I have the following function defined on $\mathbb{R}$:  $$f(x) = \begin{cases}     0 &amp; \text{if $x$ irrational} \\     1/n &amp; \text{if $x = m/n$ where $m, n$ coprime}   \end{cases}$$  I want to show that $f$ is continuous at every irrational point, and has a simple discontinuity at every rational point. I was able to show the first and partially the second (I showed that $f$ has a discontinuity at every rational point, but got stuck on showing that the discontinuity is simple).  However, I've realized that I can simply show that $\lim_{t \rightarrow x}f(t) = 0$ for every $x$, and both of the things I want to show follow from this. How can I show this?
I have to prove that a function $f:]0,1] \rightarrow \Bbb R$ : $$ f(x) = \begin{cases} \frac1q,  &amp; \text{if $x \in \Bbb Q$  with $ x=\frac{p}q$ for $p,q \in \Bbb N$ coprime} \\ 0, &amp; \text{if $x \notin \Bbb Q $} \end{cases} $$  is discontinuous in every point $x \in \ ]0,1] \cap\Bbb Q$.  And then to consider $x \in \ ]0,1] \backslash \Bbb Q$ and prove that it is continuous.  For now I learned different ways to prove continuity (epsilon-delta, sequences), but I'm never sure what would be better to use in each different case.  I wanted to prove the discontinuity by using sequences: $$\forall x_n \quad x_n\rightarrow a \quad \Rightarrow \quad f(x_n) \rightarrow f(a)$$  I tried creating a sequence $ x_n=\frac1n + a$, we know it converges to $a$ but $f(x_n)\ $ doesn't  converges to $\ f(a)$ because there would still be some points not in our set (irrational numbers that creates gaps).  But I don't think it works, so I'm asking you if you could help me solving the two questions.
So I have a 2010 15" macbook pro (MBP6,1) and I replaced the optical drive with a hard disk. I wanted to install windows 7 on the hard disk with boot camp with a USB and ISO image.  After searching I found that I need to modify my plist.info to allow my boot camp to create a bootable USB (since normally it doesn't have the option to create the USB for my model since it came with a disk drive).   After doing so and seemingly having a fine (but long) install process onto the USB. I told boot camp how much space to partition windows on and hit install.  Then the computer restarted. When it came back up it was stuck on a grey apple logo screen forever (I waited 15 minutes to ensure it wasn't just "thinking"). I force restarted and the next time it turned on it got stuck on the grey logo for a bit but then proceeded to go to a black screen that said " No bootable device - please insert a bootable device and hit any key". I tried hitting keys with he USB in but nothing happened.  I had to eventually remove the USB and reset PRAM to get it to boot into osx again  Is there any way to install windows 7 from a usb on my macbook pro using boot camp? How long can you say "the late so and so"? When you refer to the deceased, you say "the late so and so." How long can you say that? Is JFK referred to as the late John F. Kennedy? How about Abraham Lincoln? 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? ... Why are the even and odd Regge trajectories degenerate? Why is $1$ not considered a prime number?  Or, why is the definition of prime numbers given for integers greater than $1$? Words for "a bit sad" and "a bit happy"? Ways to eliminate switch in code pronoun agreement in neither nor Which one is correct?   Neither my friend nor I feel my best in the morning. Neither my friend nor I feel her best in the morning. Neither my friend nor I feel our best in the morning.    I already know that the verb agrees with the subject closer to the verb:   Neither my friends nor my brother understands me. (correct) Neither my friends nor my brother understand me. (wrong)   AND   Neither my brother nor my friends understand me. (correct) Neither my brother nor my friends understands me. (wrong) Modifying C string constants? If I am talking to somebody about a certain group of people in the third person, and then want to refer to the person I am talking with as one of those people, which do I say?     One of them were you      One of them was you. If an polynomial has a complex root, is it necessary that its conjugate is also a root? If an polynomial has a complex root, is it necessary that the conjugate of that root is also a root of the polynomial? I am not sure about this. Please comment on this. this code always returns 0 in PHP 5.2.5 for microseconds:  &lt;?php $dt = new DateTime(); echo $dt-&gt;format("Y-m-d\TH:i:s.u") . "\n"; ?&gt;   Output:  [root@www1 ~]$ php date_test.php 2008-10-03T20:31:26.000000 [root@www1 ~]$ php date_test.php 2008-10-03T20:31:27.000000 [root@www1 ~]$ php date_test.php 2008-10-03T20:31:27.000000 [root@www1 ~]$ php date_test.php 2008-10-03T20:31:28.000000   Any ideas? I use ${\dot \hat E}$ to put a hat and dot on top of an E simultaneously. However, an error occurs. Data   I want operator users on this machine to mount their own cifs shares The sudoers file already contains the /bin/mount -t cifs //*/* /media/* -o username=* command for all operators I want the users to mount a cifs share through a script typing the password only once, not twice. The sudo password and the cifs password are identical.   What I already have  This script works:  #!/bin/bash sudo 'mount -t cifs //192.168.1.1/home /media/$USER/home -o username=$USER'   ...but it requires the users to type the same password twice!     Once for sudo Once for the mount itself   This would also work:  #!/bin/bash echo -n Password:  read -s szPassword echo $szPassword | sudo -S sh -c 'echo $szPassword | mount -t cifs //192.168.1.1/home /media/$USER/home -o username=$USER'   ...but this would require me to allow all operator users to be able to sudo sh (major security problem)  Question  How to mount a cifs share in bash¹ without putting sh in the sudoers file nor creating a permanent/temporary file???  Note 1: no python, perl, C, Go, ... please? Note 2: I know I can just remove the password through the sudoers file, but I'm trying to tighten security, not loosen it, without giving up convenience... KOMA-Classes (scrbook, scrreprt, ... ): Disable chapter separation in \listoffigures and \listoftables I use \listoffigures and \listoftables. Both commands create line breaks for items from different chapters. How can you stop that?  Screenshot: It is stated that if $\psi(x_0)=\psi'(x_0)=0$ at some $x_0$, then $\psi(x)=0,\forall x$.  Hence, we cannot have the wavefunction and its derivative vanishing at the same point.  My question is, how to prove the above statement? How can I add a new value to a picklist using APEX? I have queries pulling all of the information correctly but what I can't seem to find was a way to add new options to a picklist using APEX.  I currently have   Package__c.Products__c.add(new Schema.PicklistEntry(products[j].Name));   Package__c is my custom object, and Products__c is the picklist. products[j] is pulling a list of products pulled from our products page.  And I've tried adding it as a select option, or even casting the variable type.  And this information has to go to the actual picklist within a custom object, and not within a visual force page or anything like that. How many solutions $X^{10} - I=0$ has in $M_2(\mathbb{R})$? How many solutions $X^{10} - I=0$ has in $M_2(\mathbb{R})$? Where $M_2(\mathbb{R})$ denotes the set of $2 \times 2$ real matrices.  I absolutely have no idea of where I should start from. $I$ and $-I$ solve the equation, but are there any other non-trivial solutions? Are there infinitely many solutions? How to determine an object's class? I have a number of tiff files named:    sw.001.tif sw.002.tif ...   and I want to remove the .tif at the end of each of the files.  How can I use the rename command to do this?