# A Most Curious Algebraic Identity

I recently found a very interesting Algebraic Identity: $xyz+(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)$

What's so special about it? Note that going from one side of the equality to the other, all products are switched with sums, and all sums are switched with products!

This may be seen a bit easier if I rewrite it as follows: \begin{aligned} &{~}\color{#D61F06}x \color{#D61F06}{\times} y\color{#D61F06}{\times }z \color{#FFFFFF}{)}\color{#3D99F6}{+} \color{grey}{(}x\color{#3D99F6}{+}y\color{grey}{)}\color{#D61F06}{\times}\color{grey}{(}y\color{#3D99F6}{+}z\color{grey}{)}\color{#D61F06}{\times} \color{grey}{(}z\color{#3D99F6}{+}x\color{grey}{)}\\ =&\color{grey}{(}x\color{#3D99F6}{+}y\color{#3D99F6}{+}z\color{grey}{)}\color{#D61F06}{\times}\color{grey}{(}x\color{#D61F06}{\times} y\hspace{0.9ex}\color{#3D99F6}{+}\hspace{0.9ex}y\color{#D61F06}{\times }z\color{#FFFFFF}{)}\color{#3D99F6}{+}\hspace{0.9ex}z\color{#D61F06}{\times }x\color{grey}{)} \end{aligned}

Cool!

Note by Daniel Liu
5 years, 10 months ago

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Yes; I really like this identity too. For example it can be used to prove that $r_1 + r_2 + r_3 - r = 4R$ (from Incircles and Excircles). If we substitute $x = s-a$ etc., then

$s(s-b)(s-c)+s(s-c)(s-a)+s(s-a)(s-b)-(s-a)(s-b)(s-c) = abc$

where $s$ is the semi-perimeter, and this reduces nicely using area formulas to the desired relationship.

- 5 years, 10 months ago

@Michael Ng created this problem which uses the identity.

Staff - 5 years, 10 months ago

thanks @Daniel Liu i used this to solve problems like this i wrote a solution using this identity, and i'm thinking about a problem with this identity, will post soon!

- 5 years, 9 months ago

I didn't realise that- thanks. It should inspire some good problems :)

- 5 years, 10 months ago

F B U L O U S !!!!!!

- 5 years, 2 months ago

Typo. $\color{#D61F06}{x}$ is missed in color version, line 5.

- 2 years, 5 months ago

I didn't examine. Good for making questions. However, it could have been found by people in the past.

- 5 years, 10 months ago

Yea, I'm just saying that I just noticed it. I most likely was not the person who discovered it (as seen by the comment by Michael Ng)

- 5 years, 10 months ago

I had just expanded both sides to compare. Should be correct. Do not feel disappointed by what I guessed. You could be the first person to find this. Congratulation!

- 5 years, 10 months ago

Thanks. I did not realize this very useful identity.

- 5 years, 10 months ago

Most awesome discoveries ever!Thanks,this must help a lot.

- 5 years, 9 months ago

It's following the rules of principle of duality

- 5 years, 2 months ago

No, that is not the principle of duality.

Staff - 5 years, 2 months ago

I mean l'll bit similar to that

- 5 years, 2 months ago


1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. Decomposition of tensor power of symmetric square. (1.5) Usually the conditions for µ (in Eq. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. The N-way Toolbox, Tensor Toolbox, â¦ A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components â¦. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. Symmetric tensors occur widely in engineering, physics and mathematics. CHAPTER 1. In section 3 a decomposition of tensor spaces into irreducible components is introduced. For N>2, they are not, however. Since the tensor is symmetric, any contraction is the same so we only get constraints from one contraction. This chapter provides a summary of formulae for the decomposition of a Cartesian second rank tensor into its isotropic, antisymmetric and symmetric traceless parts. 1.4) or Î± (in Eq. If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? This makes many vector identities easy to prove. Yes. : USDOE â¦ The trace decomposition theory of tensor spaces, based on duality, is presented. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? Physics 218 Antisymmetric matrices and the pfaï¬an Winter 2015 1. This decomposition, ... ^2 indicates the antisymmetric tensor product. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. Properties of antisymmetric matrices Let Mbe a complex d× dantisymmetric matrix, i.e. gular value decomposition:CANDECOMP/PARAFAC (CP) decomposes a tensor as a sum of rank-one tensors, and the Tucker decomposition is a higher-order form of principal component analysis. LetT be a second-order tensor. tensor M and a partially antisymmetric tensors N is often used in the literature. If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. THE INDEX NOTATION Î½, are chosen arbitrarily.The could equally well have been called Î± and Î²: vâ² Î± = n â Î²=1 AÎ±Î² vÎ² (âÎ± â N | 1 â¤ Î± â¤ n). An alternating form Ï on a vector space V over a field K, not of characteristic 2, is defined to be a bilinear form. Cartan tensor is equal to minus the structure coeï¬cients. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. (antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. OSTI.GOV Journal Article: DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. [3] Alternating forms. A related concept is that of the antisymmetric tensor or alternating form. There are many other tensor decompositions, including INDSCAL, PARAFAC2, CANDELINC, DEDICOM, and PARATUCK2 as well as nonnegative vari-ants of all of the above. The trace decomposition equations for tensors, symmetric in some sets of superscripts, and antisymmetric â¦ Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share â¦ Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric â¦ A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Google Scholar; 6. A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. First, the vector space of finite games is decomposed into a symmetric subspace and an orthogonal complement of the symmetric subspace. Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. ARTHUR S. LODGE, in Body Tensor Fields in Continuum Mechanics, 1974 (11) Problem. We begin with a special case of the definition. The alternating tensor can be used to write down the vector equation z = x × y in suï¬x notation: z i = [x×y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 âx 3y 2, as required.) Each part can reveal information that might not be easily obtained from the original tensor. These relations may be shown either directly, using the explicit form of f Î±Î², and f Î±Î² * or as consequences of the HamiltonâCayley equation for antisymmetric matrices f Î±Î² and f Î±Î² *; see, e.g., J. PlebaÅski, Bull Acad. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. This means that traceless antisymmetric mixed tensor $\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. The symmetry-based decompositions of finite games are investigated. In these notes, the rank of Mwill be denoted by 2n. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Viewed 503 times 7. What's the significance of this further decomposition? Full Record; Other Related Research; Authors: Bazanski, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org. Thus, the rank of Mmust be even. A tensor is a linear vector valued function defined on the set of all vectors . Active 1 year, 11 months ago. Sponsoring Org. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Decomposition. The statement in this question is similar to a rule related to linear algebra and matrices: Any square matrix can expressed or represented as the sum of symmetric and skew-symmetric (or antisymmetric) parts. Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . MT = âM. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. Sci. Cl. It is a real tensor, hence f Î±Î² * is also real. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g The result of the contraction is a tensor of rank r 2 so we get as many components to substract as there are components in a tensor of rank r 2. Antisymmetric and symmetric tensors. Contents. Polon. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 AÄ± ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. Algebra is great fun - you get to solve puzzles! The result is Antisymmetric and symmetric tensors. This is exactly what you have done in the second line of your equation. Since det M= det (âMT) = det (âM) = (â1)d det M, (1) it follows that det M= 0 if dis odd. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. There is one very important property of ijk: ijk klm = Î´ ilÎ´ jm âÎ´ imÎ´ jl. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the elasticity tensor. An alternative, less well-known decomposition, into the completely symmetric part Sof C plus the reminder A, turns out to be irreducibleunder the 3-dimensional general linear group. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: This is an example of the Youla decomposition of a complex square matrix. 1.5) are not explicitly stated because they are obvious from the context. The trace of the tensor S is the rate of (relative volume) expansion of the fluid. : Lehigh Univ., Bethlehem, Penna. For more comprehensive overviews on tensor calculus we recom-mend [58, 99, 126, 197, 205, 319, 343]. Vector spaces will be denoted using blackboard fonts. Ask Question Asked 2 years, 2 months ago. In 3 dimensions, an antisymmetric tensor is dual to a vector, but in 4 dimensions, that is not so. â What symmetry does represent?Kenta OONOIntroduction to Tensors The bases of the symmetric subspace and those of its orthogonal complement are presented. 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# Yet another proof for $\zeta (2) = \frac { { \pi }^{ 2 } }{ 6 }$

Let us consider the integral

$\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta$

Now using

$\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta \\ \\ 1-{ e }^{ i\theta } = 1-cos\theta -isin\theta = 2sin(\frac { \theta }{ 2 } )(sin(\frac { \theta }{ 2 } )-icos(\frac { \theta }{ 2 } ))\\ = 2sin(\frac { \theta }{ 2 } ){ e }^{ i(\frac { \theta }{ 2 } -\frac { \pi }{ 2 } ) }$

we have the integral as

$\displaystyle \int _{ 0 }^{ \pi }{ (i(ln(2) + ln(sin(\frac { \theta }{ 2 } )) - \left( \frac { \theta }{ 2 } -\frac { \pi }{ 2 } \right) )d\theta }$

Both its real and imaginary parts can be easily evaluated to get

$\displaystyle \frac { { \pi }^{ 2 } }{ 4 }$ (Yes as you can see the imaginary parts cancel each other)

Now, reconsider the integral

$\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta$

Let us substitute

$\displaystyle z= { e }^{ i\theta }\$/extract_itex] at $\displaystyle \theta =0 \quad z=1\\ \theta =\pi \quad z=-1\\$ also $dz\quad =\quad i{ e }^{ i\theta }$ We get the integral $\displaystyle -\int _{ -1 }^{ 1 }{ ln(1-x)\frac { 1 }{ x } dx } \\$ Here i make use of the fact that the value of a definite integral depends only on the function and not on the varriable or its past using taylor expansion we get, $\displaystyle \int _{ -1 }^{ 1 }{ (\frac { 1 }{ 1 } } +\frac { x }{ 2 } +\frac { { x }^{ 2 } }{ 3 } ...)\quad =\quad 2(\frac { 1 }{ 1 } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...)\quad =\frac { { \pi }^{ 2 } }{ 4 } (as\quad proved\quad before)\\$ $\displaystyle \zeta (2)\quad =\quad \frac { 1 }{ 1 } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...+\frac { 1 }{ 4 } \zeta (2)\quad \quad (which\quad you\quad yourself\quad can\quad check)$ $\displaystyle \frac { 3 }{ 4 } \zeta (2) = \frac { \pi ^{ 2 } }{ 8 } \quad\rightarrow \boxed{\zeta(2)= \frac { \pi ^{ 2 } }{ 6 }}$ Hence proved Do point out any flaws i might have done Entirely original, any resemblance is accidental Inspritation - Ronaks proof (just inspiration to try to prove , not copy) Note by Mvs Saketh 4 years, 11 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote  # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or \[ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
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\boxed{123} $\boxed{123}$

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FanTastic Proof. Hats off

- 4 years, 11 months ago

Thanks :)

- 4 years, 11 months ago

Nice :) ...So, how did you come up with this? Were you fiddling around with infinite series?

- 4 years, 11 months ago

Well, i recently learnt some basics of contour integration, so i was fiddling around with the little knowledge i had and stumbled upon it :)

- 4 years, 11 months ago

That's nice :) ...I had thought that it might have in part been inspired by $\int_{0}^{1}\frac{ln(1-x)}{x}dx$ (both of them have ln(1-z) )

Either ways, it's pretty cool.

Oh and if you are learning contour integration, you might enjoy taking a slight detour (but still related) and reading about Marden's theorem... It so cool (assuming you haven't already seen it) :D .

- 4 years, 11 months ago

Yes that too did inspire,

Ok , i will check it out, i havent heard of it yet,as i just looked at the basics ,thanks

- 4 years, 11 months ago

Cheers :)

- 4 years, 11 months ago

Nice proof @Mvs Saketh :)

@Ronak Agarwal Beware !! Here's some competition for you !

- 4 years, 11 months ago

thankyou but that was never the intention, i just posted it because i liked it,

- 4 years, 11 months ago

I was just kidding Saketh !!

- 4 years, 11 months ago

i know, please dont use more than one $!$ at a time, it becomes hard to know whether you are shouting or telling :P

- 4 years, 11 months ago

Sorry , it's just that I'm not used to SMS language , i have learnt all that I know here from Brilliant , so pls bear with me :)

- 4 years, 11 months ago

Cheers :)

- 4 years, 11 months ago

Me Too.

- 4 years, 11 months ago

It seems that your 100 follower question got a level 5 rating , cheers:)

- 4 years, 11 months ago

Yes but it should be level 3

- 4 years, 11 months ago

No worries , your question got rated . That's what you wanted , no ? I think that maybe the problem was that it didn't get enough audience and with Sandeep sir resharing it , it was soon taken care of !

- 4 years, 11 months ago


# Thread: Solving Exponential Equations (logs involved)

1. ## Solving Exponential Equations (logs involved)

Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case!

6^(3x)=4^(2x-3)

6^(3x)=2^(4x-6)

We can use the log laws .... And it's likely we do... Anyone give me a hand?

2. Originally Posted by mike_302
Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case!

6^(3x)=4^(2x-3)

6^(3x)=2^(4x-6)

We can use the log laws .... And it's likely we do... Anyone give me a hand?
I am assuming this is a system of two equations...

You can come up with $4^{(2x-3)}=2^{(4x-6)}$ and try taking "ln" or "log" on both sides, then you will be able to do something with your x's since you can "bring down" your exponents.

3. No, the two equations are equal... That's as far as I solved. Sorry :P

And we haven't done ln .

EDIT: I see what you mean.

This is what I come up with next.

3xlog(6)=(2x-3)log(4)

3xlog(6)=2xlog4-3log4

log(64)=2xlog(4)-3xlog(6)

What next :S ?

4. Originally Posted by mike_302
No, the two equations are equal... That's as far as I solved. Sorry :P

And we haven't done ln .

I am not quite sure what you meant by them being equal.

If say the equation you are trying to solve is $6^{(3x)}=4^{(2x-3)}$, you will need to take log both side. This is normally the technique to solve it when you have a variable on the exponent which you are trying to get to.

EDIT:
I saw you edit your post.

log(64)=2xlog(4)-3xlog(6)
well, log(a number) is still a number. so this is similar to solving: $2=2x*5-3x*4
$

CAn you take it from here?

5. Originally Posted by mike_302
Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case!

6^(3x)=4^(2x-3)

6^(3x)=2^(4x-6)

We can use the log laws .... And it's likely we do... Anyone give me a hand?
It's really the same idea as the other one you posted. For instance, the first one, if we divide by $4^{2x}$, we get

$\frac{6^{3x}}{4^{2x}} = 4^{-3}$

$\implies \left(\frac{6^3}{4^2}\right)^x = \frac{1}{64}$

Just try to isolate x, and then solve.

6. Hello, mike_302!

Okay, I'll assume you've never seen one of these before.
I'll give you a walk-through . . .

Solve for $x\!:\;\;6^{3x} \:=\:4^{2x-3}$

Take logs of both sides: . $\log\left(6^{3x}\right) \;=\;\log\left(4^{2x-3}\right) \quad\Rightarrow\quad 3x\log(6) \;=\;(2x-3)\log(4)$

. . $3x\log(6) \:=\:2x\log(4) - 3\log(4) \quad\Rightarrow\quad 3x\log(6) - 2x\log(4) \:=\:-3\log(4)$

Factor: . $x\bigg[3\log(6) - 2\log(4)\bigg] \:=\:-3\log(4)$

Therefore: . $\boxed{x \;=\;\frac{-3\log(4)}{3\log(6) - 2\log(4)}}$

This answer can be simplified beyond all recognition . . .

$\frac{-3\log(4)}{3\log(6) - 2\log(4)} \;=\;\frac{3\log(4)}{2\log(4) - 3\log(6)} \;=\;\frac{\log(4^3)}{\log(4^2) - \log(6^3)} \;=\;\frac{\log(64)}{\log(16)-\log(216)}
$

. . $= \;\frac{\log(64)}{\log(\frac{16}{216})} \;=\;\frac{\log(64)}{\log(\frac{2}{27})} \;=\;\log_{\frac{2}{27}}(64)$ . . . see what I mean?

7. Okay, just as I was posting, I read Chop Suey's post... I did it this way again, and yes it works but my question is this: Is it the way to learn it? Or is it a quick work around? Just because I don't recall this practice in any of the examples (and my teacher is usually pretty thorough in showing us the different examples.... this looks nothing like any of the ones he gave) .

Thanks for the help!

8. Lol, I hae to learn to post faster. The posts just roll in!

Thanks VERY much, I just read Soroban's post and, like I said with Chop Suey's post, I'm just confused as to which is a better way to learn it: Which is more of a work around, versus which is the more fundamental way of doing the question.

Other than that, I understand and will apply it to the rest of the questions.

Thanks!

9. Originally Posted by mike_302
Lol, I hae to learn to post faster. The posts just roll in!

Thanks VERY much, I just read Soroban's post and, like I said with Chop Suey's post, I'm just confused as to which is a better way to learn it: Which is more of a work around, versus which is the more fundamental way of doing the question.

Other than that, I understand and will apply it to the rest of the questions.

Thanks!
It is common that there are more than one way to approach a math problem. It is really up to you which method you are more comfortable with. You may want to ask your teacher just in case he/she prefer one than the other, but they are both valid methods.


# difference between scalar matrix and diagonal matrix

{ For example, $$A =\begin{bmatrix} 0\\ â3\\-1 \\1/2 \end{bmatrix}$$ is a column matrix of order 4 Ã 1. In a scalar matrix, all off-diagonal elements are equal to zero and all on-diagonal elements happen to be equal. When the order is clear from the context, we simply write it as I. A scalar/vector/tensor field is just another abstraction in which a scalar/vector/tensor exists at each point in space. For example, $$A =\begin{bmatrix} -1/2 & â5 & 2 & 3\end{bmatrix}$$ is a row matrix of order 1Â Ã 4. { A square matrix is a matrix that has the same number of rows and columns i.e. For example, $$A =\begin{bmatrix} 1\end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$ are identity matrices of order 1, 2 and 3, respectively. { Example: (2 0 0 0 − 3 0 0 0 5). 7. For example,Â $$A =\begin{bmatrix} 3 & -1 & 0\\ 3/2 & â3/2 & 1\\4 & 3 & -1\\ 7/2 & 2 & -5 \end{bmatrix}$$ is a matrix of the order 4 Ã 3. We have to find out the difference between both diagonal sums. }, Matrices are distinguished on the basis of their order, elements and certain other conditions. Our mission is to provide a free, world-class education to anyone, anywhere. Up Next. 2. Yes it is, only the diagonal entries are going to change, if at all. ] { In other words, we can say that a scalar matrix is an identity matrixâs multiple. 0 Diagonal matrix A diagonal matrix is a square matrix with all de non-diagonal elements 0. ", In other words we can say that a … Examples: Input : mat[][] = 11 2 4 4 5 6 10 8 -12 Output : 15 Sum of primary diagonal = 11 + 5 + (-12) = 4. An identity matrix is a diagonal matrix that has all diagonal elements equal to 1. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. For the following matrix A, find 2A and –1A. 6) Scalar Matrix A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. Program to swap upper diagonal elements with lower diagonal elements of matrix. "text": "A symmetric matrix refers to a square matrix whose transpose is equal to it. } ", These rows and columns define the size or dimension of a matrix. Its effect on a vector is scalar multiplication by λ. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. A scalar matrix whose diagonal elements are all 1 is called a unit matrix, or identity matrix. Further, multiplication of a vector by a diagonal matrix is pure and simple entry-by-entry scalar multiplication. ... where D is a diagonal matrix with diagonal elements holding the pivots. "mainEntity": [ "name": "Explain a scalar matrix? By using our site, you consent to our Cookies Policy. Difference order, specified as a positive integer scalar or [].The default value of n is 1.. "name": "Can we say that a zero matrix is invertible? } Basis. Basis. Scalar multiplication is easy. Join courses with the best schedule and enjoy fun and interactive classes. Question 1:Â Assertion :Â  $$A =\begin{bmatrix} 3 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 7 \end{bmatrix}$$ is a diagonal matrix. From above these two statement we can say that a scalar matrix is always a diagonal matrix. But every identity matrix is clearly a scalar matrix. A matrix consists of rows and columns. Since,Â a12Â =Â a13 =Â a21Â =Â a23Â =Â a31Â =Â a32Â =Â 0Â Thus, the given statement Â is true andÂ $$A =\begin{bmatrix} 3 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 7 \end{bmatrix}$$ is a diagonal matrix is a diagonal matrix. All the other entries will still be . Examples: This article is attributed to GeeksforGeeks.org. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all … You just take a regular number (called a "scalar") and multiply it on every entry in the matrix. Given a Boolean Matrix, find k such that all elements in kâth row are 0 and kâth column are 1. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. Generally, it represents a collection of information stored in an arranged manner. Question 4: Can we say that a zero matrix is invertible? Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. "text": "The scalar matrix is similar to a square matrix. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Mathematically, it states to a set of numbers, variables or functions arranged in rows and columns. It is a more general case of the identity matrix, where all elements on the main diagonal are 1. Up Next. "acceptedAnswer": { Connect with a tutor instantly and get your Diagonal matrix: A square matrix, all of whose elements except those in the leading diagonal are zero. The various types of matrices are row matrix, column matrix, null matrix, square matrix, diagonal matrix, upper triangular matrix, lower triangular matrix, symmetric matrix, and antisymmetric matrix. This is because its determinant is zero." Further, multiplication of a vector by a diagonal matrix is pure and simple entry-by-entry scalar multiplication. 2. A matrix is said to be zero matrix or null matrix if all its elements are zero. A matrix stores a group of related data in a structured format. In general, A = [aij]1 Ã nÂ is a row matrix of order 1 Ã n. A column matrix has only one column but any number of rows. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. A square null matrix is also a diagonal matrix whose main diagonal elements are zero. A diagonal matrix with all its main diagonal entries equal is a scalar matrix, that is, a scalar multiple λI of the identity matrix I. A diagonal matrix is a square matrix that has zeros as elements in all places, except in the diagonal line, which runs from top left to bottom right. Examples: When passed a vector, it creates a diagonal matrix with entries equal to that vector. Example: [3 0 0 0 3 0 0 0 3]. A symmetric matrix has symmetric entries with respect to the main diagonal." } So when you multiply a matrix times a scalar, you just multiply each of those entries times that scalar quantity. For the following matrix A, find 2A and –1A. "acceptedAnswer": { } 2. Given a matrix of n X n.The task is to calculate the absolute difference between the sums of its diagonal. ... where D is a diagonal matrix with diagonal elements holding the pivots. All the other entries will still be . Revise With the concepts to understand better. }, A square matrix in which all the elements below the diagonal are zero is known as the upper triangular matrix. 6. "@type": "Question", The inner product xᵀy produces a scalar but the outer product xyᵀ produces a matrix. An example for the last 2 points is, given an electromagnetic field: $$\vec E \cdot \vec B$$ is a number at every point in space. If you multiply any number to a diagonal matrix, only the diagonal entries will change. The inner product xᵀy produces a scalar but the outer product xyᵀ produces a matrix. "acceptedAnswer": { Scalar matrix: A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. A symmetric matrix and skew-symmetric matrix both are square matrices. What would be an example of the two? Cheers! This is because its determinant is zero. This topic is collectively known as matrix algebra. Thus an m Ã n matrix is said to be a square matrix if m = n and is known as a square matrix of order ânâ. Count number of islands where every island is row-wise and column-wise separated, Find a common element in all rows of a given row-wise sorted matrix, Given a matrix of âOâ and âXâ, replace âOâ with âXâ if surrounded by âXâ, Given a matrix of âOâ and âXâ, find the largest subsquare surrounded by âXâ. ... Let’s summarize the difference between a singular and non-singular n × n matrix. A matrix is said to be a column matrix if it has only one column. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n × n is said to be a scalar matrix if b ij = 0, when i ≠ j } For those of … This process continues until a 0-by-0 empty matrix is returned. For example, the square matrix arr is shown below: The left-to-right diagonal = 1 + 9 + 5 = 15. Examples : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. In general, B = [bij]m Ã 1 is a column matrix of order m Ã 1. The diag() function, when passed a matrix, extracts the diagonal elements from that matrix. You just take a regular number (called a "scalar") and multiply it on every entry in the matrix. Calculating the difference between two matrices Also, the size of the matrices also changes from m×n to n×m. Properties of matrix scalar multiplication. For example,Â $$A =\begin{bmatrix} 3 & -5 & 7\\ 0 & 4 & 0\\ 0 & 0 & 9 \end{bmatrix}$$, A square matrix in which all the elements above the diagonal are zero is known as the upper triangular matrix. "@type": "Answer", Find the largest rectangle of 1âs with swapping of columns allowed, Validity of a given Tic-Tac-Toe board configuration, Maximum size rectangle binary sub-matrix with all 1s, Maximum size square sub-matrix with all 1s. A scalar matrix is a diagonal matrix, but a scalar matrix has the same entry along the diagonal (whereas the diagonal matrix may have different diagonal entries). Answer :Â If A=[aij]nÃn is a square matrix such that aijÂ =Â 0 for iâ j,Â then AÂ is called a diagonal matrix. 3×3, 200 x 200. A matrix is said to be a rectangular matrix if the number of rows is not equal to … ... Let’s summarize the difference between a singular and non-singular n × n matrix. Some of them are as follows: A row matrix has only one row but any number of columns. 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# Bipolar coordinate system

I want to extend some work presented in a paper "Analysis of TM and TE Modes in Eccentric Coaxial Lines Based on Bipolar Coordinate System" using SageMath. Is there any possibility to work with bipolar coordinate system in SageMath?

Thanks

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Yes one can use bipolar coordinates in SageMath provided that one sets by hand the relations between bipolar and Cartesian coordinates, as follows. First, we introduce the Euclidean plane $E$ with the default Cartesian coordinates $(x,y)$:

sage: E.<x,y> = EuclideanSpace()
sage: CA = E.cartesian_coordinates(); CA
Chart (E^2, (x, y))

We then declare the bipolar coordinates $(\tau, \sigma)$ as a new chart on $E$:

sage: BP.<t,s> = E.chart(r"t:\tau s:\sigma:(-pi,pi)")
sage: BP
Chart (E^2, (t, s))
sage: BP.coord_range()
t: (-oo, +oo); s: (-pi, pi)

We set the transformation from the bipolar coordinates to the Cartesian ones, using e.g. the Wikipedia formulas. This involves $\cosh\tau$ and $\sinh\tau$. For the ease of automatic simplifications, we prefer the exponential representation of cosh and sinh:

sage: cosht = (exp(t) + exp(-t))/2
sage: sinht = (exp(t) - exp(-t))/2
sage: BP_to_CA = BP.transition_map(CA, [sinht/(cosht - cos(s)), sin(s)/(cosht - cos(s))])
sage: BP_to_CA.display()
x = (e^(-t) - e^t)/(2*cos(s) - e^(-t) - e^t)
y = -2*sin(s)/(2*cos(s) - e^(-t) - e^t)

We also provide the inverse transformation:

sage: BP_to_CA.set_inverse(1/2*ln(((x+1)^2 + y^2)/((x-1)^2 + y^2)),
....:                      pi - 2*atan(2*y/(1-x^2-y^2+sqrt((1-x^2-y^2)^2 + 4*y^2))))
sage: BP_to_CA.inverse().display()
t = 1/2*log(((x + 1)^2 + y^2)/((x - 1)^2 + y^2))
s = pi - 2*arctan(-2*y/(x^2 + y^2 - sqrt((x^2 + y^2 - 1)^2 + 4*y^2) - 1))

At this stage, we may plot the grid of bipolar coordinates in terms of the Cartesian coordinates (the plot is split in 2 parts to avoid $\tau = 0$):

sage: BP.plot(CA, ranges={t: (-4, -0.5)}) + BP.plot(CA, ranges={t: (0.5, 4)})

Let us do some calculus with bipolar coordinates. The Euclidean metric is

sage: g = E.metric()
sage: g.display()
g = dx*dx + dy*dy

From here, we declare that the default coordinates are the bipolar ones:

sage: E.set_default_chart(BP)
sage: E.set_default_frame(BP.frame())

We have then:

sage: g.display()
g = -4*e^(2*t)/(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1) dt*dt
- 4*e^(2*t)/(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1) ds*ds

Let us factor the metric coefficients to get a shorter expression:

sage: g[1,1].factor()
4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2
sage: g[2,2].factor()
4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2
sage: g.display()
g = 4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2 dt*dt
+ 4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2 ds*ds

sage: g[1,1] == 1/(cosht - cos(s))^2
True

Let us consider a generic scalar field on $E$, defined by a function $F$ of the bipolar coordinates:

sage: f = E.scalar_field({BP: function('F')(t,s)}, name='f')
sage: f.display(BP)
f: E^2 --> R
(t, s) |--> F(t, s)

The expression of the Laplacian of $f$ in bipolar coordinates is

sage: f.laplacian().expr(BP).factor()
1/4*(2*cos(s)*e^t - e^(2*t) - 1)^2*(diff(F(t, s), t, t) + diff(F(t, s), s, s))*e^(-2*t)

The gradient of $f$ is

sage: f.gradient().display()
grad(f) = -1/4*(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1)*e^(-2*t)*d(F)/dt d/dt
- 1/4*(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1)*e^(-2*t)*d(F)/ds d/ds

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Thank you very much for your prompt and extended answer. I will need some time to adapt and modify it the analysed geometry. I have to solve the Helmholtz equation - in between. Once again - thanks.

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In how many ways can we pair ourselves?

Say we have an even number of elements and we sort them into pairs in a way that every element belongs to a pair and no element belongs in two pairs.

Given $2n$ elements how many different arrangements of this sort can be made?

For example given elements named $1$, $2$, $3$ and $4$ we can do $\{\{1,2\},\{3,4\}\}$, $\{\{1,3\},\{2,4\}\}$ and $\{\{1,4\},\{2,3\}\}$ so we have $3$ different arrangements.

A wild guess I made is of the sort $\prod_{i=0}^{n-1} 2n-(1+2i)$.

The question arises form trying to figure out in how many ways can the earth population arrange themselves in couples where noone (or at most one poor human being) is left alone.

• Your guess is correct. – Empy2 Sep 7 '15 at 12:04
• It would be good exercise to put into words why it is the product of the odd numbers. – Empy2 Sep 7 '15 at 12:36
• These numbers are also called "double factorial" en.wikipedia.org/wiki/Double_factorial – user940 Sep 7 '15 at 14:11

We can start by looking at all the ways to arrange $2n$ numbers. This is $(2n)!$. Then within each of the $n$ pairs, there are $2$ ways to sort the numbers. So we want to divide our count by $2^n$. Lastly, we don't care about the order of the $n$ pairs themselves, so we further divide our count by $n!$. So the number of pairings of $2n$ numbers is

$$\dfrac{(2n)!}{2^nn!}.$$

Edit: This agrees with the OP's answer of $\;\prod_{i=0}^{n-1} 2n-(1+2i)$.

Denote the number in question by $P_{2n}$. Person number $1$ can choose his mate in $2n-1$ ways. After that there are $2n-2$ people left, which can be paired off in $P_{2n-2}$ ways. It follows that the $P_{2n}$ satisfy the recursion $$P_2=1,\qquad P_{2n}=(2n-1)\>P_{2n-2}\qquad(n>1)\ ,$$ which immediately leads to you "wild guess".

$$\frac1{n!}\binom{2n}2\binom{2n-2}2\cdots\binom42\binom22=\frac{(2n)!}{2^nn!}$$

Pick out $2$ out of all $2n$ to form a pair.

After that pick out $2$ out of remaining $2n-2$ to form a pair, et cetera.

Then every possibility has been counted $n!$ times (there are $n!$ orders for the pairs) so we must divide by $n!$ to repair this.

I appreciate all answers. To anyone interested in how I personally arrived to my guess: I imagined a string made by all elements, let's call it $S$, and noticed that any pairing can be uniquiely represented as another string where the $i$-th element is paired with the $i$-th element in $S$ (imagine one string on top of the other, the elements vertically aligned belong in the same pair). This strings however must satisfy that if element $a$ sats on top of $b$, then on top of $a$ we can only have $b$. In fact every string that obeys this rule uniquely determines a pairing. Counting them lead to the answer.

I realize that this problem has already been solved, but I got the solution in a different way and it looks a little different but plugging it in to wolfram alpha leads me to believe the previous correct solutions and this one are the same.

Background: (Note: this paragraph is how I came about the problem and you can skip it if you just want to see the answer.) This idea came up in Anthony Zee's "Quantum Field Theory in a Nutshell" book (pg 15). If you want to find certain moments of a multivariable (what is close to) Gaussian distribution, you can use a technique called Wick contraction to easily get the answer. Here is what I mean:
$$ = C_{ij}$$ $$ = C_{ij} C_{kl} + C_{il} C_{jk} + C_{ik} C_{jl}$$ $$...$$ The definition of $$C$$ does not matter, but we see we are basically pairing off the indices into however many different configurations we can. I was curious if there was a general formula for this, and here is how I got the answer.

Answer: The case of 4 indices is easy because you can just list them out like I did above--there are three. Now say we have six indices: {i, j, k, l, m, n}. Consider pairing off the first two indices to {i, j}. Now there are four more indices left to pair off, but we already know there are three different configurations for four indices. I.e. for the initial pairing of {i, j}, there are 3 "sub-configurations." Now notice there are 5 different starting pairs: {i, j}, {i, k}, {i, l}, {i, m}, {i, n}. So 6 indices gives us 5 x 3 = 15 different pairings. If you go to 8 indices, there are 7 different starting pairs which would each leave 6 indices left, which we just found out has 15 options. Therefore you have 7 x 5 x 3 options. Continuing this logic, you can see we clearly have the following:
4 indices: 3
6 indices: 5 x 3
8 indices: 7 x 5 x 3
10 indices: 9 x 7 x 5 x 3
...
Or more concisely (2n-1)!! different pairings (where !! is the double factorial).


Does the "field" over which a vector space is defined have to be a Field?

I was reviewing the definition of a vector space recently, and it occurred to me that one could allow for only scalar multiplication by the integers and still satisfy all of the requirements of a vector space.

Take for example the set of all ordered pairs of integers. Allow for scalar multiplication over the integers and componentwise vector addition as usual. It seems to me that this is a perfectly well-defined vector space.

The integers do not form a Field, which begs the question: Is there any reason that the "field" over which a vector space is defined must be a mathematical Field? If so, what is wrong with the vector field I attempted to define above? If not, what are the requirements for the scalars? (For instance, do they have to be a Group - Abelian or otherwise?)

• Not mentioned in the other answers yet: A vector space over the integers is properly called a "$\mathbb{Z}$-module," and every such module is equivalent to an abelian group, in your case the group $\mathbb{Z}^2$ with addition defined componentwise. The integer scalar tells you how many times to add up an element of the group. Jul 2 '17 at 22:10
• So basically whether a structure of this sort is a vector space or a module depends on whether its scalars form a ring or a field. So it's largely a definition thing, then? Vector space have scalar fields, and modules have scalar rings. Jul 2 '17 at 22:21
• +1 for a great question, thanks for posting! Nov 15 '20 at 19:47

If you pick the scalars from a general ring instead of insisting on a field (in particular, $\mathbb Z$ is a ring), you get a structure known as a module rather than a vector space.

Modules behave like vector spaces in certain respects, but there are also points where they are not at all as well-behaved as vector spaces. For example, a module does not necessarily have a basis, or even a well-defined dimension. This makes matrices less useful for understanding modules than they are for vector spaces. (You can still have matrices with entries in a ring; they just don't tell you everything about linear maps between the modules anymore).

• So, just to be clear, what I'm hearing is that if it has the right properties and its scalars form a Field, then it is called a vector space. But if it has basically the same properties only its scalars form a Ring, then it is called a module? Do I have that right? Jul 2 '17 at 22:15
• @Geoffrey: Correct -- the axioms are identical. Jul 2 '17 at 22:20
• @Geoffrey basically, if you it's not a field, theorems about vector spaces don't apply Jul 3 '17 at 12:52

These things are studied: they are called modules over the ring instead of vector spaces.

The main difference is that the elements of general modules do not allow a lot of the geometric intuition we have for vector spaces, so we still retain the traditional term "vector space" because it is still a useful term.

So, modules over fields (and also noncommutative fields) are called vector spaces.

While the other answers (and comments) implicitly address the question stated in the title of the OP, I thought it may be useful to include an explicit answer, as well.

Does the “field” over which a vector space is defined have to be a Field?

Yes, a vector space is defined over a field; i.e. if the scalars do not refer to a field, the resulting object is not by definition a vector space.

For completeness: as pointed out in the other answers and comments, there are objects with analogous definitions, in the case that the scalars belong to a ring (as in the example provided in the OP), and the other axioms are met, the resulting object is called a "module." As the subsection on modules on the Wikipedia vector space page says (emphasis added):

Modules are to rings what vector spaces are to fields: the same axioms, applied to a ring R instead of a field F, yield modules. The theory of modules, compared to that of vector spaces, is complicated by the presence of ring elements that do not have multiplicative inverses. For example, modules need not have bases, as the Z-module (that is, abelian group) Z/2Z shows; those modules that do (including all vector spaces) are known as free modules. Nevertheless, a vector space can be compactly defined as a module over a ring which is a field, with the elements being called vectors. Some authors use the term vector space to mean modules over a division ring.[105] The algebro-geometric interpretation of commutative rings via their spectrum allows the development of concepts such as locally free modules, the algebraic counterpart to vector bundles.


# If $g\circ f$ is injective and $f$ is surjective then $g$ is injective

Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be functions, prove that if $g\circ f$ is injective and $f$ is surjective then $g$ is injective.

I need advise or correction if something is incorrect with my proof. Thank you beforehand.

We must show that $g$ is injective, i.e for $x,y\in B, g(x)=g(y)\implies x=y$

Let $x,y\in B$ such that $g(x)=g(y)$. Because $f$ is surjective there exists $a,b \in A$ such that $f(a)=x$ and $f(b)=y$

$\implies g(f(a))=g(f(b))$

$\implies g\circ f(a)=g\circ f(b)$

$\implies a=b$ (by injectivity of $g\circ f$)

$\implies f(a)=f(b)$

$\implies x=y$

Would appreciate any correction in proof writing also!

• This is correct. – Mr.Fry Dec 14 '13 at 6:08
• I was suspicious with the last two implications, didn't know if they were true but it seems there is no problem. Thank you Faraad! – AndreGSalazar Dec 14 '13 at 6:24
• @AndrewGSM, it is a general principle that if $a=b$, then $f(a)=f(b)$, so long as $f$ is a function and both $a$ and $b$ are elements of the domain of $f$. – goblin Dec 14 '13 at 11:25

Your proof is correct. I myself would prove it exactly the same. But, I think it's useful to know more than one way, so here is an alternative solution. It's not profoundly different, but I think it's still worth mentioning.

I'm assuming that $A$ is nonempty (and, since there is a map from $A$ to $B$, $B$ is also nonempty). When $A$ is empty there's not much to prove.

The solution uses left and right inverses. A function with non-empty domain is injective iff it has a left inverse, and a function is surjective iff it has a right inverse.

So, we know that $g\circ f$ has a left inverse $h:C \to A$ and $f$ has a right inverse $k: B \to A$. We want to show that $g$ has a left inverse. Just observe that $$(f \circ h) \circ g = (f \circ h) \circ g \circ (f \circ k) = f \circ (h \circ g \circ f) \circ k = f \circ \mathrm{id}_A \circ k = f \circ k = \mathrm{id}_B,$$ so $(f \circ h)$ is a left inverse for $g$. It follows that $g$ is an injection.

PS: this solution is actually worse than your original one, because this one relies on the axiom of choice (it is used when we say that surjectivity is equivalent to having a right inverse). But it is good in the sense that we don't look at particular elements and manipulate maps as "opaque" objects.

• You write: "a function is injective iff it has a left inverse." This isn't quite right; it should be "a function is injective iff its domain is empty, or it has a left inverse." – goblin Dec 14 '13 at 11:27
• @user18921 You are right. Nice catch! I've made the correction. – Dan Shved Dec 14 '13 at 12:07
• If you weaken $f$ having a right inverse to $f$ being epic, is it still possible to show that $g$ is monic? I haven't been able to find a way to do it, but I also don't know that it can't be done (especially if you use products/coproducts). – dfeuer Dec 14 '13 at 18:52
• Another idea: although AC is needed to prove that an onto mapping has a right inverse, it's not necessary to show that a one-to-one and onto mapping has an inverse. – dfeuer Dec 14 '13 at 19:13

I need advise or correction if something is incorrect with my proof.

Would appreciate any correction in proof writing also!

To this, I would respond: its good to read different people's writing just for style. So here's my version of the proof, which is logically similar to yours but just differs on a few stylistic dimensions.

A few noteworthy points:

• You may prefer to write function arrows "backwards", as in $f : B \leftarrow A.$ See below.
• A fraction line can be used to mean "implies," see below.
• I prefer ending sentences without a big mass of symbols, using phrases like "as follows" and "below," and then putting the symbols immediately afterwards. See below.
• The word "fix" is a nice alternative to "let" when the latter has the right "basic meaning" but doesn't work grammatically. See below.
• If you're going to have a sequence of implications, I'd suggest making it as long as possible, and omitting the symbol $\implies.$ See below.

With that said, here's the proof:

Proposition. Let $g : C \leftarrow B$ denote a function and $f : B \leftarrow A$ denote a surjection. Then whenever $g \circ f$ is injective, so too is $g$.

Proof. Assume that $g \circ f$ is injective, and fix $b,b' \in B.$ The following implication will be proved. $$\frac{g(b)=g(b')}{b=b'}$$

Since $f$ is surjective, begin by fixing elements $a,a' \in A$ satisfying the equations immediately below.

$$b = f(a),\;\; b'=f(a')$$

Then each statement in the following sequence implies the next.

1. $g(b)=g(b')$
2. $g(f(a)) = g(f(a'))$
3. $(g \circ f)(a) = (g \circ f)(a')$
4. $a=a'$
5. $f(a)=f(a')$
6. $b=b'$.

Here is alternative method

note that : $$g\circ f \mbox{ injective } \implies f \mbox{ injective }$$ we have :

• $f \mbox{ is injective and surjective } \implies f \mbox{ bijective (one-to-one correspondence) }$

Since $f$ is a bijection, it has an inverse function $f^{-1}$ which is itself a bijection.

• $f^{-1} \mbox{is bijective} \implies f^{-1} \mbox{ injective }$
• $$\begin{cases} g\circ f \mbox{ injective } & \\ f^{-1} \mbox{ injective } & \\ f^{-1}(B)\subset A &\\ \end{cases} \implies g\circ f \circ f^{-1} \mbox{ injective}$$

• Since \forall x\in B \qquad \begin{align} (g\circ f)\circ f^{-1}(x)&=g\circ (f\circ f^{-1})(x)\\ &=g\circ {\rm id}_{B}(x)\\ &=g({\rm id}_{B}(x))\\ (g\circ f)\circ f^{-1}(x)&=g(x)\\ \end{align} then $$(g\circ f)\circ f^{-1}=g$$ since $(g\circ f)\circ f^{-1}$ injective then $g$ is injective

I've proved it on my own like this:

Pick two arbitrary elements of $B$, $y_1$ and $y_2$, with $g(y_1)=g(y_2)$. Since $f$ is surjective, $y_1=f(x_1)$ and $y_2=f(x_2)$ for some $x_1,x_2 \in A$. Then $g(f(x_1))=g(f(x_2))$. Since $f \circ g$ is injective, $x_1=x_2$, and so $f(x_1)=f(x_2)$, or $y_1=y_2$. Finally, $g$ is injective.

• Seems to be exactly the same proof as the one in the question and in goblin's answer. Please consider whether you're contributing something new before answering, particularly given that (1) this question already has an accepted answer, and (2) the question is 3 years old. – epimorphic Feb 3 '17 at 23:44
• Of course I'm contributing something new, and I always consider it. My proof doesn't look like the others at all... And ironically, two of the other answers recommended OP reading different styles of proofs... So yeah, I'm contributing to future people who stumble here looking for help. And, also, the three answers are basically "your proof is correct, here's an alternative proof/how I did it". Also, are you a moderator? – anon Feb 4 '17 at 15:28
• The two answers by others not prefaced with "here's my version of the proof, which is logically similar to yours but just differs on a few stylistic dimensions" each have a different sequence of implications, not just a different "style". That's what the "more than one way" in the first answer usually means, and what "I've proved it on my own like this" seems to suggest to me. I'd be willing to reverse my vote if you would explicitly say something like "here's a more prose-like style". – epimorphic Feb 4 '17 at 19:08
• To answer your last question: Primary moderation on this site is provided by the community. IIRC I found your answer through the review queues, where the only posts visible are this answer, the question, and any attached comments. And here's an answer to an old question that claims "I've proved it on my own like this" and yet looks logically identical to the proof in the OP... – epimorphic Feb 4 '17 at 19:08


# Necklace combinations with three group of beads

I have a hard question about a way how many different necklaces can be made.

Suppose that we have the following restrictions:

1. We have 3 groups of beads:

All the beads in one group are completely identical. This means that if you put two triangle beads next to each other and then switch their positions this counts as one necklace because the beads are identical

1. Necklaces are identical if they are identical under symmetric operations just as rotate them (𝑟) or turning them around (𝑠).

So if we have a necklace ordered in one way and we rotate it 180 deg or just flip a side this is count as one necklace.

1. We need to use all the 18 beads in each and every new necklace. We can not create a necklace from 17, 16 or less than 18 beads.

I read all the topics here but could not find a question about a group of identical beads. I also read Burnside lemma and Pólya_enumeration_theorem and Necklace_(combinatorics) in wikipedia, but could not find a way how to solve this and what is the correct answer.

From Burnside lemma, I found that the answer should be 57, but is this correct?

I used directly the formula from Burnside lemma, but it does not seem quite right for me, because I do not take into account that the three groups are with different numbers of beads.

$$\frac{1}{24} * (n^6 + 3 * n^4 + 12 * n^3 + 8 * n^2)$$

where n is 3 from three groups.

$$\frac{1}{24} * (3^6 + 3 * 3^4 + 12 * 3^3 + 8 * 3^2) = 57$$

However, as I said earlier despide the fact that the result looks some kind realisting I am not sure that this is the right answer, because I do not use in the formula that we have 4 triangle, 6 square and 8 circle beads.

It looks like Pólya enumeration theorem weighted version is the thing that I need. However, I am not sure how to get to the right answer

• Welcome to MSE! Pls show your work for arriving at $57$ so someone can check its correctness. You can do that by editing your question and e.g. putting your work at the end. May 6 '20 at 20:53
• Thanks @antkam, for the answer. I modified the questions and added the formula that I used and a short description why I think that despide the fact that the result looks like a good number I think that this is not quite right. May 7 '20 at 5:54
• May 7 '20 at 16:19
• Does this answer your question? Circular permutations with indistinguishable objects May 7 '20 at 16:19
• I think you should be using the word "bracelet" and not "necklace". The answer in the link provided by @Vepir gives an answer to your problem for necklaces, which means objects with a cyclic symmetry group, when you want a dihedral symmetry group. This being said you should be able to follow the same line of argument to find the formula you need. May 11 '20 at 20:04

I manage to answer the question and this is the process that I followed:

I consider the 18-bead necklace in the first part of the problem. Here are the eighteen rotations expressed in cycle form where we assume that the slots are numbered from 1 to 18 in clockwise order. The first is the identity (e: no rotation) and the second is the generator g—a rotation by a single position which, when repeated, generates all the elements of the group:

$$e = g^0 \text{ = (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18)}$$

$$g^1 \text{ = (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18) }$$

$$g^2 \text{= (1 3 5 7 9 11 13 15 17) (2 4 6 8 10 12 14 16 18)}$$

$$g^3 \text{= (1 4 7 10 13 16) (2 5 8 11 14 17) (3 6 9 12 15 18)}$$

$$g^4 \text{= (1 5 9 13 17 3 7 11 15) (2 6 10 14 18 4 8 12 16)}$$

$$g^5 \text{= (1 6 11 16 3 8 13 18 5 10 15 2 7 12 17 4 9 14)}$$

$$g^6 \text{= (1 7 13) (2 8 14) (3 9 15) (4 10 16) (5 11 17) (6 12 18)}$$

$$g^7 \text{= (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18)}$$

$$g^8 \text{= (1 9 17 7 15 5 13 3 11) (2 10 18 8 16 6 14 4 12)}$$

$$g^9 \text{= (1 10) (2 11) (3 12) (4 13) (5 14) (6 15) (7 16) (8 17) (9 18)}$$

$$g^{10} \text{= (1 11 3 13 5 15 7 17 9) (2 12 4 14 6 16 8 18 10)}$$

$$g^{11} \text{= (1 12 5 16 9 2 13 6 17 10 3 14 7 18 11 4 15 8)}$$

$$g^{12} \text{= (1 13 7) (2 14 8) (3 15 9) (4 16 10) (5 17 11) (6 18 12)}$$

$$g^{13} \text{= (1 14 9 4 17 12 7 2 15 10 5 18 13 8 3 16 11 6)}$$

$$g^{14} \text{= (1 15 11 7 3 17 13 9 5) (2 16 12 8 4 18 14 1 6)}$$

$$g^{15} \text{= (1 16 13 10 7 4) (2 17 14 11 8 5) (3 18 15 12 9 6)}$$

$$g^{16} \text{= (1 17 15 13 11 9 7 5 3) (2 18 16 14 12 10 8 6 4)}$$

$$g^{17} \text{= (1 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2)}$$

After that I found the GCD for all cycle form's with and group them in a table:

| Cycle length | Permutations | GCD with 18 |

| 1 | $$g^0$$ | GCD(0, 18)=18 |

| 2 | $$g^9$$ | GCD(9, 18)=9 |

| 3 | $$g^6$$, $$g^{12}$$ | GCD(6, 18)=GCD(12, 18)=6 |

| 6 | $$g^3$$, $$g^{15}$$ | GCD(3, 18)=GCD(15, 18)=3 |

| 9 | $$g^2$$, $$g^4$$, $$g^8$$, $$g^{10}$$, $$g^{14}$$, $$g^{16}$$ | GCD(2, 18)=GCD(4, 18)=GCD(8, 18)=GCD(10, 18)=GCD(14, 18)=GCD(16, 18)=2 |

| 18 | $$g^1$$, $$g^5$$, $$g^7$$, $$g^{11}$$, $$g^{13}$$, $$g^{17}$$ | GCD(1, 18)=GCD(5, 18)=GCD(7, 18)=GCD(11, 18)=GCD(13, 18)=GCD(17, 18)=1 |

We have 18 permutations for rotation and lets name cycle 1 with $$f_1$$, cycle 2 with $$f_2$$ .. cycle n with $$f_n$$

Than the formula for Cycling index is:

$$\frac{f_1^{18} + f_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^2 + 6f_{18}^1}{18}$$

If we solve all the possible necklaces with three colors the result should be (for the moment we do not solve for the three colors with 4, 6 and 8 beads in respective groups):

$$\frac{3^{18} + 3^9 + 2*3^6 + 2*3^3 + 6*3^2 + 6*3^1}{18} = \text{21 524 542}$$

From here because turn around is allowed we need to add and the necklace(bracelet if we follow the right terms) is with even beads we should add the symethric turn arounds.

$$\frac{f_1^{18} + f_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^2 + 6f_{18}^1 + 9f_1^2f_2^8 + 9f_2^9}{2 * 18}$$

and again for three colors without including the different weight:

$$\frac{3^{18} + 3^9 + 2*3^6 + 2*3^3 + 6*3^2 + 6*3^1 + 9 * 3^8 + 9 * 3^9}{2 * 18} = \text{10 781 954}$$

For the moment we have all the possible necklaces and bracelets with three colors. However, in order to find the necklaces and bracelets with three colors (4 reds, 6 greens and 8 blues) we need to replace:

$$f_1 = (x + y + z)$$

$$f_2 = (x^2 + y^2 + z^2)$$

$$f_3 = (x^3 + y^3 + z^3)$$

$$f_6 = (x^6 + y^6 + z^6)$$

$$f_9 = (x^9 + y^9 + z^9)$$

$$f_{18} = (x^{18} + y^{18} + z^{18})$$

and if we replace in the formula it becomes:

$$\frac{(x + y + z)^{18} + (x^2 + y^2 + z^2)^9 + 2(x^3 + y^3 + z^3)^6 + 2(x^6 + y^6 + z^6)^3 + 6(x^9 + y^9 + z^9)^2 + 6(x^{18} + y^{18} + z^{18}) + 9(x + y + z)^2(x^2 + y^2 + z^2)^8 + 9(x^2 + y^2 + z^2)^9}{36}$$

Then we need to find which expressions can expand to $$x^4y^6z^8$$.

After then by using multinominal coeficient I managed to calculate the following results

9 189 180

1260

11 340

11 340

Then I sum all of them and divide them to 36. This gives me the answer of 255 920 which is the answer of the question. We can create 255 920 bracelets with 4 red 6 green and 8 blue beads.


# Is $V$ isomorphic to direct sum of subspace $U$ and $V/U$?

Given a vector space $$V$$ and a subspace $$U$$ of $$V$$. $$V \cong U \oplus(V/U)$$ Does the above equation always hold? Where $$\oplus$$ is external direct sum. For finite dimensional vector space $$V$$, here is my attemp of prove:

Let dimension of $$U$$ be m, dimension of $$V$$ be $$n$$. Find a basis of $$U$$ : $$\{ \mathbf{ u_1, u_2, \cdots ,u_m}\}$$ and extend it to a basis for $$V$$ : $$\{ \mathbf{ u_1, u_2, \cdots ,u_m, v_1, \cdots,v_{n-m} } \}$$.

For every vector $$\mathbf{x} \in V$$, we can write $$\mathbf{x}= c_1 \mathbf{u_1}+ \cdots+ c_m \mathbf{u_m} + d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}$$ uniquely. Define a linear map $$T$$ as $$T(\mathbf{x})=(c_1 \mathbf{u_1}+ \cdots+ c_m \mathbf{u_m}, [d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}])$$ ,where $$[]$$ is used to express the equivalent class. We claim $$T$$ is an isomorphism.

Surjectivity is obvious. As for injectivity,
if $$T(\mathbf{x})=(\mathbf{0},[\mathbf{0}])$$, then $$c_1 \mathbf{u_1}+ \cdots+ c_m\mathbf{u_m}= \mathbf{0}$$
$$\Rightarrow c_1=0, c_2=0, \cdots,c_m=0$$
$$\Rightarrow x=d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}$$
Since $$[d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}] = [\mathbf{0}]$$, we have $$(d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}-\mathbf{0})\in U$$, which means $$d_1=0, d_2=0, \cdots,d_{n-m}=0$$
$$\Rightarrow \mathbf{x}=\mathbf{0}$$, so $$T$$ is injective.

Is the above proof correct? Does this mean $$V \cong \ker F \oplus (V/ \ker F) \cong \ker F \oplus\mathrm{im}F$$ for any linear map $$F$$, because $$\ker F$$ is a subspace of $$V$$ ?

The final question is about how should I prove it when the dimension of $$V$$ is infinite?

It is true that for every finite dimensional vector space $$V$$ with $$U$$ a vector subspace that $$V \cong U \oplus (V / U)$$ I think your proof is essentially correct. And yes it is true that for $$T: V \rightarrow W$$ any linear that we have $$V \cong \operatorname{ker}(T) \oplus \operatorname{Im}(T)$$ The rank-nullity theorem is a direct consequence of this.

In more technical language, we say that every "short exact sequence" of finite dimensional vector spaces over a field $$k$$ $$\textit{splits}$$. What this means is that if $$T : U \rightarrow V$$, $$S : V \rightarrow W$$ are linear maps such that $$T$$ is injective, $$\operatorname{ker}(S) = \operatorname{Im}(T)$$, and $$S$$ is surjective, then $$V \cong U \oplus W$$.

Note then that this directly gives us our result since if $$U$$ is a subspace of $$V$$, then the inclusion map $$\iota : U \rightarrow V$$ and projection map $$\pi : V \rightarrow (V / U)$$ set up exactly a short exact sequence.

In terms of whether or not this extends to infinite dimensional vector spaces, the result does hold again (assuming the axiom of choice), and the proof is essentially the same. All your proof relies on is the ability to extend a basis of a subspace to a basis of your entire space. We can do this with the axiom of choice.

• I am wondering if the following statement is false for infinite dimensional spaces. "If $V$ is a vector space and $U$ is a subspace of $V$, then there exists another subspace of $V$ called $U^\perp$ such that every element of $V$ can be uniquely expressed as the sum of an element from $U$ with an element from $U^\perp$." (I am thinking a counterexample would be if $U$ was the set of real number sequences with finite support (i.e. eventually zero) and $V$ is the set of all real number sequences.) – irchans Jan 3 at 19:03
• @irchans If we take the axiom of choice, then every subspace of a vector space has a direct sum complement (what you call the perpendicular space, but this language is typically reserved for a space equipped with some bi-linear form). The proof is pretty simple. Let $V$ be a $k$-vector space, with $U$ a vector subspace. Let $\mathcal{B}_{U}$ be a basis for $U$ and extend it to a basis $\mathcal{B}_{V}$ (using the axiom of choice) for $V$. Then let $W = \operatorname{Span}_{k}\left( \mathcal{B}_{V} \backslash \mathcal{B}_{U} \right)$. Then $V = U \oplus W$. – Adam Higgins Jan 3 at 19:13
• @irchans Perhaps the reason you think that your example is a counter example is because of the $\textit{weirdness}$ of bases of infinite dimensional vector spaces. Notice that a subset $S$ of a vector space $V$ is said to be a basis if and only if every element $v \in V$ can be written as a $\textbf{finite}$ linear combination of the elements of $S$, and that there is no finite non-trivial linear relation amongst the elements of $S$. – Adam Higgins Jan 3 at 19:19
• Thank you very much ! – irchans Jan 3 at 19:27
• This set of notes seems relevant math.lsa.umich.edu/~kesmith/infinite.pdf – irchans Jan 3 at 19:38

Your proof works. The answer to your second question is yes, that is true. For an infinite dimensional vector space, take any linear map $$F: V -> W$$. Then $$U = \ker F$$ is a subspace of $$V$$. Note that we have a short exact sequence (if you don't know what that means, don't worry, the explanation is coming) $$0\to U\to V\to V/U\to 0$$

(That is, there's an injective map $$U \to V$$ (inclusion, I'll call it $$i$$) and a surjective map $$V \to V/U$$ (the quotient map, I'll call it $$q$$) such that the image of the injection is the kernel of the surjection).

But there's also a surjective map $$V \to U$$ (projection onto $$U$$, I'll call it $$p$$), and note that for any $$u \in U$$, $$pi(u) = u$$ (since $$p$$ fixes $$u$$).

Now, we're going to show that $$V$$ is the (internal) direct sum of the kernel of $$p$$ and the image of $$i$$. First, note that it's the sum of the two: for any $$v \in V$$, $$v = (v - ip(v)) + ip(v)$$, $$ip(v)$$ is obviously in the image of $$i$$, and $$p(v - ip(v)) = p(v) - pip(v) = 0$$ (with the last equality being due to our note about $$p$$, since $$p(v)\in U$$.

And further, the intersection is trivial: if $$v \in \ker(p)\cap\mathrm{im}(i)$$ then there is some $$u\in U$$ such that $$i(u) = v$$, and $$pi(u) = p(v) = 0$$, but $$pi(u) = u$$, so $$u = 0$$, hence $$v = 0$$. Thus, $$V = \ker(p)\oplus \mathrm{im}(i)$$.

Now, it's clear that $$\mathrm{im}(i)\cong U$$ since it's the image of $$U$$ under an injective map, so we need only show that $$\ker(p)\cong V/U$$.

For that purpose, since $$q$$ is surjective, for any $$w \in V/U$$, there is some $$v \in V$$ such that $$w = q(b)$$. But since $$V = \ker(p)\oplus\mathrm{im}(i)$$, there are unique $$u \in U$$, $$x \in \ker(p)$$ such that $$v = i(u) + x$$, so $$w = q(b) = q(i(u)+x) = qi(u) + q(x)=q(x)$$ (since the image of $$i$$ is the kernel of $$q$$, in particular $$qi = 0$$). Thus, $$q|_{\ker(p)}: \ker(p)\to V/U$$ is surjective. But also, if $$q(v) = 0$$, and $$v \in \ker(p)$$. But $$\ker(q) = \mathrm{im}(i)$$, and $$p$$ fixes $$\mathrm{im}(i)$$, so the only element that it sends to $$0$$ is $$0$$ itself, so we must have $$v = 0$$, hence $$q|_{\ker(p)}$$ is also injective, so is an isomorphism.

Thus, we have $$V \cong \mathrm{im}(i)\oplus\ker(p) \cong U\oplus (V/U)$$, as required.

This is precisely a special case of the Splitting Lemma, and my proof is essentially just one part of the proof of that, translated: there are easier ways to prove it, but I thought that it would be useful to see it in a more broadly applicable form.


# Query on a Solution to the Problem: $\gcd(5a+2,7a+3)=1$ for all integer $a$.

I wish to show that the numbers $$5a+2$$ and $$7a+3$$ are relatively prime for all positive integer $$a$$.

Here are my solutions.

Solution 1. I proceed with Euclidean Algorithm. Note that, for all $$a$$, $$|5a+2|<|7a+3|$$. By Euclidean Algorithm, we can divide $$7a+3$$ by $$5a+2$$. To have

$$7a+3=(5a+2)(1)+(2a+1)$$ continuing we have,

$$5a+2=(2a+1)(2)+a$$

$$2a+1=(2)(a)+1$$

$$2=(1)(2)+0$$

Since the last nonzero remainder in the Euclidean Algorithm for $$7a+3$$ and $$5a+2$$ is 1, we conclude that they are relatively prime.

Solution 2. Suppose that $$d=\gcd(5a+2,7a+3)$$. Since $$d=\gcd(5a+2,7a+3)$$ then the following divisibility conditions follow:

(1) $$d\mid (5a+2)$$

(2) $$d\mid (35a+14)$$

(3) $$d\mid (7a+3)$$

(4) $$d\mid (35a+15)$$.

Now, (2) and (4) implies that $$d$$ divides consecutive integers. The only (positive) integer that posses this property is $$1$$. Thus, $$d=1$$ and that $$7a+3$$ and $$5a+2$$ are relatively prime.

Here are my questions:

1. Is the first proof correct or needs to be more specific? For instance cases for $$a$$ must be considered.

2. Which proof is better than the other?

Thank you so much for your help.

• I think both approaches are good. Nor are they that different really...in both cases you are trying to find smaller and smaller multiples of $d$.
– lulu
Sep 24 '18 at 15:23
• Thank you very much for the kind comment @lulu. Got a follow up question. By "you are trying to find smaller and smaller multiples of $d$, you mean the process of continously dividing the divisor to remainder so that $r$ decreases? Sep 24 '18 at 15:27
• Is there a typo in (4), i.e. shouldn't 35 be multiplied by $a$?
– user431008
Sep 24 '18 at 15:28
• I meant something less precise than that. Euclid provides a somewhat systematic way to find new numbers that $d$ divides...the second method is less systematic, but faster as you seek out convenient expressions. Working by hand, I prefer the second method...were I trying to automate the process, I'd prefer the systematic approach.
– lulu
Sep 24 '18 at 15:28
• I agree @marmot. Thank you for pointing it out. Sep 24 '18 at 15:29

In the first solution you're not using, strictly speaking, the Euclidean algorithm, but a looser version thereof:

Let $$a$$, $$b$$, $$x$$ and $$y$$ be integers; if $$a=bx+y$$, then $$\gcd(a,b)=\gcd(b,y)$$.

The proof consists in showing that the common divisors of $$a$$ and $$b$$ are the same as the common divisors of $$b$$ and $$y$$.

There is no requirement that $$a\ge b$$ or that $$y$$ is the remainder of the division. Indeed your argument actually has a weakness, because $$7a+3\ge 5a+2$$ only if $$2a\ge-1$$, so it doesn't hold for $$a\le-2$$. But $$7a+3\ge 5a+2$$ is not really needed for the argument.

Since successive application of the statement above show that $$\gcd(2a+1,2)=1$$ and the gcd has never changed in the various steps, you can indeed conclude that $$\gcd(5a+2,7a+3)=1$$.

The second solution is OK as well.

You can simplify it by noting that if $$d$$ is a common divisor of $$5a+2$$ and $$7a+3$$, then it divides also $$5(7a+3)-7(5a+2)=1$$

• Thanks for the clarification in solution 1 and by giving a simplified version for solution 2. It is very clear to me now. @egreg Sep 24 '18 at 15:35

Your first proof is correct. I would perhaps complete it, writing that\begin{align}1&=(2a+1)-2a\\&=2a+1-2\bigl(5a+2-2(2a+1)\bigr)\\&=5(2a+1)-2(5a+2)\\&=5\bigl(7a+3-(5a+2)\bigr)-2(5a+2)\\&=5(7a+3)-7(5a+2).\end{align}

Your second proof also works, but it doesn't generalize easily to other situations.

• Thank you very much for completing the proof of solution 1 Prof. I am interested in a case that solution 2 wont work. Thank you Prof. Sep 24 '18 at 15:41
• @JrAntalan Concerning the second proof, I only meant that you have to think about it in a case-by-case basis. Sep 24 '18 at 15:43
• Noted Prof. and Thank you again. Sep 24 '18 at 15:44

Here is a different rendering of the same arguments.

Let $$u=5a+2$$, $$v=7a+3$$. Then $$\pmatrix{ u \\ v} = \pmatrix{ 5 & 2 \\ 7 & 3} \pmatrix{ a \\ 1}$$ and so $$\pmatrix{ a \\ 1} = \pmatrix{ 5 & 2 \\ 7 & 3}^{-1} \pmatrix{ u \\ v} = \pmatrix{ \hphantom- 3 & -2 \\ -7 & \hphantom-5} \pmatrix{ u \\ v}$$ This gives $$1 = -7u+5v = -7(5a+2)+5(7a+3)$$ The key point here is that the matrix has determinant $$1$$ and so its inverse has integer entries.


# For any arrangment of numbers 1 to 10 in a circle, there will always exist a pair of 3 adjacent numbers in the circle that sum up to 17 or more

I set out to solve the following question using the pigeonhole principle

Regardless of how one arranges numbers $1$ to $10$ in a circle, there will always exist a pair of three adjacent numbers in the circle that sum up to $17$ or more.

My outline

[1] There are $10$ triplets consisting of adjacent numbers in the circle, and since each number appears thrice, the total sum of these adjacent triplets for all permutations of the number in the circle, is $3\cdot 55=165$.

[2] If we consider that all the adjacent triplets sum to 16 , and since there are $10$ such triplets, the sum accordingly would be $160$, but we just said the invariant sum is $165$ hence there would have to be a triplet with sum of $17$ or more.

My query

Could someone polish this into a mathematical proof and also clarify if I did make use of the pigeonhole principle.

• Yes, you used pigeonhole. Should say if all sums are $\le 16$, then the sum is $\le 160$. – André Nicolas May 25 '13 at 4:42
• @AndréNicolas noted and aha , the syntax part is exactly why I asked the question on here. – metric-space May 25 '13 at 4:44
• @nerorevenge Note: You can actually show that the 3 adjacent numbers sum to 18 or more. – Calvin Lin May 25 '13 at 8:37

Yes, you used the Pigeonhole Principle. As a very mild correction, you should say that of all sums of three consecutives are $\le 16$, then the sum is $\le 160$.

The proof (with the small correction) is already fully mathematical. Conceivably you might want to explain the $55$ further. It is clear to you and to most users of this site why $55$, but imagine the reader to be easily puzzled.

The part about "all the permutations" is vague, and technically incorrect. You are finding the sum of a consecutive triple, and summing all these sums. Permutations have nothing to do with it, we are talking about a particular fixed arrangement of the numbers.

Remark: We could use a lot of symbols. Starting at a particular place, and going counterclockwise, let our numbers be $a_1,a_2,\dots,a_{10}$. For any $i$, where $1\le i\le 10$, let $S_i=a_i+a_{i+1}+a_{i+2}$, where we use the convention that $a_{10+1}=a_1$, $a_{10+2}=a_2$, and $a_{9+2}=a_1$.

Then $S_1+S_1+\cdots+S_{10}=165$, since each of $1$ to $10$ appears in $3$ of the $S_i$, and $1+2+\cdots+10=55$.

But if all the $S_i$ are $\le 16$, then $\sum_{i=1}^{10}S_i\le 160$. This contradicts the fact that $\sum_{i=1}^{10}S_i=165$.

I prefer your proof, mildly modified.

• eh, is that all? the reason I'm asking is that, inb the past my so called 'proofs' have been criticised for being too informal – metric-space May 25 '13 at 4:45
• @nerorevenge: I am an advocate of the informal but clear. I hope that someone more formal-minded will add an answer. The only issue is that you perhaps left an easily filled gap or two. – André Nicolas May 25 '13 at 4:53
• I have added another criticism. You should really fix the wording there. – André Nicolas May 25 '13 at 4:58
• I agree.That definitely is vaguely worded. – metric-space May 25 '13 at 4:59
• Presumably the "pair" wording came with the problem. But it looks funny to call a trio a pair. – André Nicolas May 25 '13 at 5:23

We will show something stronger, namely that there exists 3 adjacent numbers that sum to 18 or more.

Let the integers be $\{a_i\}_{i=1}^{10}$. WLOG, $a_1 = 1$. Consider

$$a_2 + a_3 + a_4, a_5 + a_6 + a_7, a_8 + a_9 + a_{10}$$

The sum of these 3 numbers is $2+3 +\ldots + 10 = 54$. Hence, by the pigeonhole principle, there exists one number which is at least $\lfloor \frac{54}{3} \rfloor = 18$.

I leave it to you to show that there is a construction where no 3 adjacent numbers sum to 19 or more, which shows that 18 is the best that we can do.

• definitely a stronger proof ,and I have admit it sure is nice, I wonder why cut-the-knot.org didn't ask people to solve for what you proved. – metric-space May 25 '13 at 16:45
• @nerorevenge I used this as a problem on Brilliant.org, and most could only get to 17, and didn't understand why 18 must be true. The 18 case arises quite easily by ignoring the small values which would otherwise reduce our sum. – Calvin Lin May 25 '13 at 23:37


# To find number of real roots

Consider the equation $$x^5-5x=c$$ where c is a real number.

Determine all c such that this equation has exactly 3 real roots.

I know that between consecutive real roots of $$f$$ there is a real root of $$f'$$. Now $$f'$$ in this case is $$5x^4-5$$ which always has two real roots. So the claim should be true for all c.

But I KNOW IT IS NOT TRUE. Where am I messing up?

• Note that while it is necessary for $f'$ to have two real roots (in order for $f$ to have exactly three real roots), it is not sufficient. Of course an odd degree polynomial will always have one real root. – hardmath Dec 12 '19 at 16:12
• ok so how do I find all such c? – Angry_Math_Person Dec 12 '19 at 16:14
• Consider the graph of $p(x) = x^5 - 5x$. Changing the constant $c$ in your equation amounts to moving a horizontal line up or down across this graph. – hardmath Dec 12 '19 at 16:15

Yes, between any two roots of $$f$$, there is a root of $$f'$$. However, just because $$f'$$ has a root, that doesn't mean that $$f$$ has a root on either side. Consider $$f(x)=x^2+1$$.

As for solving this problem, the derivative has only two roots, so we can at most have three roots. For some values of $$c$$ we have three roots, for some values of $$c$$ we have a single root, and for exactly two values of $$c$$, there are two roots. The three-root region is exactly the interval between the two two-root values of $$c$$.

And finding the values of $$c$$ that gives two roots is easier than one might think. They happen exactly when one root of $$f$$ coincides with a root of $$f'$$. So find the roots of $$f'$$, and find the values of $$c$$ that make each of them a root of $$f$$, and you have found the interval of $$c$$-values that gives three roots.

As you know, the given equation has extrema at $$x=\pm1$$. These correspond to values of the polynomial

$$1-5-c$$ and $$1+5-c$$ (the RHS was moved to the left).

Hence the polynomial will grow from $$-\infty$$, reach the maximum, then the minimum and continue growing to $$\infty$$. There are three roots when $$0$$ is in the range $$(-4-c,6-c)$$.

Where am I messing up? Just look at a graph where it fails.

$$x^5-5 x -5$$

I know that between consecutive real roots of f there is a real root of f′. Now f′ in this case is 5x4−5 which always has two real roots. So the claim should be true for all c.

$$A \implies B$$ does not mean $$B \implies A$$.

The two real roots of $$5x^4 - 5$$ are the two roots at $$x = \pm 1$$.

If $$x^5 - 5x=c$$ has three roots then they will be at $$x < -1; -1 < x < 1;$$ and at $$x > 1$$ by your condition.

But there won't be three real roots if there is no root for any $$x< -1$$, or no root between $$-1$$ and $$1$$, or no root for any $$x < -1$$.

$$x=\pm 1$$ are extreme points and if one, the max, is $$>0$$ and the other $$<0$$ then there will be three real roots. But if both are "on the same side of $$0$$" there is no root between them and no root to "the other side".

$$x^5-5x -c|_{-1} = 4-c$$ and $$x^5 - 5x -c|1 = -4-c$$ so $$x =-1$$ is a max and $$x = 1$$ is a min.

If $$f(-1) = 4-c \le 0$$ is a max there will be no root for $$x < -1$$ or for $$-1 < x \le 1$$. If $$f(1) = -4-c\ge 0$$ is a min there will be no root for $$x > 1$$ or for $$-1 \le x < 1$$. So if either $$c \ge 4$$ or if $$c\le 4$$ then there are fewer than three real roots. But if $$-4 < c < 4$$ then there will be three.

Alternatively: we know what the shape of an odd polynomial $$x^5 -5x$$ looks like. It's that polynomial curve with a twisty bit in the middle, goes off to infinity as $$x \to \infty$$, goes to negative infinity as $$x \to -\infty$$. It has roots were the x-axis crosses it (or where it crosses the x-axis-- everything is relative). If we shift it up or shift it down by $$c$$ we can force the x-axis to avoid the twisty bits in the middle and have it have only one root. Or we can deliberately shift it so that the x-axis goes smack through the twisty bits and we have a maximum number of roots. So if $$c$$ is between the max and mins we maximize the number of roots and the x-axis goes through the twisty bits. If $$c$$ is beyond the max an mins we've shoved the twisty bits below or above the x-axis and there is only one root.