I know that there are infinities of different magnitudes for countable things (e.g. via Cantor's diagonal argument), and that the ratios of the sizes of two such infinities may itself be infinite. However, how do ratios of infinities work for uncountable things? And can the ratio of two infinities of uncountable things be finite? Consider an infinitely large two-dimensional plane, consisting of a point $p$ and two unit vectors $v_1$ and $v_2$ rooted at $p$. Divide the infinite plane into two regions: $A$, the region of space spanned by the &quot;wedge&quot; centered at $p$, the &quot;short way around&quot; between $v_1$ and $v_2$, and $B$, the remainder of the plane (the &quot;long way around&quot;). The areas $|A|$ and $|B|$ are infinite, however it would seem that $|A| \over |B|$ is finite, since if a circle of any nonzero radius centered at $p$ is intersected with $A$ and $B$, the ratio of the areas of the two intersected regions would always be be the constant value $\theta \over {2\pi - \theta}$, where $\theta = cos^{-1}{{|v_1 . v_2|} \over {|v_1||v_2|}}$. (Actually maybe I am mistaken, and points on a plane are countable, by extension of Cantor's argument to two dimensions...)
Recently, I was wondering how division of aleph numbers would work. First, I thought about how finite cardinality division would work. What I came up with was that the result of $A/B$ where $A$ and $B$ are both cardinalities, is the number of times that each element of $B$ had to be mapped to an element of $A$ in order to ensure that all elements of $A$ were mapped to.  Extending this to aleph numbers, specifically, I thought that $\aleph_1/\aleph_0$ would be $\aleph_1$. The reasoning behind this is that there are an infinite ($\aleph_1$) number of real numbers between any two natural numbers. As such, that mapping would need to be applied.  Can anyone validate this idea? Is this how division of aleph numbers actually works, or am I totally off base?  Thanks.
What tense should I use to describe something in the past but probably is still true? The title may be a bit confusing. See the below example:     The movie (was/is) so good.   The simplest guideline on deciding which tense to use, past tense or present tense, to describe an object comes down to whether the description was only true that that time or is true even to the present point of time.  But now we talking about literary work(movies,books,etc) of which quality tend to stay the same over time, which tense should we use? Can I print a custom paper on standsard size in word? Why Listen to Music, why not Listening Music If you have the ability to cast a spell without a spell slot, can you cast that spell using a spell slot? SR Flip-Flop: NOR or NAND? Is there a way of creating a filesystem object akin to this:  mknod files p cat file1 file2 ... fileN &gt; files   but such that it can be seeked in, as if it were a regular file? I'm currently a Combinatorics student trying to parse through this solution. I do not understand the proof currently. Any help understanding it is greatly appreciated.    Question Let the number of partitions be given by the function $p(n)$, i.e. the number of ways of writing the integer $n$ as a sum of positive integers. Prove that that if $g$ is any polynomial function in $n$, then there exists an integer $N$ such that $g(n)&lt;p(n)$ for all $n&gt;N$. Do not use the asymptotic Hardy-Ramanujan formula.    Provided Proof.    Enough to show $n^a&lt;p(n)$ for fixed positive integer $a$. (This is just saying that we want to show $p(n)$ grows faster than any degree polynomial of $n$, right?) We have that $(i_1,i_2,\ldots,i_{a+1},1^{n-i_1-\cdots -i_{a+1}})$ is a partition of $n$ as long as $i_{a+1}\leq i_a\leq \cdots i_1 \leq \lfloor\frac{n}{a+1}\rfloor$. (I really don't understand this step. What is this partition? Of the coefficients or terms of polynomial $p(n)$? And how do we know this is $\leq \lfloor\frac{n}{a+1}\rfloor$?) The number of partitions $n$ of this form is multiset binomial coefficient $$\left(\dbinom{\lfloor \frac{n}{a+1}\rfloor}{a+1}\right)=\frac{\lfloor\frac{n}{a+1}\rfloor(\lfloor\frac{n}{a+1}\rfloor+1)\cdots(\lfloor\frac{n}{a+1}\rfloor+a-1)}{(a+1)!}$$ (I'm assuming that if I can understand (2.) then I can understand this line) This means $p(n)&gt;\left(\binom{\lfloor \frac{n}{a+1}\rfloor}{a+1}\right)$ for all $n&gt;0$. Since $a$ is constant, the given fraction grows with order $n^{a+1}$, which means $p(n)&gt;n^a$ for $n$ sufficiently large. (Why does it grow with order $n^{a+1}$?)   Thanks all again in advance! Is it possible to see own 'review queue audits' results? I completed recently a few review audits, some of them I passed, some I failed.  Is there a place I could (easily) find these results? Keyboard shortcut to autocomplete the user in comments Undefined control sequence error on \fancy@reset, \f@ncyolh and \f@ncyorh when using newlfm class I updated to TeXLive 2017 and couldn't get newlfm.cls to work without errors.  Here is a minimal example:  \documentclass{newlfm} \begin{document} \begin{newlfm} \end{newlfm} \end{document}   LaTex is throwing undefined control sequence errors on \fancy@reset, \f@ncyolh, and \f@ncyorh.  What causes the errors?  What am I missing? What is the $\sqrt{i^4}$?   $i^4$ = $(i^2)^2$ So is $\sqrt{i^4}$ = $\sqrt{(i^2)^2}$ = $i^2$ = $-1$?  Or is $\sqrt{i^4}$ = $\sqrt{1}$ = $1$?  When I plug it into my TI-89 Titanium, I get $1$.  Edit: I now see why $i^4 \neq -1$. $$\sqrt{x^2} = \left|{x}\right| $$ So$$\sqrt{(i^2)^2} = \left|i^2\right| = \left|-1\right| = 1$$ 3 Statements of axiom of choice are equivalent This winter I started to study much harder the set theory and especially the axiom of choice. Unfortunately, I have problems with solving the next exercise:  Prove that the 3 statements of the axiom of choice are equivalent :  1) For any non-empty collection $X$ of pairwise disjoint non-empty sets, there exists a choice set.  2) For any non-empty collection $X$ there is a choice function.  3) For any non-empty set $X$, there exists a function $f:P(X)\setminus\{\varnothing\}\to X$ so that for any non-empty set $A\subseteq X$, $f(A) \in A$. Why xargs can't work with tail -f? I want tail -f /var/log/syslog | grep it with pattern "arpwatch" and send every line to myself via jabber: xmpp username@jabber.server using xargs  tail -f /var/log/syslog | grep arpwatch | xargs sendxmpp username@jabber.server   no working.  but tail  /var/log/syslog | grep arpwatch | sendxmpp username@jabber.server   works well.  I think it's some fundamental thing about xargs and tail -f that I don't understand. So: I have a Gantt chart created with pgfgantt where in order to fit my pages I have included  x unit=1.4cm,y unit chart=6mm   in its preamble.  This means that milestone shapes are spread out horizontally.  How can I change the vertical and horizontal units of a Gantt chart in a way which doesn't affect the aspect ratios of elements like milestones? C++ guarantees that variables in a compilation unit (.cpp file) are initialised in order of declaration. For number of compilation units this rule works for each one separately (I mean static variables outside of classes).  But, the order of initialization of variables, is undefined across different compilation units.  Where can I see some explanations about this order for gcc and MSVC (I know that relying on that is a very bad idea - it is just to understand the problems that we may have with legacy code when moving to new GCC major and different OS)? "Recent Badges" list is displayed without margins on "Home" page When the &quot;Hide Hot Network Questions&quot; preference is enabled (i.e. HNQ sidebar is hidden), we see &quot;Recent Tags&quot; and &quot;Recent Badges&quot; in the homepage instead of the HNQ. However, the &quot;Recent Badges&quot; list is displayed with no margins. ex.)   I think that the appropriate margins were set in the past. cf.) Asking the position of a person in a sequence How to ask the position(?) of a president (like 10th) of a country, correctly? How do I upgrade to a newer version of Ubuntu? What are the different ways I can use to upgrade Ubuntu from one release to another? What is the purpose of a serial number? I would like to be able to render several files one after the other, when I am not using my computer, e.g. at night. Is there a way of doing this?