Our professor gave us some practice problems, which are without solutions. One of them is to prove that the property of identity element of scalar multiplication: $1v=v$, where $1$ denotes the multiplicative identity in $\mathbb{F}$, does not follow from the other axioms of vector space. The only idea that I have is to try to find an example of a vector space that satisfies all the properties of vector space except of this one, but I have trouble finding it. I am also wondering if there is another, more elegant way, to show this.
I have two small questions that I can't solve. So, there are 8 axioms for vector spaces. One of them is that $1 \cdot a = a$ for all $a \in V$ and of course $1$ is multiplicative identity in $\mathbb{F}.$ How can I be sure this axiom cannot be derived from other 7 axioms? Also, why is there no definition of $\alpha a= a\alpha, \forall \alpha \in \mathbb{F}, \forall a \in V$? Why is multiplication of scalars and vectors not defined in commutative way?
SubView slide in animation, iphone What is the meaning of the beta-coefficient for an interaction term in a crossover study? I have asked a similar question here:  I am puzzled by the interpretation for an interaction term. In my data my Y is an interval variable with the health outcome of an experiment. I have used an interaction term in which I have interacted the condition with the predisposition of the subject considering health status at base level. They are both categorical variables (factor variables in R).  Now it gets complicated because the Condition was two treatments: in treatment A subjects got the placebo first and then the real medicine whereas in treatment B they go the real medicine first and the placebo second. All it changes is the order.   Health outcome = a + Condition * Health.Base   I have the worst state of health at the base level as my reference category I find that interaction with the Condition is statistically significant but I am not sure how to interpret this.  I use the lm() function of R (although my design looks more like an ANOVA) and in the output I get the b coefficient in an output that looks like this:  ConditionB:Health.Base.So.and.So         (Beta and p-value) ConditionB:Health.Base.Excellent         (Beta and p-value)   A statistically significant interaction term for those in Excellent health at baseline would mean that they are affected by the Condition B more than Condition A compared to the reference category people (Poor health at baseline). Is this right? What does the beta-coefficient represent?   If I would like to examine for each Health category at the base line separately without comparing to a reference category I would have to code each category as a dummy variable. However, in this case I would compare whether membership to a specific health status at the base line significantly changes between conditions compared to those who belong to the other health statuses at the base line. Is this right? Again, what does the beta-coefficient represent?  Would it be right to assume that the choice of the interaction between the Condition and the dummy variables is easier to interpret?  --- EDIT --- R output:  Call: lm(formula = HealthOutcome ~ Condition * HealthStatus,      data = datA)  Residuals:     Min      1Q  Median      3Q     Max  -1.5957 -0.5942 -0.2640  0.4423  2.4423   Coefficients:                                        Estimate Std. Error t value     Pr(&gt;|t|)     (Intercept)                            2.595652   0.053044  48.934  &lt; 2e-16 *** ConditionCondB                        -0.001449   0.077071  -0.019    0.985     HealthStatusSo.and.So                 -0.331693   0.078094  -4.247  2.35e-05 *** HealthStatusExcellent                 -0.836724   0.092692  -9.027  &lt; 2e-16 *** ConditionCondB:HealthStatusSo.and.So   0.137490   0.110612   1.243    0.214     ConditionCondB:HealthStatusExcellent  -0.199787   0.133943  -1.492    0.136     --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  Residual standard error: 0.8045 on 1059 degrees of freedom   (68 observations deleted due to missingness) Multiple R-squared:   0.16, Adjusted R-squared:  0.156  F-statistic: 40.33 on 5 and 1059 DF,  p-value: &lt; 2.2e-16 Scheduling a Round Robin tournament - 4-way games I'm looking to schedule 16 players to play a round robin tournament with each other such that there are 4 players at each table.  I'd like for each player to play with each other player exactly once over the course of 5 rounds.  I just took a look at this post:  , and I was wondering if anyone knows if the case n=m=4 has been solved, and if so what that solution is. When I want to edit a tag of a question I get following message:     Your edit will be placed in a queue until it is peer reviewed.      We welcome all constructive edits, but please make them substantial. Avoid trivial, tiny one-letter edits unless absolutely necessary.   Now the question is: are tag edits (adding a needed tag to an under-untagged, replacing a mis-used tag) substantial enough?  For example the proc-tag-wiki is just about a Ruby feature called proc - but currenlty some questions about Oracle Pro*C (a embedded SQL preprocessor) or the Linux /proc filesystem are also tagged with proc. Btw, what tag should be used for /proc?  Say I come across a Pro*C Question which is tagged with proc - is a change to proc-sql considered substantial or not? Why isn't  generic?    Seems like Collection&lt;E&gt; could have boolean remove(E o);   Then, when you accidentally try to remove (for example) Set&lt;String&gt; instead of each individual String from a Collection&lt;String&gt;, it would be a compile time error instead of a debugging problem later. For a specific package, can I find out why it is installed?  So, can I see or it's manually installed, or that it is installed as a dependency of another package? Or that it is installed as part of the distro? Can Ubuntu run on a Microsoft Surface with Windows RT? My family might be buying a Windows Surface tablet with Windows RT (Nvidia)? I was wondering if I can run Ubuntu or WUBI on it for my homework? Some time ago, we announced that . This has gone well and we're extremely thankful to everyone for their support and feedback. Because it has been successful so far, we're going to be extending the program to the majority of our graduated sites. Well, we've pretty much already done that, but there are quite a few sites that have ads enabled, yet still display nothing. The thing is, we don't really have an abundance of clients offering products that a  would find relevant, but Amazon has stand mixers, a selection of super fine chinois, deliciously sharp cutlery and enough cast iron to fill the Gulf Of Mexico with . Beginning as early as this week (week of 5 February 2018), our ad team is going to be hand selecting affiliate ads to be displayed on select sites. The use of inventory will of course depend on the availability of relevant ads to display which means:  Some sites might see very few, if any ads. Some topics simply don't come remotely close to orbiting anything worth advertising. Furthermore, while graduated, some sites simply don't have enough activity for ads to make sense.  Since these are hand-selected, it could take a while for ads to appear initially.  These would of course yield to inventory from clients if clients had products that were relevant to any given topic.   Nothing changes when it comes to the . If you're used to occasionally seeing a community promotion ad, you'll just notice that you'll also occasionally see an ad for something that is likely to be of interest to someone interested in the topic that your site covers. In fact, to be clear, this is simply a continuation of the announcement we made in November of 2016. We just wanted folks to know that we're going to be continuing the program where there's good potential to show relevant advertisements, even if we don't have clients lined up with a demand for it. Where can I give feedback about a certain ad on [site]? Open a meta post on your per-site meta and let us know. Please remember, the bulk of advertisements are an effort to get something small back from folks that hit the sites through search engines and are extremely unlikely to ever become contributors. Their soufflé stopped falling, and we'll probably never see them again. Maybe we can get a few bucks once they realized that copper ions help egg whites stiffen properly and buy a bowl or a proper bloom whisk. However, you're the experts on these topics and we're very likely going to listen to you. What sites, specifically? Are any sites exempt? If you're a former Stack Exchange 1.0 site and / or subsequently joined the network under a special charter, nothing changes - don't panic :) For the rest, we're just extending what we're currently doing with ads on graduated sites to also include those where there's an opportunity to make sensible use of affiliate ads. If a site has less than 15k (average) daily visits, or fewer than roughly 50k combined total posts - it might not yet be time to invest in selecting ads to be shown there. But, part of this exercise is also seeing what works best. I don't really have feedback, just some suggestions Ads aren't just an opportunity to sell stuff, they're an opportunity to really show people that we understand a given topic. That, of course, makes suggestions from people that do that stuff every day quite valuable. Post on your per-site meta and our community / ad critters will take a look. Questions? Comments? Kvetches? Let us know in an answer or comment below. If anything about the plan changes, we'll update this post to let folks know. How to make text aligned left/center/right in the same line? I am trying to align some text to left/center/right in the same line. For example, I want to put my phone number on the left, my name in the center, and my email on the right, how do I do that? If $A\simeq\prod A_i$ then $\sum e_i=1$ Given $A$ commutative ring with unity. If we have rings $A_1,\dots, A_n$such that $A\simeq A_1\times \dots \times A_n$ there are $e_1,\dots,e_n$ idempotent elements of $A$ such that $e_1+\dots+e_n = 1$ and $e_i e_j =0$ if $i\neq j$.  I'm not sure how general the proof should be written...  The isomorphic relation suggest that if $f$ defines the isomorphism then there is a $(a_1,\dots, a_n)\in \prod A_i$ such that $f(a_1,\dots,a_n)=1\in A$. Could it work thinking of the idemmpotent terms as $e_i=(0,\dots,0, a_i,0, \dots,0)?$ This should satisfy $e_i e_j =0$ if $i\neq j$. But I have two problems with it (1) I don't see why $e_i$ should idempotent as no assumptions in regard to the existence of idempotent elements in $A_i$ has been made. (2) The sum $\sum e_i=1$ has no reason to be true. What counter balances gravity here? Lets consider a stone tied to one end of a massless and unstretchable string... Now if i'll start rotating, it will form a conical pendulum &amp; then the equilibrium is achived when the pendulum attains horizontal circular motion.... Now to maintain that state we give a constant force to string.... Here, which force counterbalances gravity? Is it the torque due to applied force cancels out the torque due to gravity?  The angle θ changes from 0 to 90° Why isn't this sudo mv operation with wildcard working? UK Visa Refused - can I reapply immediately? How to configure extra buttons in Logitech Mouse In the neural network literature, often we encounter the word "tensor".  Is it different from a vector? And from a matrix? Have you got any specific example that clarifies its definition?  I'm a bit confused about its definition. Wikipedia doesn't help and sometimes I have the impression that its definition depends on the specific machine learning environment used (TensorFlow, Caffee, Theano). Need for using the article "the" before "United States of America"? Usually, we do not use the definite article, the, before names of countries but we use the before "United States of America".   Is there a specific reason for using the article before the country "United States of America"? Proving Pascal's Rule : ${{n} \choose {r}}={{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$ I'm trying to prove that ${n \choose r}$ is equal to ${{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$.  I suppose I could use the counting rules in probability, perhaps combination= ${{n} \choose {r}}=\frac{n!}{r!(n-r!)}$.  I want to see an actual proof behind this equation.  Does anyone have any ideas? How to show $\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$? I am able to evaluate the limit $$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$ for a given $n$ using l'Hôspital's (Bernoulli's) rule.  The problem is I don't quite like the solution, as it depends on such a heavy weaponry. A limit this simple, should easily be evaluable using some clever idea. Here is a list of what I tried:   Substitute $y = x - 1$. This leads nowhere, I think. Find the Taylor polynomial. Makes no sense, it is a polynomial. Divide by major term. Dividing by $x$ got me nowhere. Find the value $f(x)$ at $x = 1$ directly. I cannot as the function is not defined at $x = 1$. Simplify the expression. I do not see how I could. Using l'Hôspital's (Bernoulli's) rule. Works, but I do not quite like it.   If somebody sees a simple way, please do let me know.    Added later: The approach proposed by Sami Ben Romdhane is universal as asmeurer pointed out. Examples of another limits that can be easily solved this way:   $\lim_{x \to 0} \frac{\sqrt[m]{1 + ax} - \sqrt[n]{1 + bx}}{x}$ where $m, n \in \mathbb{N}$ and $a, b \in \mathbb{R}$ are given, or $\lim_{x \to 0} \frac{\arctan(1 + x) - \arctan(1 - x)}{x}$.   It sems that all limits in the form $\lim_{x \to a} \frac{f(x)}{x - a}$ where $a \in \mathbb{R}$, $f(a) = 0$ and for which $\exists f'(a)$, can be evaluated this way, which is as fast as finding $f'$ and calculating $f'(a)$.  This adds a very useful tool into my calculus toolbox: Some limits can be evaluated easily using derivatives if one looks for $f(a) = 0$, without the l'Hôspital's rule. I have not seen this in widespread use; I propose we call this Sami's rule :). Questions with mathjax don't fit in the box This bug is intermittent, only showing up on some answers, but not others.  Sometimes, posts that have mathjax equations expand the math, but don't resize the text box, cutting off some text on the bottom.  Note that this is on the IOS public beta, so that may partially have caused the problem.     App Version: 1.5.3.5 Device: iPhone 5s (GSM) OS Version: Version 10.0 (Build 14A5322e) A ring without the Invariant Basis Number property