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Multiplicative inverse questions This is my first time posting on Math exchange. I have been self-teaching myself Mathematics and recently started learning some Algebra. I was posed the following questions, in which I would like to answer before I proceed, but having some trouble finding proper solutions. So the format will be question, and my answer.      What name do we give an element $a \in \mathbb{Z}_n$ that has a multiplicative inverse?   A modular multiplicative inverse of an integer $a$ is an integer $x$ such that $ax \cong 1 (\mod m)$. This means that the remainder after dividing $ax$ by integer $m$ is equal to $1$.      Give an example, for some $n \in \mathbb{N}$, of an element of $\mathbb{Z}_n$, that does not have a multiplicative inverse?   This question is throwing me off because - does the $n$ we speak of matter for both $n \in \mathbb{N}$ and $\mathbb{Z}_n$ ot solely $n$? My guess here would be $0$ because we cannot have an inverse of $0$.      Prove that multiplicative inverse are unique; an element $a \in \mathbb{Z}_n$ cannot have two multiplicative inverses.    Suppose that an element $a\in\mathbb{Z}_n$ has an inverse $a'$. Next, suppose that $a^*$ is also an inverse of $a$. Then, $a'$ is a solution to $a \cdot_n x=1$, and similarly $a^*$ has a solution to $a \cdot_n x=1$.   Lemma: Suppose $a'$ is multiplicative inverse of $a$ in $\mathbb{Z}_n$, then for any $b \in \mathbb{Z}_n$ the equation $a \cdot_n x=b$ has a unique solution, such that $x = a' \cdot_n b$.   By Lemma, equation $a \cdot_n x = 1$ has a unique solution, namely $x=a'=a^*$. Therefore, an element $a\in \mathbb{Z}_n$ cannot have two multiplicative inverses.    Thank you for the help. I would love any constructive feedback for these. If you can help me elaborate on some ideas, that would be really helpful also.  Thank you very much! Prove $2^n-1$ is divisible by $(2^m-1)^2$ iff... Prove that for any $m,n\in\mathbb{Z}^+$, the number $2^n-1$ is divisible by $(2^m-1)^2$ iff the number $n$ is divisible by $m(2^m-1)$.  My work so far:  It's an if and only if statement, so the common approach is to split the proof into two "if then" proofs ($A\rightarrow B \text{ and } B\rightarrow A$).  Starting with $A\rightarrow B$, for a direct proof, I would assume $A$ and prove $B$. I can use the definition of divisibility:  Assume $n=m(2^m-1)k$ for some integer $k$  Then I need to prove that $2^n-1=(2^m-1)^2h$ for some integer $h$  So I'm thinking I need to start with the expression $2^n-1$ and transform it to $(2^m-1)^2h$, or vice versa, and I need to use the assumption $n=m(2^m-1)k$ somehow.  I notice that the LS of the equality contains the variable $n$ while the RS contains $m$, and the assumption relates $n$ to $m$, so that could be a way forward. For example, simple substitution gives $2^{m(2^m-1)k}-1$, but I'm not sure what I could do from here. Perhaps there's a divisibility theorem that could help?  Other things I notice: $2^n-1$ and $(2^m-1)^2$ are always odd. Also, the divisors in both cases are multiples of $2^m-1$.  Any help/hints appreciated. 19.10: Deactivate Alt Key From Triggering Applications' Menu Bar How can I stop the Alt key from triggering the menu bar of applications in Ubuntu 19.10?  I know that there are these , , and , as well as  but those solutions are no longer applicable. I did try searching for a similar option to the one in the answer of those questions but with no success. This is a question of the book Linear Algebra of Kenneth Hoffman.  Describe explicitly all inner products on $\mathbb{R}$ and on $\mathbb{C}$.  I think that if $&lt;,&gt;$ is one inner product on $\mathbb{R}$ (or $\mathbb{C}$) then $&lt; ,&gt;=K[,]$, where $[,]$ is the standard inner product. Is that correct? For a graph $G$, let $W$ be the (random) vertex occupied at the ﬁrst time the random walk has visited every vertex. That is, $W$ is the last new vertex to be visited by the random walk. Prove the following remarkable fact:      For the   random walk on an $n$-cycle, $W$ is uniformly distributed over all vertices different   from the starting vertex. Can FFMPEG extract images from a video with timestamps? How to split the output and store it in an array? This is the output:  3,aac-lc, 93.8, aaclc, 77.3, h.264, 1024.6, h.264, 1029.1, 31, 31, 0,0,0.000000,31,31,0,0,0.000000,7,0,0,0.000000,30,1280 720,10,0,0,0.000000,30,1280 720   I tried with 2 scenario:   Storing in an array    @arr=split(',',$stats);   echo "statistics: $stats"  Storing in a variable   echo $stats | cut -d ',' -f | read s1  echo $s1    But neither scenario is working. How to decide whether to replace or repair? Need hint for $\lim_{x\to 0} \frac{(x+1)^\frac{1}{x}-e}{x}$ Please give me some hints or solution for this limit. I had it on my exam and I'm curious how to solve it. $$ \lim_{x\to 0} \frac{(x+1)^\frac{1}{x}-e}{x} $$ The Medibuntu Project has come to an end - what do I do now? Show that a finite cyclic group of order $n$ has exactly one subgroup of order $d$ dividing $n$ and these are all the subgroups it has My attempt  Note- Addition is under $mod$ $d$  Let $G = (a^0,a^1,a^2.....a^{n-1})$  $n=kd$  Let $H_d = (a^{0k},a^{1k},a^{2k}......a^{{(d-1)}k})$  For $a^{z_1k},a^{z_2k} \in H_d$  $a^{z_1k}a^{z_2k} = a^{{(z_1+z_2)}k} \in H_d$  $H_d$ is closed  For $a^{zk} \in H_d$  $a^{zk} (a^{zk})^{-1}=a^0$  $(a^{zk})^{-1} = (a^{-1})^{zk}$  Since $a^{-1}$ is $a^{n-1}$ we can rewrite it as   $(a^{zk})^{-1} = (a^{n-1})^{zk}=a^{z(n-1)k} \in H_d$  Since $(a^{zk})^{-1} \in H_d$ and $H_d$ is closed, $H_d$ is a subgroup.  I can easily prove the 3rd part by using Lagrange's theorem.  I have no idea how to prove that only $1$ subgroup of order $d$ exists. Maybe by proving that the elements of the $2$ groups have to be the same?  I would really appreciate if someone could verify my proof so far and give some hints to complete the proof. What the pointer size in 64 bits computer in C++? Drag and drop not working inside Nautilus on Ubuntu 17.10 I'm having a strange problem on Ubuntu 17.10: when I try to drag and drop something inside Nautilus it simply doesn't work, it doesn't even select the item, it seems like it's ignoring the action... Also, I cannot select things inside nautilus by drawing rectangles with the mouse.  All of these actions work only on the Desktop. Here is a demo:    I'm using the Xorg session with nVidia proprietary drivers. I also noticed that Nautilus seems to hang when I try to copy, paste or cut a file...  EDIT 1: I solved the problem for the slow copy/paste by disabling the clipboard sync plugin of KDEConnect. I also discovered that if I open Nautilus as root, I can drag and drop or select whatever I want... Uniform continuity of $f(x) = x \sin{\frac{1}{x}}$ for $x \neq 0$ and $f(0) = 0.$ Log color bar/meta data in PGFPLOT I need to find a way to import some .json models into Blender, and it seems I can't just import them the same way I can import a .obj model. So is there some way to convert a .json model to a .obj model, or another file type Blender can import? I've tried using Blockbench for this, but I think it only works with Minecraft recourse pack .json models, as those are the only models that will ever actually load in it. Thank you to anyone who can help me figure this out. How far can I refactor the code in someone else's question? I am working on setting up a Minecraft server on a machine I have at home. This particular machine has two quad-core CPUs running at 2.0Ghz so I would like to take advantage of the number of cores and run a server with 4 or 6 of the cores. My server of choice is CraftBukkit, the latest version of which is craftbukkit-1.4.7-R1.0 and launched with a start-up script of:  "C:\Program Files\Java\jre7\bin\java.exe" -Xms1G -Xmx8G -jar craftbukkit-1.4.7-R1.0.jar"   The  says:     Minecraft servers, as of 1.1.0, can use multiple cores, so now they will not sit idle.   but does not really expand on this.  My questions, then:   Is it possible to set the number of cores the server uses? How do I do it - via start-up options, or other methods? Is it really worth it and would it make that much of a difference with a lot of players? I would like to install  chat platform on my Ubuntu 12.04 LTS system. Please, could anyone provide me the steps to successfully install this software?. I've been trying to install this software using wine but I couldn't succeed it. For example,   If this van's a-rockin', don't come a-knockin'  (song title)   (song title)   I have a few related questions:   What is the "a-" or "-a" called? Is there a function beyond a rhythmic one? Why is "a-" attached to a word at all -- and why attach to a word at all?