I want to prove that $\mathbb N\times \mathbb N$ is countable set using cantor first diagonal method:  where every-time we count the elemnts on the digonal with the direction of the arrow ($(1,1)\mapsto1,(2,1)\mapsto 2,(1,2)\mapsto3,(3,1)\mapsto4$ etc). I know that if I had same problem with $\mathbb N \cup{0}\times \mathbb N \cup{0}$ I could define cantor's function by $f(x,y)=\frac {(x+y+1)(x+y)}{2}+y$  which I'm not sure its' bijective. How can I define a function according to my scheme? EDIT: The question is NOT about proving $\mathbb N\times \mathbb N$ is countable but on writing appropriate bijective function for the described diagonal method.
I'm examining a proof I have read that claims to show that the Cartesian product $\mathbb{N} \times \mathbb{N}$ is countable, and as part of this proof, I am looking to show that the given map is surjective (indeed bijective), but I'm afraid that I can't see why this is the case. I wonder whether you might be able to point me in the right direction?  Indeed, the proof begins like this:  "For each $n \in \mathbb{N}$, let $k_n, l_n$ be such that $n = 2^{k_n - 1} \left(2l_n - 1 \right)$; that is, $k_n - 1$ is the power of $2$ in the prime factorisation of $n$, and $2 l_n - 1$ is the (necessarily odd) number $\frac{n}{2^{k_n - 1}}$."  It then states that $n \mapsto \left(k_n , l_n \right)$ is a surjection from $\mathbb{N}$ to $\mathbb{N} \times \mathbb{N}$, and so ends the proof.  I can intuitively see why this should be a bijection, I think, but I'm not sure how to make these feelings rigorous?   I suppose I'd say that the map is surjective since given any $\left(k_n , l_n \right) \in \mathbb{N} \times \mathbb{N}$ we can simply take $n$ indeed to be equal to $2^{k_n - 1} \left(2l_n - 1 \right)$ and note that $k_n - 1 \geq 0$ and thus $2^{k_n - 1}$  is both greater or equal to one so is a natural number (making the obvious inductive argument, noting that multiplication on $\mathbb{N}$ is closed), and similarly that $2 l_n - 1 \geq 2\cdot 1 - 1 = 1$ and is also a natural number, and thus the product of these two, $n$ must also be a natural number. Is it just as simple as this?  I suppose my gut feeling in the proving that the map is injective would be to assume that $2^{k_n - 1} \left(2 l_n - 1 \right) = 2^{k_m - 1} \left(2 l_m - 1 \right)$ and then use the Fundamental Theorem of Arithmetic to conclude that $n = m$. Is this going along the right lines? The 'implicit' definition of the mapping has me a little confused about the approach.    On a related, but separate note, I am indeed aware that if $K$ and $L$ are any countable sets, then so is $K \times L$, so trivially, taking the identity mapping we see trivially that this map is bijective and therefore that $\mathbb{N}$ is certainly countable (!), and thus so is $\mathbb{N} \times \mathbb{N}$. Hence, it's not really the statement that I'm interested in, but rather the exciting excursion into number theory that the above alternative proof provides.
bash aliases equivalent for powershell? By default my Windows PowerShell starts in C:\Users\Santosh, my XAMPP installation is in D:\ so the htdocs folder is located at D:\xampp\htdocs. If I have to edit something in htdocs folder then I have to type full cd D:\xampp\htdocs\ (autocompletion is not so kind) then edit that file.  If this PowerShell were a Bash I would do this in .bash_aliases file:  alias htdocs='cd D:\xampp\htdocs'  Is it possible to maintain Bash aliases like file and alias any command in PowerShell? Let $A$ be a $n\times n$ matrix with entries $a_{ij}=i+j $ . Calculate rank of $A$.  My work : I noticed that A is symmetric . Hence all of its eigen vectors are real .. That is all i have got .   Your help will be highly appreciated .Thank you . How can I run Debian stable but install some packages from testing? Say you're running a server and you don't want to upgrade to Testing (Squeeze) from Stable (Lenny) to just install a required package or two.  What's the best way of installing only certain packages from Testing? Matrix of Linear Transformation by right multiplication I am trying to solve the following problem:   Let $A$ be an $n\times n$ matrix, and let $V$ denote the space of $n$-dimensional row vectors. What is the matrix of the linear operator ‘‘right multiplication by $A$’’ with respect to the standard basis of $V$?  I am not sure where to begin with this problem. I have a bunch of client point of sale (POS) systems that periodically send new sales data to one centralized database, which stores the data into one big database for report generation.  The client POS is based on PHPPOS, and I have implemented a module that uses the standard XML-RPC library to send sales data to the service. The server system is built on CodeIgniter, and uses the XML-RPC and XML-RPCS libraries for the webservice component. Whenever I send a lot of sales data (as little as 50 rows from the sales table, and individual rows from sales_items pertaining to each item within the sale) I get the following error:  Fatal error: Allowed memory size of 134217728 bytes exhausted (tried to allocate 54 bytes)   128M is the default value in php.ini, but I assume that is a huge number to break. In fact, I have even tried setting this value to 1024M, and all it does is take a longer time to error out.  As for steps I've taken, I've tried disabling all processing on the server-side, and have rigged it to return a canned response regardless of the input. However, I believe the problem lies in the actual sending of the data. I've even tried disabling the maximum script execution time for PHP, and it still errors out. Auto revision summary has wrong plural: "added 1 characters in body" As seen at:           Update: I'm 101% sure the down votes were due to the lack of free hands.    Update 2: likes me, like me not, like me, likes me not... yeah ! How to quickly send commands to nested tmux sessions? I have the following in my .tmux.conf  set -g prefix M-j bind-key j  send-prefix   I need to press (Atl+J) + (J) + bound-key to send something to the nested tmux session. I feel it is rather slow. Is there any better way? For example, I would love to be able to do (Alt+J) + (2x bound-key) to do stuff in the nested session. I constantly execute commands in the top tmux session instead of executing them in the nested one. Also, how come everybody binds prefix to C-a? I find it awfully slow and unpleasant to type this combination. Am I missing something? I want to dynamically create a sequence of strings by manipulate an array of elements and create some arithmetic procedure.  for name in FIRST SECOND THIRD FOURTH FIFTH; do     $name = $(( $6 + 1 ))     $name = "${$name}q;d"     echo "${$name}"; printf "\n" done   The desire outcome would be the below for $6 equals 0.  1q;d 2q;d 3q;d 4q;d 5q;d   But I get this error  reel_first_part.sh: line 18: FIRST: command not found reel_first_part.sh: line 19: ${$name}q;d: bad substitution reel_first_part.sh: line 18: FIRST: command not found reel_first_part.sh: line 19: ${$name}q;d: bad substitution reel_first_part.sh: line 18: FIRST: command not found reel_first_part.sh: line 19: ${$name}q;d: bad substitution   I guess it's something simple. It used to work when I did something like   FIRST=$(( $6 + 1 )) FIRST="${FIRST}q;d" When I'm looking down the front page on , I often wish I could tell at a glance what happened to the question to bump it there. Sometimes it is obvious from looking at the number of answers, other times you have to follow a link to find out if it has been edited or if an answer has been added or updated.  I'm guessing this is just as applicable on other SE sites - this is the kind of thing I'm suggesting:    A couple of suggestions from the comments that might be good:   To reduce clutter the word 'ago' could be removed at the same time. The new text could be visible only to established users - ie users who are likely to be reviewing/editing and would benefit from seeing at a glance what kind of change has occurred. Understanding slice notation How should a question which quotes a question be punctuated? Suppose there is a sentence in question form, like the one appearing below:     Where do you belong to?   Now, if a speaker refers to that question, he will frame another sentence and will place the question in the quotation marks in that sentence, like what is appearing below:     They said, "Where do you belong to?"   Now, if the speaker wants to convert this sentence too in question form, he will put the original question in another question, like what is appearing below:     What did they mean by saying "Where do you belong to?"   Now, if the speaker wants to rephrase this sentence, it would become as follows:     By saying "Where do you belong to?" what did they mean?    My question is how the 4th sentence is to be punctuated. It has two question marks in it. Does the sentence correctly take two question marks? If so, why will the 3rd sentence not take two question marks − one, just before the closing inverted commas and, the second, just after those commas? The 4th sentence and the 3rd sentence are all same, except being a bit different in placement of the words. Will they take two different sets of punctuation marks? I'm trying to prove commutativity of addition for vector spaces, using the axioms for vector spaces. Apparently commutativity can be proven! Im having trouble getting a good feel for what is allowed and what is not. Here's my work so far:  $u+v+u+v = 2(u+v) = 2u + 2v = u+u+v+v = u+(u+v)+v$  Here I just wanna claim that $u+(v+u)+v = u+(u+v)+v$   $\Rightarrow -u+u+(v+u)+v+(-v) = -u+u+(u+v)+v+(-v)$ : here im just adding -u to the right, and -v to the left. Question: is this "adding to both sides" really legit in this context? Why?   Quick help proof: $-v+v = (-1)v+(1)v = (-1+1)v = 0v = 0 = v-v$  And another: $ 0+v = v+(-v) + v = (1)v + (-1)v + v = (1-1)v + v = v = v+0$  We have $0 + (u+v) + 0 = 0+(v+u)+0 \Rightarrow u+v = v+u$  This feels ugly and not at all elegant, especially the great leap "add -u to both sides" feels completely out of place. Do I need more lemmas? Is there a more elegant way?   //not homework or anything, just for my own pleasure, feel free to provide theory, as it is more insightful than solutions. :) Thanks!   EDIT: corrected notation a little. What is the benefit of review a scientific paper as a referee? Divisibility by 7 What is the fastest known way for testing divisibility by 7? Of course I can write the decimal expansion of a number and calculate it modulo 7, but that doesn't give a nice pattern to memorize because 3 is a primitive root. I'm looking for alternative ways that can help you decide when a number is divisible by 7 by hand.  I'm sorry if this is a duplicate question, but I didn't find anything similar on the site. I've noticed that in comments, when you create `code region` with backticks, the backslash will still work as an escape character within that region (but it should not, IMHO).  To clarify:   In comments, `\\` will be rendered as \, but it should be rendered as \\ In the post body itself, `\\` will be rendered as \\ (i.e., correctly)   This is especially confusing when you enter regular expressions into a comment, since backslash-escaping is very common there, and confusion increases if you are dealing with a regex that is double-escaped for representation in a string.   At some point you lose track how many backslashes you must make so the comment renders correctly, or you don't notice fast enough (or at all) and the comment contains wrong code where the same markup in a question or answer would come out as perfectly correct.   It's just that comments can't be corrected anymore after the grace period, leaving you with the choice of either deleting the comment, thus crippling the conversation, or engaging in a trial-and-error game of writing another comment with even more backslashes to explain the issue.  My question is two-fold:   Will this error be corrected? If yes, existing comments should probably be searched and "fixed" by a routine that tries to do the right thing (this is probably somewhere between hard and impossible) Or, can I rest assured that never ever the behavior of the backslash in comments is going to be changed, so I can start working around this issue by hand, knowing that they won't start displaying incorrectly at some point in the future? "work for" / "work in" / "work at" I would like to know what usually follows after each proposition in sentences with the word &quot;work&quot;. These are what I understand so far: I work for my boss. ( usually for certain person ) I work in a accounting corporation. (usually for business area ) I work at Samsung. ( usually for specific name of company ) Could you please tell me if I understand rightly? comparing the first column of two files and printing the entire row of the second file if the first columns match I have two files in these formats:   file1 :  air smell hand dude road cat   file 2:   air,4,21,01,13,3,2 smell,21,4,2,5,6 dude,1,31,42,1 road,1,4,2,1,4 cat,1,5,6,3,1 hand,1,4,2,1,6 mouse,1,3,5,6,2   what I want to do is print the entire row of file 2, if the first string in column 1 of file 2 is found in file 1, and I want to keep the order of file 1.   expected output:   air,4,21,01,13,3,2 smell,21,4,2,5,6 hand,1,4,2,1,6 dude,1,31,42,1 road,1,4,2,1,4 cat,1,5,6,3,1 How to Install Octave in El Capitan? newcommand vs. DeclareMathOperator As far as I understand, whatever can be done using DeclareMathOperator can also be achieved using newcommand, but not the other way around. If this is the case, what are the circumstances where DeclareMathOperator is the better choice? When should I take solely newcommand?  I found a question regarding , and I wonder what other aspects should be taken into consideration when it comes to choosing which method to use when defining math operators. Why would a single celled organism evolve to be multi-celled? I read a story this week on Richard Lenski who has been 'evolving' E. coli for more than 50,000 generations now.  One comment I read was from someone who doesn't accept Evolution who pointed out that we haven't seen a single celled organism 'evolve' into a multi-celled organism.  Another person responded and said that a bacteria is not going to evolve into something that isn't a bacteria.  So, if Evolution created single celled organisms and then multi-celled organisms how might that change have happened?  And is it possible to recreate that set of driving forces to make a bacteria something other than a bacteria?  To that end, what advantage does being multi-cellular have over being unicellular (if that's even a word)?