Friday, September 27, 2002

Experience Mathematics # 15 : OR and AND

Suppose that your mom says that you can have either an Apple or a Banana. Can you have both? One of the most fundamental rules of logic says that the expression either $A$ or $B$ is true only if one of $A$ or $B$ is true. That is to say, you cannot have both the Apple and the Banana (assuming you wish to obey your mom.)

What if your mom says you can have an Apple or a Banana? In this case, you can have both.

Suppose your mom asks you if you have had an Apple or a Banana. Can you honestly say yes if you have had an Apple and an Orange? The answer is yes. If she asks you if you have had an Apple and a Banana, you can answer yes only if you have had both.

Suppose your mom insists that you should not have an Apple. Is it OK to have a Banana? How about Baked Beans? It depends. The alternatives to an Apple allowed by your mom depend on the context. For example, if the alternatives allowed consist of the other fruits in the house, you cannot have baked beans instead of the Apple, but you could have a Banana. However, if the context of discussion is the five servings of fruits and vegetables that you must have every day, then Baked Beans are allowed. In Mathematics, when we refer to a set $A$, then we must specify the universal set $U$ from where the elements of $A$ are picked. Then the complement of $A$ is the set of all the elements that are in $U$ but not in $A$. Then there is no confusion when we claim: $a$ is not an element of $A$. By this statement we mean that $a$ is an element of the complement of $A$.

Tuesday, September 17, 2002

Experience Mathematics #14 - Uncountable sets

Many of the infinite sets we have encountered are countable. Even numbers, the set of prime numbers, integers and rational numbers, all have the same number of elements as $N$, the set of natural numbers. Are there any infinite sets that have more elements than the natural numbers?

This question was answered by Cantor, who showed that the real numbers outnumber the natural numbers. All the real numbers between $0$ and $1$ have a decimal expansion such as $x=0.13212987\dots$. Cantor showed that all numbers of this form cannot be put into one-to-one correspondence with the set of natural numbers. To be able to understand his proof, find a number that differs from $x$ in the first decimal place. Take any number $y$ with $2$ in the first decimal place. Since $2$ is different from $1$, $y$ differs from $x$ in the first decimal place.

To return to Cantor’s proof, suppose that you are able to find a one-to-one correspondence between the natural numbers and all the real numbers in the interval $(0,1)$. Let us denote by $x_1$ the number corresponding to $1$; $x_2$, the number corresponding to $2$, and so on. Now consider a number $y$ (between $0$ and $1$) that is different from $x_1$ in the first place after the decimal; different from $x_2$ in the second place after the decimal; and so on. Clearly, $y$ cannot appear in the list, since it is different from all the $x$’s. Thus we have found a real number between $0$ and $1$ that is not in the above correspondence. This contradiction shows that no such correspondence is possible. In other words, the real numbers are uncountable in number.

Are there any infinite sets that have more elements than $N$ but less elements than the set of real numbers?

Thursday, September 12, 2002

Experience Mathematics #13 - The cartesian society in Hilbert Hotel


A set S is countable if it can be put in one-to-one correspondence with $N$. Suppose that you are the manager of the Hilbert Hotel, a hotel with a countable number of rooms. Now, even though the hotel is full, when $200$ new guests arrive after lunch at the Restaurant at the End of the Universe, you can accommodate them. All you have to do is to move the guests in Room $1$ to Room $201$, the guests in Room $2$ to Room $202$, etc. In other words, you will move the guests in Room $n$ to Room $200+n$.

Now, suppose the hotel is empty. The members of the Cartesian Society (Motto: We Think, Therefore We Exist!) decide to have a convention. Each member of the Cartesian Society has an identification mark of the form $(a, b)$, where $a$ and $b$ are natural numbers. The chairman of the society is $(1,1)$ and for any two numbers $a$ and $b$, there is a member corresponding to the ordered pair $(a, b)$. Don’t confuse $(2,3)$ with $(3,2)$: they are quite different people. There are many, many members in this society. But all of them can be accommodated in the Hilbert Hotel. Can you find a one-to-one correspondence of the members of the Cartesian Society with the natural numbers?

First write down the ID numbers of all the members of the Cartesian Society in the form of a table. For example, put $(3,5)$ in the third row and the fifth column of the table. Now find a way to “count” them. In other words, assign a natural number to each of them in a systematic fashion.

Friday, September 06, 2002

Experience Mathematics #12 -- A part can be equal to the whole!

A set S is countable if it can be put in one-to-one correspondence with $N$. For example, if we take the set of even numbers, we can establish a one-to-one correspondence as follows. $1$ corresponds to $2$, $2$ to $4$, $3$ to $6$, and so on. This shows that the number of even numbers is equal to the number of natural numbers.

This contradictory idea—that a part of an object is equal to the whole—troubled many philosophers. But they got over it, and began to compare the concept of infinity with the concept of God. However, when Biologists made it possible to clone human beings, they have stopped approving of the idea.

The famous mathematician Hilbert told the story of a hotel with an infinite number of rooms. Suppose the Hilbert Hotel is full, but the hotel manager wants to accommodate a guest who arrives suddenly. How does he manage that? Well, he asks the guest in room number $1$ to move to Room $2$, the guest in Room $2$ to move to Room $3$, and so on. Room $1$ becomes empty and is readied for the new guest. Can you figure out how to accommodate $30$ guests, even if the hotel is full? 

What if a travel agent calls the manager, and says she is sending groups of tourists to the hotel. The numbers of people in the groups are: $3$, $7$, $11$, $15$, and so on. Help the manager come up with the required one-to-one correspondence in these cases, so that he can accommodate all the guests.

Friday, August 30, 2002

Experience Mathematics # 11 -- Counting

How can you tell if there are enough chairs in your classroom so that every student has a chair? One way is to count the chairs and the number of students in your class. A simpler way is to ask each student to sit down. If all the students are able to sit down, and no chair is left over, the number of students and chairs is equal.

This fundamental idea is at the heart of mathematics, because it deals with counting. Just like there is a weight that represents a kilogram, there is a unit in mathematics that represents a number $n$. Let $I(n)$ be the set containing the first $n$ natural numbers. The set $I(n)$ is the “unit” that represents the number $n$. For example, the set $A$ with elements $a, b, c, \dots, z$, has $26$ elements because this set can be put in one to one correspondence with $I(26)$. 

But what about infinite sets? The set $N$ of natural numbers is a unit for infinite sets. We say a set $S$ is countable if it can be put in one to one correspondence with $N$. It is remarkable that the following subsets of $N$ are countable. Can you find the one to one correspondence between $N$ and these sets?

1. The set of all even numbers.

2. The set with elements $1, 4, 7, 10, \dots$

3. The set of all rational numbers.

Can you show that the set of prime numbers is countable?

When it comes to infinite sets, a part can be equal to the whole.

Wednesday, August 21, 2002

Experience Mathematics #10 -- Sets

A fundamental object in mathematics is a set. You can think of a set as a collection of objects. We are familiar with the set $N$ of natural numbers. The empty set is a set with no elements.

We say that a set $B$ is a subset of a set $A$ if each element of $B$ is also in $A$. Two sets are equal if they are subsets of each other. The empty set is a subset of every set. For example the set of even numbers $\left\{ 2, 4, 6, \dots\right\}$ is a subset of $N$.

Q1. How many subsets does the empty set have? (Hint: There is atleast one subset.)

The members of the set are called its elements.

Let $I(n)$ denote the set containing the first $n$ natural numbers. The set $I(n)$ has n elements.

Q2. List all the subsets of the set $I(1)$. How many subsets does $I(1)$ have?

Q3. List all the subsets of the set $I(2)$. How many subsets does $I(2)$ have?

Q4. List all the subsets of the set $I(3)$. How many subsets does $I(3)$ have?

Suppose you delete $3$ from each subset of $I(3)$. What do you get? Use this idea to do Q5.

Q5. List all the subsets of the set $I(4)$. How many subsets does $I(4)$ have?

Q6. How many subsets does $I(n)$ have? Guess the answer from your experiments above.

If you cannot find a pattern, you must have made a mistake in listing the sets earlier.

Friday, August 16, 2002

Experience Mathematics #9 -- Prime numbers

The prime numbers are numbers that only have $1$ and themselves as factors. The first few prime numbers are $2, 3, 5, 7, 11, 13, 17, 19, \dots$. One of the most important ideas in the theory of numbers is that a number can be written in a unique way as a product of prime powers. For example,
$4=2^2$, $15$ is $3$ times $5$, and $20$ is $2\times 5$.

The uniqueness of the prime factorization is used to show that the square root of $2$ is not a rational number. A rational number is a number in the form $p/q$, where both $p$ and $q$ are integers, and $q$ is not equal to zero. By doing the following questions, you can prove that the square root of two is an irrational number. The proof is by contradiction. We assume that the square root of $2$ can be written as a fraction, and then show our assumption implies a false statement.

Q1. Suppose the square root of $2$ equals $ p/q$ for some integers $p$ and $q$. Show that $2q^2=p^2$.

Q2. Show that the power of $2$ in the prime factorization of the number $2q^2$ is an odd number.

Q3. Show that the power of $2$ in the prime factorization of the number $p^2$ is an even number.

The above two statements are contradictory. Thus our assumption, that the square root of $2$ is a rational number, must be false.

Can you generalize this proof to prove that the square root of $3$ and $5$ are also irrational numbers?

Friday, August 09, 2002

Experience Mathematics #8 -- Finding patterns

Mathematics is all about finding patterns. Begin learning to spot patterns in numbers today, and maybe one day you can solve a big mathematical problem—like Maninder Agarwal, Neeraj Kayal, and Nitin Saxena, computer scientists at IIT, Kanpur. These mathematical stars have just announced a new algorithm, by which they can tell whether a number is a prime number. An implementation of this algorithm will mean that everyone will have to re-look at computer programs used to keep messages transferred on the Internet confidential.

In today’s activity, find the next three terms in the following sequences.

Q1. $1, 3, 5, 7,\dots $

Q2. $2, 4, 6, 8, \dots $

Q3. $1, 4, 9, 16, \dots $

Q4. $2, 3, 5, 7, 11, 13, 17, \dots $

Q5. $1, 2, 4, 8, 16, \dots $

Q6. $1, 4, 1, 4, 2, \dots $

Q7. $3, 6, 9, 12, 15, 18, \dots $

Q8. $0, 3, 8, 15, 24, \dots $

Q9. $0, 1, 1, 2, 3, 5, 8, \dots $

Q10. $1, 3, 6, 10, 15, 21, \dots $

Q11. $2, 8, 20, 40, 70, \dots $

Q12. $1, 5, 14, 20, 55, \dots $

Friday, August 02, 2002

Experience Mathematics # 7 -- A continued fraction

We all know that the square root of $2$ is an irrational number. That is to say, it cannot be written as a fraction $p/q$. Here $p$ and $q$ are integers, and $q$ is not zero. But it can be written in the form of a continued fraction.

Here is how you can discover the continued fraction representation of the square root of $2$. You will need a calculator to do the calculations. Carefully understand the following calculations.
Note that the $.41421\dots $ starts repeating, and we get a fraction that looks like



You can chop off the fraction at any point and get a fraction that is approximately equal to the square root of $2$.

As for this week’s activity, do a similar calculation (using a calculator) for the square root of $3$ and the square root of $5$ and find the continued fraction representation of these irrational numbers.

Friday, July 26, 2002

Experience Mathematics #6 - A special triangle


Once again Sarthak Parikh (Class VIII, Sardar Patel) gave interesting stories, and made nice pictures in response to our last column. This week’s column is about Pascal’s Triangle and the patterns that can be found in it. Pascal’s triangle has an infinite number of rows, and it begins as follows. 

1
1  1
1   2   1
1   3   3   1
1   4   6   4   1
1   5   10   10   5   1

Each row has one more entry than the previous row. The row begins and ends with 1. The other entries in the middle are the sum of the two entries above. For example, in the fourth row: $4=1+3; 6= 3+3; 4=3+1$.

Pascal’s triangle is a storehouse of patterns. Make 16 rows of Pascal’s triangle, and find as many patterns as you can. For example, can you find $1, 2, 3, 4,\dots$ in the triangle? What is the sum of entries in each row? If we replace all the even numbers in the triangle by $0$, and all the odd entries by $1$, we get the Fractal known as Sierpinski Triangle. This computer-generated picture shows this pattern. 



Thursday, July 18, 2002

Experience Mathematics # 5 -- Mathematical stories

This week’s column is about word problems. Many children find it difficult to solve word problems in mathematics. But can you do the reverse? Can you make word problems based on mathematical statements? Also, make a picture that represents the given statement.

In each of the following questions, you have to write the given mathematical statement in words. Then you have to make a story around it, as in the example below.

Example: $10+2=12$. Ten when added to two is equal to twelve.

Story: I was going along the road when I saw two owls. I was surprised, since owls usually sleep during the day. Then I thought that they must be owls from the Harry Potter books, come to deliver letters. Suddenly I saw 10 more owls. They were carrying a big parcel. In all, it was a strange day, where I saw 12 owls in all.

We leave the picture for you.

Q1.  $8-3=5$.

Q2. $2-6=-4.$

Q3. $6 \times 3=18 $

Q4. $16=4^2$

Q5. $x+20= 37$, where $x$ is some number.


Thursday, June 27, 2002

Experience Mathematics #4 -- Division by zero!

Both Koonal Sharda (Class VI, DPS, Mathura Road) and Sarthak Parikh (Class VIII, Sardar Patel) gave the correct solutions with good explanations. What is creditable is that students are expected to answer these questions only when they reach Class XI. However, they say that Gauss found these patterns when he was only 10 years old! But anyway, well done.

Assuming the pattern that you observed continues forever—and it does—the sum $O$ of the first $n$ odd numbers is given by the formula: $O=1+3+…+2n-1=n^2$. This fact can be proved using Mathematical Induction. Adding $n$ is the same thing as adding $1$ $n$ times. So we obtain: $E=2+4+…+2n=n^2 +n$. From this it is easy to find the sum of the first $n$ numbers. We divide each term by $2$ to get $$1+2+3+…+n=(n^2 +n)/2 = n(n+1)/2.$$

Once again Sarthak asked some deep questions on which today’s experiments are based. He asked: What meaning can be given to division by 0?

At the outset, let me say that division by 0 is not allowed.

The reason is the confusion caused by the following argument: We know that $0=0.$ This implies that $0$ times $1$ is equal to $0 $ times $2$. So $0 \times  1 = 0 \times 2$.  Now canceling $0$ from both sides (by dividing both sides with $0$) we obtain $1=2$, which clearly is false. And there is a theorem in mathematics that says that a false proposition implies any proposition. Which implies that Santa Claus exists. And also implies that Santa Claus does not exist.
Now you can see why mathematicians have very sensibly banned division by $0$.

However, this does not mean that mathematicians do not try to give division by $0$ some meaning. Try the following (you may need a calculator or a computer for some of these experiments):
1. Make a table of the function $y=1/x$. In the first column place numbers very close to $0$ that are positive. For example $0.1, 0.01, 0.001, \dots, 0.0000000001$. In the second column, replace $x $ in $1/x$ by this number and see what values come out for y. For example, when we replace $x$ by $.01$, we get $y=100. $ Describe the results of your experiments.

2. Now replace $x$ by negative numbers that come closer and closer to $0$ and find out what value $ y$ takes. Describe the results of your experiments.

3. Try the above with the function $y=x^2/x$. Does the value come closer and closer to a number? 

Describe the results of your experiments.

We say a function (depending on the variable $x$) approaches the limit $+\inf$  (read plus infinity) when $x$ approaches a number $a$ from the left, if given any large number $M$ (any large number you can think of), we can find a number a little smaller than $a$, so that the corresponding value of the function becomes larger than $M$. (The ‘left’ refers to the number being to the left of $a$ on the number line.) Similarly, we can define the limit approaching from the right, and the limit approaching minus infinity.
Intuitively, it is easier to understand the limit of a function if it approaches a number $q$, when $x$ approaches a number $a$. We replace $x$ by numbers close to $a$, and find the value of the function. If the value comes close to a number $q$, the limit is $q$. Try the following experiments and get a feel for the definition.

4. Multiply $(1-x)$ in turn by $(1+x),$ $(1+x+x^2),$ $(1+x^2+x^3)$ and simplify. Can you generalize the pattern?

5. What is the limit of $(1-x^2)/(1-x),$ $(1-x^3)/(1-x),$ $\dots$, $(1-x^n)/(1-x),$ as $x$ approaches $1$. (Hint: Replace $x$ by numbers close to $1$, like $1, 1.1, 0.9, 1.01, 0.99, 1.001, 0.999,$ $\dots,$ and guess the answer in each case.)

Tuesday, June 18, 2002

Experience Mathematics # 3 -- The sum of the first n odd numbers


Akarsh Gupta (Class VII, A.P.J., Noida), Sarthak Parikh (Class VIII, Sardar Patel) and Richa Sharma (Class IX, Swami Vivekananda Sarvodaya School) all gave excellent explanations/solutions to the questions asked in the previous column. The answers (very briefly) are:

Q1. The numbers that have $0, 2, 4, 6, 8$ in the last digit are divisible by $2$.

Q2. The numbers that have $0$ or $5$ in the last digit are divisible by $2$.

Q3. If the sum of the digits of a number is $9$ or a multiple of $9$, then that number is divisible by $9$.

Q4. If the sum of the digits of a number is a multiple of $3$ then that number is divisible by $3$.

Sarthak Parikh also asked an interesting question. Sarthak asked: How many lines of symmetry does a parallelogram have? What about the square? Rhombus? Rectangle? Octagon? Cut these shapes out of paper and label the vertices. Experiment by drawing various lines, and flipping the shape. If you get back the same shape (but with different labels) you have discovered a line of symmetry. For example, a square with vertices labeled ABCD, you may get a square labeled BACD (see the top two squares in the picture). If you notice, by flipping across a line you have rotated the square by 180 degree. You should also try to find other rotational symmetries. When you rotate a square ABCD clockwise by 90 degrees, you get the square DABC (bottom square in the picture). After doing some experiments, perhaps you would like to answer Sarthak’s query?



Here is another activity involving squares—the squares of natural numbers.
Complete the following table. For n going from 1 to 20 write the square of n in the second column, and the sum of the first n odd numbers in the third column. The odd numbers are (1, 3, 5, …).

 $n$
$n^2$
Sum of first $n$ odd numbers
$1$
$1$
$1=1$
$2$
$4$
$1+3= 4$
$3$
$9$
$1+3+5=9$


$20$



Make a picture showing the pattern above. (Hint: see the picture below)


  1. Find a formula for the sum $O$ of the first $n$ odd numbers: $O = 1+3+5+7+\cdots +(2n-1).$
  2. Use the formula for $O$ to find a formula for the sum $E$ of the first $n$ even numbers: $E=2+4+\cdots+2n .$
  3. Use the formula for $E$ to find a formula for the sum of the first $n$ natural numbers. 

Check your work by putting $n=1, 2, 3, 4, 5$ for each formula you find.

Thursday, June 13, 2002

Experience Mathematics # 2 - Divisibility


Among all the students who tried to answer the questions presented last week, Anirudh Matange (Class VI, Ahlcons School) gave the best explanations. The answers are: 

Q1. The Commutative Law for multiplication. 

Q2. The Prime Numbers $(2, 3, 5, 7, 11, 13, 17, 19)$ give rise to only two rectangles: 

Q3. The largest number of rectangles arise from $12, 18,$ and $20.$ 

Q4. The rectangles with side $2$ can be formed by the Even Numbers: $2, 4, 6, \dots, 20.$ 

Q5. All the Multiples of $3$ (namely: $3, 6, \dots, 18$) can form rectangles with side $3.$ 

Here is another mathematical activity on the concept of divisibility. When a number $n$ divides a number $m$ evenly, we say that $m$ is divisible by $n$, or that $n$ is a factor of $m$. There are a number of tests that determine whether a given number is divisible by $2, 3, 5,$ or $9.$ By doing the following experiments, you can discover these tests for yourself. 

1. Consider the numbers: $2030, 4201, 89782, 129083, 124, 5435, 67656, 9087, 8888, 90919.$ For each number, you have to tell the last digit when the given number is multiplied by $2$. 

2. For each of the numbers in Activity 1, you have to tell the last digit when the given number is multiplied by $5$. 

3. Make a table with 2 columns. In the first column, place the multiples of $9$ less than $100$ ($9, 18, 27, 36,\dots , 81, 90$). In the second column, note the sum of their digits. 

4. Make a table with two columns. In the first column, put any $10$ multiples of $3$ less than $100$—such as $12, 15, 63, 51$—that are not all multiples of $9$. In the second column, note the sum of their digits. 

Now use your experiments and guess the answer of the following: 

Q1. Which of the following numbers is divisible by $2$: $100000, 1201, 2342, 9083, 2124, 21245, 1906, 6757, 1978, 9879.$ 

Q2. Which of the following numbers is divisible by $5$: $100000, 1201, 2342, 9083, 2124, 21245, 1906, 6757, 1978, 9879.$ 

Q3. What is the rule for checking whether a number is divisible by $9$? 

Q4. Which of the following numbers is divisible by $3$: $101010, 1201, 20112, 2124, 21223, 1906, 6757, 1978, 9879, 1080.$ 

Bright students should verify, and then prove that their guesses are correct. But the proofs, while easy, are not covered in school syllabi. For the moment, it is enough to know the tests of divisibility without knowing why they work.

Sunday, June 02, 2002

Experience Mathematics # 1 -- Mathematical activities

Introduction

Children learn by doing. By doing these activities, they will experience interesting mathematical ideas. They will also gain experience in thinking mathematically. This will help them understand concepts easily, and better their performance in exams.
It is also very important to remember that the encouragement of parents and grandparents motivates a child a lot. Praise them when they show their intelligence by doing mathematical activities successfully. This will make the children work hard to become better at mathematics.
Why do children find mathematics difficult? The most important reason is that they find mathematics removed from their daily existence. However, it is not too difficult to give mathematical experiences to children. In this column, we will give an activity for children. It is a good idea for parents to help the child with the activity, if the child is studying in class I-V. For older children, show them this column and challenge them to explain the answers of the questions below.
*** 


This activity requires 20 round pegs. The round pieces of a Carrom Board set will do nicely. You may also use 20 buttons of the same size. 

Pick a number—say 15. Take 15 pegs and make as many rectangles as you can out of them. Each time you are able to make a rectangle, reproduce the rectangle in a drawing book and note the dimensions. (You can use dots to make rectangular arrays.) For example, using 15 pegs, you can draw the following rectangles. 




Repeat this activity for each number from 1 to 20. Now try to answer the following questions:
  1. When you rotate a rectangle by 90 degrees, you get another rectangle with the same number of dots. What is this law called? 
  2. List the numbers that give rise to only 2 rectangles? 
  3. Which number (or numbers) lead to the largest number of rectangles? 
  4. What are the dimensions of all the rectangles with one side consisting of 2 dots? 
  5. What are the dimensions of all the rectangles with one side consisting of 3 dots? 
It is not necessary that you will be able to answer all the questions. But it is important to try hard. In the next column, we will give explanations of the mathematics underlying this activity.