text:
Question

# In an isosceles triangle $ABC$, with $AB = AC$, the bisectors of $\angle B = \angle C$ intersect each other at $O$.Join $A$ to $O$. Show that $(i) OB = OC$ and $(ii) AO$ bisects $\angle A$.

Hint: Join $A$ to $O$. Apply the given condition to the isosceles triangle and use similarity criterion.
According to given data we have 3 conditions,
$AB = AC \to (1)$
$OB$ is the bisector of $\angle B$
So,$\angle ABO = \angle OBC = \dfrac{1}{2}\angle B \to (2)$
$OC$ is the bisector of$\angle C$
So,$\angle ACO = \angle OCB = \dfrac{1}{2}\angle C$$\to (3)$
Case-1
So, here we have to prove $OB = OC$
Proof:
Now by using condition (1) we can say that,
$AB = AC$
From this condition we say that
$\Rightarrow \angle ACB = \angle ABC$ [ Where we know that Angles opposite to equal sides are equal]
$\ \Rightarrow \dfrac{1}{2}\angle ACB = \dfrac{1}{2}\angle ABC \\ \\ \$
$\Rightarrow \angle OCB = \angle OBC$ [From (2) and (3)]
Hence,
$OB = OC$ [Sides opposite to equal angles are equal]
Hence proved that $OB = OC$.
Case-2
We have to prove that $\angle OAB = \angle OAC$
By using case (1) we know that $OB = OC$
And also from $\Delta ABO$ and$\Delta ACO$, we have
$\Rightarrow AB = AC$ (Given)
$\Rightarrow AO = OA$ (Common)
$\Rightarrow OB = OC$(From (case 1))
$\therefore \Delta ABO \cong \Delta ACO$ (By SSS Congruence rule)
$\Rightarrow \angle OAB = \angle OAC$ (CPCPT Theorem)

Hence we have proved that$\angle OAB = \angle OAC$.

CASE - 1 CASE - 2
NOTE: In this problem given construction is mandatory to prove the given statements so join $A$ and $O$ points.


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# lwarp to html issues with termcal

I am trying to convert my LaTeX document into HTML for accessibility reasons. Unfortunately, I have not found a package that correctly creates an accessible document (for images and tables), and so I was told to convert it to HTML, and that way, it can be accessible. I have tried various techniques that are documented in the Stack Exchange and found that lwarp has the best conversion to the display of my PDF output. However, it seems that lwarp has an issue when using the package termcal. After I compiled and ran the code lwarpmk html, I end up with an error ! Misplaced alignment tab character &. <inserted text> ...ut \scshape \centering Tuesday& \struct \scshape \centering.... I have tried going into the code of termcal and based on the lwarp documentation enclosing the line of code in error with \StartDefiningTabulars and \EndDefiningTabulars did not work. It did create an HTML document, but the calendar was missing. Any help with this will be greatly appreciated.

I am running the code on MacBook Pro 2019 using a program called Texpad with pdfLaTeX.

Below is my MWE from Brian Hall's sample syllabus that I have changed to use termcal:

% This syllabus template was created by:
% Brian R. Hall
% Associate Professor, Champlain College
% www.brianrhall.net

% Document settings
\documentclass[11pt]{article}
\usepackage{lwarp}
\usepackage[margin=1in]{geometry}
\usepackage[pdftex]{graphicx}
\usepackage{multirow}
\usepackage{setspace}
\usepackage[pdftex,                                     % For accessibility requirements
pdftitle={CSCI 133 SPRING 2020},
pdflang={en-GB}]{hyperref}
\usepackage{termcal} % Works with the calendar display
\pagestyle{plain}
\setlength\parindent{0pt}

\begin{document}

% Course information
\begin{tabular}{ l l }
\multirow{3}{*}{%InsertImage
}
& \LARGE Course Code \\\\
\mrowcell & \LARGE Course Title \\\\
\mrowcell & \LARGE Day(s), Time, Place \\\\
\end{tabular}
\vspace{10mm}

\newpage

% Course Outline
\newcommand{\TRClass}{%
\skipday % Monday (no class)
\calday[Tuesday]{\classday} % Tuesday
\skipday % Wednesday (no class)
\calday[Thursday]{\classday} % Thursday
\skipday % Friday
\skipday\skipday % weekend (no class)
}

\newcommand{\Holiday}[2]{%
\options{#1}{\noclassday}
\caltext{#1}{#2}
}

\textbf {\large Tentative Course Outline}:

The weekly coverage might change as it depends on the progress of the class.  However, you must keep up with the reading assignments.

\begin{center}
\begin{calendar}{01/27/2020}{17} % Semester starts on 01/27/2020 and last for 17
% weeks, including finals week
\setlength{\calboxdepth}{.3in}
\TRClass
% schedule
\caltexton{1}{Intro to course and data structures}
\caltextnext{C++ review}
\caltextnext{C++ review, cont.}

% ... and so on

% Holidays
\Holiday{2/14/2020}{Lincoln's Birthday Holiday}
\Holiday{2/17/2020}{Presidents' Day Holiday}
\Holiday{4/6/2020}{Spring Break}
\Holiday{4/7/2020}{Spring Break}
\Holiday{4/8/2020}{Spring Break}
\Holiday{4/9/2020}{Spring Break}
\Holiday{4/10/2020}{Spring Break}

% ... and so on

%\options{12/11/2017}{\noclassday} % finals week
%\options{12/12/2017}{\noclassday} % finals week
%\options{12/13/2017}{\noclassday} % finals week
%\options{12/14/2017}{\noclassday} % finals week
%\options{12/15/2017}{\noclassday} % finals week

%\caltext{5/15/2017}{\textbf{Final Exam: 8:00am - 10:30am}}
\end{calendar}
\end{center}

\end{document}

• I can compile your document with TeX4ht (when I remove lwarp and explicit pdftex option for hyperref). It complains about \mrowcell, but so does pdflatex. The output seems OK. May 29, 2020 at 7:00
• @michal.h21 I tried again, and yes the output is ok. I realized that the .css file kept the formating of it to look similar to the PDF output. However, I will still prefer lwarp due to its accessible features. Thank you for your comment and help. May 29, 2020 at 17:08
• Have you seen Creating a PDF/A PDF/X and PDF/UA document? I do not like mass page-type invitations because I have to typwrite them first into my calendar. It's better to send Outlook invitations so they can go directly into your personal electronical calendar. This works for the blind the deaf the lazy and the forgetful. Jun 9, 2020 at 16:29

format: plain


text:
The completed SDSS-IV extended Baryon Oscillation Spectroscopic Survey: BAO and RSD measurements from anisotropic clustering analysis of the quasar sample in configuration space between redshift 0.8 and 2.2

Jiamin Hou, Ariel G. Sánchez, Ashley J. Ross, Alex Smith, Richard Neveux, Julian Bautista, Etienne Burtin, Cheng Zhao, Román Scoccimarro, Kyle S. Dawson, Arnaud De Mattia, Axel De La Macorra, Hélion Du Mas Des Bourboux, Daniel J. Eisenstein, Héctor Gil-Marín, Brad W. Lyke, Faizan G. Mohammad, Eva Maria Mueller, Will J. Percival, Graziano RossiMariana Vargas Magaña, Pauline Zarrouk, Gong Bo Zhao, Jonathan Brinkmann, Joel R. Brownstein, Chia Hsun Chuang, Adam D. Myers, Jeffrey A. Newman, Donald P. Schneider, M. Vivek

Research output: Contribution to journalArticlepeer-review

3 Scopus citations

Abstract

We measure the anisotropic clustering of the quasar sample from Data Release 16 (DR16) of the Sloan Digital Sky Survey IV extended Baryon Oscillation Spectroscopic Survey (eBOSS). A sample of 343 708 spectroscopically confirmed quasars between redshift 0.8 < z < 2.2 are used as tracers of the underlying dark matter field. In comparison with DR14 sample, the final sample doubles the number of objects as well as the survey area. In this paper, we present the analysis in configuration space by measuring the two-point correlation function and decomposing it using the Legendre polynomials. For the full-shape analysis of the Legendre multipole moments, we measure the baryon acoustic oscillation (BAO) distance and the growth rate of the cosmic structure. At an effective redshift of zeff = 1.48, we measure the comoving angular diameter distance DM(zeff)/rdrag = 30.66 ± 0.88, the Hubble distance DH(zeff)/rdrag = 13.11 ± 0.52, and the product of the linear growth rate and the rms linear mass fluctuation on scales of $8 \, h{-1}\, {\rm Mpc}$, f?8(zeff) = 0.439 ± 0.048. The accuracy of these measurements is confirmed using an extensive set of mock simulations developed for the quasar sample. The uncertainties on the distance and growth rate measurements have been reduced substantially (?45 and ?30 per cent) with respect to the DR14 results. We also perform a BAO-only analysis to cross check the robustness of the methodology of the full-shape analysis. Combining our analysis with the Fourier-space analysis, we arrive at $D{{\bf c}}-{\rm M}(z-{\rm eff})/r-{\rm drag} = 30.21 \pm 0.79$, $D{{\bf c}}-{\rm H}(z-{\rm eff})/r-{\rm drag} = 13.23 \pm 0.47$, and $f\sigma-8{{\bf c}}(z-{\rm eff}) = 0.462 \pm 0.045$.

Original language English (US) 1201-1221 21 Monthly Notices of the Royal Astronomical Society 500 1 https://doi.org/10.1093/mnras/staa3234 Published - Jan 1 2021

All Science Journal Classification (ASJC) codes

• Astronomy and Astrophysics
• Space and Planetary Science


text:
Question

# A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was $₹1500$ after $4$ year of service and $₹1800$ after $10$ years of service, what was his starting salary and what is the annual increment?

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Solution

## Let $₹x$ be the starting salary of a man and $₹y$ be the fixed annual increment.Then,Salary after $4$ years of service $=₹\left(x+4y\right)$Salary after $10$ years of service $=₹\left(x+10y\right)$As per the question, given that,$\left(x+4y\right)=₹1500$…………eq.$1$$\left(x+10y\right)=₹1800$………..eq.$2$On subtracting equation ($1$) from equation ($2$), we get,$6y=300$$y=\frac{300}{6}=50$Substituting $y=50$ in equation ($1$), we get,$x+4×50=1500$$x+200=1500$$x=1500-200$$x=1300$Therefore, the starting salary of a man was $₹1300$ and the annual increment is $₹50$.

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# Complex number algebra

#### Jimi2404

##### New member
Hi ive recently had a question that i can't get my head across.

The equation is: (4+2i)(A+B)=6+7i

The question asks to multiply out the let hand side of the complex equation, obtain equations for A and B and thus find the values for them.

Because it involves complex numbers I'm finding it hard to solve. Any help will be fantastic

Thank you

#### Subhotosh Khan

##### Super Moderator
Staff member
Hi ive recently had a question that i can't get my head across.

The equation is: (4+2i)(A+B)=6+7i

The question asks to multiply out the let hand side of the complex equation, obtain equations for A and B and thus find the values for them.

Because it involves complex numbers I'm finding it hard to solve. Any help will be fantastic

Thank you
Did you follow the first hint - i.e. -

"multiply out the let hand side of the complex equation"

What did you get?

#### Dr.Peterson

##### Elite Member
The equation is: (4+2i)(A+B)=6+7i

The question asks to multiply out the let hand side of the complex equation, obtain equations for A and B and thus find the values for them.

Because it involves complex numbers I'm finding it hard to solve.
The main idea will be that after expanding the LHS you will have ____ + ____i = 6 + 7i. In order for these to be equal (the blanks being real numbers), the real and imaginary parts have to be equal: ____ = 6, ____ = 7. Then you can solve the system of equations.

Now go do it, and let us see what you get.

#### pka

##### Elite Member
Hi ive recently had a question that i can't get my head across.
The equation is: (4+2i)(A+B)=6+7i
The question asks to multiply out the let hand side of the complex equation, obtain equations for A and B and thus find the values for them.
To Jimi2404, Please double check your post. As posted the equation has no definite solution. See here.
I suppect it should be $$\displaystyle (4+2{\bf{i}})(A+B{\large\bf{i}})=6+7\bf{i}$$

#### Dr.Peterson

##### Elite Member
To Jimi2404, Please double check your post. As posted the equation has no definite solution. See here.
I suppect it should be $$\displaystyle (4+2{\bf{i}})(A+B{\large\bf{i}})=6+7\bf{i}$$
Yes, I missed that because I didn't try to actually solve it. Assuming A and B are supposed to be real, there is no solution; if not, then there are infinitely many possibilities for each of A and B, though A + B would be fixed.

#### Jimi2404

##### New member
Hi everyone

yeah i forgot to write in the i for B which makes the equation - (4+2i)(A+Bi)=6+7i

#### Jimi2404

##### New member
Did you follow the first hint - i.e. -

"multiply out the let hand side of the complex equation"

What did you get?

when i did that i got

4A+4Bi+2Ai+2bi^2=6+7i

i know that 2 imaginary numbers equal a real number when multiplied but i didn't know where to go from there.

#### Dr.Peterson

##### Elite Member
when i did that i got

4A+4Bi+2Ai+2bi^2=6+7i

i know that 2 imaginary numbers equal a real number when multiplied but i didn't know where to go from there.
First, carry out that multiplication: What real number is 2bi^2 equal to?

Then, carry out the addition: What are the real and imaginary parts on the left hand side?

Set those equal to the corresponding parts on the right hand side.

Staff member
Remember:

i^2 = -1

Staff member

#### pka

##### Elite Member
when i did that i got
4A+4Bi+2Ai+2bi^2=6+7i
i know that 2 imaginary numbers equal a real number when multiplied but i didn't know where to go from there.
Because $$\displaystyle {\bf{i}}^2=-1$$ we now have $$\displaystyle (4A-2B)+(2A+4B){\bf{i}}=6+7\bf{i}$$
Setting real parts equal & imaginary parts equal.
$$\displaystyle 4A-2B=6\\2A+4B=7$$

Solve for $$\displaystyle A~\&~B$$

#### Otis

##### Senior Member
Otis swiped my red ink.......
I expect Denis is awash in red ink; he'd probably be happy to give you some. :cool:


text: dc.contributor.advisor Omre, Karl Henning nb_NO dc.contributor.author Lindberg, David nb_NO dc.date.accessioned 2014-12-19T13:58:39Z dc.date.available 2014-12-19T13:58:39Z dc.date.created 2010-09-30 nb_NO dc.date.issued 2010 nb_NO dc.identifier 354281 nb_NO dc.identifier ntnudaim:5735 nb_NO dc.identifier.uri http://hdl.handle.net/11250/258753 dc.description.abstract In this thesis, we solve the seismic inverse problem in a Bayesian setting and perform the associated model parameter estimation. The subsurface rock layers are represented by categorical variables, which depends on some response variables. The observations recorded appear as a convolution of these response variables. We thus assess the categorical variables' posterior distribution based on a prior distribution and a convolved likelihood distribution. Assuming that the prior model follows a Markov chain, the full model becomes a hidden Markov model. In the associated Posterior-Prior deconvolution algorithm, we approximate the convolved likelihood in order to use the recursive forward-backward algorithm. The prior and likelihood distributions are parameter dependent, and two parameter estimation approaches are discussed. Both estimation methods make use of the marginal likelihood distribution, which can be computed during the forward-backward algorithm.In two thorough test studies, we perform parameter estimation in the likelihood. Approximate posterior models, based on the respective parameter estimates, are computed by Posterior-Prior deconvolution algorithms for different orders. The signal-to-noise ratio, a ratio between the observation mean and variance, is found to be of importance. The results are generally more reliable for large values of this ratio. A more realistic seismic example is also introduced, with a more complex model description. The posterior model approximations are here more poor, due to under-estimation of the noise parameter. nb_NO dc.language eng nb_NO dc.publisher Institutt for matematiske fag nb_NO dc.subject ntnudaim:5735 no_NO dc.subject SIF3 fysikk og matematikk no_NO dc.subject Industriell matematikk no_NO dc.title Parameter estimation in convolved categorical models nb_NO dc.type Master thesis nb_NO dc.source.pagenumber 106 nb_NO dc.contributor.department Norges teknisk-naturvitenskapelige universitet, Fakultet for informasjonsteknologi, matematikk og elektroteknikk, Institutt for matematiske fag nb_NO


text:
zbMATH — the first resource for mathematics

On the numerical stability of algorithmic differentiation. (English) Zbl 1238.65013
Summary: In contrast to integration, the differentiation of a function is an ill-conditioned process, if only an oracle is available for its pointwise evaluation. That is, unrelated small variations in the value of the composite function are allowed at nearly identical arguments. In contrast, we show here that, if the function is defined by an evaluation procedure as a composition of arithmetic operations and elementary functions, then automatic, or algorithmic differentiation is backward stable in the sense of Wilkinson. More specifically, the derivative values obtained are exact for a perturbation of the elementary components at the level of the machine precision. We also provide a forward error analysis for both the forward and reverse mode. The theoretical analysis is confirmed by numerical experiments.

MSC:
 65D25 Numerical differentiation 65G50 Roundoff error 65Y04 Numerical algorithms for computer arithmetic, etc.
Software:
MAPM; Matlab; mctoolbox; XBLAS
Full Text:
References:
 [1] Berz M, Bischof C, Corliss G, Griewank A (eds) (1996) Computational differentiation: techniques, applications, and tools. Proceedings series. SIAM · Zbl 0857.00033 [2] Bücker M, Corliss G, Hovland P, Naumann U, Norris B (eds) (2005) Automatic differentiation: applications, theory, and tools. In: Lecture notes in computational science and engineering, vol 50. Springer, Berlin [3] Cacuci D, Weber C, Oblow E, Marable J (1980) Sensitivity theory for general systems of nonlinear equations. Nucl Sci Eng 88: 88–110 [4] Christianson B (1991) Reverse accumulation and accurate rounding error estimate for Taylor series coefficients. Optim Method Softw 1: 81–94 · doi:10.1080/10556789208805508 [5] Corliss, G, Faure, C, Griewank, A, Hascoet, L, Naumann, U (eds) (2002) Automatic differentiation of algorithms–from simulation to optimization. Springer, Berlin · Zbl 0983.68001 [6] Corliss, G, Griewank, A (eds) (1991) Automatic differentiation: theory, implementation, and application. Proceedings series. SIAM, Philadelphia [7] Griewank A, Utke J, Walther A (2000) Evaluating higher derivative tensors by forward propagation of univariate Taylor series. Math Comput 69(231): 1117–1130 · Zbl 0952.65028 · doi:10.1090/S0025-5718-00-01120-0 [8] Griewank A, Walther A (2008) Principles and techniques of algorithmic differentiation, 2nd edn. SIAM, Philadelphia · Zbl 1159.65026 [9] Higham NJ (2002) Accuracy and stability of numerical algorithms, 2nd edn. SIAM, Philadelphia · Zbl 1011.65010 [10] Iri M (1991) History of automatic differentiation and rounding error estimation. In: Griewank A, Corliss GF (eds) Automatic differentiation of algorithms: theory, implementation, and application. SIAM, Philadelphia, pp 1–16 · Zbl 0782.65028 [11] Iri M, Tsuchiya T, Hoshi M (1988) Automatic computation of partial derivatives and rounding error estimates with applications to large-scale systems of nonlinear equations. Comput Appl Math 24:365–392. Original Japanese version appeared in J Inf Process 26:1411–1420 (1985) · Zbl 0677.65015 [12] Lefèvre V, Muller JM, Tisserand A (1998) Toward correctly rounded transcendentals. IEEE Trans Comput 47(11): 1235–1243 · Zbl 05106162 · doi:10.1109/12.736435 [13] Li XS, Demmel JW, Bailey DH, Henry G, Hida Y, Iskandar J, Kahan W, Kapur A, Martin MC, Tung T, Yoo DJ (2002) Design, implementation and testing of extended and mixed precision blas. ACM Trans Math Softw 28:152–205. http://crd.lbl.gov/$$\sim$$xiaoye/XBLAS · Zbl 1070.65523 [14] Linnainmaa S (1976) Taylor expansion of the accumulated rounding error. BIT (Nordisk Tidskrift for Informationsbehandling) 16(1): 146–160 · Zbl 0332.65024 [15] MathWorks T. Matlab. http://www.mathworks.com/ [16] Michelsen M (1982) The isothermal flash problem. Part I: stability. Fluid Phase Equilibria 9(1): 1–19 · doi:10.1016/0378-3812(82)85001-2 [17] Miller W, Wrathall C (1980) Software for Roundoff analysis of matrix algorithms. Academic Press, New York · Zbl 0474.65017 [18] Muller JM, Brisebarre N, de Dinechin F, Jeannerod CP, Lefèvre V, Melquiond G, Revol N, Stehlé D, Torres S (2010) Handbook of floating-point arithmetic. Birkhäuser, Boston. ACM G.1.0; G.1.2; G.4; B.2.0; B.2.4; F.2.1. ISBN:978-0-8176-4704-9 · Zbl 1197.65001 [19] Ogita T, Rump S, Oishi S (2005) Accurate sum and dot product. SIAM J Sci Comput 26(6): 1955–1988 · Zbl 1084.65041 · doi:10.1137/030601818 [20] Ring M (2001) MAPM, a portable arbitrary precision math library in C. C/C++ Users J. http://www.tc.umn.edu/$$\sim$$ringx004/mapm-main.html [21] Rump S (2001) Rigorous and portable standard functions. BIT 41(3): 540–562 · Zbl 0994.65017 · doi:10.1023/A:1021971313412 [22] Rump S, Böhm H (1983) Least significant bit evaluation for arithmetic expressions. Computing 30: 189–199 · Zbl 0503.65031 · doi:10.1007/BF02253892 [23] Society IC (2008) IEEE standard for floating-point arithmetic. Tech rep, Microprocessor Standards Committee of the IEEE Computer Society [24] Speelpenning B (1980) Compiling fast partial derivatives of functions given by algorithms. PhD thesis, University of Illinois at Urbana, Champaign [25] Wilkinson J (1971) Modern error analysis. SIAM Rev 13: 548–568 · Zbl 0243.65018 · doi:10.1137/1013095 [26] Ziv A (1991) Fast evaluation of elementary mathematical functions with correctly rounded last bit. ACM Trans Math Softw 17: 410–423 · Zbl 0900.65038 · doi:10.1145/114697.116813
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
Non commutative ring without unity.

Find order of smallest non-commutative ring.

According to me, the order should be 4. 2×2 matrix with 2 non-zero entries over finite field Z2 and bottom row is zero.This matrix has no unity.

• I’m not sure what your ring elements are. Are these in the ring? What is $\begin{pmatrix} 1 & 1\\ 0& 0\end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}?$ – Dan Robertson Nov 29 '17 at 13:55
• Also what is your definition of a (noncommitative) ring? You can’t have a noncommutative ring with no unity if your definition requires the existence of a unit. Often a ring with no unit is called a rng (a ring with no ‘i’). – Dan Robertson Nov 29 '17 at 13:58
• Non commutative ring means it is not commutative under multiplication.I'm talking about the 2×2 matrix with only 2 non zero entries at any place and element are from finite field Z2. – Seeker Nov 29 '17 at 14:03
• @user378511 This ring has a unit, but is not closed under addition. – lisyarus Nov 29 '17 at 14:07
• @lisyarus take bottom row of the matrix zero. – Seeker Nov 29 '17 at 14:08

Yes, your example (pieced together from the original post and the comments) $\begin{bmatrix}F_2&F_2\\0&0\end{bmatrix}$ is a noncommutative rng without identity of order $4$.

Since $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ acts as an identity on the left, it would have to be equal to any candidate two-sided identity, if it existed, but $\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\neq \begin{bmatrix}0&1\\0&0\end{bmatrix}$

Obviously you can't have such a ring with $1$ or $2$ elements.

If such a ring had $3$ elements, say $\{0,a,b\}$, all distinct, then $a+b$ must be somewhere in this set. $a+b\in \{a,b\}$ creates a contradiction, so $a+b=0$ necessarily, so that $b=-a$. But multiplication in $\{-a,0,a\}$ necessarily commutes.

What you gave is actually a nice representation for one of my favorite examples of semigroup rings. You take the semigroup $S=\{a,b\}$ with multiplication defined by $a^2=ba=a$ and $b^2=ab=b$, and make the semigroup ring $F_2[S]$. I think this is the same ring as if we had taken $a=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $b=\begin{bmatrix}1&1\\0&0\end{bmatrix}$.

• So the question should be asked what is the order of smallest non-commutative "rng"? Right? – Seeker Nov 29 '17 at 14:41
• @user378511 "rng=ring possibly without identity". If I was given free reign with the question I would pose it as "what's the smallest ring that isn't commutative and doesn't have identity?" but that's just me. How you wrote it is understandable. – rschwieb Nov 29 '17 at 14:44
• Oh yeah!This sounds more understandable to me. Next time I'll be more precise with the language ;) – Seeker Nov 29 '17 at 14:49


text:
# Find the 500-th term of an arithmetic sequence with {eq}a_1 = 6.9 \text{ and } d = 0.3 {/eq}.

## Question:

Find the 500-th term of an arithmetic sequence with {eq}a_1 = 6.9 \text{ and } d = 0.3 {/eq}.

## Arithmetic Sequence

An arithmetic sequence is a sequence in which each term and its previous term have the same difference. This difference is known as the common difference. Since we know the first term and the common difference in the question we can find the nth term of the sequence by using the formula,

{eq}a_n = a_1 + (n-1)d {/eq}, where {eq}a_n {/eq} is the {eq}n^{th} {/eq} term and {eq}d {/eq} is a common difference.


text:
+ 2 like - 0 dislike
1164 views

In the heart of the original derivation of the AdS/CFT is the decoupling limit. From the Supergravity point of view ($g_s N \gg 1$), say for D3 branes, this amounts to taking the near horizon limit $r/R \ll 1$ with $R$ the AdS radius.  (See the well known review)

• Is there any paper that discusses different ways of taking a decoupling limit? (I am aware of the non-conformal case of other D-branes with running dilatons) ex: http://arxiv.org/pdf/hep-th/9802042.pdf
• For example, if one considers more elaborate constructions of D-brane configurations -ex compact say spherical or toroidal (Say we put a stack of N spherical D-branes together) one can in principle form a lot of dimensionless ratios that one can tune, is the obstacle finding SUGRA solutions with these geometric properties - or are in general these solutions  unstable?
edited May 28, 2014

+ 1 like - 0 dislike

Well, i totally agree that i did not give a very well posed problem, but I was just curious if other people had specific examples...(I have not yet tried to work out this problem myself)

Anyway, what I get from Lubos answer is that in principle one might be able to do this but of course he needs to choose a specific configuration that is stable and then discuss if this is interesting...

Personally, my original motivation (these are my thoughts so they might be totally wrong..) was that when one takes the decoupling limit the radius of AdS and of the sphere are the same and one cannot tune their ratio.... Thus we need an extra parameter to do this and I was wondering whether it can come from other brane configurations...

For me as far as I understand AdS/CFT  branes are non perturbative objects but their effects scale like $e^{\frac{1}{g_s}}$ , so gravitationally they are not "enough" to create bh as single objects...

So we put a lot of them N and when we both take N and g to be large then they collapse and form the black hole.

Apparently, since we want a stable configuration etc we take an extremal bh and the near horizon limit is the well known $AdS_5 \times S^5$.

Now if one could cook up a stable configuration having an extra parameter (this is what I do not know how to do and what role do different configurations play in their near horizon geometry) I believe that in principle one can get a near horizon limit with a small sphere compared to ads.......(so this parameter to tune the relative size of the two)

answered May 28, 2014 by anonymous

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#### Lower Bounds for Approximate Knowledge Compilation

##### Alexis de Colnet, Stefan Mengel

Knowledge compilation studies the trade-off between succinctness and efficiency of different representation languages. For many languages, there are known strong lower bounds on the representation size, but recent work shows that, for some languages, one can bypass these bounds using approximate compilation. The idea is to compile an approximation of the knowledge for which the number of errors can be controlled. We focus on circuits in deterministic decomposable negation normal form (d-DNNF), a compilation language suitable in contexts such as probabilistic reasoning, as it supports efficient model counting and probabilistic inference. Moreover, there are known size lower bounds for d-DNNF which by relaxing to approximation one might be able to avoid. In this paper we formalize two notions of approximation: weak approximation which has been studied before in the decision diagram literature and strong approximation which has been used in recent algorithmic results. We then show lower bounds for approximation by d-DNNF, complementing the positive results from the literature.

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This documentation is not using the current rendering mechanism and will be deleted by December 31st, 2020. The extension maintainer should switch to the new system. Details on how to use the rendering mechanism can be found here.

# 1.5.9 TYPO3 LTS 4.5¶

* TYPO3 LTS 4.5 support version
* Security fix


text:
## B.6 Named Vectors

### B.6.1 Creating Named Vectors

Vectors in R can be named – each element can be assigned a string label.

x <- c(20, 40, 99, 30, 10)
names(x) <- c("a", "b", "c", "d", "e")
x # a named vector
##  a  b  c  d  e
## 20 40 99 30 10

Other ways to create named vectors include:

c(a=1, b=2, c=3)
## a b c
## 1 2 3
structure(1:3, names=c("a", "b", "c"))
## a b c
## 1 2 3

For instance, the summary() function returns a named vector:

summary(x) # NAMED vector, we don't want this here yet
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
##    10.0    20.0    30.0    39.8    40.0    99.0

This gives the minimum, 1st quartile (25%-quantile), Median (50%-quantile), aritmetic mean, 3rd quartile (75%-quantile) and maximum.

Note that x is still a numeric vector, we can perform various operations on it as usual:

sum(x)
## [1] 199
x[x>3]
##  a  b  c  d  e
## 20 40 99 30 10

Names can be dropped by calling:

unname(x)
## [1] 20 40 99 30 10
as.numeric(x) # we need to know the type of x though
## [1] 20 40 99 30 10

### B.6.2 Subsetting Named Vectors with Character String Indices

It turns out that extracting elements from a named vector can also be performed by means of a vector of character string indices:

x[c("a", "d", "b")]
##  a  d  b
## 20 30 40
summary(x)[c("Median", "Mean")]
## Median   Mean
##   30.0   39.8


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# $p$-Splittable Integers

Let $$p$$ be a positive integer. For each nonnegative integer $$k$$, write $$[k]$$ for the set $$\{0,1,2,\ldots,k\}$$. Also, we define $$[-1]:=\emptyset$$. We say that an integer $$k\geq -1$$ is $$p$$-splittable if there is a partition of $$[k]$$ into $$p$$ subsets $$A_1$$, $$A_2$$, $$\ldots$$, $$A_p$$ such that $$\sum_{x\in A_1}\,x=\sum_{x\in A_2}\,x=\ldots=\sum_{x\in A_p}\,x$$ (i.e., these sets have the same sum). Such a partition $$\left\{A_1,A_2,\ldots,A_p\right\}$$ is called a $$p$$-splitting of $$[k]$$.

What are all $$p$$-splittable integers for a given $$p$$? How many $$p$$-splittings of $$[k]$$ are there for each of these available $$k$$'s? If the exact number of $$p$$-splittings of $$[k]$$ is not easily computable, then what is the asymptotic answer?

Clearly, $$k=-1$$ and $$k=0$$ are $$p$$-splittable. We can also ignore the trivial case $$p=1$$. We know that, if $$p=2$$, then all $$p$$-splittable numbers are of the forms $$4t-1$$ and $$4t$$, where $$t$$ is a nonnegative integer. If $$p$$ is an odd prime, then all $$p$$-splittable numbers are integers of the forms $$tp-1$$ and $$tp$$, where $$t\in\{0,2,3,4,\ldots\}$$.

For $$p=2$$, we can show that the number of $$2$$-splittings of a $$2$$-splittable integer $$k$$ is given by the coefficient of $$x^{\frac{k(k+1)}{4}}$$ in the expansion of $$\prod_{r=1}^k\,\left(1+x^r\right)$$. For example, if $$k=3$$, there are two $$2$$-splittings of $$[3]$$, namely, $$\big\{\{0,1,2\},\{3\}\big\}$$ and $$\big\{\{1,2\},\{0,3\}\big\}$$, whereas $$\prod_{r=1}^3\,\left(1+x^r\right)=1+x+x^2+2x^3+x^4+x^5+x^6$$ whose coefficient of $$x^{\frac{k(k+1)}{4}}=x^3$$ is also $$2$$. Similarly, there are $$2$$, $$8$$, and $$14$$ $$2$$-spittings of $$[k]$$ for $$k=4$$, $$k=7$$, and $$k=8$$, respectively. I do not know if there is any closed form for this coefficient for an arbitrary $$2$$-splittable $$k$$.

I conjecture the following:

(1) If $$p$$ is odd, then, for any $$j\in\{-1,0,1,2,\ldots,p-2\}$$ such that $$p\mid j(j+1)$$, every integer of the form $$tp+j$$, where $$t\in\{2,3,4,\ldots\}$$, is $$p$$-splittable, and nothing else (except $$-1$$ and $$0$$) is $$p$$-splittable.

(2) If $$p$$ is even, then, for any $$j\in\{-1,0,1,2,\ldots,2p-2\}$$ such that $$2p\mid j(j+1)$$, every integer of the form $$2tp+j$$, where $$t$$ is a positive integer, is $$p$$-splittable, and nothing else (except $$-1$$ and $$0$$) is $$p$$-splittable.

This conjecture is true, at least, if $$p$$ is a prime power (where $$j=-1$$ and $$j=0$$ are the only possible choices of $$j$$). If you can show that (1) holds for $$t=2$$ and for $$t=3$$, and that (2) holds for $$t=1$$, then you are done. It is worth noting that, if $$k$$ is $$p$$-splittable, then $$k+2p$$ is $$p$$-splittable.

P.S. I include $$k=0$$ and $$k=-1$$ for the sake of completeness. There is nothing subtle about these numbers.

• Just curious, but is there a reason you went for combi-number-theory rather than combinatorial-number-theory? Is the latter tag not allowed by the system due to its length? – pjs36 Dec 5 '15 at 19:48
• Yes, there is a 25-character limit. – Batominovski Dec 5 '15 at 19:48
• Would arithmetic-combinatorics work better as a tag, or does that have a different flavor to it? Admittedly I don't really know how that area defines itself - just asking :-) – Jyrki Lahtonen Dec 8 '15 at 7:45
• I don't feel like this is an arithmetic combinatorics problem, if my understanding of arithmetic combinatorics is not far off. I could be wrong. – Batominovski Dec 8 '15 at 7:53
• Ok. I believe you! – Jyrki Lahtonen Dec 8 '15 at 7:53

likely last edit, formatting for clarity:

We say $n$ is $p$-splittable if there is a partition $A_1,...,A_p$ of $\{1,...,n\}$ with $\sum A_i := \sum_{a \in A_i} a = \sum A_j$ for all $i,j\leq p$

We call $n$ uniformly $p$-splittable if there is such a partition with $|A_i|=|A_j|$ for all $i,j \leq p$

We call such a partition a $p$-split of $n$

Let $s_p(n)$ $(\bar{s}_p(n))$ denote the number of (uniform) $p$-splits of $n$

Some truths:

$\bar{s}_p \leq s_p$

If $n$ is $p$-splittable then $p| \frac{(n+1)n}{2}$

If we have equality $s_p(n)=1$ (there is exactly one $p$-split of $n$)

$2p$ and $2p-1$ are $p$-splittable, $2p$ uniformly.

If $n$ is $p$-splittabe and $p'|p$ then $n$ is $p'$-splittable and we obtain a lower bound of $s_{p'}(n)\geq \frac {p!}{(\frac{p}{p'}!)^{p'}}$ (the number of ways to make a $p$-split into a $p'$-split by melting groups of $\frac{p}{p'}$ together)

(it may be possible to get a bound in terms of $s_p(n)$ if one can argue away double counting)

If $n$ is uniformly $p$-splittable then $mn$ is uniformly $p$-splittable for all m$\in \mathbb{N}$

If $n$ is $p$-splittable and $k$ is uniformly $p$-splittable, then $m+k$ is $p$-splittable

If $m,n$ are uniformly $p$-splittable then $m+n$ is uniformly $p$-splittable

The $p$-splittable numbers are then a finite union of (affine) copies of the uniformly $p$-splittables and are generated by finitely many primitive $p$-splittable numbers, a trivial upper bound on the count of primitives is the smallest non-zero uniformably $p$-splittable number $2p$. A better bound is achieved by $\# \{j \in \{-1,0,...,2p-2\} \; with \; 2p|j(j+1)\}$

It is conjectured this bound is exact and the primitive $p$-splittables are of the form $2p+j$ for $j$ in this set. (This is numerically proven for $p\leq 184$)

If $n$ is uniformly $p$-splittable then $n$ must be a multiple of $p$. I will give a full categorization:

Let $p \in \mathbb{N}$ then the uniformly $p$-splittables are $p \mathbb{N} \backslash p$ if $p$ is odd and $2p \mathbb{N}$ if p is even (or $\mathbb{N}$ for $p=1$)

proof: It is apparent that $2p$ is uniformly $p$-splittable and $p$ is not. It suffices to show that for odd $p$ $3p$ is uniformly $p$-splittable and for even $p$ no odd multiple is.

Let $p$ be even and $k$ odd. Then $\frac{kp(kp+1)}{2}$ is no multiple of $p$ (it is a odd multiple of $\frac p 2$)

Let $p$ be odd:

let $[i]$ denote $i \; mod \; p$

Then a uniform $p$-split of $3p$ is given by $A_i = \{ 1+[i], (p+1)+[\frac{p-1}{2}+i], 2p+1+[p-1-2i]\}$ for $i=0,...,p-1$ Obviously each $A_i$ has 3 elements and each number $\leq 3p$ is represented. We need to show $[i]+[\frac{p-1}{2}+i]+[p-1-2i]$ is independent of $i$.

This is true because for $i \leq (p-1)/2$ all the arguments are in $(0,...,p-1)$ and for $(p+1)/2 \leq i \leq p-1$ we have $[\frac{p-1}{2}+i]=\frac{p-1}{2}+i-p$, $[p-1-2i]=p-1-2i+p$ in each case the sum is equal to $\frac{3(p-1)}{2}\square$

$\bar{s}_p(p^2) \geq (p-1)!$ (by construction of $(p-1)!$ uniform $p²$-splits)

Things to look into:

• The original conjecture. It can be achieved by

1. simply constructing $p$-splits for $2p+j$ for the given $j$ (note $j=-1,0$ are trivial) or

2. Constructing an algorithm for it. Problems here lie in proving the algorithm finishes, this leads into study of finding how to split a set into heaps of (differing) given sizes (splits would be the special case where all stacks have equal size)

3. Note that the requirement on $j$ is equivalent to $j \equiv -1$ or $0 \mod p_i$ for all prime powers $p_i|p$ (note this proves the conjecture for prime $p$ after giving a split for $3p-1$ ($p$ odd) like here)

• Said study of "fitting" a set into a given tuple of integers $(d_1,...,d_n)$ (ie finding a partition with $\sum A_i = d_i$) Here the sets of the form $\{1,...,n\}$ are of a special importance for proving things about splittability

• Study of splittability for sets $A\subset \mathbb{N} \neq \{1,...,n\}$

• Study of $s_p$ and/or $\bar{s}_p$, I didn't look into the number of splits except when a bound just fell out of a proof.

• Bounty award goes to you due to your significant attempt, despite the incomplete answer. – Batominovski Dec 15 '15 at 21:09
• My critique would be that the work needs cleaning up. For example, by $2pN\setminus p$ for odd $p$, I think you meant $p\mathbb{N}\setminus\{p\}$. Also, $\mod{}$ (with TeX command \mod) and $\lim$ (with TeX command \lim) are better than $mod$ and $lim$ – Batominovski Dec 15 '15 at 21:11
• I'll likely do some major reformatting tomorrow and may or may not put this to rest depending on whether i make any progress on the current working conjecture of "if n>2p and there is a p-split of n+2p then for every partition of $\{n+1,...,n+2p\}$ into p sets of size two there is a p-split of n+2p over that partition. (currently backed by numerical evidence for n=5, p=3 and by the requirement becoming weaker the bigger n becomes wrt p) proving that would essentially reduce the initial conjecture to calculating out all the n with p<n<3p (most of them are trivial because 2p|n(n+1)) any ideas? – obstkuchen Dec 15 '15 at 22:55


text:
Definitions
Nearby Words

# Biquandle

## Biracks

In mathematics, biquandles and biracks are generalizations of quandles and racks. Whereas the hinterland of quandles and racks is the theory of classical knots, that of the bi-versions, is the theory of virtual knots.

Biquandles and biracks have two binary operations on a set $X$ written $a^b$ and $a_b$. These satisfy the following three axioms:

1. $a^\left\{b\left\{c_b\right\}\right\}= \left\{a^c\right\}^\left\{b^c\right\}$

2. $\left\{a_b\right\}_\left\{c_b\right\}= \left\{a_c\right\}_\left\{b^c\right\}$

3. $\left\{a_b\right\}^\left\{c_b\right\}= \left\{a^c\right\}_\left\{b^c\right\}$

These identities appeared in 1992 in reference [FRS] where the object was called a species.

The superscript and subscript notation is useful here because it dispenses with the need for brackets. For example if we write $a*b$ for $a_b$ and $a**b$ for $a^b$ then the three axioms above become

1. $\left(a**b\right)**\left(c*b\right)=\left(a**c\right)**\left(b**c\right)$

2. $\left(a*b\right)*\left(c*b\right)=\left(a*c\right)*\left(b**c\right)$

3. $\left(a*b\right)**\left(c*b\right)=\left(a**c\right)*\left(b**c\right)$

For other notations see .

If in addition the two operations are invertible, that is given $a, b$ in the set $X$ there are unique $x, y$ in the set $X$ such that $x^b=a$ and $y_b=a$ then the set $X$ together with the two operations define a birack.

For example if $X$, with the operation $a^b$, is a rack then it is a birack if we define the other operation to be the identity, $a_b=a$.

For a birack the function $S:X^2-> X^2$ can be defined by

$S\left(a,b_a\right)=\left(b,a^b\right).,$

Then

1. $S$ is a bijection

2. $S_1S_2S_1=S_2S_1S_2 ,$

In the second condition, $S_1$ and $S_2$ are defined by $S_1\left(a,b,c\right)=\left(S\left(a,b\right),c\right)$ and $S_2\left(a,b,c\right)=\left(a,S\left(b,c\right)\right)$. This condition is sometimes known as the set-theoretic Yang-Baxter equation.

To see that 1. is true note that $S\text{'}$ defined by

$S\text{'}\left(b,a^b\right)=\left(a,b_a\right),$

is the inverse to

$S ,$

To see that 2. is true let us follow the progress of the triple $\left(c,b_c,a_\left\{bc^b\right\}\right)$ under $S_1S_2S_1$. So

$\left(c,b_c,a_\left\{bc^b\right\}\right) to \left(b,c^b,a_\left\{bc^b\right\}\right) to \left(b,a_b,c^\left\{ba_b\right\}\right) to \left(a, b^a, c^\left\{ba_b\right\}\right).$

On the other hand, $\left(c,b_c,a_\left\{bc^b\right\}\right) = \left(c, b_c, a_\left\{cb_c\right\}\right)$. Its progress under $S_2S_1S_2$ is

$\left(c, b_c, a_\left\{cb_c\right\}\right) to \left(c, a_c, \left\{b_c\right\}^\left\{a_c\right\}\right) to \left(a, c^a, \left\{b_c\right\}^\left\{a_c\right\}\right) = \left(a, c^a, \left\{b^a\right\}_\left\{c^a\right\}\right) to \left(a, b_a, c_\left\{ab_a\right\}\right) = \left(a, b^a, c^\left\{ba_b\right\}\right).$

Any $S$ satisfying 1. 2. is said to be a switch (precursor of biquandles and biracks).

Examples of switches are the identity, the twist $T\left(a,b\right)=\left(b,a\right)$ and $S\left(a,b\right)=\left(b,a^b\right)$ where $a^b$ is the operation of a rack.

A switch will define a birack if the operations are invertible. Note that the identity switch does not do this.

## Biquandles

A biquandle is a birack which satisfies some additional structure, as described by Nelson and Rische. It should be noted that the axioms of a biquandle are "minimal" in the sense that they are the weakest restrictions that can be placed on the two binary operations while making the biquandle of a virtual knot invariant under Reidemeister moves.

In Preparation

In Preparation

In Preparation

## References

• [FJK] Roger Fenn, Mercedes Jordan-Santana, Louis Kauffman Biquandles and Virtual Links, Topology and its Applications, 145 (2004) 157-175
• [FRS] Roger Fenn, Colin Rourke, Brian Sanderson An Introduction to Species and the Rack Space, in Topics in Knot Theory (1992), Kluwer 33-55
• [K] L. H. Kauffman, Virtual Knot Theory, European J. Combin. 20 (1999), 663--690.
Search another word or see Biquandleon Dictionary | Thesaurus |Spanish


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# Leki Replacement Expander - Pair

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£3.49

## Quick Overview

Standard Poles:

• 18mm for the upper section

• 16mm for the lower section

Ultralite Poles:

• 16mm for the upper section

• 14mm for the lower section

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## Details

If your old poles are slipping, or not holding at all, these are the parts that sit on a thread on the end of the poles inside. To access, simply loose, then pull the sections fully apart, then they can be replaced.

These will fit all Leki poles unless they have an 'SLS' or 'ELS' graphic at the top or bottom of the pole. Sizes are generally as follows, but should be double checked before ordering. Also make sure the existing expander is red with 4 prongs at the top.

Standard Poles:

• 18mm for the upper section
• 16mm for the lower section

Ultralite Poles:

• 16mm for the upper section
• 14mm for the lower section

## Additional Information

Description If your old poles are slipping, or not holding at all, these are the parts that sit on a thread on the end of the poles inside. To access, simply loose, then pull the sections fully apart, then they can be replaced. These will fit all Leki poles unless they have an 'SLS' or 'ELS' graphic at the top or bottom of the pole. Sizes are generally as follows, but should be double checked before ordering. Also make sure the existing expander is red with 4 prongs at the top. Standard Poles: 18mm for the upper section 16mm for the lower section Ultralite Poles: 16mm for the upper section 14mm for the lower section 880 Leki Trekking Poles

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To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free)
Top

Surface Area of a Triangular Prism Calculator
Top
Surface Area of a Triangular prism Calculator is used find the surface area. The parameters to find the surface area perimeter, the sides, height of the prism and the base area.

Step by Step Calculation

Back to Top
Step 1 :

Formula:

Surface Area = $\frac{1}{2}$(b * h)2 + (S1 + S2 + S3)H

Surface Area = (b * h) + (S1 + S2 + S3)H

Where

b = base

h = height

H = Prism height

S1, S2, S3 = Sides of triangle

Step 2 :

Put the values in the formula and calculate it further.

Example Problems

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1. Find the surface area of triangular prism whose sides are equilateral with 4cm, the prism height is 6 cm and base height is 3.4 cm.

Step 1 :

Given: The surface area of triangular prism whose sides are equilateral with 4 cm.

S1, S2, S3 = 4 cm

b = 4 cm

h = 3.4 cm

H = 6 cm

Formula:

Surface Area = $\frac{1}{2}$(b * h)2 + (S1 + S2 + S3)H

Surface Area = (b * h) + (S1 + S2 + S3)H

Step 2 :

Put the values in the formula

Surface Area = (4 * 3.4) + (4 + 4 + 4)*6

Surface Area = 85.6 cm2

Answer  :

Surface Area = 85.6 cm2

2. Find the surface area of triangular prism whose sides s1 = 5 m, s2 = 4 m and s3 = 3 m. The prism height is 7 cm and base height is 4.5 cm.

Step 1 :

Given: The surface area of triangular prism whose sides are equilateral with 4 cm.

S1 = 5 m

S2 = 4 m

S3 = 3 m

b = 4 m

h = 7 m

H = 4.5 m

Formula:

Surface Area = $\frac{1}{2}$(b * h)2 + (S1 + S2 + S3)H

Surface Area = (b * h) + (S1 + S2 + S3)H

Step 2 :

Put the values in the formula

Surface Area = (4 * 4.5) + (5 + 4 + 3)*7

Surface Area = 102 m2

Answer  :

Surface Area = 102 m2

*AP and SAT are registered trademarks of the College Board.


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# What do E1 and E2 reactions have in common?

Both consist of a step where the leaving group leaves, and one where a proton is abstracted to form a $\pi$ bond.


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Three of the four vertices of a rectangle in the xy-coordinate plane

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Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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10 Jun 2015, 04:19
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Three of the four vertices of a rectangle in the xy-coordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?

(A) (–4.5, 2.5)
(B) ( –4, 5)
(C) (6, –2 )
(D) (7, –3 )
(E) (10, 1)

Kudos for a correct solution.

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Re: Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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10 Jun 2015, 07:59
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Bunuel wrote:
Three of the four vertices of a rectangle in the xy-coordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?

(A) (–4.5, 2.5)
(B) ( –4, 5)
(C) (6, –2 )
(D) (7, –3 )
(E) (10, 1)

Kudos for a correct solution.

ALTERNATE

Length of Diagonals are same in a Rectangle

i.e. Distance Between ( –5, 1) and (8, 0) should be same as Distance between ( –4, 4) and 4th Vertex (x, y)

i.e. $$(0-1)^2+(8+5)^2 = (y-4)^2+(x+4)^2$$

i.e. $$1+169 = (y-4)^2+(x+4)^2$$

Check Options:

4th Vertex can only be in either 1st or 4th Quadrant so Option A and B are already out of sync

Option (C) - (6, –2 ): (y-4)^2+(x+4)^2 = (-6)^2 + (10)^2 = 136 so INCORRECT
(D) (7, –3 ): (y-4)^2+(x+4)^2 = (-7)^2 + (11)^2 = 49+121 = 170 so CORRECT

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Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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10 Jun 2015, 07:51
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Bunuel wrote:
Three of the four vertices of a rectangle in the xy-coordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?

(A) (–4.5, 2.5)
(B) ( –4, 5)
(C) (6, –2 )
(D) (7, –3 )
(E) (10, 1)

Kudos for a correct solution.

ALTERNATIVE

Slope of Line joining points A( –5, 1) and B( –4, 4) = (4-1) / [(-4)-(-5)] = 3/1 = m1

Slope of Line joining points B( –4, 4) and C( 8, 0) = (0-4) / [(8)-(-4)] = -4/12 = -1/3 = m2

This also suggests that 4th Point must be in Quadrant IV i.e Option C or D only can be true

i.e. AB and BC lines are perpendicular as m1 * m2 = -1

Slope of Line joining points A(–5, 1) and D( x, y) = (y-1) / (x+5) = m3

Slope of Line joining points C(8, 0) and D( x, y) = (y-0) / (x-8) = m4

Now m3 * m4 = -1

i.e. [(y-1) / (x+5)] * [(y-0) / (x-8)] = -1

i.e $$y(y-1) = -(x+5)*(x-8)$$

Let's Check Option C (6, -2)
LHS = (-2)(-2-1) = 6

RHS = (6+5)(6-8) = -22

INCORRECT

Let's Check Option D (7, -3)
LHS = (-3)(-3-1) = 12

RHS = -(7+5)(7-8) = 12

CORRECT

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Re: Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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13 Jun 2015, 06:43
Draw it on the paper, it takes a few seconds to "guess" the correct answer.

Then we could check the result with the formula for the Distance between two points.

Distance btw (-5;1) and (-4;4) is Sqrt(10)

then the correct point must be distant from (8;0) by Sqrt(10)
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Re: Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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13 Jun 2015, 08:15
3
Actually this is very easy to be solved if you draw on a paper.

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Re: Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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15 Jun 2015, 02:46
1
3
Bunuel wrote:
Three of the four vertices of a rectangle in the xy-coordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?

(A) (–4.5, 2.5)
(B) ( –4, 5)
(C) (6, –2 )
(D) (7, –3 )
(E) (10, 1)

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Your GMAT scratchpad has a grid; use it to plot the diagram to scale.
Attachment:

2015-06-15_1342.png [ 13.23 KiB | Viewed 9612 times ]

“Eyeball” solution: The corner at ( –4, 4) looks like a right angle, so complete the rectangle with the dashed lines shown. The 4th point must be located approximately where the bigger dot is drawn. The closest answer choice is the point (7, –3 ). Alternatively, you could plot all of the answer choice points and see which one “works” with the three given points.

Alternatively, we could solve by checking the slopes of the solid lines we drew between the given points to prove those lines are perpendicular, that is, to prove we have drawn two sides of the rectangle correctly.

The slope of the long solid line = (4- 0)/(-4 - 8) = -1/3.
The slope of the short solid line = (4 - 1)/(-4 - (-5)) = 3.

The product of these slopes is -1/3*3 = -1, proving that the lines are perpendicular.

Compute the location of the 4th point, using the fact that the short sides have the same slope. The known short side connects the points ( –5, 1) and ( –4, 4). In other words, the bottom left corner is 1 to the left and 3 down from the top left corner. The unknown bottom right corner should therefore be 1 to the left and 3 down from the top right corner, or x = 8 – 1 = 7 and y = 0 – 3 = –3, corresponding to the point (7,–3).

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Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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28 Jun 2015, 12:19
Bunuel wrote:
Bunuel wrote:
Three of the four vertices of a rectangle in the xy-coordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?

(A) (–4.5, 2.5)
(B) ( –4, 5)
(C) (6, –2 )
(D) (7, –3 )
(E) (10, 1)

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Your GMAT scratchpad has a grid; use it to plot the diagram to scale.
Attachment:
2015-06-15_1342.png

“Eyeball” solution: The corner at ( –4, 4) looks like a right angle, so complete the rectangle with the dashed lines shown. The 4th point must be located approximately where the bigger dot is drawn. The closest answer choice is the point (7, –3 ). Alternatively, you could plot all of the answer choice points and see which one “works” with the three given points.

Alternatively, we could solve by checking the slopes of the solid lines we drew between the given points to prove those lines are perpendicular, that is, to prove we have drawn two sides of the rectangle correctly.

The slope of the long solid line = (4- 0)/(-4 - 8) = -1/3.
The slope of the short solid line = (4 - 1)/(-4 - (-5)) = 3.

The product of these slopes is -1/3*3 = -1, proving that the lines are perpendicular.

Compute the location of the 4th point, using the fact that the short sides have the same slope. The known short side connects the points ( –5, 1) and ( –4, 4). In other words, the bottom left corner is 1 to the left and 3 down from the top left corner. The unknown bottom right corner should therefore be 1 to the left and 3 down from the top right corner, or x = 8 – 1 = 7 and y = 0 – 3 = –3, corresponding to the point (7,–3).

Hi Bunuel, please explain why the rectangle cant be such that one side joins the points (-5,1) and (8,0), and the opposite side joins the points (-4,4) and the unknown? Thats what I did initially, but it led to a wrong answer. What (apparently obvious) clues did I miss?

My approach was this: two parallel lines have equal slopes so slope for the line joining (-5,1) and (8,0) = slope for the line joining the points (-4,4) and the unknown. For the latter part (slope involving unknown point), x coordinate would be one unit more than (8,0), since the difference b/w x coordinates is 1 unit here: (-5,1) and (-4,4). Then solved for y (the only unknown variable left now).

TIA!
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Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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28 Jun 2015, 21:59
mrish7 wrote:
Bunuel wrote:
Bunuel wrote:
Three of the four vertices of a rectangle in the xy-coordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?

(A) (–4.5, 2.5)
(B) ( –4, 5)
(C) (6, –2 )
(D) (7, –3 )
(E) (10, 1)

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Your GMAT scratchpad has a grid; use it to plot the diagram to scale.
Attachment:
2015-06-15_1342.png

“Eyeball” solution: The corner at ( –4, 4) looks like a right angle, so complete the rectangle with the dashed lines shown. The 4th point must be located approximately where the bigger dot is drawn. The closest answer choice is the point (7, –3 ). Alternatively, you could plot all of the answer choice points and see which one “works” with the three given points.

Alternatively, we could solve by checking the slopes of the solid lines we drew between the given points to prove those lines are perpendicular, that is, to prove we have drawn two sides of the rectangle correctly.

The slope of the long solid line = (4- 0)/(-4 - 8) = -1/3.
The slope of the short solid line = (4 - 1)/(-4 - (-5)) = 3.

The product of these slopes is -1/3*3 = -1, proving that the lines are perpendicular.

Compute the location of the 4th point, using the fact that the short sides have the same slope. The known short side connects the points ( –5, 1) and ( –4, 4). In other words, the bottom left corner is 1 to the left and 3 down from the top left corner. The unknown bottom right corner should therefore be 1 to the left and 3 down from the top right corner, or x = 8 – 1 = 7 and y = 0 – 3 = –3, corresponding to the point (7,–3).

Hi Bunuel, please explain why the rectangle cant be such that one side joins the points (-5,1) and (8,0), and the opposite side joins the points (-4,4) and the unknown? Thats what I did initially, but it led to a wrong answer. What (apparently obvious) clues did I miss?

My approach was this: two parallel lines have equal slopes so slope for the line joining (-5,1) and (8,0) = slope for the line joining the points (-4,4) and the unknown. For the latter part (slope involving unknown point), x coordinate would be one unit more than (8,0), since the difference b/w x coordinates is 1 unit here: (-5,1) and (-4,4). Then solved for y (the only unknown variable left now).

TIA!

Hi mrish7,

Considering your argument, the slope of line joining points (-5,1) and (8,0) = -1/13

Now the third point (-4,4) must be joined with (-5, 1) or with (8, 0) to make the other side of rectangle and the new line must have slope of 13 because Product of slopes of two perpendicular lines = -1

Case 1: (-4,4)is joined with (-5, 1), the slope of this line = 3/1

Case 2: (-4,4)is joined with (8, 0) the slope of this line = 4/(-12) = -1/3

Both cases rejected. Therefore this assumption is INCORRECT

I hope it helps!
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Re: Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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15 Oct 2016, 20:09
Isnt there is a formula to do the same?
x coordinate (x1+x3-x2) and y coordinate (y1+y2-y3)??
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Re: Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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15 Oct 2016, 20:14
1
Just derived - The correct formula is
x coordinate- ( x1+x3-x2) and y coordinate is (y1+y3-y2).
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Re: Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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15 Oct 2016, 20:49
AmritaSarkar89 wrote:
Just derived - The correct formula is
x coordinate- ( x1+x3-x2) and y coordinate is (y1+y3-y2).

Although your previous score is already pretty decent but if you are considering taking GMAT again to improve it further then a small piece of advice is "Do away with formulas" as much as possible.

A person getting the score as your profile shows can't be illogical and logical people should understand maths logically. It rewards them with greater score always

In Coordinate geometry, ALWAYS DRAW THE FIGURE while solving questions... It makes your understanding very smooth and easy
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Re: Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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15 Oct 2016, 21:18
GMATinsight wrote:
AmritaSarkar89 wrote:
Just derived - The correct formula is
x coordinate- ( x1+x3-x2) and y coordinate is (y1+y3-y2).

Although your previous score is already pretty decent but if you are considering taking GMAT again to improve it further then a small piece of advice is "Do away with formulas" as much as possible.

A person getting the score as your profile shows can't be illogical and logical people should understand maths logically. It rewards them with greater score always

In Coordinate geometry, ALWAYS DRAW THE FIGURE while solving questions... It makes your understanding very smooth and easy

Yes I am planning on retaking to hit something close to 750. The problem is since our childhood we had been pushed into memorizing formulas and till date I am struggling to get away with that canopy on my thought process.
Honestly yes, I am trying to logically breakdown every problem
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Re: Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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29 Apr 2018, 10:47
Three of the four vertices of a rectangle in the xy-coordinate plane are ( –5, 1), ( –4, 4), and (8, 0). What is the fourth vertex?

(A) (–4.5, 2.5)
(B) ( –4, 5)
(C) (6, –2 )
(D) (7, –3 )
(E) (10, 1)

Since diagonals of rectangles are Equal ,we can calculate the distance between the diagonals and other two points of vertex must also satisfy that distance .
Points ( –5, 1), and (8, 0) have a distance of sq root of 170 and similarly point ( –4, 4) third vertex and fourth vertex points mentioned in the option must satisfy this distance which is option D(7,-3).
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Re: Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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13 May 2018, 09:34
As it’s a rectangle, opposite sides are equal and parallel.
So the slope of the opposites is going to be same m1=m2= 0-4/8+4 =-1/3
As we have the slope and (-5,1) eq of line obtained: 3y+2x=-2.
As it’s already clear that the point will lying in the fourth quadrant, so eliminate options and substitue it on the eq of the line.
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Re: Three of the four vertices of a rectangle in the xy-coordinate plane  [#permalink]

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13 May 2018, 14:57
Without solving, we have y = 1, 4, 0. 1-4 = -3; hence, we need |3| as the difference in y which implies that the last y must be 3 or -3. Only D makes this true so D.
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Re: Three of the four vertices of a rectangle in the xy-coordinate plane &nbs [#permalink] 13 May 2018, 14:57
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text:
# The purpose of expansionary fiscal policy is to _____. A slow the growth of the GDP B prevent hyperinflation C increase the separation between government and private industry D increase output

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B prevent hyperinflation
C increase the separation between government and private industry
D increase output

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text:
# Probability-Equally likely or not?

Probability

Question. "$11$ identical balls are distributed in $4$ distinct boxes randomly. Then the probability that any $3$ boxes will together get a greater number of balls than the remaining one is:"

I am confused about the distribution of balls. Are all the cases i.e. $\{5,5,0,1\}$ and $\{2,3,4,2\}$ equally likely or not?

• No, those two cases are not equally likely (according to the most straightforward reading of the problem). Each ball is equally likely to be put into any of the four boxes, but it is not equally likely for all eleven to be put into box 1, for example, than it is for two to be put into box 1, and three each to be put into the other boxes. The question is asking you to identify what the distribution of number in boxes is, and to use that in your answer. – Brian Tung Sep 17 '17 at 20:04
• By the way, "i.e." means "that is"; this would imply that your two examples are all the cases. You probably mean "e.g.," which means "for example." (It's a trivial issue, but sometimes mathematics turns on trivialities!) – Brian Tung Sep 17 '17 at 20:05

The point is to find all combinations of natural numbers $(a,b,c,d)$ such that $\max\{a,b,c,d\}\le 5$ and $a+b+c+d=11$. Without loss of generality, assume $a\ge b\ge c\ge d$. Notice that $\max\{a,b,c,d\}$ can only be $3$, or $4$, or $5$.

• If $\max\{a,b,c,d\}=3$, then the only possible combination is $(3,3,3,2)$;
• If $\max\{a,b,c,d\}=4$, then the possible combinations are $(4,4,3,0)$, $(4,4,2,1)$, $(4,3,3,1)$,$(4,3,2,2)$;
• If $\max\{a,b,c,d\}=5$, then possible combinations include $(5,5,1,0)$,$(5,4,2,0)$,$(5,4,1,1)$,$(5,3,3,0)$, $(5,3,2,1)$,$(5,2,2,2)$.

Together, since four boxes are distinct, for each combination above, the number of ways to rearrange is

• For $(5,4,2,0)$ and $(5,3,2,1)$: $A^4_4$;
• For $(4,4,3,0)$, $(4,4,2,1)$,$(4,3,3,1)$,$(4,3,2,2)$,$(5,5,1,0)$,$(5,4,1,1)$, $(5,3,3,0)$: $A^4_4/A^2_2$;
• For $(3,3,3,2)$ and $(5,2,2,2)$:$A^4_4/A^3_3$

Since the total number of events are $4^{11}$, the probability is $$\frac{2A^4_4+7A^4_4/A^2_2+2A^4_4/A^3_3}{4^{11}}.$$

• That number seems too low. I think you must also multiply each box configuration by the ways the balls can end up in that configuration. For example, for $(5,4,2,0)$ the first $5$ balls could go to box 1, or they could be the $4$ first ones and the sixth, and so on. Simulation gives the answer somewhere around $0.86$. – ploosu2 Sep 18 '17 at 9:09

The solution given by OnoL has the correct configurations for the ball amounts in the boxes but it is missing the possible ways the balls can be placed to the boxes. Because we are considering all the ways $11$ balls can be placed to $4$ boxes ($4^{11}$) these multipliers must be considered. They are also different for different configuration (this answers your second question: the different cases have different probabilities). Luckily it doesn't depend on the order of the boxes so we can still only consider the ordered configurations (and then multiply by the number of permutations of that configuration).

The way $11$ balls can be placed to form $(a, b, c, d)$ is

$${{11}\choose{a}} {{11-a}\choose{b}} {{11-a-b}\choose{c}}$$

(The balls for the first box can be chosen freely, then choose from the rest $11-a$ balls $b$ balls for the second and so on. The last (fourth) box has to have the remaining balls so that coefficient ${{11-a-b-c}\choose{d}} = {{d}\choose{d}} = 1$, since $a+b+c+d=11$.)

The number of ball placings that satisfy the claim "any $3$ boxes will together get a greater number of balls than the remaining one" is given by the sum over all possible configurations (these were found by OnoL)

$(3,3,3,2), (4,4,3,0), (4,4,2,1), (4,3,3,1), (4,3,2,2), (5,5,1,0), (5,4,2,0), (5,4,1,1), (5,3,3,0), (5,3,2,1), (5,2,2,2)$

where the summand is the product

$($number of permutations of that configuration$)$ $\times$ $($the number of ways to place the balls to form that configuration$)$.

I calculated this with a computer and got $3618384$. So the wanted probability is

$$\frac{3618384}{4^{11}} = \frac{226149}{262144} \approx 0.86269.$$


text:
# The invariant 3-form on a compact Lie group

Let $G$ be a compact Lie group. We have the well-known Maurer-Cartan left-invariant and right-invariant $1$-forms $\theta$ and $\bar\theta$ in $\Omega^1(G, \mathfrak{g})$, probably discussed in every Lie theory lectures.

However the canonical bi-invariant closed $3$-form $\chi = \frac{1}{12} (\theta, [\theta, \theta]) = \frac{1}{12} (\bar\theta, [\bar\theta, \bar\theta])$ in $\Omega^3(G)$ may be a little bit less-known. And when I heard of it I had some questions in mind...

1) Are there canonical invariant $5$-forms and higher invariant forms on a $G$?

2) How are they related to Lie algebra cohomology and equivariant cohomology?

3) The construction looks a little bit like $tr(A \wedge dA)$, if we regard $\theta$ as a connection $A$ on the frame bundle of $G$, and use the Maurer-Cartan equation $d\theta = -\frac{1}{2}[\theta, \theta]$.

Now there is another famous $3$-form: the Chern-Simons form $tr(A \wedge dA + \frac{2}{3} A \wedge A \wedge A)$, and I wonder if these two are somehow related. Does the $tr(A \wedge A \wedge A)$ part vanish here, due to Jacobi identity?

(and, note there are Chern-Simons 5-forms etc.)

Thank you very much.

-

The relation between the three dimensional Chern-Simon term and the invariant 3-form is a special case of the BRST bicomplex which acts on the Lie algebra valued equivariant forms of a principal bundle. Locally, these forms depend polynomially on the gauge connection of the principal bundle and on the Maurer-Cartan one form of the Lie group of the gauge transformations.

One starts from the Chern-Simons term (in any dimension), and through a series of descent equations (involving the BRS operator and the exterior derivative) reaches an invariant odd degree form of the group manifold. At every stage, the descent equations (Sometimes called the Stora-Zumino descent equations) increase the polynomial degree of the Maurer-Cartan by 1 and decreases the degree of the gauge connection by one.

Here is a reference with the explicit construction starting from the five dimensional Chern-Simons term.

The intermediate terms in the descent are actually cocycles of an Abelian extentsion of the gauge group Lie algebra known as the Fadeev-Mickelsson extension.

Please, see also the following Quantum field theory books (first), (second) treating this subject.

-

The answers to your questions form part of a somewhat long and beautiful story which is explained, e.g., in Greub, Halperin and Vanstone's Connections, Curvature, and Cohomology. Vol. 2: Lie Groups, Principal Bundles, and Characteristic Classes. There is no Google Books preview available, but you can at least read a review here.

-

The 3-form in question is the extension to de Rham cohomology of the group of the canonical semisimple Lie algebra 3-cocycle $\langle - [-,-] \rangle$.

A Lie algebra cocycle is the same as a closed element in the Chevalley-Eilenberg algebra. If any Lie algebra cocycle is in transgression with an invariant polynomial on the Lie algebra, then this is witnessed by the existence of a corresponding Chern-Simons element in the Weil algebra. The detailed construction is recalled here.

From this simple cohomological construction result most of the phenomena in higher Chern-Simons theory relating these algebraic entities. I have once written up some notes on this with Jim Stasheff and Hisham Sati (from section 6.3 on it discusses what I just mentioned).

-


text:
A Fractions can be defined as any part of a whole number, a small part or even a piece or amount. A fraction is made up of a numerator and a denominator, where the numerator would tell us how many parts of the whole there are, a while the denominator would give us an idea of how many parts are in the whole.

When and if the numerators are found to be smaller than the denominator then the fraction is considered to be a proper fraction.When and if the numerators are found to be greater than the denominators then the fraction is labelled as an improper fraction.When and if a fraction is accompanying a whole number then it is considered as a mixed number. In the recent past, the usage of fraction was done to describe shares of objects or group objects, but now these fractions have been replaced by decimals and the calculations are often completed with computers or calculators.

## What is a Fraction?

A fraction is a number representing a part of a whole. The whole may be a single object or a group of objects and the parts have to be equal.

The general form of a fraction is a/b , where a and b are integers and b ? 0. Here a is called the numerator and b is called the denominator.

In the following figure, 4 out of 6 sectors are shaded. Hence the shaded region represents $\frac{4}{6}$.

### Examples on Fractions

In the figure (i) $\frac{3}{7}$ is obtained when we divide a whole into 7 equal parts and take 3 parts of it.

In figure (ii) $\frac{1}{8}$ we divide a whole into 8 equal parts and take 1 part out of it.

Now, $\frac{5}{12}$ is a fraction and we read it as "five-twelfths", where 12 is number of parts into which a whole has been divided and 5 is number of equal parts which have been taken out.
Here 5 is called the Numerator and 12 is called the Denominator.

## Types of Fractions

Depending on the relative values of the numerator and the denominator, fractions are classified as:

a) Proper fractions
b) Improper fractions
c) Mixed fractions

Proper fractions

Consider the fraction $\frac{(1)}{(2)}$ , $\frac{(3)}{(8)}$ , $\frac{(2)}{(9)}$ and $\frac{(10)}{(35)}$ . We can see that in all these fractions, the numerator is lesser than the denominator. Hence the value of these fractions is lesser than 1. Such fractions are called proper fractions.

Improper fractions

Unlike proper fractions, the numerator of an improper fraction will be greater than its denominator, and so the value of these fractions will be more than 1. Examples of improper fractions are $\frac{(4)}{(3)}$ , $\frac{(5)}{(2)}$ , $\frac{(29)}{(5)}$ , $\frac{(101)}{(11)}$ etc.

Mixed fractions

These types of fractions are a combination of a whole number and a part; that is a whole number and a proper fraction. Some examples are 5$\frac{(1)}{(2)}$ , 10$\frac{(3)}{(8)}$ , 9$\frac{(11)}{(13)}$ etc.

Adding fractions is an important concept in fractions. To add two fractions, we have to make sure the denominator is the same for both the fractions and then we have to add the numerator with out disturbing the denominator.

Given below are some examples for adding fractions

Find the sum $\frac{1}{5}$ + $\frac{2}{5}$

Solution:

Step 1: Add the numerators: 1 + 2 = 3
Step 2: Retain the common denominator, 5
Step 3: Sum of the fractions = $\frac{(Sum\ of\ the\ Numerators)}{(Common\ denominator)}$ =$\frac{3}{5}$

Based on this the addition of fractions is divided into two types:
• Adding fractions with like denominators
• Adding fractions with unlike denominators

## Subtracting Fractions

Subtracting fractions is an important concept. To subtract two fractions, we have to make sure the denominator is same for both the fractions and then we have to subtract the numerator with out disturbing the denominator.

Example for Subtracting a Fraction

Example for subtracting fraction:

Find $\frac{7}{11}$ - $\frac{3}{11}$

Solution:

$\frac{7}{11}$ - $\frac{3}{11}$ = $\frac{7-3}{11}$ = $\frac{4}{11}$

Based on this the subtraction of fractions divided into two types,
• Subtracting fractions with like denominators
• Subtracting fractions with unlike denominators

## Subtracting Negative Fractions

According to rule, a subtraction signs will applies to fraction as well. To subtract a negative fraction the denominator must be same. After getting common denominator we subtract numerators.

### Examples on Subtracting Negative Fractions

Below you could see example based up on subtracting negative fractions

Example: $-\frac{6}{4}$ - $(-\frac{3}{5})$

Solution: Given $-\frac{6}{4}$ - $(-\frac{3}{5}$)

Step 1: Denominators are not same, so find the LCM of the denominator, LCM of 4,5 =20.
Step 2: Express the fractions as equivalent fractions with the denominator as the LCM.

$-\frac{6}{4}$ = $-\frac{6\times5}{4\times5}$ = $-\frac{30}{20}$

$(-\frac{3}{5})$ = $(-\frac{3\times4}{5\times4})$ = $(-\frac{12}{20})$

Step 3: Now subtract negative fractions

$-\frac{30}{20}$ - $(-\frac{12}{20})$ = $-\frac{18}{20}$

Step 4: Simplify the fraction

$-\frac{18}{20}$ = $-\frac{9}{20}$

## Subtracting Fractions with Like Denominators

Subtract fractions with like denominators. We subtract the numerator with out disturbing denominators in this part.

In order to find the difference between two fractions with common denominators, use the following steps:
• Step 1: Find the difference between the numerators.
• Step 2: Retain the common denominator.
• Step 3: Difference between the fractions = $\frac{Difference\ between\ the\ Numerators}{The\ Common\ Denominator}$

### Examples of Subtracting Fractions with Like Denominators

Given are some examples based on subtracting fractions with like denominators

Find $\frac{3}{5}$ - $\frac{2}{5}$

Solution:

Step 1: Difference between the numerators = 3 -2 = 1.
Step 2: Retain the common denominator, 5.
Step 3: Difference between the fractions = $\frac{Difference\ between\ the\ Numerators}{The\ Common\ Denominator}$ = $\frac{1}{5}$

## Subtracting Fractions with Unlike Denominators

In order to subtract a fraction from another fraction with different denominators, we first find equivalent fractions with the same denominator. The common denominator is the least common multiple (LCM) of the denominators.

The difference of fractions with different denominators can be obtained through the following steps:
• Step 1: Find the LCM of the denominators
• Step 2: Express the fractions as equivalent fractions with the denominator as the LCM
• Step 3: Subtract the smaller numerator from the bigger one
• Step 4: Retain the common denominator (LCM)
• Step 5: Difference of fractions = Difference of numerators from step 3 / Common denominator (LCM)

### Examples for Subtracting Fractions with Unlike Denominators

Given are some examples based on subtracting fractions with unlike denominators

1. Subtract $\frac{3}{4}$ from $\frac{5}{6}$

Solution: We have to find; $\frac{5}{6}$ - $\frac{3}{4}$

Step 1: Find the LCM of the denominator, LCM of 4,6 = 12
Step 2: Express the fractions as equivalent fractions with the denominator as the LCM
$\frac{5}{6}$ =$\frac{5\times2}{6\times2}$ = $\frac{10}{12}$ , and $\frac{3}{4}$ = $\frac{3\times3}{4\times3}$ = $\frac{9}{12}$
Step 3: Subtract the smaller numerator from the bigger one;
10-9 = 1
Step 4: Retain the common denominator, 12
Step 5: Difference of the fractions= Difference of numerators from step 3 / Common denominator (LCM) = $\frac{1}{12}$

2. Find $\frac{49}{8}$ - $\frac{11}{4}$

Solution:

The LCM of the denominators, 8 and 4 is 8

So we have $\frac{11}{4}$ = $\frac{11\times2}{4\times2}$ = $\frac{22}{8}$

Hence, $\frac{49}{8}$ - $\frac{11}{4}$ = $\frac{49}{8}$ - $\frac{22}{8}$

= $\frac{49-22}{8}$

= $\frac{27}{8}$

## Multiplying Fractions

In order to multiply two fractions, we multiply the numerators together and then multiply the denominators.

The product of two fractions can be obtained through the following steps.
Step 1: Multiply the numerators
Step 2: Multiply the denominators
Step 3: The product of the fractions = $\frac{Product\ of\ the\ Numerators}{Product\ of\ the\ Denominators}$
Step 4: Write the product in its simplest form
We can reduce the fractions into the simplest form by dividing both the numerator and the denominator by the greatest common factor.

### Examples for Multiplying Fractions

Below are some examples based on multiplying fractions

1. Find the product of $\frac{2}{5}$ and $\frac{3}{7}$

Solution:

Step 1: Multiply the numerators, 2 × 3=6
Step 2: Multiply the denominators, 5 × 7= 35
Step 3: The product of the factors = $\frac{Product\ of\ the\ Numerators}{Product\ of\ the\ Denominators}$
= $\frac{6}{35}$
Here we cannot reduce this fraction further as the greatest common factor of
the numerator and denominator is 1.

2. Find the product of $\frac{2}{7}$ and $\frac{1}{14}$

Solution:

Let us follow the steps to find the product,

$\frac{2}{7}$ × $\frac{1}{14}$ = $\frac{2\times1}{7\times14}$

= $\frac{2}{98}$

= $\frac{1}{49}$ , by dividing the numerator and denominator by 2

## Multiplying Fractions with Variables

When multiplying two or more fractions, the numerators are all multiplied together along with the denominators.

This is made possible by multiplying fractions that contains variables.

One method to carry out this operation is to reduce the multiplying fractions before the actual multiplication.

When we multiply a fraction, we have to look out for common numerator and denominator in the fractions and need not be in same fraction.

Example for multiplication of fraction with variables:

Let us multiply $\frac{10}{21}$ , $\frac{14}{35}$ , and $\frac{42}{32}$

As this would finally result into a huge value, so these are reduced further so that this operation becomes manageable.

The common factor for the numerator and the denominator work out and gives us a result of $\frac{1}{4}$ .

$\frac{10}{21}$ X $\frac{14}{35}$ X $\frac{42}{32}$ = $\frac{2\times2\times2}{32}$ = $\frac{1}{4}$

## Dividing Fractions

A fraction obtained by interchanging the numerator and denominator of a given fraction is called the Reciprocal of the fraction.

We can divide a fraction by another fraction through the following steps:
• Step 1: Find the reciprocal of the divisor.
• Step 2: Multiply the dividend by the fraction from step 1.
• Step 3: Write the reduced form of the fraction obtained.

Thus in general, we can write,

$\frac{a}{b}$ ÷ $\frac{c}{d}$ = $\frac{a}{b}$ × $\frac{d}{c}$ where a, b, c, and d are integers.

### Examples for Dividing Fractions

Some examples based on dividing fractions

1. Find $\frac{2}{15}$ ÷ $\frac{1}{3}$

Solution:

Step 1: Find the reciprocal of the divisor, reciprocal of $\frac{1}{3}$ is $\frac{3}{1}$

Step 2: Multiply the dividend by the fraction from step 1, $\frac{2}{15}$ × $\frac{3}{1}$ = $\frac{6}{15}$

Step 3: Write the reduced form of the fraction obtained,

25$\frac{6}{15}$ = $\frac{2}{5}$, by removing the greatest common factor, 3, of the numerator and the denominator

Hence, $\frac{2}{15}$ ÷ $\frac{1}{3}$ = $\frac{2}{5}$

2. Divide $\frac{6}{17}$ by $\frac{24}{51}$

Solution:

$\frac{6}{17}$ ÷ $\frac{24}{51}$ = $\frac{6}{17}$ × $\frac{51}{24}$

= $\frac{3}{4}$ , by removing the greatest common factor, 6 × 17 , of the numerator and the denominator.

## Comparing Fractions

Two fractions $\frac{a}{b}$ and $\frac{c}{d}$ can be compared as follows

Cross multiply as shown:

(i) If a $\geq$ d = b $\geq$ c, then $\frac{a}{b}$ = $\frac{c}{d}$

(ii) If a $\leq$ d = b $\leq$ c, then $\frac{a}{b}$ = $\frac{c}{d}$

(iii) If a = d = b = c, then $\frac{a}{b}$ = $\frac{c}{d}$

Note:

(i) If the numerator and the denominator of a fraction are both multiplied or divided by the same non-zero number, then the value of the fraction remains unchanged.

(ii) Every whole number is a fraction since every whole number can be written as $\frac{a}{1}$

(iii) A given fraction and the fraction obtained by multiplying (or dividing) both of the numerator and the denominator by the same non-zero number are called equivalent fractions.

### Examples for Comparing Fractions

Given below are some examples on comparing fractions:

1. Which of the given fractions $\frac{3}{4}$ , $\frac{5}{7}$ is greater ?

Solution: Cross multiply the given fractions as directed above

We get: 3 x 7 = 21 and 4 x 5 = 20

Clearly, 21 > 20

Therefore, $\frac{3}{4}$ is greater.

2. Compare the fractions $\frac{7}{8}$ , $\frac{8}{9}$ ?

Solution: Cross multiply the given fractions as directed above $\frac{7}{8}$ , $\frac{8}{9}$

We get: 9 x 7 = 63 and 8 x 8 = 64

Clearly, 64 > 63

Therefore $\frac{7}{8}$ = $\frac{8}{9}$

## Ordering Fractions

Arranging of fractions either in ascending (increasing) order or descending (decreasing) order is called the ordering of fractions.

If we pick up two like fractions the one with the greatest numerator is greater and the one with the least denominator is the smallest.

### Examples for Ordering Fractions

Given below are some examples on Ordering Fractions

1. Arrange the given fractions in ascending order of magnitude.

$\frac{7}{12}$ , $\frac{5}{12}$ , $\frac{17}{12}$ , $\frac{3}{12}$ and $\frac{13}{12}$

Solution:

The given fractions are $\frac{7}{12}$ , $\frac{5}{12}$ , $\frac{17}{12}$ , $\frac{3}{12}$ and $\frac{13}{12}$

Since the denominators of the fractions are the same, they are called like fractions

To arrange the fractions in the ascending order we see the values of the numerators of the fractions from the smallest to the greatest.

Therefore, the ascending order is $\frac{3}{12}$ , $\frac{5}{12}$ , $\frac{7}{12}$ , $\frac{13}{12}$ and $\frac{17}{12}$

2. Arrange the given fractions in descending order of magnitude.

$\frac{28}{140}$ , $\frac{60}{140}$ , $\frac{98}{140}$ , $\frac{65}{140}$ and $\frac{13}{140}$

Solution:

The given fractions are $\frac{28}{140}$ , $\frac{60}{140}$ , $\frac{98}{140}$ , $\frac{65}{140}$ and $\frac{13}{140}$

Since the denominators of the fractions are the same, they are called like fractions
To arrange the fractions in a descending order we see the values of the numerators of the fractions from the greatest to the smallest

Therefore, the ascending order is $\frac{13}{140}$ , $\frac{28}{140}$ , $\frac{60}{140}$ , $\frac{65}{140}$ and $\frac{98}{140}$

## Fraction to Decimal

Fractions can be represented as decimals. The accurate and the sure shot way of converting a fraction into decimal is by long form division. The quotient gives the required decimal.

Another method of converting fractions to decimals

For some fractions we can find a number such that if the denominator is multiplied by this number it will result in a power of 10. Such fractions can be converted to decimals by the following steps:

Step 1: Identify a number such that if the denominator of the fraction is multiplied by this number will result in a power of 10.

Step 2: Multiply the numerator and the denominator of the fraction with the same number as in Step 1.

Step 3: Place the decimal point such that there are the same number of digits to the right of the decimal point as in the denominator.

### Examples for converting fraction into decimal

Below are some examples based on converting fractions into decimals

1. Convert $\frac{2}{3}$ to a decimal number.

Solution:

We know that $\frac{2}{5}$ = 2 ÷ 5

Let us perform the long form division.

Hence, $\frac{2}{5}$ = 0.4

2. Convert $\frac{7}{8}$ to a decimal.

Solution:

We know that $\frac{7}{8}$ = 7 ÷ 8

Hence , $\frac{7}{8}$ = 0.875

In the above examples, the division resulted in zero remainder after a finite number of steps. But here are cases where the remainder does not become zero. This results in non-terminating decimals.

3. Convert $\frac{1}{3}$ to a decimal.

Solution:

Performing long term division we get,

The remainder does not become zero.

Hence , $\frac{1}{3}$ = 0.333...

The decimal does not terminate. Here 3 repeats in the decimal. This is represented as $\frac{1}{3}$ = 0.3

4. Convert $\frac{3}{5}$ to a decimal.

Solution:

Step 1: Identify a number such that if the denominator of the fraction is multiplied by this number it will result in a power of 10.

We can see that if we multiply the denominator by 2 we get a power of 10 in the denominator.

Step 2: Multiply the numerator and the denominator of the fraction with the same number in the Step 1.

= $\frac{3}{5}$ = $\frac{(3\times2)}{(5\times2)}$

= $\frac{6}{10}$

Step 3: Place the decimal point such that there are the same number of digits to the right of the decimal point as in the denominator.

We get, = $\frac{3}{5}$ = $\frac{6}{10}$ = 0.6

## Fraction to Percent

Percent literally means out of 100. A percent can be represented as a fraction and as a decimal number. We have learned how to write fractions as decimal numbers. The accurate and the sure shot way of converting a fraction to a decimal is by long form division. The quotient gives the required decimal. We have to multiply this decimal number by 100 and place the percent symbol with the answer.

If the denominator of the fraction is 100, we can write the equivalent percent by removing the denominator and place the percent symbol with the numerator.

Consider the following diagram

In this diagram, 3 out of 4 boxes are shaded. We write this as $\frac{3}{4}$ of the diagram is shaded.

We know $\frac{3}{4}$ = $\frac{(3\times25)}{(4\times25)}$ , writing an equivalent fraction by multiplying the numerator and denominator by 25

Here we made the denominator 100, by multiplying it by 25

Hence we get, $\frac{3}{4}$ = $\frac{75}{100}$

So we can say $\frac{75}{100}$ of the diagram is shaded. In other words 75% of the diagram is shaded.

### Examples on Fraction to Percent

Given are some examples on converting fraction to percent

1. Write the fraction; $\frac{4}{5}$ as a percent

Solution:

Performing long division, we get,

$\frac{4}{5}$ = 0. 8

Multiplying this number by 100, 0.8 100= 80.0

Hence 0.8 =80%

$\frac{4}{5}$ = 80%

Alternate method:

We make the denominator of the given fraction as 100.

$\frac{4}{5}$ = $\frac{(4\times20)}{(5\times20)}$ , multiplying the numerator and denominator by 20

$\frac{4}{5}$ = $\frac{80}{100}$

= 80%

2. Convert the fraction; $\frac{7}{8}$

Solution

By performing long division, we get,

$\frac{7}{8}$ = 0.875

0.875 x 100 = 87.5

Therefore, $\frac{7}{8}$ = 87.5%

## Simplifying Fractions

Simplifying fractions could also result in reducing the fractions to its lowest possible equivalent.

The steps for simplifying fractions

Step 1: Start with the lowest term which can divide both the numerator and the denominator until you cannot proceed further.

Step 2: Divide the fraction term by the greatest common factor.

### Examples for Simplifying Fractions

Given below are some examples on simplifying fractions

1. Simplify the fraction $\frac{4}{14}$

Solution :
Given fraction is $\frac{4}{14}$

4 = 2 × 2
14 = 2 × 7
Greatest common factor of 4 and 14 is 2.

$\frac{4}{14}$ × $\frac{2}{2}$ = $\frac{2}{7}$

The simplest form of the given fraction is $\frac{2}{7}$

2. Simplify the fraction $\frac{55}{66}$

Solution :
Given fraction is $\frac{55}{66}$

55 = 11 × 5
66 = 11 × 6
The greatest common factor of 55 and 66 is 11.

$\frac{55}{66}$ × $\frac{11}{11}$ = $\frac{5}{6}$

The simplest form of the given fraction is $\frac{5}{6}$


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# E91 protocol's error rate

E91 protocol detects the eavesdropper by using CHSH inequality. If the value of CHSH inequality is equal to 2sqrt(2), then there is no eve. On the other hand, if the value is between -sqrt(2) and sgrt(2), there is eve. The question is, if the value is neither between -sqrt(2) and sgrt(2) nor equal to 2sqrt(2), then what happens to the eve's presence.


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# Calculus: Derivatives: Product and Quotient Rule – #474

Question: Find $$f'(x)$$ when $$f(x)=\frac{5{{x}^{2}}+4}{{{x}^{2}}+10}$$.