My friend is playing a wizard/sorcerer (in a campaign I'm not in) and he's claiming this (irect quote from discord):  you can learn any spell by either finding a scroll or copying it from another wizard you can be a level 2 wiz, 8 sorc and learn level 5 wizards spells  However, from what I'm reading...  You determine what spells you know and can prepare for each class individually, as if you were a single classed member of that class. (PHB 164)  and then also,  When you find a wizard spell of 1st level or higher, you can add it to your spellbook if it is of a level for which you have spell slots... (PHB 114)  It doesn't mention how the spellbook works with multiclassing specifically, but based on these rules wouldn't a 2 wiz / 8 sorc only be able to copy level one spells into their spellbook?
The &quot;Your Spellbook&quot; sidebar for the wizard includes the following line:  Copying a Spell into the Book. When you find a wizard spell of 1st level or higher, you can add it to your spellbook if it is of a level for which you have spell slots and if you can spare the time to decipher and copy it.  Does this mean that a Wizard 1/Cleric 19 can copy any wizard spell they find into their spellbook? Is there anything preventing them from casting these spells?
I  as unclear what you're asking after receiving flags on it in the moderator queue.   Once I did that, the text for the question title in the moderator queue changed from:     How to use handler to display images from database without passing ID in querystring?    to     How to use handler to display images from database without passing ID in querystring? [undefined more votes needed]    The text on the main page has stayed consistent with the new 'on hold' verbiage, however.  Here's a screenshot: Is the sorting algorithm used by .NET's `Array.Sort()` method a stable algorithm? Is the sorting algorithm used by .NET's Array.Sort() method a  algorithm? In The Dark Knight Rises, at the beginning of the film Bane kidnapped a guy from the plane, then he drew blood from the guy he kidnapped. And then throughout the movie I didn't see any news regarding this kidnap scene. But, before the end of the movie we find out that the guy kidnapped by Bane was Dr Pavel. Bane needed him to convert the fusion project into a nuclear bomb and Dr Pavel was the only guy who could have done it.   But I don't understand, why did he draw blood from Dr Pavel? How to move boot and root partitions to another drive The difference between log and ln $$\dfrac{1}{2}\ln(x+7)-(2 \ln x+3 \ln y)$$  Our professor let's us solve this but i do not understand how $\ln$ works. He says it has same properties with $\log$ but i still don't get it. What's the difference of the two? How do I install Adobe Acrobat Reader from the repository? How do I install Adobe Acrobat Reader in Ubuntu 14.04 using a Trusty apt repository?  This is NOT a duplicate of . In Synaptic &gt; Repositories &gt; Other software I enabled "Canonical Partners", but I still cannot install acroread as it is not available on my 64-bit system.  Here's the output from the console:   root@liv-inspiron:/home/liv# apt-get install acroread Reading package lists... Done Building dependency tree        Reading state information... Done E: Unable to locate package acroread root@liv-inspiron:/home/liv# apt-get install acroread:i386 Reading package lists... Done Building dependency tree        Reading state information... Done E: Unable to locate package acroread I'm trying to prove that the converse of Lagrange's theorem is true for finite abelian groups (i.e. "given an abelian group $G$ of order $m$, for all positive divisors $n$ of $m$, $G$ has a subgroup of order $n$"). This is an exercise from a book, and it is in the section on finite abelian groups, so I know I have to use the fundamental theorem of finite abelian groups. I have come up with a proof, but it seems a bit messy, and I'm not entirely sure if it's correct. It is given below.     Let the order of $G$ be $m$ = $p_1^{\alpha_1} \ldots p_k^{\alpha_k}$. It is known that $G$ is a direct product of $p$-groups, say:      $$G = G_1 \times \ldots \times G_k$$      where each $G_i$ is a $p_i$-group. By the fundamental theorem of finite abelian groups, each $G_i$ is isomorphic to a direct product of cyclic groups of the form      $$\mathbb{Z}_{{p_i}^{\beta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\beta_l}},$$      where $\beta_1, \ldots, \beta_l$ are positive integers such that $\sum_{j=1}^l \beta_j = \alpha_i$. Now if $n$ divides $m$, then we must have      $$n = p_1^{\gamma_1} \ldots p_k^{\gamma_k}$$      for some $\gamma_1, \ldots, \gamma_k$ with $0 \leq \gamma_i \leq \alpha_i$.      Claim: Each $G_i$ has a subgroup of order $p_i^{\gamma_i}$      Proof: As above, we have that       $$ G_i \cong \mathbb{Z}_{{p_i}^{\beta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\beta_l}} $$      where $\beta_1, \ldots, \beta_l$ are positive integers such that $\sum_{j=1}^l \beta_j = \alpha_i$.      Now since $0 \leq \gamma_i \leq \alpha_i$, we can find   $l$ numbers $\delta_1, \ldots , \delta_l$ such that $\gamma_i = \sum_{j=1}^l \delta_j$, and $0 \leq \delta_j \leq \beta_j$. (This choice of numbers is not necessarily unique).      Then $p_i^{\delta_j} | p_i^{\beta_j}$ for each $j = 1, \ldots , l$. Hence, for each factor $\mathbb{Z}_{{p_i}^{\beta_j}}$, there exists a subgroup of order $p_i^{\delta_j}$, namely $\mathbb{Z}_{{p_i}^{\delta_j}}$ (using the fact that the converse of Lagrange's theorem is true for finite cyclic groups). Taking the direct product of each of these subgroups, we get a new subgroup $G_i'$ of $G_i$:      $$G_i' \cong \mathbb{Z}_{{p_i}^{\delta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\delta_l}}$$      The order of this subgroup is $p_i^{\delta_1} \times \ldots \times p_i^{\delta_l} = p_i^{\delta_1 + \ldots + \delta_l} = p_i^{\gamma_i} $. So we have found a subgroup of $G_i$ of order $p_i^{\gamma_i}$, as required.      So each factor $G_i$ in the product $G = G_1 \times \ldots \times G_k$ has a subgroup $G_i'$ of order $p_i^{\gamma_i}$.       Therefore, $G$ has a subgroup   $$G_1' \times G_2' \times \ldots \times G_k'$$       of order $p_1^{\gamma_i}...p_k^{\gamma_k} = n$, which completes the proof.   I have two questions about this: firstly, does this proof seem to work? Secondly, is there a way to make the proof more concise (e.g. a way to prove the statement without using all these indices)? I have several input and option elements on my page, each (well almost) have an event attached to update some text on the page once they change. I use jQuery which is really really cool :)   I also use Microsofts  framework, utilizing the UpdatePanel. The reason why I do that is that certain elements are created on the page based on some server-side logic. I don't really want to explain why I use the UpdatePanel - even if it could (it can with quite some effort) be rewritten to use only  I still want that UpdatePanel.  You probably guessed it - once I have a postback on the UpdatePanel the jQuery events stops working. I actually was expecting this, since the "postback" is not really a new postback so my code in document.ready that binds the events won't be fired again. I also confirmed my suspicion by reading up on it in the jQuery help libraries.  Anyway I'm left with the problem of rebinding my controls after the UpdatePanel is done updating the . I preferably need a solution that does not require adding more .js files (jQuery plug-ins) to the page but something as simple as being able to catch the UpdatePanel's 'afterupdating' where I can just call my method to rebind all the form elements. Matrices commute if and only if they share a common basis of eigenvectors? Here is an interesting piece of code that my fellow team members were just having a slightly heated discussion about...    Dim fred As Integer    If True Then fred = 5 : fred = 3 : fred = 6 Else fred = 4 : fred = 2 : fred = 1   After executing the above code snippet, what is the value of fred?  Try not to cheat and debug the code.  This is a highly contrived code example that started out as an example of using the colon with an If statement, but then someone decided to take it upon themselves to proffer a result for fred.  UPDATE: I would not normally write code like this and this snippet only serves as an example. As it so happens, this question originated from a discussion involving the creation of a coding standards document for our team. When is the expected stable release date of Ubuntu Phone and Ubuntu Tablet? Is there an expected stable release date of Ubuntu Phone and Ubuntu Tablet?  If so, what are their dates? Many C++ books contain example code like this...  std::cout &lt;&lt; "Test line" &lt;&lt; std::endl;   ...so I've always done that too. But I've seen a lot of code from working developers like this instead:  std::cout &lt;&lt; "Test line\n";   Is there a technical reason to prefer one over the other, or is it just a matter of coding style? What type of path in shebang is more preferable? Choosing C++ debugger on linux How to inject keystrokes via a shell script? I'm trying to inject keystrokes into the input daemon so as to simulate typing from a Bash script. Is this possible, and if so, how can I make it happen? How to add D-link dwm-156 to usb switch mode for automatic eject of driver CD / switch to modem mode? I am running Ubuntu 13.10, 32-bit.  I had every time need to go terminal to "eject" the windows driver CD,nor gone to disk &amp; eject windows driver CD to make the d-link dwm-156 gsm modem functioning through network manager.  It seems that usb mode switch is not working.  But on my friend's computer with ubuntu 14.04 ,usb mode switch is working and i can connect d-link dwm-156 to mobile broadband through network manager as soon as i get in.  Is there any way to make usb mode switch to eject the windows driver CD of d-link dwm-156 in my ubuntu 13.10,32-bit at start up?   The lsusb output is Bus 002 Device 013: ID 2001:a706 D-Link Corp before ejecting the CD (modem as storage device of windows driver).  The lsusb output is Bus 002 Device 005: ID 2001:7d01 D-Link Corp after ejecting the CD (as modem).   I don't want to use the driver in the CD,as i need to send sms &amp; find the data usage. Cannot remove any bluetooth device How do I configure swappiness? How does a fork bomb work? Is there some sort of classification of all minimal non-cyclic groups? Does there exist some sort of classification of all minimal non-cyclic groups (non-cyclic groups, such that all their proper subgroups are cyclic)  I know the following classes of such groups:  1) $C_p × C_p$, where $p$ is a prime  2) $Q_8$  3) $\langle a,b | a^p = b^{q^m} = 1, b^{−1}ab = a^{r}\rangle$, where $p$ and $q$ are distinct primes and $r ≡ 1 \pmod q$, $r^q ≡1 \pmod p$.  (These three classes completely cover the case, when our group is finite: )  4)$C_{p^{\infty}}$, where $p$ is a prime  5)$(\{ \frac{n}{p^m}| m, n \in \mathbb{Z} \}, +)$, where $p$ is a prime  (These two classes completely cover the case, when our group is infinite abelian: )  6)Infinite non-abelian groups, such that all their nontrivial proper subgroups are isomorphic to $C_{p}$ for a fixed prime $p$ (Tarski monster groups)  7)Infinite non-abelian groups, such that all their nontrivial proper subgroups are isomorphic to $C_{\infty}$ ().  However, I do not know, whether there exists anything, that does not fall into these classes. I only know, that if such groups exist, they have to be infinite non-abelian.