More specifically, I have to find a $q$ in $A$ such that $q$ is larger than any $p$ in $A$.  The only thing I can think of is using (${2 - {p^2}}$) somehow, or ${(p+x)^2}&lt;2$, but other than that I don't know how to proceed. This exercise is taken from the very first chapter of Walter Rudin's Principles of Mathematical Analysis. Quoting the exercise:     To do this, we associate with each rational $p &gt; 0$ the number:     $$q = p - \frac{{{p^2} - 2}}{{p + 2}} = \frac{{2p + 2}}{{p + 2}}$$   But he didn't explain how he came up with that number.   The only thing I've thought of is the following procedure:  $$\eqalign{   &amp; {p^2} + (2 - {p^2}) = 2 \Rightarrow {p^2} + (\frac{{2 - {p^2}}}{2}) &lt; 2,{\text{ and since }}2 - {p^2}{\text{ is always positive,}}  \cr    &amp; {p^2} &lt; \left[ {{p^2} + (\frac{{2 - {p^2}}}{2})} \right] &lt; 2  \cr    &amp; {\text{Therefore, let }}q = \sqrt {{p^2} + (\frac{{2 - {p^2}}}{2})} ,{\text{ then:}}  \cr    &amp; p &lt; q &lt; \sqrt 2  \cr} $$  However, my result doesn't necessarily belong to ${\Bbb Q}$. I wonder how did he obtain the value of $q$ in the book.  EDIT: Feel free to point out if there are more appropriate tags for this question.
First, my apologies if this has already been asked/answered.  I wasn't able to find this question via search.  My question comes from Rudin's "Principles of Mathematical Analysis," or "Baby Rudin," Ch 1, Example 1.1 on p. 2.  In the second version of the proof, showing that sets A and B do not have greatest or lowest elements respectively, he presents a seemingly arbitrary assignment of a number $q$ that satisfies equations (3) and (4), plus other conditions needed to show that $q$ is the right number for the proof.  As an exercise, I tried to derive his choice of $q$ so that I may learn more about the problem.  If we write equations (3) as $q = p - (p^2 - 2)x$, we can write (4) as   $$ q^2 - 2 = (p^2 - 2)[1 - 2px + (p^2 - 2)x^2]. $$  Here, we need a rational $x &gt; 0$, chosen such that the expression in $[...]$ is positive.  Using the quadratic formula and the sign of $(p^2 - 2)$, it can be shown that we need   $$ x \in \left(0, \frac{1}{p + \sqrt{2}}\right) \mbox{ for } p \in A, $$  or, for $p \in B$, $x &lt; 1/\left(p + \sqrt{2}\right)$ or $x &gt; 1/\left(p - \sqrt{2}\right)$.  Notice that there are MANY solutions to these equations! The easiest to see, perhaps, is letting $x = 1/(p + n)$ for $n \geq 2$.  Notice that Rudin chooses $n = 2$ for his answer, but it checks out easily for other $n$.  The Question:  Why does Rudin choose $x = 1/(p + 2)$ specifically?  Is it just to make the expressions work out clearly algebraically?  Why doesn't he comment on his particular choice or the nature of the set of solutions that will work for the proof?  Is there a simpler derivation for the number $q$ that I am missing?
Catenatives followed by infinitives and gerunds Is it possible to remove a terminal command? For example, if there was a command call lookup (not an alias), how would I go about removing it? The AC adapter supplied with my laptop is designed to accept voltage in the range of 100-240VAC at 50-60Hz, but the AC cable that plugs into the adapter unit itself is only rated to 125VAC 7.5A 2.5A.  I know the adapter can accept 220V power, but is it safe to use this cable with the AC adapter to plug into a 220V socket?  Edit: The part in question is a 90W HP Smart AC Adapter.  The laptop does not cause the AC adapter to draw more than about 80 watts from the socket (as measured with a Kill A Watt meter).  Edit 2: The amperage rating of the cable as marked on the attached label is 2.5A, not 7.5A. By the time, many files are still on my system and I don't need them anymore, so how to delete all files that are one year old at least? How do I install an older version of an app on an iPhone? When I've worked a bit with my source code, I did my usual thing commit and then I pushed to a remote repository. But then I noticed I forgot to organize my imports in the source code. So I do the amend command to replace the previous commit:  &gt; git commit --amend   Unfortunately the commit can't be pushed back to the repository. It is rejected like this:  &gt; git push origin To //my.remote.repo.com/stuff.git/  ! [rejected]        master -&gt; master (non-fast forward) error: failed to push some refs to '//my.remote.repo.com/stuff.git/'   What should I do? (I can access the remote repository.) How do I create RAW TCP/IP packets in C++? I'm a beginning C++ programmer / network admin, but I figure I can learn how to do this if someone points me in the right direction. Most of the tutorials are demonstrated using old code that no longer works for some reason.   Since I'm on Linux, all I need is an explanation on how to write raw Berkeley sockets. Can someone give me a quick run down? Should I declare chocolate as "food" when entering the US? It is being customary we bring a few authentic Russian chocolate bars/sweets to the US but we are always struggling whether we should check that "food" checkbox in the "Customs Declaration" form?  It is under the following entry in the form (yeah, together with insects, right):     I am (We are) bringing:      fruits, vegetables, plants, seeds, food, insects   I've been always checking it but that just gets us into more lines and triggers extra questions after getting to the baggage check. Encrypting a folder I'm familiar with bitlocker, having been using it for a while, but I am looking at just encrypting a folder instead. But I just don't understand the options it's giving me.  When I right-click a folder > properties > advanced > [x] encrypt contents to secure data > OK &lt; apply > ...  I get:  () apply changes to this folder () apply changes to this folder, sub-folder and files.   This doesn't make sense to me at all, if I encrypt a folder then the whole folder, everything within it is encrypted, then why should I be given these options, how can I have a folder encrypted and files within that folder unencrypted, that seems stupid to me, it doesn't make sense at all.  Can anybody explain why these options exist and how can I simply encrypt a folder? Overleaf does not compile after using Biblatex Bohr's model of an atom doesn't seem to have overcome the drawback of Rutherford's model Is there a pronoun I can use as a gender-neutral pronoun when referring back to a singular noun phrase?  Each student should save his questions until the end. Each student should save her questions until the end. Why was the previous Code of Conduct FAQ (and all answers) removed? Problems with character input using scanf() Embedding $S_n$ into $A_{n+2}$ How can I calculate the angle and proper turn direction between two 2D vectors? I am working on some movement AI where there are no obstacles and movement is restricted to the XY plane. I am calculating two vectors, v, the facing direction of ship 1, and w, the vector pointing from the position of ship 1 to ship 2.  I am then calculating the angle between these two vectors using the formula  arccos((v · w) / (|v| · |w|))   The problem I'm having is that arccos only returns values between 0° and 180°.  This makes it impossible to determine whether I should turn left or right to face the other ship.  Is there a better way to do this? How do I make Chromium use Flash from Google Chrome? Now, that Adobe has stopped issuing new releases for Linux desktops, the only option for users is to use Google Chrome, for those who want to use the latest Flash updates. I use Chromium. Is there any way to get Chromium use the Flash from Google Chrome? The reason I as this question is that I have noticed both these browsers run in the background, so technically, the browsers' plugins are available to use. Single Variable Calculus Reference Recommendations In $\mathbb{Z}/(n)$, does $(a) = (b)$ imply that $a$ and $b$ are associates? [Update: Based on the hints provided by @zcn and @whacka, I believe I have found a solution. See my answer below.]  Below, $R$ is a commutative ring with $1$.  In John J. Watkins' Topics in Commutative Ring Theory, the author observes that if $a$ and $b$ are associates in $R$, then they generate the same ideal: $(a) = (b)$. This is clearly true, because $a = ub$ and $b = va$ for some units $u$ and $v$, and so $a \in (b)$ and $b \in (a)$, whence $(a) \subset (b)$ and $(b) \subset (a)$. Indeed, we didn't even require $u$ and $v$ to be units to reach this conclusion.  The author then claims that if $R$ is not a domain, then the converse may not be true: we may have $(a) = (b)$ even if $a$ and $b$ are not associates. As a "counterexample" he gives $R = \mathbb{Z}/(6)$, with $a = 2 + (6)$ and $b = 4 + (6)$. Clearly these elements generate the same ideal: $(a) = (b) = \{0, 2 + (6), 4 + (6)\}$. The author asserts that $a$ and $b$ are not associates, but this is false: $5 + (6)$ is a unit, and $(2 + (6))(5 + (6)) = 4 + (6)$, so $a$ and $b$ actually are associates.  I figured that the author botched the example but that surely the basic fact must be true: if $R = \mathbb{Z}/(n)$ then there may be nonassociates $a,b \in R$ with $(a) = (b)$. However, I was not able to find a counterexample and eventually resorted to writing a computer program and found that there are no counterexamples in $\mathbb{Z}/(n)$ for $n \leq 100$.  So now I'm starting to think that there is no counterexample in $\mathbb{Z}/(n)$ at all, but I have not yet been able to prove this.  My questions:   If I am correct that $(a) = (b)$ in $\mathbb{Z}/(n)$ implies that $a$ and $b$ are associates, I would appreciate a nudge in the right direction toward a proof. (But if the proof is elementary, please give only a hint, not a solution :-) I'll post my solution once I have one. In either case, could you please point me to a genuine counterexample? If there is one in $\mathbb{Z}/(n)$ for some $n$, that would be ideal (no pun intended), otherwise a counterexample in any commutative ring with $1$ would be great. Why does part of my model disappear when I zoom in on it in the 3D Viewport? I'm having an odd issue - When I try to zoom in on a detail using my MMB, the "camera" in the 3D view seems to be cutting off the exterior surface of my object and showing me the inside instead. I'd expect this behavior, but only when I was much more zoomed in.  If I zoom out a bit (one click on the scroll wheel) then I can see the exterior as I intend.  Is there any way to adjust this behavior?  Thanks!