Friday, December 20, 2002

Experience Mathematics #23 -- Ambigrams (by Punya Mishra)

Symmetry is important in mathematics and in art. Today we will look at a special kind of wordplay based on ideas of symmetry and figure and ground. Consider the word below: 

Can you read it? Now turn the page you are holding upside down and try reading it that way. The word stays the same. This image/word has rotational symmetry—essentially, it stays the same when rotated 1800.

Such visual wordplays are called ambigrams. The word “ambigram” was first coined by the cognitive scientist Douglas Hofstadter. Here is an ambigram of the word ambigram itself.

Ambigrams can be of many different kinds. For instance consider the word “logical” below.

This word has reflection symmetry i.e. it will read the same even when reflected in a mirror.

Some ambigrams are not about symmetry as much as they are about reading words in multiple ways. Here is one titled “good-evil” Can you see both words? Look carefully. This is similar to figure-ground paintings by M. C. Escher.

Creating ambigrams is great fun. Why don’t you try creating some yourself? If you want to see more examples of such wordplay you can search on Google or go to my wordplay gallery:

This guest column has been written by Professor Punya Mishra, College of Education, Michigan State University, USA. You can email him at

Friday, December 13, 2002

Experience Mathematics # 22: The mobius strip

Take a long, thin strip of paper, give it a half twist, and paste the two ends together. What you get is a mobius strip (see picture).

Compare the mobius strip with the cylinder, which you get if you don’t give a half twist. A mobius strip has only one surface. Can you see why? Draw a line along the edge and keep going on. Eventually, you will arrive at the starting point. A cylinder, on the other hand, has an inside and an outside surface. The artist Escher portrayed this idea in Mobius Strip II (woodcut, 1963) (see picture).

If you cut the paper cylinder in half, you will get two cylinders. However, if you cut the mobius strip, you will get something very similar to a mobius strip. How many half-twists does the cut mobius strip have?

The mobius strip is one of the many surfaces that appear in Topology, a branch of mathematics. Topologists have been described as the mathematicians who cannot tell the difference between a coffee mug and a doughnut. This is because a mug can be “continuously deformed” to become a doughnut. A doughnut (which is a tyre tube, topologically speaking) cannot be transformed into a sphere. So, according to topologists, the tyre tube is not the same as a sphere, but a coffee mug is homotopic to a doughnut.

Pic credits: Both were stolen off the web, I don't know from where. 

Thursday, December 05, 2002

Experience Mathematics # 21 -Euclid's fifth axiom

Euclid’s fifth axiom says that given a line $l$ and a point $P$ not on the line, there is exactly one line parallel to $l$ passing through the point $P$. For centuries people thought that Euclid’s fifth axiom was “obvious”. But some mathematicians did not find it obvious. 

Finally, Reimann and Lobachevsky, both modified the axiom and tried to derive a new geometry. 

Reimann began with the axiom: Given a line $l$ and a point $P$ not on the line, there is no line parallel to $l$ passing through the point $P$. Reimann derived many geometrical theorems that are applicable on the surface of a sphere. For example, he showed that the sum of angles of a triangle is always greater than 180 degrees. Try drawing a triangle on a sphere and see why this has to be true.

Similarly, if we take a hyperbola ($y=1/x$) and rotate it around the y-axis, then we obtain a surface where Lobachevsky’s geometry holds. Lobachevsky’s geometry contains the axiom: Given a line $l$ and a point $P$ not on the line, there is more than one line parallel to $l$ passing through the point $P$. In this geometry, the sum of angles of a triangle is always less than $180$ degrees.

There is a property of the surface (known as curvature) that determines the geometry. Only surfaces with curvature zero follow the Euclidean geometry. Another example of a surface is that of a saddle (of a horse). Can you tell which geometry is applicable on this surface?

Friday, November 15, 2002

Experience Mathematics # 20 -- The sum of angles in a triangle

Euclidean or plane geometry begins with notions of points and lines, and the notion that a point lies on a line. Think of lines as sets, and a point as an element belonging to a set. Points and lines satisfy certain axioms. In Euclidean Geometry (or plane geometry), the axioms are based on Euclid’s original axioms. From these axioms, we can use the rules of logic to derive theorems (or propositions) that can be regarded as truthful statements that apply to the plane. Here a plane is a model, or a mini-universe where those axioms and theorems hold.

For example, consider the theorem: The sum of angles in a triangle is $180$ degrees. The various terms in this theorem (angle, triangle etc.) are constructs in the plane that we wish to study. The theorem itself is a property that will hold in our mini-universe. The proof should proceed from the axioms, use the definitions of the various constructs, and follow the rules of logic.

Even though the theorem is true, it does not imply that the sum all triangles is $180$ degrees. For example, consider the surface of the earth. Draw a triangle with a right angle at the North Pole. Suppose the two sides of this angle go down to the equator, and the third side of the triangle is the equator. The sum of angles of this triangle—made on the surface of the earth—is $270$ degrees!

In fact, in this non-euclidean geometry, the sum of angles in a triangle is always greater that $180$ degrees.

Can you find a surface where a sum of angles in a triangle is always less than $180$ degrees?

Friday, November 08, 2002

Experience Mathematics #19 -- Euclid's axioms

Just like elements and sets, Points and Lines are undefined notions.

We can think of a line as a set of points. These satisfy certain axioms, such as: Given a line $l$ and a point $P$ not on the line, there is only one line that is parallel to $l$ containing the point $P$. Axioms are considered to be self-evident truths.

However, several gaps were found in Euclid’s axioms. For example, consider Euclid’s proof that the base angles of an isosceles triangle are equal. Suppose we have an isosceles triangle $ABC$, where the side $AB$ is equal to the side $AC$. Drop a perpendicular $AD$ from a vertex to the side $BC$. There is nothing in Euclid’s axioms that says that the point $D$ is between the points $B$ and $C$. Nevertheless, Euclid proves that the triangles $ABD$ and $ACD$ are congruent. From this it is easy to see that the base angles of an isosceles triangle are equal.

The great mathematician Hilbert completed Euclid’s work by listing a few more axioms. These included the betweenness axioms. For example, given three points $A, B and C$, one of the axioms said either $B$ is between $A$ and $C$, or $C$ is between $A$ and $B$ or $A$ is between $C$ and $B$.

To return to Euclid’s proof, some steps need to be added to show that $D$ is between $B$ and $C$. 

But that is not all. We could consider a geometry where given a line $l$ and a point $P$ not on the line, there are no lines parallel to $l$ containing the point $P$. Such a non-euclidean geometry exists on the surface of the Earth. So one of Euclid’s axioms cannot be considered to be a self-evident truth after all.

Friday, October 25, 2002

Experience Mathematics #18 - All about itself

Russel’s Paradox shows that considering sets that contain themselves (or even asking whether they contain themselves or not) can lead to contradictory situations. But Real Life has many such self-referential situations. In this column, we will collect together many amusing (and not!) statements, such as this one.

“All Cretans are Liars”, said the Cretan Epimenides. Did Epimenides tell the truth? How can he, since he is a Cretan, and hence a liar? But if he lied, maybe he is telling the truth!

What about: This sentence is false. Is it true or false? Go through each sentence in this column and evaluate whether it is true or false.

This sentence has four words. This one, however, has six words. This one has one too too many words.

This sentence has no comma. This sentence does not describe itself.

This article is written by the author of this article. In other words, the author of Experience Mathematics writes Experience Mathematics. It is self-referential, since it refers to itself. In fact, the article refers to itself several times—but only once does the article refer to itself twice in one sentence. The author of this article is careful not to write self-referential statements.

Is this a question or not. How about this statement?

The above two statements beg the question. But what is the question? Was that the question? Does this answer the question?

The sentence below is false. The above sentence is true.

Lets not say any more, and end.

Friday, October 11, 2002

Experience Mathematics # 17 -- If it is, then it is not

A set can be thought of as a collection of objects. But what is it, really? The above sentence does not say: A set is a collection of objects. So is a set a collection of objects, or can it only be thought of as a collection of objects?

Sets can be of two types: those that contain themselves, and those that do not. For example, consider the set $F$ of fruits in your home. This set is not a fruit, so cannot contain itself. Now consider the set $A$. The set $A$ contains all sets that can be described in less than sixteen words. The above sentence has only $15$ words and describes $A$, so $A$ must be a member of itself.

Now consider the set $R$ of all sets that do not contain themselves as a member. In particular, $F$ is a member of $R$. The question is: Is $R$ a member of itself?

Well, if it is, then by definition $R$ consists of sets that do not contain themselves as a member. So $R$ is not a member of $R$. In short, if it is, then it is not.

Conversely, suppose $R$ is not a member of itself. Then since $R$ contains all sets that are not members of themselves, $R$ must be an element of $R$. Thus, if it is not, it is!

This paradox—pointed out the famous philosopher, Bertrand Russell—led to the formalization of set theory. Formally speaking, a ‘set’ and the relation ‘is an element of’ are undefined notions that satisfy certain axioms. However, we can continue to think of a set as a collection of objects. Just make sure that we consider only well defined sets—where we can decide whether any given object is an element of the set or not. That saves us from all Russellian disasters.

Thursday, October 03, 2002

Experience Mathematics #16 -- An apple a day

If you study mathematics, then you will have to deal with many statements that contain expressions of the form: If $A$ then $B$  (or, $A$ implies $B$).

Suppose it is true that if you have an Apple a day, then you keep the doctor away. Is it true that if you did not visit the doctor, then you must have had an Apple everyday? Not necessarily. In other words: “if $A$ then $B$” is a true statement, then “if $B$, then $A$” may be false. The statement “if $B$, then $A$” is the converse of “if $A$ then $B$”.

The converse is not to be confused with the contrapositive of the statement. The contrapositive of “if $A$ then $B$” is: “if not $B$ then not $A$”. Unlike the converse, if a statement is true, its contrapositive is true too. Indeed, either they are both true, or they are both false. For example, suppose that it is true that an Apple a day keeps the doctor away. Now if the doctor comes to visit you, you must not have had an Apple some day. Mathematics contains axioms (that may be regarded as “truths”) together with chains of implications—statements of the form “$A$ implies $B$”, where $A$ and $B$ are mathematical expressions. Suppose your axioms say:

1. An Orange contains the daily requirement of Vitamin C.

2. Having your daily requirement of Vitamin C will keep you healthy.

3. If you are healthy, the doctor will stay away

Then, logic dictates that an Orange a day will keep the doctor away. Unfortunately, an Apple does not contain a lot of Vitamin C.

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?