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How to disable captive portal detection? How to remove exclamation mark on Wi-Fi or signal icon in Lollipop? When connected to the Internet, the system checks if it's in a walled garden, which is simply trying to connect to Google. If it fails, in Android 5, there'll be an exclamation mark on the WiFi icon or signal icon.   In some situations, this is annoying. Say, in China, Google is blocked, the exclamation mark will always be there, which is meaningless.  How can I disable this check, and make the system assume, connected WiFi/mobile data is connected Internet? Applying for a Canadian ETA after previous visits without one Let $R \simeq S$ be isomorphic commutative rings with unity. Prove the following:  a). If $R$ is an integral domain then $S$ is an integral domain.  For this, I said that If I let $f: R \to S$ be an isomorphism and $(R, + , \circ), (S, \oplus, \ast)$, then let $x, y\in R$,  such that $x \circ y = 0_R$, because $f$ is an isomorphism, $f(x \circ y) = f(x) \ast f(y) = f(0_R)$. If I let $R$ be an integral domain then, as per the definition $R$ is commutative, has unity, and no zero divisors.   So $f(x \circ y) = f(x) \ast f(y) = f(0_R)$ where we have either $f(x \circ y) = f(0_R) \ast f(y) = f(0_R)$ or $f(x \circ y) = f(x) \ast f(0_R) = f(0_R)$.   Since $f$ is an isomomorphism, let $s = f(x), t = f(y)| s,t \in S$. Then $s \ast t = f(0_R)$. By definition, $S$ also has unity so $0_S$ exists. If we map $0_R \to 0_S$. Then either $s = 0_S$ or $t = 0_S$.   So then, since $S$ has unity, is commutative, and has a no zero divisors as we showed above, it is also an Integral domain.  I'm not sure I have this correctly.  b). If $R$ is a field then $S$ is a field. I don't really understand how to do this one. Since $S$ is an integral domain, it has a $0_S$. So I need to show that $S$ distributes $\ast$ over $\oplus$.  Any help with these is appreciated How to find and replace using sed on records starting with Specific number What do the supercharges in extended supersymmetry do? What do the supercharges in  do?  In ${\cal N}=1$ supersymmetry there are a certain number of fermions and and equal number of bosons. You can transform all fermions to the bosons (and vice versa) in a 1 to 1 fashion using a single supercharge, $Q$.  So what happens when you have, for example, ${\cal N}=2$ supersymmetry with 8 supercharges? Since $Q$ is a generator of supersymmetry transformations, is it a linear combination of these supercharges that act on the particles? In which case could one particle be acted on by two separate linear combinations of $Q$? Or is it strictly one linear combination of $Q$ per fermion/boson?  Also, what does ${\cal N}$ mean physically? What difference does ${\cal N}=2$ have to ${\cal N}=1$ other than more supercharges? Or is that the only difference between the two theories? What does [param: NotNull] mean in C#? It seems that I'm missing something about this.  First of all, the series is convergent: $\lim_{n\rightarrow\infty}\frac{2^{-n-1} (n+1)^2}{2^{-n} n^2}=\frac{1}{2}$ (ratio test)  What I tried to do is to find a limit of a partial sum $\lim_{n\rightarrow\infty}S_n$ as follows: $S_n=\frac{\frac{1}{6} n (n+1) (2 n+1)}{\frac{1-\left(\frac{1}{2}\right)^n}{2 \left(1-\frac{1}{2}\right)}}$. Still, the limit is $\infty$ and I'm clearly doing something wrong. Summarizing flatness of county with statistic Elevation variance per square mile I need to classify every county in the US as either hilly, moderately hilly, or flat. Accuracy is not critical, just a rough classification. I used R and followed the advice in  answer to import elevation and counties and calculated the zonal variance. This was inherently biased towards larger counties since they have more surface for hills, so I need to divide by area. If I divide variance by square miles does that reasonably represent hilliness? The preliminary results seem decent, with the highest values being places like Colorado, Hawaii, Montana, etc. Disk space usage doesn't add up with df & du It has been said that everyone lives by selling something.  Can I substitute the bold part with &quot;it is said&quot;? I think since it is kinda like a saying, using present simple is more appropriate. But I also feel &quot;it is said&quot; can also mean this sentence was said before, and the impact on the present is that now people generally accept that. And present perfect I think carry the implication that people now may or may not say this anymore. Modular exponentiation by hand ($a^b\bmod c$) How do I efficiently compute $a^b\bmod c$:  When $b$ is huge, for instance $5^{844325}\bmod 21$? When $b$ is less than $c$ but it would still be a lot of work to multiply $a$ by itself $b$ times, for instance $5^{69}\bmod 101$? When $(a,c)\ne1$, for instance $6^{103}\bmod 14$?  Are there any other tricks for evaluating exponents in modular arithmetic?  This is being asked in an effort to cut down on duplicates, see  and . Convergence of Ratio Test implies Convergence of the Root Test It is currently said that MD5 is partially unsafe. Taking this into consideration, I'd like to know which mechanism to use for password protection.  This question,   suggests that hashing multiple times may be a good idea, whereas  suggests using salt.  I'm using PHP. I want a safe and fast password encryption system. Hashing a password a million times may be safer, but also slower. How to achieve a good balance between speed and safety? Also, I'd prefer the result to have a constant number of characters.   The hashing mechanism must be available in PHP It must be safe It can use salt (in this case, are all salts equally good? Is there any way to generate good salts?)   Also, should I store two fields in the database (one using MD5 and another one using SHA, for example)? Would it make it safer or unsafer?  In case I wasn't clear enough, I want to know which hashing function(s) to use and how to pick a good salt in order to have a safe and fast password protection mechanism.  Related questions that don't quite cover my question: (exercise from Tao's analysis book) Proof of a lemma relating to power set of X I'm stuck at one exercise from chapter of sets from Terence Tao's analysis book. I need to proof the lemma:  Lemma: Let $X$ be a set. Then the set $\{Y : Y \:\text{is a subset of}\: X\}$ is a set.  Note: The set $\{Y : Y \:\text{is a subset of}\: X\}$ is known as the power set of $X$, defined as  $2^X $  I can understand why the lemma is true, but I have no clue how to prove it. Furthermore, the author gave a hint that I found very confusing, it follows:  Hint: start with the set $\{0, 1\}^X$ and apply the replacement axiom, replacing each function $f$ with the object $f^{-1}(\{1\})$. Evaluate the following integral:  $$\int \frac{\sqrt{\sin ^4x+\cos ^4x}}{\sin ^3x \cos x }dx$$ where $x \in \big(0,\frac{\pi}{2} \big)$  Could some give me hint as how to approach this question?  I tried to use the fact that $\sin ^4x+\cos ^4x=1-\frac{sin^22x}{2}$ but it didn't help. How should I proceed? How to get a file's Media Type (MIME type)? Can Iranian-Canadians enter on arrival to the US? Dimensions of symmetric and skew-symmetric matrices I am using latest version of VirtualBox (VirtualBox 4.3.6 for Windows hosts) and after installing any OS and clicking inside of it mouse disappear. It is not visible until I press right CRTL. Anyone know what can cause this problem. I have also installed extension pack. Proof of an inequality by induction: $(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n$ Let $n \in \mathbb N^+$. Show that if $x_1, x_2, ... , x_n$ are $n$ real numbers such that $-1 \le x_i \le 0$ for each $1 \le i \le n$, then $$(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n$$  I am using a proof by induction, but I am unable to complete it. I would appreciate any tips on how I could finish the proof.   Incomplete Proof Let $P(n)$ be the proposition that $(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + \sum_{i = 1} ^ n x_i$. When $n = 1$, LHS = RHS = $1 + x_i$. Clearly $1 + x_i \ge 1 + x_i$ and so $P(1)$ is true. Assume that $P(k)$ is true, i.e. $(1 + x_1)(1 + x_2)...(1 + x_k) \ge 1 + \sum_{i=1}^k x_i$. Case 1 Suppose $x_{k + 1}  = -1$. Then $1 + x_{k + 1} = 0$, hence LHS = $(1 + x_1)(1 + x_2)...(1 + x_k)(1 + x_{k+1}) = 0$. RHS = $1 + x_1 + x_2 + ... + x_k - 1 = x_1 + x_2 + ... + x_k$ which is always nonpositive since each $x_i$ is nonpositive. Hence RHS $\le 0$ and so LHS $\ge$ RHS. Case 2 Suppose $x_{k+1} = 0$. Then $1 + x_{k+1} = 1$, hence multiplying it to LHS does not change anything. Also, adding $x_{k+1}$ to RHS does not change anything. Therefore, LHS $\ge$ RHS still holds. Case 3 Suppose $-1 &lt; x_{k+1} &lt; 0$. Then $0 &lt; 1 + x_{k+1} &lt; 1$. Adding $x_{k+1}$ to RHS makes the RHS smaller since $x_{k+1}$ is negative. If LHS $= 0$, then multiplying $1 + x_{k+1}$ to LHS does not change anything, and so LHS $\ge$ RHS holds. If LHS > $0$, then multiplying $1 + x_{k+1}$ to LHS makes the LHS smaller since $0 &lt; 1 + x_{k+1} &lt; 1$. In this case LHS $\ge$ RHS does not necessarily hold? ...