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# Monthly Archives: August 2011

## Video: Documentary on Proof of Fermat’s Last Theorem

Image via Wikipedia

Not a standard post! This post is a tribute to one of the greatest mathematicians in world history, Pierre de Fermat on his 410th birthday. New to his name? Read this wikipedia article to discover about Pierre Fermat.
His Last Theorem is very famous amongst mathematicians because of its unsolvability. It was unsolved for almost 4 centuries. Learn about Fermat’s Last Theorem here. As a tribute, below is a youtube video created for BBC Horizon programme. This video is Simon Singh‘s moving documentary of Andrew Wiles‘ extraordinary search for the most elusive proof in number theory, i.e., proof of Fermat’s last theorem.

## NPTEL: An Innovation in Visual and Online Learning

First of all, Happy Independence Day to all my Indian Friends and followers. This post is about an Indian pioneer in online learning, namely NPTEL.
National Programm on Technology Enhanced Learning (NPTEL) provides E-learning through online Web and Video courses in Engineering, Science and humanities streams. The mission of NPTEL is to enhance the quality of Engineering education in the country by providing free online courseware. All videos of NPTEL include the lectures of Indian professors in IITs and IISc and they can be found either on NPTEL homepage or at their YouTube page. On Youtube, the Video Courses are organised as PLAYLISTS under the following Categories:
1. Core Sciences
2. Civil Engineering
3. Computer Science and Engineering
4. Electrical Engineering
5. Electronics and Communication Engineering
6. Mechanical Engineering

.
And on their website, these are arranged in a order of Subjects.
$\textnormal {Visit NPTEL Website}$
$\textnormal{Watch Videos On YouTube.com}$
Bellow are two Course Videos, as a demo, one on Semiconductors and other on Artificial Intelligence. It would be better to watch them, before you go for whole.
(more…)

## A Problem On Several Triangles

A triangle $T$ is divided into smaller triangles such that any two of the smaller triangles either have no point in common, or have a vertex in common, or actually have an edge in common. Thus no two smaller triangles touch along part of an edge of them.
For an illustration let me denote the three vertices of T by 1, 2 and 3. Now number each of the vertices of the small triangles by 1, 2, 3. Do this in an arbitrary way, but such that vertices lying on an edge of T must not be numbered by the same number as the vertex of T opposite to that edge.

Show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3.

# Solution

To show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3, we show that the number of small triangles whose vertices are labeled with $1,2,3$ is odd and thus actually $>0$!

We enumerate all small triangles in the picture as $T_1$, $T_2, \ldots, T_n$ and denote by $a_i$ the number of edges with endpoints $1$ and $2$ in each triangle $T_i$. Thus, if say the vertices of $T_i$ are labeled by $1,1,2$, then $a_i=2$, and so on …

Observe now that obviously we have

$\displaystyle a_1+a_2+a_3+\cdots +a_n= A+2B,$

where $A$ is the number of triangles whose vertices are labeled $1,2,3$, while $B$ is the number of those triangles labeled by $1,1,2$ or $1,2,2$. (Actually it is easily seen that $a_i=2$ for such triangles, while $a_i=1$ if the vertices of $T_i$ are $1,2,3$ and $a_i=0$ otherwise.) All we have to show is that $A$ is odd.

Let $C$ denote the number of $12$-edges lying inside the original triangle $T$ and let $D$ be the number of $12$-edges lying on the boundary of $T$. Every interior $12$-edge lies in two triangles $T_i$ and thus it is counted twice in the sum $a_1+a_2+a_3+\cdots +a_n$, while every boundary $12$-edge is counted only once. In conclusion we get

$\displaystyle a_1+a_2+a_3+\cdots +a_n= 2C+D,$

which yields

$\displaystyle A+2B=2C+D.$

Hence $A$ is odd if and only if $D$ is odd. It is therefore enough to show that $D$ is odd.

According to the hypothesis of the problem, edges labeled $12$ or $21$ can occur only on the $12$-edge of the large triangle $T$. We start walking along the edge $12$ of the triangle $T$ starting at the vertex $1$ toward the vertex $2$. Now, only when we first pass an edge labeled $12$ will we arrive at the first vertex labeled $2$. A number of vertices labeled $2$ may now follow, and only after we have passed a segment $21$ do we reach a label $1$, and so on. Thus after an odd number of segments $12$ or $21$ we arrive at vertices labeled $2$, and after an even number of such segments we arrive at vertices labeled $1$. Since the last vertex we will reach is the vertex $2$ of the big triangle $T$, it follows that the total number of segments $12$ or $21$ lying on the side $12$ of the big triangle $T$ must be odd! The same reasoning applies for each of the other edges of the big triangle $T$, so we deduce that $D$, the total number of $12$ or $21$-edges lying on the boundary of $T$, must be odd. Proved

# Graphical Proof

It is obvious. As a result of this numbering we get following diagram:

Problem Image

## Problem1: Smallest Autobiographical Number:

A number with ten digits or less is called autobiographical if its first digit (from the left) indicates the number of zeros it contains,the second digit the number of ones, third digit number of twos and so on.

For example: 42101000 is autobiographical.

Find, with explanation, the smallest autobiographical number.

Solution of Problem 1

## Problem 2: Fit Rectangle:

A rectangle has dimensions $39.375$ cm $\times 136.5$ cm.

• Find the least number of squares that will fill the rectangle.
• Find the least number of squares that will fill the rectangle, if every square must be the same size and Find the largest square that can be tiled to completely fill the rectangle.

Solution of Problem 2

## Solutions of Problem 1:

The restrictions which define an autobiographical number make it straightforward to find the lowest one. It cannot be 0, since by
definition the first digit must indicate the number of zeros in the number. Presumably then, the smallest possible autobiographical number will contain only one 0.If this is the case, then the first digit must be 1. 10 is not a candidate because the second digit must indicate the number of 1s in the number–in this case, 1. So If the
number contains only one zero, it must contain more than one 1.
(If it contained one 1 and one 0, then the first two digits
would be 11, which would be contradictory since it actually contains two 1s).
Again, presumably the lowest possible such number will contain the lowest
possible number of 1s, so we try a number with one 0 and two 1s. It will be of the form: 12-0–..
Now, there is one 2 in this number, so the first three digits must be 121. To meet all the conditions discussed above, we can simply take a 0 onto the end of this to obtain 1210, which is
the smallest auto-biographical number.

## Solution of Problem 2:

We solve the second and third parts of the question
first:

We convert each number to a fraction and get a common denominator, then find the gcd (greatest common divisor) of the numerators.

That is, with side lengths $39.375$ cm and $136.5$ cm , we convert those numbers to fractions (with a common
denominator):
$39.375 = \dfrac{315}{8}$.

$136.5 = \dfrac{273}{2} = \dfrac{1092}{8}$.

Now we need to find the largest common factor of 315 and 1092.
Which is 21. So $\dfrac{21}{8}=2.625$ is the largest number that divides evenly into the two numbers $39.375$ and $136.5$.
There will be $\dfrac{1092}{21} \times \dfrac{315}{21} = 52 \times 15 = 780$ squares, each one a $2.625$ cm $\times 2.625$ cm square needed to fill the rectangle (52 in each row,with 15 rows).

Now we shall solve the first part.
Number of squares lengthwise is 52 and breadthwise is 15. Now we will combine these squares in order to find least number of squares to fill the rectangle. First three squares would be of
dimension 15 by 15. In this way length of 45 units is utilized. Now the rectangle which is left with us excluding three squares is 7 by 15. Again in the same way we can make two squares of dimension 7 by 7. In this way breadth of 14 units is utilized.
Now we are left with the rectangle of dimension 7 by 1.
These can further be subdivided into seven squares each of
dimension 1 by 1. In this way the least number of squares to fill the
rectangle is 3 + 2+ 7 = 12. The required answer is 12.
Note that the three numbers 3, 2, and 7 are involved in the Euclidean Algorithm for finding the g.c.d.!

Source: Internet

## On the Rope Boys are Tau and Girls Are Pi

Cartoon of

Today, in the morning when I was teaching my sister about numbers, viz. rational numbers and irrational numbers, this one was created. My sister, Kavita Tiwari, is in Vth grade and mainly study Math ,Science and Social Science. On last wednesday we visited a Circus show. We all liked that very much. And she became a fan of a girl acrobat. She always says that female acrobats are far better than male acrobats in circus-shows having rope-walk show. Male acrobats look like they are about to fall, but female ones are frank in this work. When I discussed with her about some constants like $\pi, \, \tau, \, e$ etc., she asked me to stop for a while, and started making this cartoon. Her thoughts were mingled with math. I was surprized by the meaning this cartoon had. I really liked her stuff, and thought to post it on the web.

Notes:

• I have searched everywhere, but couldn’t find same topic/idea like this.
• For those who don’t know:

[ My english is also weaker than her. ]

## Blog of the Month -August 2011

I announced that I shall chose a blog from the education blogsphere as Blog of the Month. To complete this task, I googled for days, read them, analysed them and now I have the winner of ‘Blog of the Month’.
This is the first month of this series and discussing article is made in hurry, so one can feel an emptiness and lack of interest in it. But believe, Blog of the monthwas not selected in hurry. I took quick looks on about 500 blogs and thousands of posts. I created a list of all blogs I read and rated them on behalf of their qualities, visitors, content, language etc. From the list of 513 blogs, the shortlisted blogs were:

1. What’s New (math)
2. Gödel’s Lost Letter and P=NP(Math and Computer Science)
3. Peter Cameron’s Blog(math)
4. Let’s Play Math(math)
5. Unapologetic Mathematician(math)
6. Cock Tail Party Physics(Physics)
7. WordPress Tips(Blogging)
8. Honglang Wang’s Blog (Math and Programming)
9. The GeomBlog(CS)
10. Republic Of Mathematics (Math and Media)

I count a lot of things that there’s no need to count. Just because that’s the way I am. But I count all the things that need to be counted.

And Yes! The blog of the month is Peter Cameron’s Blog with useful content, interactive language and multidimensional approach to mathematics.

Peter Cameron is a professor of mathematics in London and he writes about math, media and education at http://cameroncounts.wordpress.com. He mingles everything with math, like poetry – media – fun and internet. His blog is full of Expositories, Problems and Results, Posts about doing – playing and learning mathematics, Poetry, Events Talks and Conferences, Typesettings and Mathematics in Media. A list of categorized posts can be found here.

# Reviews

Rating: 8.9/10
View: 7.0/10
Content: 9.5/10
Interaction: 9.0/10
Language: 9.5/10
Frequency of Posts: 8.5/10
Content Management: 10/10

[Last Updated: 20:03 IST 2011/08/05]

What are you views and thoughts on this selection? Rate Peter Cameron’s blog on the base of 10. Your comments are heartly welcomed.

## Chess Problems

1. In how many ways can two queens, two rooks, one white bishop, one black bishop, and a knight be placed on a standard $8 \times 8$ chessboard so that every position on the board is under attack by at least one piece?
Note: The color of a bishop refers to the color of the square on which it sits, not to the color of the piece.
2. Can you attack every position on the board with fewer than seven pieces?