Sending JSF textbox value to a managed bean function as a parameter
How to send form input values and invoke a method in JSF bean
Is there an alternative to Cantor's cardinalities that makes proper subsets smaller than their sets? Cantor defined an infinite set as a set whose subset can be placed in a one-to-one correspondence with its subset. That is, take the set of all natural numbers: {0, 1, 2, 3, 4,...}. From that set, you can form a subset of all even numbers: {0, 2, 4, 6,...}. There is a one-to-one correspondence between both these sets. After all, you can multiply each member in the set of all natural numbers by 2, and you’ll get the set of all even numbers. Because of this one-to-one correspondence, the two sets have the same cardinality, despite the set of even numbers being a subset of the set of natural numbers. This is counter-intuitive to me. How can a subset of a set be as big as the set itself? To make the intuition more concrete, consider the set of multiples of 1,000,000: {0, 1,000,000, 2,000,000,...}. There is a one-to-one correspondence between this set and the set of all natural numbers. So they have the same cardinality. This is really counter-intuitive.  From as far as I can tell, there are basically two axioms at play here:  A1: If all members of set A can be placed in a one-to-one correspondence with all the members of set B, then both sets have the same cardinality.  A2: If set A is a proper subset of set B, then A has a smaller cardinality than set B.  How Cantor defined infinite sets is that he accepted A1 and rejected A2. That is counter-intuitive to me. But mathematicians might say that my intuition is just wrong. So my question is whether we can define infinite sets in another way, i.e. accept A2 but reject A1. That is, the set of even numbers is smaller than the set of all natural numbers, even though both sets have a one-to-one correspondence. If you think this is counter-intuitive because sets with one-to-one correspondence with each other should have the same cardinality, and thus think this is wrong, why can’t I say the same when Cantor accepted A1 and rejected A2? Either way, there is something counter-intuitive and paradoxical.   So, my questions are:   Can we accept A2 and reject A1 for the definition of an infinite set? What arguments, if any, are given for accepting A1 and rejecting A2? Does Java have buffer overflows? Does Java have buffer overflows? If yes can you give me scenarios? How can infinite information be theoretically encoded or stored in a single qubit? I've just gotten started with Nielsen and Chuang's text, and I'm a little stuck. They mention that theoretically, it would be possible to store an infinite amount of information in the state of a single qubit. I'm not sure I completely comprehend this.    Here's how I rationalized it: You take all the information you want to store, put it in binary form, and make it the real component of $\alpha $ or $\beta$ (the coefficients of the computational basis states).   Now I'm not sure if I've understood it right, but since it's still fuzzy in my head it would be great to get some kind of ELI5 explanation or possibly a more detailed picture of how this would, even theoretically, be possible.    Apologies if the question doesn't meet standards. I'm new to the forum and would be open to feedback regarding asking questions or answering them. Proving that $\sum\limits_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$ I have to prove that:     $$\sum_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$$   I tried to open up the right side with Pascal's definition that: $$ { n \choose k} = {n-1 \choose {k}} + {n-1 \choose {k-1}}$$  Here is what I came up with, and I am sure it is wrong because it does not equal the left side:  $$ {m+n+1 \choose m+1} = {m+n \choose m+1} + {m+n \choose m} = ... ={m+n \choose m+1} + {m+n-1 \choose m} + {m+n-2 \choose m-1} + ... + {m \choose m+1} = \sum_{k=0}^{n} {m+k \choose m+k-n+1 }  $$  Which, again, probably is wrong because it is not equal $\sum_{k=0}^{n} { m+k \choose m}$. Any help is appreciated How to use host names rather than ip addresses on home network? I've a small home network up and running and managed to set up ssh. Can anyone tell me what I need to do to refer to one computer on the network from another by host name rather than IP address (which is what I'm doing now)? How can I draw a graph like the diagram below? I'm trying, but something is wrong with my code in LaTeX: \begin{asy} pair dynkinMark(real x, real y, string s = &quot;&quot;, pen p = defaultpen, pair B = dir(45)) { if (s != &quot;&quot;) { dot(&quot;$&quot;+s+&quot;$&quot;, (x,y), B, p); }     else { dot((x,y), p); }     return (x,y); }  dynkinMark(0,1, &quot;1&quot;, dir(135)); dynkinMark(0,2, &quot;2&quot;, dir(135)); dynkinMark(0,3, &quot;3&quot;, dir(135)); dynkinMark(0,4, &quot;4&quot;, dir(135)); dynkinMark(7,1, &quot;1'&quot;); dynkinMark(7,2, &quot;2'&quot;); dynkinMark(7,3, &quot;3'&quot;); dynkinMark(7,4, &quot;4'&quot;); path p1 = (0,1)..(2,2.0)..(5,3.5)..(7,4); path p2 = (0,2)..(2,1.2)..(5,2.5)..(7,3); path p3 = (0,3)..(2,2.8)..(5,1.5)..(7,1); path p4 = (0,4)..(2,3.8)..(5,3.5)..(7,2); draw(p1, blue); draw(p2, heavycyan); draw(p3, mediumcyan); draw(p4, lightblue); dot(&quot;$12$&quot;, IP(p1, p2), dir(90), heavygreen); dot(&quot;$13$&quot;, IP(p1, p3), dir(90), heavygreen); dot(&quot;$23$&quot;, IP(p2, p3), dir(90), heavygreen); dot(&quot;$14$&quot;, IP(p1, p4), dir(90), heavygreen); dot(&quot;$24$&quot;, IP(p2, p4), dir(90), heavygreen); real y = 0.8; label(&quot;$s_1$&quot;, (IP(p1,p2).x, y), heavygreen); label(&quot;$s_2$&quot;, (IP(p1,p3).x, y), heavygreen); label(&quot;$s_1$&quot;, (IP(p2,p3).x, y), heavygreen); label(&quot;$s_3$&quot;, (IP(p1,p4).x, y), heavygreen); label(&quot;$s_2$&quot;, (IP(p2,p4).x, y), heavygreen); \end{asy} No \xrightrightarrow[a]{b}? I know only the very basics of TikZ, having only used it to produce a pretty flow chart for one of my documents. I recently learned that you can use Inkscape and export to TikZ.  What are the limitations of the Inkscape->TikZ workflow? Is there some compelling reason to take the time to learn TikZ (which I'm told has a fairly steep learning curve) rather than simply exporting to TikZ from Inkscape? I had a look at Knuth's The Art of Computer Programming, book 1. In chapter 1, section 1.2.2, exercise 25, he presents the following algorithm for calculating logarithm:  given $x\in[1,2)$, do the following:  L1: [Initialize] Set $y\leftarrow0$, $z\leftarrow x/2$, $k\leftarrow1$.  L2: [Test for end] If $x=1$, stop.  L3: [Compare] If $x-z&lt;1$, go to L5.  L4: [Reduce values] Set $x\leftarrow x-z$, $z\leftarrow x/2^k$, $y\leftarrow y+\log_b(2^k/(2^k-1))$, and go to L2.  L5: [Shift] Set $z\leftarrow z/2$, $k\leftarrow k+1$, and go to L2.  (The algorithm needs an auxiliary table which stores $\log_b2$, $\log_b(4/3)$, and so on, to as many values as the precision of the computer.  Then Knuth concludes: this exercise is to explain why the above algorithm will terminate and why it computes an approximation of $y=\log_b x$.  Well, I see that it works, but I am not able to explain why... Is the 9-speed Sora crankset compatible with 105 11-speed chain? I have heard that there are some Pokémon that are region specific.  Could someone please clarify which Pokémon are region specific and in which regions that they can be found in? Lighter in CPU/Memory Usage: Lubuntu or Xubuntu I am looking for an Ubuntu version that consumes less Memory and CPU. I have read both Lubuntu and Xubuntu (The homepages, wikipedia, phoronix and other sites comparing both). But from experience, which one uses less memory and is less CPU intensive. I need to install them in very old hardware and want to persuade the owner of the hardware of the benefits of Ubuntu.  in this case I want to install 11.10 or 12.04 when it comes out. How are each behaving in those versions?  The 2 PCs I will be installing either Xubuntu or Lubuntu are:  Granpa PC:  CPU - Pentium 2 450Mhz RAM - 64MB DIMM Video - 16MB Used for - Documents and Internet. No listening to music, no looking at videos. Just using it for document writing.  The other old meat:  CPU - Pentium 3 550Mhz RAM - 128MB DIMM Video - 16MB Used for - Documents and Internet also but they want.. or maybe they are wishing for it to use it to see movies and listen to music. This one has internet. The other one does not.  UPDATE  Wanted to update this question with the test results for over 8 months worth of using Lubuntu and Xubuntu. When it comes to very old hardware, am talking about 64 MB, 128 MB or even 256 MB of RAM computers, Lubuntu is better than Xubuntu when comparing Memory usage. When it comes to CPU usage, both are the same.  This will not improve I/O Disk performance which will be the bottleneck for old computers. Also the difference in performance between Xubuntu and Lubuntu is about 10% to 20% percent in memory usage. Again, CPU usage is the same.  Now, when it comes to how responsive the Desktop is, LXDE (Lubuntu) has a much better response time than XFCE (Xubuntu). I would say about 50% or just simply noticeable to the eye. Lubuntu is similar to how Windows XP looks while Xubuntu is more like how Gnome 2 looked.  The computers (The old ones mentioned here) and 2 more are still working in excellent conditions thanks to the Ubuntu family. I also want to add I am using 13.10 for both distros. Suppose $n$ divides $3^n + 4^n$. Show that $7$ divides $n$. Suppose $n \geq 2$ and $n$ is a divisor of $3^n + 4^n$. Prove that $7$ is a divisor of $n$.  My work so far:  I had a hypothesis that if $n| 3^n + 4^n$, then $n = 7^k$ for some $k\in\mathbb{N}$. But this is not necessarily so. Take $n = 7⋅379$, where $3^7 + 4^7 = 7^2⋅379$. Then, $3^7+4^7$ divides $3^n + 4^n$, and since $n$ divides $3^7+4^7$, we must have $n|3^n+4^n$. Running Synaptic I get the following error message:  E: The package hl1440lpr needs to be reinstalled, but I can't find an archive for it. E: Internal error opening cache (1). Please report.   Upon accepting the message, Synaptic quits. The  defines multiple padding schemes for signature generation/verification ( and ),  and encryption/decryption ( and the less safe ).   Which attacks are possible on signature generation/verification and encryption/decryption if no padding mechanisms are applied, i.e. if raw RSA is used over the plaintext or hash? How is $\dfrac1{(1-x)^5}=\sum_{n\geq0}{n+4\choose4}x^n$ Can someone please explain to me how is $$\dfrac1{(1-x)^5}=\sum_{n\geq0}{n+4\choose4}x^n$$  Thanks! Closed form for $\sum_{k=0}^{m} {\binom {m}{k}} a^{k} (b+ck)^N$ Is there a closed form for the following?   $$\sum_{k=0}^{m}  {\binom {m}{k}} a^{k} (b+ck)^N$$      how about a pretty limit for large $b$.   I have tried using the binomial expansion for the $(b+ck)^N$ to convert the above expression to double sum and then tried to simplify the double sum which was not successful. Any idea if I have find a closed form for it? Why do we use this definition of "algebraic integer"? A number is an &quot;algebraic integer&quot; if it is the root to a monic polynomial with integer coefficients. Artin says (Algebra, p. 411):  The concept of algebraic integer was one of the most important discoveries of number theory. It is not easy to explain quickly why it is the right definition to use, but roughly speaking, we can think of the leading coefficient of the primitive irreducible polynomials $f(x)$ as a &quot;denominator.&quot; If $\alpha$ is the root of an integer polynomial $f(x)=dx^n+a_{n-1}x^{n-1}...$ then $d\alpha$ is an algebraic integer, because it is a root of the monic integer polynomial $x^n + a_{n-1}x^{n-1} + ... + d^{n-1}a_0$. Thus we can &quot;clear the denominator&quot; in any algebraic number by multiplying it with a suitable integer to get an algebraic integer.  When I first learned of algebraic integers, I looked online and saw some hints that maybe they were used to prove the Abel-Ruffini theorem. So I put off questioning their usage for a while; I now think I understand one proof of this theorem (the one at the end of Artin's Algebra) and it has nothing to do with algebraic integers (that I can tell). So basically: why is it important if a number is an algebraic integer? I think I understand what he's saying about the relationship between roots of integer polynomials and algebraic integers, but I fail to see why this is &quot;one of the most important discoveries of number theory.&quot; I'm a little confused about the output I'm seeing in my routing table, particularly the 'metric' column:  $ route -n Kernel IP routing table Destination     Gateway         Genmask         Flags Metric Ref    Use Iface 0.0.0.0         192.168.0.1     0.0.0.0         UG    0      0        0 wlan0 172.16.35.0     0.0.0.0         255.255.255.0   U     0      0        0 vmnet1 192.168.0.0     0.0.0.0         255.255.255.0   U     9      0        0 wlan0 192.168.82.0    0.0.0.0         255.255.255.0   U     0      0        0 vmnet8   According to the man page, metric indicates the 'distance' to the target. I was a little confused about what exactly "target" was. I assumed it was my router (it goes to gateway 0.0.0.0, which then goes to my router at 192.168.0.1). Thus, I expected the metric to be a single hop to my router. However, it's 9! Why is this so number high? How to modify an invalid '/etc/sudoers' file? How do I edit an invalid sudoers file? It throws the below error and it's not allowing me to edit again to fix it.  Here is what happens:  $ sudo visudo &gt;&gt;&gt; /etc/sudoers: syntax error near line 28 &lt;&lt;&lt; sudo: parse error in /etc/sudoers near line 28 sudo: no valid sudoers sources found, quitting