain $\Omega$ then $f$ is constant? I am not able to  find out what property of holomorphic map I need to apply. please help.Thank you.  $f(z)=u(x,y)+iv(x,y)$, $\bar{f}(z)=u(x,y)-iv(x,y)$ Can you negate the Frightened condition by not looking at the source of your fear? As the title says, if you cause a creature to be Frightened of you, as in the Frightened condition, but they can somehow manage to not see you, do they stop rolling with disadvantage to attacks/ability checks?  Below are scenarios to illustrate my question.  Scenarios:   If a creature is frightened of you, and they turn their back against you to hit someone else, would they no longer have disadvantage to attack rolls/ability checks? If a creature is frightened of you and they have Blindsight, can they close their eyes to stop seeing you, and hence stop rolling with disadvantage? If a creature has 60 ft of movement and is frightened of you, and there is a corner they can reach in 15 ft that will block you from sight, can they run to that corner to stop being frightened, then run back to attack without disadvantage? Assume they would not provoke OA to do this, although the OA is immaterial to the question. Marker shadows in Google Maps v3