Let $R$ a set with $+$ and $\cdot $ s.t.  1) $(R,+)$ is a group (not necessarily comutative)  2) $\cdot $ is associative, and distributive for $+$, i.e. $a\cdot (b\cdot c)=(a\cdot b)\cdot c$, $a\cdot (b+c)=a\cdot b+a\cdot c$ and $(a+b)\cdot c=a\cdot c+b\cdot c$  3) there is $1\in R$ s.t. $1\cdot x=x\cdot 1=x$.  Show that $R$ is a ring.  I really don't know how to did it. I think that I have to show that $(R,+)$ is commutative, but I didn't success.
I have a simple question answer to which would help me more deeply understand the concept of (non)commutative structures. Let's take for example (our teacher's definition of) a ring:  Let $R\neq \emptyset$ be a set, let $\oplus:A\times A \to A$ and $\bullet :A\times A \to A$ be binary operations. Moreover, let $(R, \oplus)$ be a commutative group, $(R, \bullet)$ be a monoid and following property holds for all $a, b, c\in R$: $$a\bullet(b\oplus c) = (a\bullet b)\oplus(a \bullet c)$$ $$(b\oplus c)\bullet a = (b\bullet a)\oplus(c \bullet a)$$ Then ordered triple $\mathbf R = (R, \oplus, \bullet \mathbf)$ is called a (unitary) ring.  Moreover, we call ring $\mathbf R$ commutative iff $(R, \bullet)$ is a commutative monoid. Commutativity of a ring is always a matter of its multiplicative operation because the additive operation is always assumed to be commutative.  Could anyone explain me the bold part? Why do we even in non-commutative rings (and fields etc.) assume the addition to be always commutative? Is there some serious reason? Would it make any trouble? Or studying of structures with non-commutative addition just doesn't give us anything new so we can take addition as commutative simply because of our comfort?
contractive property A topological property $\mathcal{R}$ is contractive if $(X,\tau)$ has property $\mathcal{R}$ and if $\tau^{\prime} \subset \tau$, then (A, $\tau^{\prime}$) has property $\mathcal{R}$.  A topological space is $T_B$ if each compact subset  is closed.  A topological space is called a US-space provided that each convergent sequence has a unique limit.  A topological space is called a C- C-space provided that  the  compact subsets  are precisely the closed sets.     Is property C-C , US contractive,  hereditary?      Is property $T_B$  contractive? Is there a formula for $(1+i)^n+(1-i)^n$? I'm wondering if there is a formula for the value of $(1+i)^n+(1-i)^n$?  I calculated the first terms starting with $n=1$ to be, in order, $2$, $0$, $-4$, $-8$, $-8$, $0$, $16$, $\dots$  So it seems to be some sequence of positive and negative powers of $2$ with $0$s thrown in. Is there a more explicit formulation of what $(1+i)^n+(1-i)^n$ is, based on $n$?   With the binomial theorem, I get it equal to $$ \sum_{k=0}^n\binom{n}{k}i^{n-k}(1+(-1)^{n-k}). $$ Can this be made nicer? Thanks. What are the technical obstacles that doesn't allow OSX to be installed in a PC? Prove that if prime $p$ divide $a_{2k}-2$, then $p$ divide also $a_{2k+1}-1$. Intuition behind the definition of linear transformation I have studied that given vector spaces $V_1$ and $V_2$, a function $T:V_1 \rightarrow V_2$ is called a linear transformation of $V_1$ into $V_2$, if following two properties are true for all $u, v \in V_1$ and scalar $c$:  $(1)$:  $T(u+v) = T(u) + T(v)$ and $(2):  T(cu) = c T(u) $.  My questions are   $1$: What is the geometrical interpretation of properties $1$ and $2$ which says that $T$ preserves additivity and scalar multiplication. I am not able to see this geometrically. What is the meaning of preserving additivity and scalar multiplication.  $2$: At some place I have studied that a linear transformation will be linear if it sends each line to line and planes to planes and so on. How can we interpret this based on these two properties.  I need help to understand this.   Thank you very much for your time.. Is it possible to get total render time via Python code when render’s done? I installed kali linux OS on my laptop.  The installation went well but when I booted in kali linux the screen was backwards. I signed in as root and the screen flipped back but the mouse is reverse (x &amp; y axis).  When I move my mouse left it moves right when I move it down it moves up. If I want to press a button at the upper right I need to press it at the down left.  The screen is acting backwordes (x &amp; y axis)but I see the graphics straight and the mouse is acting backwordes.  I installed kali linux again with a fresh installation. I have the same problem as before.  I installed kali linux again with a fresh installation. I have the same problem as before.   I searched the web And I don't find information on a problem like this Formatting OGR VRT file to convert CSV to KML? I am attempting to get a nice KML for simple review of POI/geocoding from a CSV. I a using ogr2ogr and a VRT. You can see the VRT file below, with two attempts to get this right.  Basically, the issue is using ogr2ogr -f KML ... with the VRT w/o renaming results in a lot of "SimpleData" tags in the KML.   I'd like to get some CSV fields as the KML Placemark "Name" and "Description", concatenated fields, actually. -dsco NameField=RegionName option in ogr2ogr, but I don't see a way to concatenate fields into name, e.g Street, City, State..  In the .vrt below, the SrcSQL field results in no records. It does open, but with layer info. The Field tag (when not commented out) does result all records with in the new alias, but the "name" is the only field that comes through, it drops all other fields.  Is there a way to concatenate fields using ?  Of course, perhaps ogr is not capable of concatenation ...  &lt;OGRVRTDataSource&gt;     &lt;OGRVRTLayer name="roofing-poi2"&gt;         &lt;SrcDataSource&gt;roofing-poi2.csv&lt;/SrcDataSource&gt;         &lt;GeometryType&gt;wkbPoint&lt;/GeometryType&gt;         &lt;LayerSRS&gt;WGS84&lt;/LayerSRS&gt;         &lt;GeometryField encoding="PointFromColumns" x="Longitude" y="Latitude"/&gt; &lt;!--         &lt;Field src="StreetAddress" name="name" /&gt; --&gt;         &lt;SrcSQL&gt;SELECT StreetAddress as name&lt;/SrcSQL&gt;     &lt;/OGRVRTLayer&gt; &lt;/OGRVRTDataSource&gt; $P[X=x]=0$ when $X$ is continuous variable I know that for continuous variable $P[X=x]=0$.   But i can't visualize that if $P[X=x]=0$, there is infinite number of possible $x$'s. And also why do their probabilities get infinitely small ? Possible Duplicate:                 What is the difference between does have and has? For example, compare she does have a car and she has a car. Convergence in measure and almost everywhere In a finite measure space, let $\{f_{n}\}$ be a sequence of measurable functions. Show that $f_{n}  \rightarrow f$ in measure if and only if every subsequence $\{f_{n_{k}}\}$ contains a subsequence $\{f_{n_{k_{j}}}\}$, that converges almost everywhere to f.   I can prove from converge in measure to converge almost everywhere, but I don't know how to write it down for the other direction. checking apt-get update lock file How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers?    I have this idea:  Let $d=\gcd(m,n)$. Then there exist positive integers $x,y$ such that $mx-ny=d$ (WLOG). We shall find $G=\gcd(a^m+1,a^n+1)$.  If $m,n$ are odd, then $d$ is odd, therefore one and only one of $x,y$ is even. We have:  $$a^{ny}(a^d+1)=a^{mx}+a^{ny}=(a^{mx}-1)+(a^{ny}+1)=(a^{mx}+1)+(a^{ny}-1).$$  If $x$ is even and $y$ is odd, then $a^{m}+1\mid a^{mx}-1$ and $a^{n}+1\mid a^{ny}+1$, therefore $G\mid a^{ny}(a^d+1)$. If $x$ is odd and $y$ is even, then $a^{m}+1\mid a^{mx}+1$ and $a^{n}+1\mid a^{ny}-1$, thus $G\mid a^{ny}(a^d+1)$. However, since $\gcd(a^m+1,a^{ny})=\gcd(a^n+1,a^{ny})=1$, so $\gcd(G,a^{ny})=1$, hence $G\mid a^d+1$. We also have $a^d+1\mid a^{m}+1$ and $a^d+1\mid a^{n}+1$, so $a^d+1\mid G$. Thus $G=a^d+1$.  If $v_2(m)=v_2(n)=v_2(d)=k&gt;1$, then there exist some odd numbers $m_1,n_1,d_1$ such that $m=2^km_1,n=2^kn_1,d=2^kd_1$. We shall have $m_1x-n_1y=d_1$, so one and only one of $x,y$ is even, and we can use the same argument when $m,n$ are odd, so $G=a^d+1$.  However, if $v_2(m) \neq v_2(n)$, I cannot find any solutions for this. I think that $G \in \{1,2\}$, but I cannot prove or disprove it. How can I find $G=\gcd(a^m+1,a^n+1)$ if $v_2(m) \neq v_2(n)$ ? Moreover, is there anything from my arguments that can be improved ? In the decimal system: $...-3, -2, -1, 0, 1, 2, 3...$ In the binary system: $...-11, -10, -1, 0, 1, 10, 11...$ In the unary system: $-111, -11, -1, \text{ ??? }, 1, 11, 111...$   I was only able to find a related question but none with an answer so far: Can Kingsoft Office be installed? Can Kingsoft Office be installed in Ubuntu? Why is negative times negative = positive? Someone recently asked me why a negative $\times$ a negative is positive, and why a negative $\times$ a positive is negative, etc.  I went ahead and gave them a proof by contradiction like so:  Assume $(-x) \cdot (-y) = -xy$  Then divide both sides by $(-x)$ and you get $(-y) = y$  Since we have a contradiction, then our first assumption must be incorrect.  I'm guessing I did something wrong here. Since the conclusion of $(-x) \cdot (-y) = (xy)$ is hard to derive from what I wrote.  Is there a better way to explain this?  Is my proof incorrect?  Also, what would be an intuitive way to explain the negation concept, if there is one? How many Naruto Shippuden dubbed episodes are there? I am watching Naruto and I am currently on episode 164. How many episodes of Shippuden do they have dubbed, and are they still dubbing it? Does a (half) beheaded Ettin take bleeding damage of some sort? Let's say a character cuts off one of an Ettin's heads, but the other head is still intact, and it's as such still alive. Obviously, there's a lot of blood. Does the Ettin take continuous damage from naturally bleeding out? Or does it not work that way? Are there any non-decreasing securities on the stock market? Are there any securities traded on the stock market (i.e. ETPs) that never decrease in value?    As an aside (in case anyone mentions bonds):  I'm aware there's ETFs for AAA-rated government bonds, but even these dip significantly.  For example, the  has dipped by 9% in a 85-day period before (21/08/2016 to 13/11/2016). Alright Ubuntu wizards, I jacked up my system a bit unnecessarily, which is now giving me the chance to learn more about how to fix it...a vicious cycle. Anyway, I have it mostly up and running again to my liking, though I am receiving this (photo attached) prompt when my system boots:  libkmod: ERROR ../libkmod/libkmod-config.c:686 kmod_config_parse: /etc/modprobe.d/blacklist.conf line 56: ignoring bad line starting with 'rt2800usb'   Other users had success with just deleting the bad line from their blacklist.conf file, which I have done. Line 56 no longer exists, but I'm still receiving the same error. So am I missing something? Is there some follow-up command I need to give Ubuntu after saving the conf file?