I saw a demonstration that there was an irrational number that raised to another irrational number resulted in a rational number. The demonstration was as follows: We know that $\sqrt{2}$ is irrational. If $\sqrt{2}^\sqrt{2}$ is rational, we are done. Otherwise, we have $(\sqrt{2}^\sqrt{2})^\sqrt{2} =  \sqrt{2}^2 = 2$, which is rational, and we are done. It's a interesting demonstration, because it is not necessary to determine the numbers in order to show that they exist. However, I would like to know if, after all, it is possible to determine whether $\sqrt{2}^\sqrt{2}$ is rational or irrational. I don't know how could I aproache this problem.
I know  that uses $x={\sqrt{2}}^{\sqrt{2}}$ to prove that there must exist an irrational to an irrational power that evaluates to a rational. But I don't know if $x$ itself is known to be irrational or rational.  Could someone enlighten me? Thanks!
Is there a way to detach matplotlib plots so that the computation can continue? After these instructions in the Python interpreter one gets a window with a plot:  from matplotlib.pyplot import * plot([1,2,3]) show() # other code   Unfortunately, I don't know how to continue to interactively explore the figure created by show() while the program does further calculations.  Is it possible at all? Sometimes calculations are long and it would help if they would proceed during examination of intermediate results. List sorting by custom field I dont know how to sort a list of SObject Case.  I have a List&lt;Case&gt; newCases which i iterate over in for cycle, where i save Cases to my new List&lt;Case&gt; casesList.  for (Case newCase : newCases) {             if(newCase.Case_Content__c == '111' &amp;&amp; newCase.Status == '24'){                 String cuid = newCase.CUID__c;                 cuidSet.add(cuid);                 casesList.add(newCase);                 casesList.sort(); //TODO sorting by CUID__c             }         }   What i need to do is, to sort this new list by field CUID__c. So the Cases with same CUID are together.  Thank you for any help :) Let $G$ be a finite abelian group, then $G$ has a subgroup of order $n$ if and only if $n\mid G$.  Proof: by Lagrange if $H\leq G$ then $|H|$ divides $|G|$ so this proves one of the implications.  We prove the other implication by strong induction on the order of $G$, for primes this is just Cauchy, so we can take these like base cases. Now let $n$ be a proper divisor of $|G|$, take $p$ a prime in the factorization of $|G|$ (instead of $|G|$ is should be $n$, I had made a mistake when typing the first time), then by cauchy there is an element $x$ of $G$ of order $p$, since $G$ is abelian $\langle x\rangle$ is normal in $G$ so we can look at the quotient group $G/\langle x\rangle$, this group has order $\frac{|G|}{p}$, now take a prime $p_1$ in the factorization of $\frac{|G|}{n}$. By induction the quotient group has a subgroup $E$ of order $G/pp_1$ now look at the preimage of $E$ under the natural homomorphism from $G$ to $G/\langle x\rangle$, it is easy to see this is a subgroup of $G$, we call it $I$, consider the restriction of the natural projection to $I$, this is a surjective homomorphism from $I$ to $E$ with kernel $\langle x\rangle$ so by the first isomorphism theorem $I/x\cong E$ so $\frac{|I|}{|\langle  x\rangle|}=|E|\implies  |I|=\frac{|G|}{pp_1}{p}=\frac{|G|}{p_1}$ Notice $\frac{|G|}{p_1}$ is a multiple of $n$ since $p_1$ was in the fatorization of $\frac{|G|}{n}$. So by induction $I$ has a subgroup of order $n$ which is also a subgroup of $G$. GetFeatureInfo from Image WMS I am an Indian citizen studying in Italy on an Italian visa that expires 19 November, 2015. I have permesso di soggiorno. I want to visit the United Kingdom for my vacations for a week during March. I know the procedures already, however I am a little bit confused about the amount that I must show as a proof of support with my bank statements. I will stay for a week with my friends who are also students in Italy and will be in London for 5 days. What would be the amount in Euros that I need to show for my visit to UK? Let $E$ be measurable and define $f:E\rightarrow\mathbb{R}$ such that $\{x\in E : f(x)&gt;c\}$ is measurable for all $c\in\mathbb{Q}$, is $f$ measurable?  There are a number of equivalent definitions for the measurability of a function and the most obvious one would be to show that $\{x\in E : f(x)&gt;c\}$ is measurable for all $c\in\mathbb{R}$.  Thus my strategy has been to consider an arbitrary irrational $y$ and use the density of the rationals in the reals to show that there exists some open set $O$ such that $m^*(O-\{x\in E : f(x)&gt;y\})&lt; \varepsilon$.  I would do this by choosing some rational $q&lt;y$, close enough to $y$, such that the set of values in $E$ which get sent under $f$ to values strictly greater than $q$ and less than or equal to $y$ is small enough.  But that doesn't in general seem possible since I could keep sending intervals of some fixed positive length to smaller and smaller regions just beneath $y$.  However conversely, such a function would not be enough for a counter example, since finding such a function only rules out this strategy, but doesn't necessarily entail non-measurability.  Is there a better definition of the measurability of a set I should use? or does the statement actually not imply measurability?  Thanks. How can I compute $\mathrm{Aut}(\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z})$? Persistent Dummy Output The soundcard I want to use is detected and works (it only recently switched to dummy output and didn't allow me to go back). HDMI is no longer listed under Audio and Video - KDE Control Module; everything under audio module is set to dummy output and its the only 'soundcard' on the list now. I can view my previous sound cards in alsamixer, but changing the settings from there does nothing. How can I get my real soundcards to work again? I have already tried the following answers without success:  EDIT: name@name-desktop:~$ lspci -vvv | grep -A8 Audio 00:01.1 Audio device: Advanced Micro Devices, Inc. [AMD/ATI] Kaveri HDMI/DP Audio Controller         Subsystem: ASRock Incorporation Kaveri HDMI/DP Audio Controller         Control: I/O+ Mem+ BusMaster+ SpecCycle- MemWINV- VGASnoop- ParErr- Stepping- SERR- FastB2B- DisINTx+         Status: Cap+ 66MHz- UDF- FastB2B- ParErr- DEVSEL=fast &gt;TAbort- &lt;TAbort- &lt;MAbort- &gt;SERR- &lt;PERR- INTx-         Latency: 0, Cache Line Size: 64 bytes         Interrupt: pin B routed to IRQ 43         Region 0: Memory at ff764000 (64-bit, non-prefetchable) [size=16K]         Capabilities: &lt;access denied&gt;         Kernel driver in use: snd_hda_intel         Kernel modules: snd_hda_intel -- 00:14.2 Audio device: Advanced Micro Devices, Inc. [AMD] FCH Azalia Controller (rev 01)         Subsystem: ASRock Incorporation FCH Azalia Controller         Control: I/O- Mem+ BusMaster+ SpecCycle- MemWINV- VGASnoop- ParErr- Stepping- SERR- FastB2B- DisINTx-         Status: Cap+ 66MHz- UDF- FastB2B- ParErr- DEVSEL=slow &gt;TAbort- &lt;TAbort- &lt;MAbort- &gt;SERR- &lt;PERR- INTx-         Latency: 32, Cache Line Size: 64 bytes         Interrupt: pin A routed to IRQ 16         Region 0: Memory at ff760000 (64-bit, non-prefetchable) [size=16K]         Capabilities: &lt;access denied&gt;         Kernel driver in use: snd_hda_intel  Could this image possibly help?:  There is another soundcard but I'm unsure of what its for because it never outputs sound, but it does have a lot of settings: that sound card is labelled as &quot;HD-Audio Generic&quot;.   So I went into Software &amp; Updates under the Additional drivers tab and I see this:  It was on Do not use the device so I changed it and applied the settings but unfortunately the soundcard isn't back even after reboot. Using Georeference information stored in tfwx (from ArcGIS) in QGIS? I have georeferenced some images (ArcGIS -> Update Georeferencing) which creates a *.tfwx file with the new parameters for georeferencing.   I renamed it to *.tfw but QGIS is not able to read the georeferencing information.   Does anybody has any useful tips how to fix this? Binding energy and mass I’ve been told that a greater binding energy means the nucleus is more tightly bound, and therefore that decreases the mass of the nucleus with respect to its nucleons when separated. But why does a tighter nucleus mean a smaller mass? Hash character in LaTeX I want to check if points layer that I have intersect with polygons layer that I have, as Boolean column in the points dataframe. I have two GeoPandas dataframes, the first one is many points and looks like this: &gt;&gt;&gt;ID  geometry 0  12  POINT (5.0279 7.4547) 1  45  POINT (6.6539 12.139) ...  and the second dataframe is one layer of many different polygons  that looks something like this: &gt;&gt;&gt;name     code   geometry 0  Desert   12     POLYGON ((5.52013 13.8902, 5.5265 13.892,...) 1  Water    24     POLYGON ((5.53756 13.88472, 5.5291 13.8791,...) ...  I'm trying to check if there is any intersection between the points layer and the regions layer. For that I have determined the crs and then used intersects : regions=regions.to_crs({'init': 'epsg:4326'}) points=points.set_crs({'init': 'epsg:4326'}) inter=points.geometry.intersects(regions.geometry)  A The script runs with the following warnings:  FutureWarning: '+init=:' syntax is deprecated. ':' is the preferred initialization method. When making the change, be mindful of axis order changes: https://pyproj4.github.io/pyproj/stable/gotchas.html#axis-order-changes-in-proj-6 return _prepare_from_string(&quot; &quot;.join(pjargs)) /opt/conda/lib/python3.8/site-packages/geopandas/base.py:39: UserWarning: The indices of the two GeoSeries are different. warn(&quot;The indices of the two GeoSeries are different.&quot;)  Then when I check the results the only value is False, like all th epoints do not intersect: inter.unique().tolist() &gt;&gt;&gt;[False]  *I saw on QGIS that the there are points that intersect and there are points that do not so there is no way this result is true *I have checked the dtypes - each one of my geodataframes has one column that is geometry and called geometry. My end goal: to add new column in the points geodataframe that will tell if it intersects the regions or not. Let $a,b,c&gt; 0$ be three real numbers such that $a+b+c=1$. I want to prove that   $$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+a^3+b^3}\le\frac{1}{5abc}.$$  My attempt: Using AM-GM on each denominator gives (here, LHS denotes the left-hand side) $$LHS\le\frac{1}{3a^{\frac12}bc}+\frac{1}{3b^{\frac12}ac}+\frac{1}{3c^{\frac12}ab}.$$  However, I think that my attempt doesn't work because the last expression can get larger than $\frac{1}{5abc}$. In fact, if we multiply with $3abc$ then the original inequality is: $$\sqrt a+\sqrt b+\sqrt c\le\frac35$$  However, as $a,b,c&lt;1$, $$\sqrt a +\sqrt b + \sqrt c&gt; a+b+c=1&gt;\frac35.$$ So my upper bound is always larger than the given one. iMac with 8GB RAM -- memory always almost full Intuition behind the definition of a measurable set This week I saw the definition of a measurable set for an outer measure.     Let $\mu^*$ be an outer measure on a set $X$. We call $A \subseteq X$ measurable if $$\mu^*(E) = \mu^*(A\cap E) + \mu^*(A^c\cap E)$$ for every $E \subseteq X$.   This is not the first time I've seen this definition. Unlike most other things in mathematics, over time I have gained absolutely no intuition as to why this is the definition.   The only explanation I've ever seen is that a set is measurable if it 'breaks up' other sets in the way you'd want. I don't really see why this is the motivation though. One reason I am not comfortable with it is that you require a measurable set to break up sets which, according to this definition, are non-measurable; why would you require that? Of course, you can't say what a non-measurable set is without first defining what it means to be measurable so I suppose no matter what your condition is, it will have to apply to all subsets of $X$.     Is there an intuitive way to think about the definition of measurable sets? Is there a good reason why we should use this definition, aside from "it works"? Does MariaDB or MySQL implement the VALUES (expression) table value constructor? I am just wondering if MariaDB or MySQL implements the &lt;table value constructor&gt; in the SQL Spec. In  and ?  SELECT * FROM ( VALUES (1) ) AS t(x);  x  ---  1 (1 row)   (syntax from ). I need to amend my return but I already received check from IRS with my refund. Should I cash it out now or wait until I amend my return? I need to amend my return, but I have already received a check from IRS (they approved my return that I need to amend). Should I cash out this check and then amend my return, or should I dispose this check and wait for a check from an amended return? Update from 14.04 to 16.04 failed with GPG error Is $L^{p}$ space with alternate norm Banach? Often (via Google Now, or when searching for a solution to some problem), I'll open a question from a non-favourite site in the app. Sometimes, I'd like to see what else is going on in the site where the question was posted. Currently, I have one of two ways:   Via tag icon:   Click one of the tag icons beneath the title Delete the [tag] text (at which point, thankfully, the sorting changes to Active from Votes).  Via manually searching for the site, which is more cumbersome. I type in hard and Hardware Recommendations is the fourth entry (below Ask Different, Raspberry Pi and Arduino)! This more annoying for meta sites, where I have to pick Meta from the dropdown menu.   Can I pretty please have a way to head to the site directly from the question? Maybe:   the site name in the top bar could become clickable and lead to the site, or an option in the More menu?     This would be especially useful for Meta sites, where questions posted around the same time could be related. Interestingly enough, most questions that Google Now are for metas of sites I visit on occasion, but haven't pinned to the list.