Let $A = \{p \in \mathbb{Q}^+ : p^2 \lt 2\} $ and $B = \{p \in \mathbb{Q}^+ : p^2 \gt 2\} $. We want to prove that $A$ has no maximum and $B $ has no minimum. I spent a good hour on this without being able to do it. This is how it was done in the book:     Associate with every rational $p \gt 0$ the number      (3) $$q=p- \frac{p^2-2}{p+2}$$        Then      (4) $$q^2-2 = \frac{2 (p^2-2)}{(p+2)^2} $$        If $p $ is in $A $ then $p^2-2 \lt 0$, (3) shows that $q \gt p $, and (4) shows that $q^2 \lt 2$. Thus q is in $A $.  (and analogous for $B $).   Ok I can see it works. But how did he find that $q $? Why divide $p^2-2$ by $p+2$ and then subtract it to $p $? What process does one go to find the expressions that work?
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?
Twin Paradox, calculating spacetime intervals from both perspectives I've very recently started to try to understand special relativity. I'm want to get a decent understanding of the twin paradox. I'll post what I've done so far and highlight what's gone wrong for me.  The situation is that Alice and Bob both begin at point $x_1$ in Alice's coordinate system $(x,t)$ (we have orientated the axis so that y and z will not matter). Bob then instantaneously moves off with velocity $v$ in the positive $x$ direction. Bob's coordinate system is now $(x',t')=(\gamma (x-vt), \gamma (t-v\frac{x}{c^2}))$. In Alice's coordinates, Bob reaches point $x_2$ after having moved distance $d$, then instantly turns around and travels back to Alice at $-v$. I want to show that Alice's space time interval $\Delta s_A^2$ is greater than Bob's $\Delta s_B^2$, as each path's space time interval is proportional to the proper time passed along that path.    According to Alice: $\Delta s_A^2=-c^2\Delta t^2 + \Delta x_A^2$ where $\Delta t=\dfrac{2d}{v}$ and $\Delta x_A=0$ as she didn't move, but the time elapsed is the time it took bob to travel the distance and then come back. Thus $\Delta s_A^2=-\dfrac{4c^2d^2}{v^2}$  Still according to Alice, Bob's $\Delta s_B^2=-c^2\Delta t^2+\Delta x_B^2$, where the change in time is the same, but now bob has moved distance $d$ twice, thus $\Delta x_B=2d$ ($\Delta x$ refers to total distance traveled rather than displacement which is 0 in this case). Now we have $\Delta s_B^2=-\dfrac{4c^2d^2}{v^2}+4d^2$.     The |size| of Bob's spacetime interval now definitely smaller than Alice's, and this would be all OK, except that when I do the calculations in Bob's frame, they don't agree. This is contradictory to the fact that $\Delta s^2$ is conserved under Lorentz transformations.    According to Bob: $\Delta {s'}_A^2=-c^2\Delta t'^2 + \Delta {x'}_A^2$ where $\Delta t'=\dfrac{2d'}{v}$. I'm not sure if I'm right in saying that $v$, the relative velocity of the two frames, is the only velocity upon which both Alice and Bob will agree on, other than the speed of light. Anyway, $d=\gamma d'$ as lengths contract by $\gamma$ i.e. $d'$ is smaller than $d$ by factor $\gamma$. Also, $\Delta {x'}_A=2d'$, so we have $\Delta {s'}_A^2=-\dfrac{4c^2d'^2}{v^2}+4d'^2=-\dfrac{4c^2d^2}{\gamma ^2 v^2}+\dfrac{4d^2}{\gamma ^2}=4d^2\dfrac{v^2-c^2}{\gamma^2 v^2}\neq \Delta s_A^2$  Although I haven't included them here, my calculations for $\Delta {s'}_B^2$ and $\Delta s_B^2$ agree. Sorry for the ultra-long post, but any help would we well appreciated! I thought of this question right after I posted a tweet about a service upgrading me to a free student account since I am in college. I said "That really helps a broke college student out." I actually paused for a second while I was writing that to decide if I should say     That really helps out a broke college student.   or     That really helps a broke college student out.   Are there any prescriptive rules about splitting phrasal verbs like this? I know this breaks the "don't put a preposition at the end of a sentence" rule, but that "rule" has an exception for phrasal verbs.  To clarify:   Is there any rule that says phrasal verbs can't be split, even if it is just an imposed, prescriptive rule? Does splitting a phrasal verb to put a preposition at the end of a sentence fall under the phrasal verb exception to the prepositions at the end of a sentence rule? Centering in tabularx and X columns How can I center the column when using X in tabularx environment like in this example?  \begin{tabularx}{\textwidth}{@{}lXXXXX@{}}  1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\    \end{tabularx} Using the quadratic formula, we have either 0, 1, or 2 solutions. I wonder if we could use it this formula for congruence? Or is there a formula to solve quadratic equation for congruence?  Edit Assume that $ax^2 + bx + c \equiv 0 \pmod{p}$, where $p$ is prime with $(a, p) = 1$, then is there a formula for this equation?    Thanks, What is the difference between OpenOffice and the newly created LibreOffice? What is the difference between OpenOffice and LibreOffice? I understand that LibreOffice is another fork that is going to be developed. I have used OpenOffice since v1 and am wondering if there is a definite advantage to uninstalling and trying LibreOffice. Solving Keplerian two-body problem numerically? I'm trying to solve the two-body problem numerically, setting up G, m1 and m2 =1.0.  The masses are placed at the positions -10 and 10 respectively along the x-axis and gave them both 0 on the y-axis. I am having some real issue with the initial conditions fitting for a keplerian orbit with a given eccentricity (e=0.9)! Also how would the initial condition change for different mass ratios? I've been trying to understand how the second order derivative "formula" works:  $$\lim_{h\to0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}$$  So, the rate of change of the rate of change for an arbitrary continuous function. It basically feels right, since it samples "the after $x+h$ and the before $x-h$" and the $h^2$ is there (due to the expected /h/h -> /h*h), but I'm having trouble finding the equation on my own.  It's is basically a derivative of a derivative, right? Newtonian notation declares as $f''$ and Leibniz's as $\frac{\partial^2{y}}{\partial{x}^2}$ which dissolves into:  $$(f')'$$ and $$\frac{\partial{}}{\partial{x}}\frac{\partial{f}}{\partial{x}}$$  So, first derivation shows the rate of change of a function's value relative to input. The second derivative shows the rate of change of the actual rate of change, suggesting information relating to how frequenly it changes.  The original one is rather straightforward:  $$\frac{\Delta y}{\Delta x} = \lim_{h\to0} \frac{f(x+h) - f(x)}{x + h - x} = \lim_{h\to0} \frac{f(x+h) - f(x)}{h}$$  And can easily be shown that $f'(x) = nx^{n-1} + \dots$ is correct for the more forthcoming of polynomial functions. So, my logic suggests that to get the derivative of a derivative, one only needs to send the derivative function as input to finding the new derivative. I'll drop the $\lim_{h\to0}$ for simplicity:  $$f'(x) = \frac{f(x+h) - f(x)}{h}$$  So, the derivative of the derivative should be:  $$f''(x) = \lim_{h\to0} \frac{f'(x+h) - f'(x)}{h}$$  $$f''(x) = \lim_{h\to0} \frac{  \frac{ f(x+2h) - f(x+h)}{h} - \frac{ f(x+h) - f(x)}{h}  }{h}$$  $$f''(x) = \lim_{h\to0} \frac{  \frac{ f(x+2h) - f(x+h) - f(x+h) + f(x)}{h}  }{h}$$  $$f''(x) = \lim_{h\to0} \frac{  f(x+2h) - f(x+h) - f(x+h) + f(x)  }{h^2}$$  $$f''(x) = \lim_{h\to0} \frac{  f(x+2h) - 2f(x+h) + f(x)  }{h^2}$$  What am I doing wrong? Perhaps it is the mess of it all, but I just can't see it. Please help. get values from ANY type Object $e^{i\theta}=\cos\theta + i\sin \theta$  $e^{i\sin^{-1}x}=\cos(\sin^{-1}x)+i\sin(\sin^{-1}x)$  $i\sin^{-1}x=\ln|\sqrt{1-x^2} + ix|$  $\sin^{-1}x=-i\ln|\sqrt{1-x^2} + ix|$  Now from here I'm kind of lost, since it seems like this should be the definition, but when I look it up, the definition of inverse hyperbolic sine is:  $\sinh^{-1}x=\ln(\sqrt{1+x^2} + x)$  So although they're very similar, I guess I just don't know how to handle the logarithm and anything to the ith power or drop off the absolute value. Creating an index of quality from multiple variables to enable rank ordering Viruses at this period of time do not fit the current definition of life.   Much of the reasoning behind this is that we currently believe that all life must be made up of cells.  Also, many biologist/scientists say that viruses are not living because all they are is just chemicals carrying out their predefined chemistry.   Has anyone every considered the fact that cells, the supposed building blocks of life, themselves simply only carry out predefined chemical actions? All actions cells carry out are chemically driven, not consciously or spiritually driven. Cells do not make choices, they follow chemical directions present in their DNA and RNA. Viruses act very similarly, following the action transcribed in their DNA or RNA.   How can we really even draw a defined line of what life is? Question regarding an inequality How to prove that $$ \frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\cdots+\frac{x_n}{1+x_1^2+\cdots+x_n^2}&lt;\sqrt{n} $$ knowing that $(x_n)$ is a positive sequence ? I looked up all kinds of inequalities such AM-GM, Chebyshev, Cauchy-Schwarz, but I couldn't manage to obtain anything useful.. Can anyone help ? 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I want to avoid that because I have links to directories full of media files that would involve copying hundreds of gigabytes I don't need to back up. Fearing (but not certain) that rsync -a would copy all those media files I instead added the --no-links flag.  This does not seem to be behavior I want.  It just ignores copying any link which is problematic because I do have links I want copied (e.g. links to common header files from different project directories). Assuming #1 above (without the --no-links flag) is what I really want and it just copies the link without copying the linked-to files, will the links break when they are backed up?  For example I may rsync source directory /home/me/projects/misc to /media/extdrive/backup/home/me/projects/misc.  In this case I assume rsync is not smart enough nor does it try to correct the contents of symlinks for the relative directory changes.  Is this correct?  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Show: $\sum_{k=1}^{\infty}\mu(\left\{f\geq k\right\})\leq\int f\, d\mu\leq\sum_{k=0}^{\infty}\mu(\left\{f>k\right\})$ Show that für $f\colon (\Omega,\mathcal{A})\to\mathbb{R}_{\geq 0}$ measurable, it is     $$ \sum_{k=1}^{\infty}\mu(\left\{f\geq k\right\})\leq\int f\, d\mu\leq\sum_{k=0}^{\infty}\mu(\left\{f&gt;k\right\}). $$      I do not know if one needs it here, but we defined the integral of a non-negative measurable function $f\colon (\Omega,\mathcal{A})\to (\overline{\mathbb{R}},\overline{\mathcal{B}})$ by $$ \int_{\Omega}f\, d\mu=\sup\left\{\sum_{k=1}^{n}(\inf\left\{f(x): x\in A_k\right\})\mu(A_k): n\in\mathbb{N}, A_1,\ldots,A_n\in\mathcal{A}\mbox{ disjoint },\Omega=\bigcup_{k=1}^{n}A_k\right\}. $$    My first idea was to use this definition, but I did not come along with it...  Maybe you can help me to show that?  (I think, maybe it has something to do with monotone convergence theorem or majorated convergence?)