Small exercice about finite groups

Posted by Mathoman, Sunday 30 October 2011 at 12:01 - Maths for mathematicians - Tags

The theory of finite groups is full of little, weird exercises. Here is one of them. It seems quite simple... Try it out!

Is it true that every automorphism of a finite group, which sends at least half of the elements of the group to their inverses, is an involution?

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Some solutions to Arnold's Trivium

Posted by Mathoman, Monday 3 January 2011 at 12:05 - Maths for mathematicians - Tags

Nearly twenty years ago Vladimir Arnol'd has published a collection of exercises which he called Trivium but many of which are not trivial at all. In the french version of this blog you can find most solutions. You are invited to contribute solutions if you wish. Exercises no. 27, 41, 51, 58, 68, 69, 70, 73, 74 are still unsolved here.

Since the answers are posted in comments not following the original order of Arnold's exercises, the best way to find a solution on that page is to use Ctrl+F and look for "no.54", for example. The numbering of the exercises follows this french version which I found on the web and can be slightly different to the english, russian or german version.

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A convergent series

Posted by Mathoman, Friday 29 October 2010 at 23:39 - Riddles And Exercises - Tags

To find out for which $\alpha>0$ the series $\sum_{k=1}^\infty\,\frac1{k^\alpha}$ converges, one can use the following inequality which is valid for all $n\in\llbracket2,\infty\llbracket$ ,

$\int_2^n\frac{{\rm d}x}{x^\alpha}\;<\;\sum_{k=2}^n\,\frac1{k^\alpha}\;<\;\int_1^n\frac{{\rm d}x}{x^\alpha}\:.$

This inequality is easily proved (for example with the help of a drawing) and implies that

$\begin{align*} \frac1{1-\alpha}\left(\frac1{n^{\alpha-1}}-\frac1{2^{\alpha-1}}\right)\;&<\;\sum_{k=2}^n\,\frac1{k^\alpha}\;<\;\frac1{1-\alpha}\left(\frac1{n^{\alpha-1}}-1\right)\qquad\text{ for }\alpha\neq1\:,\\ \ln (n) - \ln (2)\;&<\;\sum_{k=2}^n\,\frac1{k}\;<\;\ln (n) - \ln (1)\:. \end{align*}$

By letting n go to infinity we conclude that the series converges if $\alpha>1$ and diverges to infinity if $\alpha\leq1\,.$ Having this in mind, the following question is natural.

Exercise:

Let $\sum_{k=1}^\infty\,u_k$ be a convergent series. Is it true that the series $\sum_{k=1}^\infty\,u_k^3$ converges too?

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Compression of a soccer ball

Posted by Mathoman, Sunday 11 July 2010 at 00:58 - No Maths - Tags

Today is the finale of the soccer world championship and the last opportunity ta ask my reader the following question:

When a player receives a ball on his head, the ball is compressed for a very short moment before it is bounced back. Have a guess, what is the maximal deformation of the ball (in cm)?

Here are some measures about the soccer ball according to the rules of the FIFA: it is of spherical shape, made with leather or similar material and has, at the beginning of the match, a circonference between 68 and 70 cm, a weight between 410 and 450 g and a pressure between 1,6 and 2,1 atm (1600 - 2100 g/cm²).

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Compute an approximate value of an integral

Posted by Mathoman, Wednesday 16 June 2010 at 20:05 - Riddles And Exercises - Tags

A friend sent me a nice collection of exercises about which I will talk soon on this blog. One of the questions is simply:

Calculate the mean value of sin¹⁰⁰(x) with a precision of 10%.

I suppose that one must understand calculate the mean value on an interval having the length of a period (for example between 0 and pi).
According to the author of the list of exercises a student who is unable to solve this problem within five minutes does not master mathematics at all... What about you? :-)

To close this entry here are two nice math sentences:

To speak algebraically, Mr. M. is execrable, but Mr. G. is (x+1)ecrable. — Edgar Alan Poe

Even the strongest number needs its zeros: 100000000.
— Zarko Petan

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Theorem about a circle, three chords and a midpoint

Posted by Mathoman, Friday 29 January 2010 at 13:36 - Riddles And Exercises - Tags

Here is a nice exercise in plane geometry. As often in mathematics the statement is quite simple — but the proof is not!

Let $\scr{C}$ be a circle, A,B two distinct points on $\scr{C}$ and M be the midpoint of the chord [AB]. Take two other chords, [PQ] and [SR], that pass through M. Let C (resp. D) be the intersection of [AB] with [PS] (resp. [RQ]).
Prove that M is the midpoint of the chord [CD].

Chords of a circle, midpoint of segments, plane geometry, a theorem in euclidean geometry, butterfly theorem

Suprising! If M is the midpoint of [AB] then also of [CD].

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Christmas Riddle

Posted by Mathoman, Friday 25 December 2009 at 13:57 - Riddles And Exercises - Tags

X-mas is the time to crack nuts. At least that's what normal people do. But mathematicians enjoy cracking very peculiar nuts, like the following one. Show that the equation below is true for every positive integer.

$\sum_{k=0}^n\left$\begin{array}{c}2n+1\\2k+1\end{array}\right$(2k+1)\:=\:2^{2n-1}(2n+1).$

As always in maths, the slogan is short is beautiful, i.e., you should find a solution that is as short and elegant as possible, without using much computations. For this exercise ou may get inspiration from Santa Claus who has to distribute gifts into ChristmasStockings...

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An exercise about groups

Posted by Mathoman, Sunday 20 December 2009 at 12:45 - Riddles And Exercises - Tags

A topological group is a set G with a group structure and a topology such that the binary law

$G \times G \rightarrow G ,\;\; (x,y) \rightarrow xy,$

and the map of the inverse

$G \rightarrow G ,\;\; x \rightarrow x^{-1},$

are continuous. In other words both structures, the algebraic and the topological one, are linked in a natural way. So the following question is of interest.

Question

Are there topological groups which are isomorphic as groups and homeomorphic as topological spaces but not isomorphic as topological groups?

Answer

Yes. The proof goes in three stages:

Let G and H be dense subsets of $\mathbb{R}$ and $f :\, G \rightarrow H$ a monotonic bijection. Then f is a homeomorphism.

Without restriction of generality we can suppose that f is increasing. We prove that it is continuous. Let $x_0\in G$ and $\epsilon>0$ . Since H is dense in $\mathbb{R}$ we have $H\cap\,]f(x_0)-\epsilon,f(x_0)[\,\neq\emptyset$ . Thus there exists $y_1\in H\cap\,]f(x_0)-\epsilon,f(x_0)[\,$ . Similarly there exists $y_2\in H\cap\,]f(x_0),f(x_0)+\epsilon[\,$ . Since f is onto we can write $y_k=f(x_k)$ with $x_k\in G, k=1,2$ . Let $\delta=\min(x_0-x_1,x_2-x_0)$ . Then for all x in G
$\begin{align*}x_0-\delta<x<x_0+\delta \;\;\;\Longrightarrow\;\;\;& f(x_0-\delta)<f(x)<f(x_0+\delta)\\ \Longrightarrow\;\;\;&y_1=f(x_1)\leq f(x)\leq f(x_2)=y_2\\ \Longrightarrow\;\;\;&f(x_0)-\epsilon<f(x)<f(x_0)+\epsilon\,, \end{align*}$
which proves that f is continuous in $x_0$ . The proof of the continuity of the inverse function $f^{-1}$ is the same.
Let G and H be countable and dense subsets of $\mathbb{R}$ . Then they are homeomorphic.

First we write
$\begin{align*} G&=\{x_0,x_1,x_2,\ldots\}\;\;\;\;\;(*)\,,&H&=\{y_0,y_1,y_2,\ldots\}\;\;\;\;(**)\,. \end{align*}$
We will now list G and H in a new way, and The aim is to make a monotonic bijection (then it is automatically a homeomorphism). For this we proceed as follows.
- k=0. We take $x'_0=x_0,\;y'_0=y_0$
- k=1. We take $x'_1=x_1$ . To choose $y'_1$ we look at the order of $x'_0$ and $x_1'$ .
  If $x'_1<x'_0$ then we take as $y'_1$ any element in H which is smaller than $y'_0$ .
  If $x'_1>x'_0$ then we take as $y'_1$ any element in H which is greater than $y'_0$ .
- k=2. We take as $y'_2$ the first element of $H\setminus\{y'_0,y'_1\}$ listed in (**). To choose $x'_2$ we look at the order of $y'_0,y'_1,y'_2$ .
  If $y'_2$ is smaller than $y'_0$ and $y'_1$ we take as $x'_2$ any element in G which is smaller than $x'_0$ and $x'_1$ .
  If $y'_2$ is greater than $y'_0$ and $y'_1$ we take as $x'_2$ any element in G which is greater than $x'_0$ and $x'_1$ .
  If $y'_2$ is between $y'_0$ and $y'_1$ we take as $x'_2$ any element in G which is between $x'_0$ and $x'_1$ .
- k=3. We take as $x'_3$ the first element of $G\setminus\{x'_0,x'_1,x'_2\}$ listed in (*). To choose $y'_3$ we look at the order of $x'_0,x'_1,x'_2,x'_3$ . There are 24 possible orders for these four numbers.
  If $x'_3<x'_0<x'_1<x'_2$ we take as $y'_3$ any element in H which is smaller than $y'_0,y'_1,y'_2$ .
  If $x'_2<x'_3<x'_0<x'_1$ we take as $y'_3$ any element in H which is bewteen $y'_2$ and $y'_0$ .
  And so forth.
The topological groups $G=\mathbb{Q}+\mathbb{Q}\sqrt2$ and $H=\mathbb{Q}+\mathbb{Q}\sqrt3$ fulfill our wishes.

After what has been said before, G and H are homeomorphic as topological spaces. Clearly they are isomorphic groups. But they are not isomorphic topological groups. In fact, suppose that there exists an isomorphism of topological groups $f :\, G \to H$ . One proves easily by induction that f(n)=nf(1) for every integer n, and then that f(r)=rf(1) for every rational r. Therefore by continuity
$f(\sqrt2)=\sqrt2f(1),\;\;\;\;\;\lightning$
a contradiction in $H=\mathbb{Q}+\mathbb{Q}\sqrt3$ .

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Cargo-cult science training

Posted by Mathoman, Friday 4 December 2009 at 14:59 - Maths And Society - Tags

Here is very interesting article about "modern ideas" of teaching, written by João Magueijo a physics professor at Imperial College London. Unfortunately, his observations are so true. It seems that the same problem, i.e., the rise of bureaucrats and educationalists, occurs not only in the UK but also in Germany and France.

As one commentator puts it so rightly: The whole development is part of a broader crisis of our societies becoming increasingly unable to discern the values on which they rest so comfortably and to protect their vulnerable creators.

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An exercise about the temperature distribution on the earth

Posted by Mathoman, Friday 23 October 2009 at 14:55 - Riddles And Exercises - Tags

Heute mal ein bisschen climatology! The following question, funny and useless, is dedicated to my friend A. Wirth, who has left pure mathematics in order to use his mind for applied sciences like oceanography and meteorology ;-)

Exercise: Suppose that the earth is a perfect ball and that the temperature on the earth's surface is a continuous function. Prove that there are infinitely many pairwise disjoint sets {A,B} with A and B points on the earth surface such that the temperature in A and B is the same and such that the distance between A and B is 1000 km.

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Mixing wine and beer

Posted by Mathoman, Saturday 22 August 2009 at 11:33 - No Maths - Tags

Before summer I wrote a blog entry about mixing wine and beer. french friends told me that such a mixture can't exist, putting beer into wine is a sacrilege... It is okay to drink one after the other, but never, never simultaneously! The order though, wine first or beer first, depends on the contry:

Germany: Wein auf Bier das rat ich dir, Bier auf Wein das lasse sein.
France: Bière sur vin est venin, vin sur bière est belle manière.
England: Beer after wine, and you’ll feel fine, wine after beer and you’ll feel queer.

So I was very suprised when during summer brazilian friends introduced me to espanhola, a drink that they like to mix on the beach and which contains exactly wine and beer! The taste depends on the proportions (mostly more beer than wine), the color looks rather dirty and flakes swim around in the glass...


Those brazilians mix...	...wine with beer!

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Comatrix and adjugate matrix

Posted by Mathoman, Tuesday 14 July 2009 at 13:38 - Riddles And Exercises - Tags

The comatrix com(M) of a n x n-matrix M is the n x n-matrix whose entry in (l,k) is $\small{(-1)^{l+k}}$ times the determinant of the matrix which you get by deleting in M the line l and the row k.
It is the transpose of the comatrix which is of interest to us; it is called adjugate matrix, and it is shown in every lecture on linear algebra that it has the following fundamental property:

$^t\text{com}(M)\:M\;=\;M\:^t\text{com}(M)\;=\;\det(M)\:I\:.$

If you work with matrix coefficients in a ring R, it follows that the matrix M is invertible in the matrix ring precisely if the scalar det(M) is invertible in the ring R. For example over $\small\mathbb{Z}$ the invertible matrices are those whose determinant is 1 oder -1 ist.

Exercise: Prove that com preserves matrix multiplication, i.e.,

com(I) = I et com(MN) = com(M) com(N).

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Dimension of the Commutator Space

Posted by Mathoman, Thursday 9 July 2009 at 14:02 - Riddles And Exercises - Tags

Here is a beautiful exercise from matrix theory.

Let A be a n x n matrix. Show that the dimension of the commutator space of A (i.e., the set of all matrices that commute with A) is at least n.

There is a proof that works on any field. But in the special cases of real or complex numbers there are alternative proofs.

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A sunday child

Posted by Mathoman, Sunday 5 July 2009 at 20:08 - Riddles And Exercises - Tags

Here is a personalised birthday gift for my father, in form of a mathematical riddle.

Today, sunday 5th of july 2009, is my father's birthday. He was born on a sunday of a leap year. What is his age today?

In order to solve that exercise you can use the fact that I am older than twenty-three, that my father was older than twenty-three when he took upon him the responsibility of becoming my father and finally that he isn't hundred years old.

Anyway, happy birthday, Daddy!

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Deserve your dessert!

Posted by Mathoman, Wednesday 1 July 2009 at 11:30 - Riddles And Exercises - Tags

We all know those little riddles that entertain mathematicians at the end of lunch at the cafeteria. Here is one of them:

Connect the following nine points by four straight lines, without lifting your pencil.

°             °             °

°             °             °

°             °             °

Easier said than done. The solution in the video below shows that you should not get contained in your thinking habits...

MathOMan connects 9 points with 4 lines

Smiling Bin is even more witty and takes her revenge by challenging me with the following task, which must be impossible for any specialist on connectivity questions in topology.

Without lifting your pencil draw a circle and its center (but not more).

Here is the video where she shows us her tricky solution:

Bin draws a circle and its midpoint

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Categories

Small exercice about finite groups

Some solutions to Arnold's Trivium

A convergent series

Compression of a soccer ball

Compute an approximate value of an integral

Theorem about a circle, three chords and a midpoint

Christmas Riddle

An exercise about groups

Cargo-cult science training

An exercise about the temperature distribution on the earth

Mixing wine and beer

Comatrix and adjugate matrix

Dimension of the Commutator Space

A sunday child

Deserve your dessert!

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