Which CMS written in Ruby or Python do you recommend me to choose for my blog? The requirements:   being able to upload the images without having them cut their size comments (not really necessary as I can use discuss) themes free
This is a general, community wiki  question to address non-specific "I need a CMS or Wiki that does x, y, and z..." questions.  If your question was closed as a duplicate of this question and you feel that the information provided here does not provide a sufficient answer, please open a discussion on .    I have a list of features that I want for my website's Content Management System (CMS) - where can I find a [free] script that includes all of them?
I noticed that I have three warnings when I submitted my Sitemap. They all say this:      When we tested a sample of URLs from your Sitemap, we found that some   URLs redirect to other locations. We recommend that your Sitemap   contain URLs that point to the final destination (the redirect target)   instead of redirecting to another URL.   I'd like to try and resolve this issues but I'm not sure what to do.  Sitemap url for reference: This is a weird phenomenon.  Consider the following code:  \documentclass{article} \begin{document}   $$Z^* \frac{Z^*}{Z^*}$$ \end{document}   It produces the following picture:    The asterisks appear in various places in comparison to the Z. This looks particularly bad in the denominator. Any idea what's behind this? Can we interchange the sentences " What's the date today?" and "what day is it today?"  In grammar is it correct to use these sentences interchangeably?  For example "What's the day today?" and "what date is it today?" How to make existing public files private? Maximum length of a USB Cable? Video produced from images only plays in VLC but no other players Permission Subscribe to Dashboards: Add Recipients depends on permission(s): Subscribe to Dashboards Is there a CSS selector for elements containing certain text? It seems that a number of the statistical packages that I use wrap these two concepts together. However, I'm wondering if there are different assumptions or data 'formalities' that must be true to use one over the other. A real example would be incredibly useful. Using old version of numpy even with a newer version installed Proving $\sqrt2$ is irrational I used the method of contradiction by assuming that $\sqrt 2$ is a rational number. Then, by the definition of rational number, there exist two integers $p$ and $q$ whose ratio equals $\sqrt 2$. Thus, $$\frac pq = \sqrt2\tag{x}$$  Squaring both sides, $$p^2/q^2 = 2\tag{a}$$  or $$p^2 = 2q^2\tag{b}$$ This means that $p^2$ is an even number, implying that $p$ is even.  Now, any even integer can be written as $2^kf$ where $f$ is any odd integer and $k$ is some positive integer (the minimum value of $k$ and $f$ is $1$ since even numbers start from $2 = 2^1\cdot1$). For odd numbers, $k=0$.  For example, $8 = 2^3 \cdot 1$, $4= 2^2 \cdot 1$, $18= 2^1 \cdot 9$, $24 = 2^3 \cdot 3$, $-12= 2^2 \cdot(-3)$, etc.  Now, from $(b)$, $q^2$ can be even or odd.  Case 1: $q^2$ is even (thus meaning $q$ is even). Then $p= 2^{k_1} \cdot f_1$ (say) and $q = 2^{k_2} * f_2$ (say). Note here in this case, both conditions $k_1=k_2$ and $f_1=f_2$ can't hold simultaneously since that would mean $p/q =1$ and here $p/q =\sqrt2$ (from $(x)$)). Let's consider the condition when $k_1=k_2=k$ but $f_1\ne f_2$. Then $$\frac{p^2}{q^2}=\frac{(2^kf_1)^2}{(2^kf_2)^2} = \frac{f_1^2}{f_2^2} =2$$ (from $(a)$), i.e.: odd/odd ($f_1$ and $f_2$ are odd) can never equal $2$.  Now lets consider $f_1=f_2=f$ but $k_1\ne k_2$, thus $(2^{k_1}f)^2/(2^{k_2}f)^2= 2^{2(k_1-k_2)}= 2$ meaning $k_1 - k_2 = 0.5$ but $k_1,k_2$ are integers, so their difference can't be $0.5$. (Also here, $k_1-k_2$ must be greater or equal to $1$ since $2^{k_1-k_2}=\sqrt 2$, $k_1-k_2$ can't be negative since $\sqrt 2&gt;1$, but $k_1-k_2\ge1$ can't satisfy $(a)$ since the minimum value of $p/q$ in this case will be $2$ which is surely greater than $\sqrt2$.)  Now lets consider $f_1\ne f_2$ and $k_1\ne k_2$, thus $\frac{(2^{k_1} f_1)^2}{(2^{k_2} f_2)^2}= 2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal $2$ since $\frac{f_1^2}{f_2^2}$ is either odd or "odd/odd" ie: it doesn't contain 2 as a factor and $2^{2(k_1-k_2)}$ is a power of $2$. So in $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$, there's no chance of cancellation of the odd factor $\frac{f_1^2}{f_2^2}$.  Case 2: $q^2$ is odd (thus meaning $q$ is odd). $q=f'$ (say, here $k=0$ for $q$ but not for $p$), thus from $(a)$ $2^{2k}\frac{f^2}{f'^2} = 2$ but this isn't possible since a power of $2$ multiplied with odd factor can't equal $2$.  Thus, both case 1 and 2 suggest that for any possible combination of $k_1,k_2,f_1$ and $f_2$, $p/q\ne\sqrt2$, i.e.: for no value of integers $p$ and $q$, $p/q = \sqrt2$. Thus, this contradicts our assumption that $\sqrt2$ is rational. Therefore it must be irrational.  PLS NOTE that i would like to clarify that $f_1^2$/$f_2^2$ is either an odd integer or a fraction of odd/odd or 1/odd form, thus not containing  2 as a factor,for any $2^n$, if it has to be reduced to 2 , must be multiplied with $1/ 2^(n-1)$ but $f_1^2$/$f_2^2$ can't be of $1/2^[n-1]$ form since for odd nos, in $2^k$*f notation, k is 0 , so 2 vanishes as $2^k$ becomes 1 in this case thus f1/f2 can't be represented as  1/ $2^(n-1)$ in which n has to be existent.  therefore $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal 2 as to further explain the 3rd condition of case 1  Difficulty: is my approach correct? This proof which I thought is different from proofs found on the internet or books since I have used $p$ and $q$ as any integers, which may have a common factor. So I am not sure if I am on right track. Will someone please check this out and make kind comments? Assigning CRS to Lat/Lon in R giving error: "Geographical CRS given to non-conformant data" Why do some statistics symbols have a "squared", e.g. Variance $\sigma^2$, "R squared" $R^2$ or heritability $H^2$ Why does a bigger sensor lead to a shallower depth of field? I no longer use the service that I originally used to register, and would like to change the service I use to log in, or at least add a second account.  Is this possible? How do I change my login provider or associate a backup method to log in?     For more information, see "" in the . How to Start Web Mapping? I want to start working on a web map at work to map some of our statistics by city.    I have an idea, but I'm not exactly sure where to start.   Are there any definitive resources that provide step by step guidance on how to do this? Alternatives to Dmitri Bertsekas' Introduction to Probability and Sheldon Ross' A First Course on Probability I have Ross' A First Course on Probability and Bertseka's Introduction to Probability book. However these two books do not exactly give me what I am looking for. The problem is  Bertsekas' book is introductory, while Ross' book explains expression $E[\ E[X\mid Y] \ ]$ just by briefly saying it is a function of $y$ if I remember correctly. I need a probability book which will make me feel comfortable with the symbols and letters rather than ready made formulas. I want to be able grasp when a random variable is a constant or when it is not. I want  to be able to read and understand equations involving both random variables and constants and when are interpreted as a rv. when they are not etc. Why do we need to use `int main` and not `void main` in C++? Why do we need to use int main and not void main in C++? How to get the EXIF data from a file using C# I would like to write a small program in C# which goes through my jpeg photos and, for example, sorts them into dated folders (using MY dating conventions, dammit...).   Does anyone know a relatively easy way to get at the EXIF data such as Date And Time or Exposure programatically? Thanks! I'm trying to prove that $(ma, mb) = $|$m$|$(a, b)$ , where $(ma, mb)$ is the greatest common divisor between $ma$ and $mb$. My thoughts: If $(ma, mb) = d$ , then $d$|$ma$ and $d$|$mb$ → $d$|$max + mby$ → $d$|$m(ax+by)$. This implies that $d$|$m$ or $d$|$(ax+by)$. This is the same as  $d$|$m$ or $d$|$a$ and $d$|$b$, so $d$|$m$ or $d$|$(a,b)$. This is the same as $d$|$m|$ or $d|(a,b)$, so $d$|$|m|(a,b)$. I don't know what to do.