Color edges of $K_6$, there is a cycle of length 4 If we color the edges of $K_6$ (complete graph on $6$ vertices) red or blue, how to prove that there is a cycle of length $4$ with monochromatic edges ?
Color the edges of $K_6$ red or blue. Prove that there is a cycle of length 4 with monochromatic edges. Color the edges of $K_6$ red or blue. Prove that there is a cycle of length 4 with monochromatic edges.  Attempt: I know that i have to...  prove that there must be TWO vertices with “red-degree” at least 3, or two with "blue-degree" at least 3. Call these U and V. (We'll assume we're in the red case; the blue case is similar.) Now, ask if they’re connected by a red edge, and how many red-adjacent neighbors they share. If you analyze this, there should be only a few cases to consider – and each should be relatively straightforward. For each case, you’ll show either that that case can’t actually happen in $K_6$, or that it gives rise to a monochromatic 4-cycle.
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. Get name of caller function in PHP? Is there a PHP function to find out the name of the caller function in a given function? What does the term "Extended Partition" mean? Is it safe to use this type of partition? In the image below, what does Extended 103 GB mean? When I have .txt files or .py files, I can Quick Look those files in Finder. I have .jl files that are codes written in the Julia Language. Of course they are text files. They will open in a text editor like Sublime Text. How can I Quick Look these .jl files in Finder? On every use profile page the search bar on the top of the page already contains user:&lt;id&gt; text to simplify search for post of the currently visible user. But this doesn’t work correctly for the  user:        Instead of searching for posts owned by Community user I’ve got Jeff Atwood posts:        In this case minus sign was just ignored and instead of (-1, i.e. Community user), the (+1, i.e. Jeff Atwood) is used. Same happened for any other negative id for the “user:” option in the search box. Locally bounded additive function is linear Let $T:\mathbb{R}^k\to\mathbb{R}^m$ is additive, i.e. $T(X_1+X_2)=T(X_1)+T(X_2)$. $T$ is locally bounded. Prove that $T$ is a linear transformation.  I know there is a relation between boundedness, convexity and scalar multiplicity. But not sure where to start or how to tie things together. Add a "flag" button in the revision list If $11z^{10}+10iz^9+10iz-11 = 0$. Then possible value of $\mid z \mid,$ is If $11z^{10}+10iz^9+10iz-11 = 0$. Then possible value of $\mid z \mid,$ is  $\bf{My\; Try::}$ Given $11z^{10}+10iz^9+10iz-11 = 0\Rightarrow \displaystyle z^9 = \frac{11-10iz}{11z+10i}.$  Now Put $z = x+iy\;,$ we get $\displaystyle (x+iy)^9 = \frac{11-10i(x+iy)}{11(x+iy)+10i} = \frac{(11+10y)-10ix}{11x+i(10+11y)}$  Now i did not understand how can i solve it,  Help Required  Thanks How do I automate a key press in AppleScript? Is it possible to automate the key press of a key (such as "a" or "&lt;") in Mozilla Firefox, Google Chrome or Safari (in this order)? I'd like to automate the key press of "a" for example in Mozilla Firefox for 100 times every x seconds. With x being a value between 1 second and 10 seconds chosen randomly at every key press. What Applescript could make the trick?  Also, where can I find a list of all codes for each key of the keyboard? A situation when two or more people speak at the same time The word or phrase I am looking for is quite opposite to that of the situation when people, such as students or sports players, sing their national anthem with the same tone and words.  I want a single word or even phrase that can describe the situation when two or more people speak at the same time with different tone and words . . . and at the end, the listener cannot get what they are saying. This situation can be found while debating, telling a news about any incident or even while complaining.     Due to _____ , the teacher could not get what they are saying.   If it's not feasible to find a noun or phrase for such situation, then here I request to have an adverb standing alone to describe how they are speaking, for instance, what is the adverb for "at the same time"?.  Here is how, I want to convert sentence 1 into sentence 2.        They spoke at the same time, and the teacher could not get what they are saying.   They spoke _____ , and the teacher could not get what they are saying. find the number of injections/surjections I've been banging my head on this problem for some time now, and could really use help. Bear in mind I'm not very good at this sort of thing and am struggling to get by in class.  Problem:     Given $ \alpha : A \to B       \;\;\;\;\;   |A|=p \;\;\;\;|B|=q $      Find:         Number of Injections: $A \to B$   Number of Surjections: $ A \to B$    Prove that ∃ a bijection $A \to B$       $\Rightarrow$ p = q   find number of bijections: $A \to B$         I apologize for dropping such a big problem on you all, but I just can't get anywhere with this. Help?  EDIT: These answers are really helpful and I believe I've got down the number of Injections, but  we haven't gone through permutations or combinatorics in class so my understanding of it is really hamstrung. I'm unclear on the Stirling theorem as discussed here: , specifically the part where inclusion-exclusion comes into play. I'm not sure how to apply that, since right now I'm letting p = 4 and q = 3, so I don't know where inclusion-exclusion would be used (if at all!) since I don't know how elements would be missing. Import a python module without the .py extension How do you concatenate characters in java? Concatenating strings would only require a + between the strings, but concatenating chars using + will change the value of the char into ascii and hence giving a numerical output. I want to do System.out.println(char1+char2+char3... and create a String word like this.  I could do   System.out.print(char1); System.out.print(char2); System.out.print(char3);   But, this will only get me the characters in 1 line. I need it as a string. Any help would be appreciated.  Thanks Changing my Apple ID's email: address is not available My Custom Skin won't Change Seeking Python script for creating .mxd files? Regular expression to match DNS hostname or IP Address? Does anyone have a regular expression handy that will match any legal DNS hostname or IP address?  It's easy to write one that works 95% of the time, but I'm hoping to get something that's well tested to exactly match the latest RFC specs for DNS hostnames. How would you implement an LRU cache in Java? Please don't say EHCache or OSCache, etc. Assume for purposes of this question that I want to implement my own using just the SDK (learning by doing). Given that the cache will be used in a multithreaded environment, which datastructures would you use? I've already implemented one using  and , but I'm curious if any of the new concurrent collections would be better candidates.  UPDATE: I was just reading through  when I found this nugget:     If you need constant-time access and want to maintain the insertion order, you can't do better than a LinkedHashMap, a truly wonderful data structure. The only way it could possibly be more wonderful is if there were a concurrent version. But alas.   I was thinking almost exactly the same thing before I went with the LinkedHashMap + Collections#synchronizedMap implementation I mentioned above. Nice to know I hadn't just overlooked something.  Based on the answers so far, it sounds like my best bet for a highly concurrent LRU would be to extend  using some of the same logic that LinkedHashMap uses. How to implement Bitcount using only Bitwise operators? Transformation of Random Variable $Y = X^2$ I'm learning probability, specifically transformations of random variables, and need help to understand the solution to the following exercise:      Consider the continuous random variable $X$ with probability density function $$f(x) = \begin{cases} \frac{1}{3}x^2 \quad  -1 \leq x \leq 2, \\ 0 \quad \quad \text{elsewhere}. \end{cases}$$ Find the cumulative distribution function of the random variable $Y = X^2$.   The author gives the following solution:   For $0 \leq y \leq 1: F_Y(y) = P(Y \leq y) = P(X^2 \leq y) \stackrel{?}{=} P(-\sqrt y \leq X \leq \sqrt y) = \int_{-\sqrt y}^{\sqrt y}\frac{1}{3}x^2\, dx = \frac{2}{9}y\sqrt y.$  For $1 \leq y \leq 4: F_Y(y) = P(Y \leq y) = P(X^2 \leq y) \stackrel{?}{=} P(-1 \leq X \leq \sqrt y) = \int_{-1}^{\sqrt y}\frac{1}{3}x^2\, dx = \frac{1}{9} + \frac{1}{9}y\sqrt y.$  For $y &gt; 4: F_{Y}(y) = 1.$    Previous to this exercise, I've managed to follow the solutions of two similar (obviously simpler) problems for a strictly increasing and strictly decreasing function of $X$, respectively. However in this problem, I don't understand the computations being done, specifically:    How does the three intervals $0 \leq y \leq 1$, $1 \leq y \leq 4$ and $y &gt; 4$ are determined? In the two previous problems I've encountered, we only considered one interval which was identical to the interval where $f(x)$ was non-zero. In the case where $0 \leq y \leq 1$, why does $P(X^2 \leq y) = P(-\sqrt y \leq X \leq \sqrt y)$ and not $P(X \leq \sqrt y)$? I have put question marks above the equalities that I don't understand.   I think I have not understand the theory well enough. I'm looking for an answer that will make me understand the solution to this problem and possibly make the theory clearer.