Category Archives: pure mathematics

Set Theory, Relations, Functions Preliminaries: II

Relations:

Concept of Order:

Let us say that we create a “table” of two columns in which the first column is the name of the father, and the second column is name of the child. So, it can have entries like (Yogesh, Meera), (Yogesh, Gopal), (Kishor, Nalin), (Kishor, Yogesh), (Kishor, Darshna) etc. It is quite obvious that “first” is the “father”, then “second” is the child. We see that there is a “natural concept of order” in human “relations”. There is one more, slightly crazy, example of “importance of order” in real-life. It is presented below (and some times also appears in basic computer science text as rise and shine algorithm) —-

Rise and Shine algorithm:

When we get up from sleep in the morning, we brush our teeth, finish our morning ablutions; next, we remove our pyjamas and shirt and then (secondly) enter the shower; there is a natural order here; first we cannot enter the shower, and secondly we do not remove the pyjamas and shirt after entering the shower. 🙂

Ordered Pair: Definition and explanation:

A pair $(a,b)$ of numbers, such that the order, in which the numbers appear is important, is called an ordered pair. In general, ordered pairs (a,b) and (b,a) are different. In ordered pair (a,b), ‘a’ is called first component and ‘b’ is called second component.

Two ordered pairs (a,b) and (c,d) are equal, if and only if $a=c$ and $b=d$. Also, $(a,b)=(b,a)$ if and only if $a=b$.

Example 1: Find x and y when $(x+3,2)=(4,y-3)$.

Solution 1: Equating the first components and then equating the second components, we have:

$x+3=4$ and $2=y-3$

$x=1$ and $y=5$

Cartesian products of two sets:

Let A and B be two non-empty sets then the cartesian product of A and B is denoted by A x B (read it as “A cross B”),and is defined as the set of all ordered pairs (a,b) such that $a \in A$, $b \in B$.

Thus, $A \times B = \{ (a,b): a \in A, b \in B\}$

e.g., if $A = \{ 1,2\}$ and $B = \{ a,b,c\}$, tnen $A \times B = \{ (1,a),(1,b),(1,c),(2,a),(2,b),(2,c)\}$.

If $A = \phi$ or $B=\phi$, we define $A \times B = \phi$.

Number of elements of a cartesian product:

By the following basic counting principle: If a task A can be done in m ways, and a task B can be done in n ways, then the tasks A (first) and task B (later) can be done in mn ways.

So, the cardinality of A x B is given by: $n(A \times B)= n(A) \times n(B)$.

So, in general if a cartesian product of p finite sets, viz, $A_{1}, A_{2}, A_{3}, \ldots, A_{p}$ is given by $n(A_{1} \times A_{2} \times A_{3} \ldots A_{p}) = n(A_{1}) \times n(A_{2}) \times \ldots \times n(A_{p})$

Definitions of relations, arrow diagrams (or pictorial representation), domain, co-domain, and range of a relation:

Consider the following statements:

i) Sunil is a friend of Anil.

ii) 8 is greater than 4.

iii) 5 is a square root of 25.

Here, we can say that Sunil is related to Anil by the relation ‘is a friend of’; 8 and 4 are related by the relation ‘is greater than’; similarly, in the third statement, the relation is ‘is a square root of’.

The word relation implies an association of two objects according to some property which they possess. Now, let us some mathematical aspects of relation;

Definition:

A and B are two non-empty sets then any subset of $A \times B$ is called relation from A to B, and is denoted by capital letters P, Q and R. If R is a relation and $(x,y) \in R$ then it is denoted by $xRy$.

y is called image of x under R and x is called pre-image of y under R.

Let $A=\{ 1,2,3,4,5\}$ and $B=\{ 1,4,5\}$.

Let R be a relation such that $(x,y) \in R$ implies $x < y$. We list the elements of R.

Solution: Here $A = \{ 1,2,3,4,5\}$ and $B=\{ 1,4,5\}$ so that $R = \{ (1,4),(1,5),(2,4),(2,5),(3,4),(3,5),(4,5)\}$ Note this is the relation R from A to B, that is, it is a subset of A x B.

Check: Is a relation $R^{'}$ from B to A defined by x<y, with $x \in B$ and $y \in A$ — is this relation $R^{'}$ *same* as R from A to B? Ans: Let us list all the elements of R^{‘} explicitly: $R^{'} = \{ (1,2),(1,3),(1,4),(1,5),(4,5)\}$. Well, we can surely compare the two sets R and $R^{'}$ — the elements “look” different certainly. Even if they “look” same in terms of numbers, the two sets $R$ and $R^{'}$ are fundamentally different because they have different domains and co-domains.

Definition : Domain of a relation R: The set of all the first components of the ordered pairs in a relation R is called the domain of relation R. That is, if $R \subseteq A \times B$, then domain (R) is $\{ a: (a,b) \in R\}$.

Definition: Range: The set of all second components of all ordered pairs in a relation R is called the range of the relation. That is, if $R \subseteq A \times B$, then range (R) = $\{ b: (a,b) \in R\}$.

Definition: Codomain: If R is a relation from A to B, then set B is called co-domain of the relation R. Note: Range is a subset of co-domain.

Type of Relations:

One-one relation: A relation R from a set A to B is said to be one-one if every element of A has at most one image in B and distinct elements in A have distinct images in B. For example, let $A = \{ 1,2,3,4\}$, and let $B=\{ 2,3,4,5,6,7\}$ and let $R_{1}= \{ (1,3),(2,4),(3,5)\}$ Then $R_{1}$ is a one-one relation. Here, domain of $R_{1}= \{ 1,2,3\}$ and range of $R_{1}$ is $\{ 3,4,5\}$.

Many-one relation: A relation R from A to B is called a many-one relation if two or more than two elements in the domain A are associated with a single (unique) element in co-domain B. For example, let $R_{2}=\{ (1,4),(3,7),(4,4)\}$. Then, $R_{2}$ is many-one relation from A to B. (please draw arrow diagram). Note also that domain of $R_{1}=\{ 1,3,4\}$ and range of $R_{1}=\{ 4,7\}$.

Into Relation: A relation R from A to B is said to be into relation if there exists at least one element in B, which has no pre-image in A. Let $A=\{ -2,-1,0,1,2,3\}$ and $B=\{ 0,1,2,3,4\}$. Consider the relation $R_{1}=\{ (-2,4),(-1,1),(0,0),(1,1),(2,4) \}$. So, clearly range is $\{ 0,1,4\}$ and $range \subseteq B$. Thus, $R_{3}$ is a relation from A INTO B.

Onto Relation: A relation R from A to B is said to be ONTO relation if every element of B is the image of some element of A. For example: let set $A= \{ -3,-2,-1,1,3,4\}$ and set $B= \{ 1,4,9\}$. Let $R_{4}=\{ (-3,9),(-2,4), (-1,1), (1,1),(3,9)\}$. So, clearly range of $R_{4}= \{ 1,4,9\}$. Range of $R_{4}$ is co-domain of B. Thus, $R_{4}$ is a relation from A ONTO B.

Binary Relation on a set A:

Let A be a non-empty set then every subset of $A \times A$ is a binary relation on set A.

Illustrative Examples:

E.g.1: Let $A = \{ 1,2,3\}$ and let $A \times A = \{ (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}$. Now, if we have a set $R = \{ (1,2),(2,2),(3,1),(3,2)\}$ then we observe that $R \subseteq A \times A$, and hence, R is a binary relation on A.

E.g.2: Let N be the set of natural numbers and $R = \{ (a,b) : a, b \in N and 2a+b=10\}$. Since $R \subseteq N \times N$, R is a binary relation on N. Clearly, $R = \{ (1,8),(2,6),(3,4),(4,2)\}$. Also, for the sake of completeness, we state here the following: Domain of R is $\{ 1,2,3,4\}$ and Range of R is $\{ 2,4,6,8\}$, codomain of R is N.

Note: (i) Since the null set is considered to be a subset of any set X, so also here, $\phi \subset A \times A$, and hence, $\phi$ is a relation on any set A, and is called the empty or void relation on A. (ii) Since $A \times A \subset A \times A$, we say that $A \subset A$ is a relation on A called the universal relation on A.

Note: Let the cardinality of a (finite) set A be $n(A)=p$ and that of another set B be $n(B)=q$, then the cardinality of the cartesian product $n(A \times B)=pq$. So, the number of possible subsets of $A \times B$ is $2^{pq}$ which includes the empty set.

Types of relations:

Let A be a non-empty set. Then, a relation R on A is said to be: (i) Reflexive: if $(a,a) \in R$ for all $a \in A$, that is, aRa for all $a \in A$. (ii) Symmetric: If $(a,b) \in R \Longrightarrow (b,a) \in R$ for all $a,b \in R$ (iii) Transitive: If $(a,b) \in R$, and $(b,c) \in R$, then so also $(a,c) \in R$.

Equivalence Relation:

A (binary) relation on a set A is said to be an equivalence relation if it is reflexive, symmetric and transitive. An equivalence appears in many many areas of math. An equivalence measures “equality up to a property”. For example, in number theory, a congruence modulo is an equivalence relation; in Euclidean geometry, congruence and similarity are equivalence relations.

Also, we mention (without proof) that an equivalence relation on a set partitions the set in to mutually disjoint exhaustive subsets.

Illustrative examples continued:

E.g. Let R be an equivalence relation on $\mathbb{Q}$ defined by $R = \{ (a,b): a, b \in \mathbb{Q}, (a-b) \in \mathbb{Z}\}$. Prove that R is an equivalence relation.

Proof: Given that $R = \{ (a,b) : a, b \in \mathbb{Q}, (a-b) \in \mathbb{Z}\}$. (i) Let $a \in \mathbb{Q}$ then $a-a=0 \in \mathbb{Z}$, hence, $(a,a) \in R$, so relation R is reflexive. (ii) Now, note that $(a,b) \in R \Longrightarrow (a-b) \in \mathbb{Z}$, that is, $(a-b)$ is an integer $\Longrightarrow -(b-a) \in \mathbb{Z} \Longrightarrow (b-a) \in \mathbb{Z} \Longrightarrow (b,a) \in R$. That is, we have proved $(a,b) \in R \Longrightarrow (b,a) \in R$ and so relation R is symmetric also. (iii) Now, let $(a,b) \in R$, and $(b,c) \in R$, which in turn implies that $(a-b) \in \mathbb{Z}$ and $(b-c) \in \mathbb{Z}$ so it $\Longrightarrow (a-b)+(b-c)=a-c \in \mathbb{Z}$ (as integers are closed under addition) which in turn $\Longrightarrow (a,c) \in R$. Thus, $(a,b) \in R$ and $(b,c) \in R$ implies $(a,c) \in R$ also, Hence, given relation R is transitive also. Hence, R is also an equivalence relation on $\mathbb{Q}$.

Illustrative examples continued:

E.g.: If $(x+1,y-2) = (3,4)$, find the values of x and y.

Solution: By definition of an ordered pair, corresponding components are equal. Hence, we get the following two equations: $x+1=3$ and $y-2=4$ so the solution is $x=2,y=6$.

E.g.: If $A = (1,2)$, list the set $A \times A$.

Solution: $A \times A = \{ (1,1),(1,2),(2,1),(2,2)\}$

E.g.: If $A = \{1,3,5 \}$ and $B=\{ 2,3\}$, find $A \times B$, and $B \times A$, check if cartesian product is a commutative operation, that is, check if $A \times B = B \times A$.

Solution: $A \times B = \{ (1,2),(1,3),(3,2),(3,3),(5,2),(5,3)\}$ whereas $B \times A = \{ (2,1),(2,3),(2,5),(3,1),(3,3),(3,5)\}$ so since $A \times B \neq B \times A$ so cartesian product is not a commutative set operation.

E.g.: If two sets A and B are such that their cartesian product is $A \times B = \{ (3,2),(3,4),(5,2),(5,4)\}$, find the sets A and B.

Solution: Using the definition of cartesian product of two sets, we know that set A contains as elements all the first components and set B contains as elements all the second components. So, we get $A = \{ 3,5\}$ and $B = \{ 2,4\}$.

E.g.: A and B are two sets given in such a way that $A \times B$ contains 6 elements. If three elements of $A \times B$ are $(1,3),(2,5),(3,3)$, find its remaining elements.

Solution: We can first observe that $6 = 3 \times 2 = 2 \times 3$ so that A can contain 2 or 3 elements; B can contain 3 or 2 elements. Using definition of cartesian product of two sets, we get that $A= \{ 1,2,3\}$ and $\{ 3,5\}$ and so we have found the sets A and B completely.

E.g.: Express the set $\{ (x,y) : x^{2}+y^{2}=25, x, y \in \mathbb{W}\}$ as a set of ordered pairs.

Solution: We have $x^{2}+y^{2}=25$ and so

$x=0, y=5 \Longrightarrow x^{2}+y^{2}=0+25=25$

$x=3, y=4 \Longrightarrow x^{2}+y^{2}=9+16=25$

$x=4, y=3 \Longrightarrow x^{2}+y^{2}=16+9=25$

$x=5, y=0 \Longrightarrow x^{2}+y^{2}=25+0=25$

Hence, the given set is $\{ (0,5),(3,4),(4,3),(5,0)\}$

E.g.: Let $A = \{ 1,2,3\}$ and $B = \{ 2,4,6\}$. Show that $R = \{ (1,2),(1,4),(3,2),(3,4)\}$ is a relation from A to B. Find the domain, co-domain and range.

Solution: Here, $A \times B = \{ (1,2),(1,4),(1,6),(2,2),(2,4),(2,6),(3,2),(3,4),(3,6)\}$. Clearly, $R \subseteq A \times B$. So R is a relation from A to B. The domain of R is the set of first components of R (which belong to set A, by definition of cartesian product and ordered pair)  and the codomain is set B. So, Domain (R) = $\{ 1,3\}$ and co-domain of R is set B itself; and Range of R is $\{ 2,4\}$.

E.g.: Let $A = \{ 1,2,3,4,5\}$ and $B = \{ 1,4,5\}$. Let R be a relation from A to B such that $(x,y) \in R$ if $x. List all the elements of R. Find the domain, codomain and range of R. (as homework quiz, draw its arrow diagram);

Solution: Let $A = \{ 1,2,3,4,5\}$ and $B = \{ 1,4,5\}$. So, we get R as $(1,4),(1,5),(2,4),(2,5),(3,4),(3,5),(4,5)$. $domain(R) = \{ 1,2,3,4\}$, $codomain(R) = B$, and $range(R) = \{ 4,5\}$.

E.g. Let $A = \{ 1,2,3,4,5,6\}$. Define a binary relation on A such that $R = \{ (x,y) : y=x+1\}$. Find the domain, codomain and range of R.

Solution: By definition, $R \subseteq A \times A$. Here, we get $R = \{ (1,2),(2,3),(3,4),(4,5),(5,6)\}$. So we get $domain (R) = \{ 1,2,3,4,5\}$, $codomain(R) =A$, $range(R) = \{ 2,3,4,5,6\}$

Tutorial problems:

1. If $(x-1,y+4)=(1,2)$, find the values of x and y.
2. If $(x + \frac{1}{3}, \frac{y}{2}-1)=(\frac{1}{2} , \frac{3}{2} )$
3. If $A=\{ a,b,c\}$ and $B = \{ x,y\}$. Find out the following: $A \times A$, $B \times B$, $A \times B$ and $B \times A$.
4. If $P = \{ 1,2,3\}$ and $Q = \{ 4\}$, find the sets $P \times P$, $Q \times Q$, $P \times Q$, and $Q \times P$.
5. Let $A=\{ 1,2,3,4\}$ and $\{ 4,5,6\}$ and $C = \{ 5,6\}$. Find $A \times (B \bigcap C)$, $A \times (B \bigcup C)$, $(A \times B) \bigcap (A \times C)$, $A \times (B \bigcup C)$, and $(A \times B) \bigcup (A \times C)$.
6. Express $\{ (x,y) : x^{2}+y^{2}=100 , x, y \in \mathbf{W}\}$ as a set of ordered pairs.
7. Write the domain and range of the following relations: (i) $\{ (a,b): a \in \mathbf{N}, a < 6, b=4\}$ (ii) $\{ (a,b): a,b \in \mathbf{N}, a+b=12\}$ (iii) $\{ (2,4),(2,5),(2,6),(2,7)\}$
8. Let $A=\{ 6,8\}$ and $B=\{ 1,3,5\}$. Let $R = \{ (a,b): a \in A, b \in B, a+b \hspace{0.1in} is \hspace{0.1in} an \hspace{0.1in} even \hspace{0.1in} number\}$. Show that R is an empty relation from A to B.
9. Write the following relations in the Roster form and hence, find the domain and range: (i) $R_{1}= \{ (a,a^{2}) : a \hspace{0.1in} is \hspace{0.1in} prime \hspace{0.1in} less \hspace{0.1in} than \hspace{0.1in} 15\}$ (ii) $R_{2} = \{ (a, \frac{1}{a}) : 0 < a \leq 5, a \in N\}$
10. Write the following relations as sets of ordered pairs: (i) $\{ (x,y) : y=3x, x \in \{1,2,3 \}, y \in \{ 3,6,9,12\}\}$ (ii) $\{ (x,y) : y>x+1, x=1,2, y=2,4,6\}$ (iii) $\{ (x,y) : x+y =3, x, y \in \{ 0,1,2,3\}\}$

More later,

Nalin Pithwa

A leaf out of Paul Erdos’ biography: My Brain is Open: by Bruce Schechter

Reference:

My Brain is Open: The Mathematical Journeys of Paul Erdos by Bruce Schechter, a TouchStone Book, Published by Simon and Schuster, New York.

https://www.amazon.in/My-Brain-Open-Mathematical-Journeys/dp/0684859807/ref=sr_1_1?ie=UTF8&qid=1526794050&sr=8-1&keywords=my+brain+is+open

Chapter One: Traveling.

The call might come at midnight or an hour before dawn — mathematicians are oddly unable to handle the arithmetic of time zones. Typically, a thickly accented voice on the other end of the line would abruptly begin:

“I am calling from Berlin. I want to speak to Erdos.”

“He’s not here, yet.”

“Where is he?”

“I don’t know.”

“Why don’t you know!” Click!

Neither are mathematicians always observant of social graces.

For more than sixty years mathematicians around the world have been roused from their abstract dreams by such calls, the first of the many disruptions that constituted a visit from Paul Erdos. The frequency of the calls would increase over the next several days and would culminate with a summons to the airport, where Erdos himself would appear, a short, frail man in a shapeless old suit, clutching two small suitcases that contained all of his worldly possessions. Stepping off the plane he would announce to the welcoming group of mathematicians, “My brain is open!”

Paul Erdos’ brain, when open, was one of the wonders of the world, an Ali Baba’s cave, glittering with mathematical treasures, gems of the most intricate cut and surpassing beauty. Unlike Ali Baba’s cave, which was hidden behind a huge stone in a remote desert, Erdos and his brain were in perpetual motion. He moved between mathematical meetings, universities, and corporate think tanks, logging hundreds of thousands of miles. “Another roof, another proof,” as he liked to say. “Want to meet Erdos?” mathematicians would ask. “Just stay here and wait. He’ll show up.” Along the way, in borrowed offices, guest bedrooms, and airplane cabins, Erdos wrote in excess of 1600 papers, books and articles, more than any other mathematician who ever lived. Among them are some of the great classics of the twentieth century, papers that opened up entire new fields and became the obsession and inspiration of generations of mathematicians.

The meaning of life, Erdos often said, was to prove and conjecture. Proof and conjecture are the tools with which mathematicians explore the Platonic universe of pure form, a universe that to many of them is as real as the universe in which they must reluctantly make their homes and livings, and far more beautiful. “If numbers aren’t beautiful, I don’t know what is,” Erdos frequently remarked. And although, like all mathematicians, he was forced to make his home in the temporal world, he rejected worldly encumbrances. He had no place on earth called home, nothing resembling a conventional year-round, nine-to-five job, and no family in the usual sense of the word. He arranged his life with only one purpose, to spend areas many hours a day as possible engaged in the essential, life-affirming business of proof and conjecture.

For Erdos, the mathematics that consumed most of his waking hours was not a solitary pursuit but a social activity, a movable feast. One of the greatest mathematical discoveries of the twentieth century was the simple equation that two heads are better than one. Ever since Archimedes traced his circles in sand, mathematicians, for the most part, have laboured alone — that is, until some forgotten soul realized that mathematics could be done anywhere. Only paper and pencil were needed, and those were not strictly essential. A table-cloth would do in a pinch, or the mathematician could carry his equations in his head, like a chessmaster playing blindfolded. Strong coffee, and in Erdos’ case even more powerful stimulants, helped too. Mathematicians began to frequent the coffeehouses of Budapest, Prague, and Paris, which led to the quip often attributed to Erdos:”A mathematician is a machine for turning coffee into theorems.” Increasingly, mathematical papers became the work of two, three, or more collaborators. That radical transformation of how mathematics is created is the result of many factors, not the least of which was the infectious example set by Erdos.

Erdos had more collaborators than most people have acquaintances. He wrote papers with more than 450 collaborators —- the exact number is still not known, since Erdos participated in the creation of new mathematics until the last day of his life, and his collaborators are expected to continue writing and publishing for years. The briefest encounter could lead to a publication — for scores of young mathematicians a publication that could become the cornerstone of their life’s work. He would work with anyone who could keep up with him, the famous or the unknown. Having been a child prodigy himself, he was particularly interested in meeting and helping to develop the talents of young mathematicians. Many of the world’s leading mathematicians owe their careers to an early meeting with Erdos.

Krishna Alladi, who is now a mathematician at the University of Florida, Gainesville, is one of the many young mathematicians whom Erdos helped. In 1974, when Alladi was an undergraduate in Madras, India, he began an independent investigation of a certain number theoretic function. His teachers could not help Alladi with his problem, nor could his father, who was a theoretical physicist and head of Madras Institute of Mathematics. Alladi’s father told some of his knowledgeable friends about his son’s difficulty, and they suggested that he write to Erdos.

Because Erdos was constantly on the move, Alladii sent a letter to the Hungarian Academy of Sciences. In an astonishingly short time, Alladi heard from Erdos, who said he would soon be lecturing in Calcutta. Could Alladi come there to meet him? Unfortunately, Alladi had examinations and could not attend, so he sent his father in his place to present the results of his research. After his father’s talk, Alladi recounts, “Erdos walked up to him and told him in very polite terms that he was not interested in the father but in the son.” Determined to meet with the promising young mathematician, Erdos, who was bound for Australia, rerouted his trip to stop briefly in Madras, which lies about 860 miles south of Calcutta.

Alladi was astonished that a great mathematician should change his plans to visit a student. He was nervous when he met Erdos at the airport, but that soon passed. “He talked to me as if he had known me since childhood,” Alladi recalls. The first thing Erdos asked was, “Do you know my poem about Madras?” And then he recited:

The home of the curry and the dhal,

Where Iyers speak only to Iyengars

And Iyengars speak only to God.

The Iyers and Iyengars are two Brahmin sects. The Iyers worship Shiva the Destroyer but will also worship in the temples of the Iyengars, who worship only Lord Vishnu, the Protector. Erdos explained that this was his variation on the poem about Boston and the pecking order among the Lowells, the Cabots, and God. Having put Alladi at ease, Erdos launched into a discussion of mathematics. Erdos was so impressed with Alladi, who was applying to graduate schools in the United States, that he wrote a letter on his behalf. Within a month, Alladi received the Chancellor’s Fellowship at the University of California, Los Angeles.

A celebrated magazine article about Erdos was called, “The Man Who Loved Only Numbers.” While it is true that Erdos loved numbers, he loved much more. He loved to talk about history, politics, and almost any other subject. He loved to take long walks and to climb towers, no matter how dismal the prospective view, he loved to play ping-pong, chess, and Go, he loved to perform silly tricks to amuse children and to make sly jokes and thumb his nose at authority. But, most of all, Erdos loved those who loved numbers, mathematicians. He showed that love by opening his pocket as well as his mind. Having no permanent job, Erdos also had little money, but whatever he had was at the service of others. If he heard of a graduate student who needed money to continue his studies, he would sent a cheque. Whenever he lectured in Madras, he would send his fee to the needy widow of the great Indian mathematician Srinivasa Ramanujan; he had never met Ramanujan or his wife, but the beauty of Ramanujan’s equations had inspired Erdos as a young mathematician. In 1984, he won the prestigious Wolf prize, which came with a cash reward of $at 50000, easily the most money Erdos had ever received at one time. He gave$30000 to endow a postdoctoral fellowship in the name of his parents at the Technion in Haifa, Israel, and used the remainder to help relatives, graduate students, and colleagues:”I kept only \$720,” Erdos recalled.

In the years before the internet, there was Paul Erdos. He carried a shopping bag crammed with latest papers, and his brain was stuffed with the latest gossip as well as an amazing database of the world of mathematics. He knew everybody: what they were interested in; what they had conjectured, proved, or were in the midst of proving; their phone numbers; the names and ages of their wives, children, pets; and, much more. He could tell off the top of his head on which page in which obscure Russian journal a theorem similar to the one you were working on was proved in 1922. When he met a mathematician in Warsaw, say, he would immediately take up the conversation where they had left it two years earlier. During the iciest years of the Cold War Erdos’s fame allowed him freely to cross the Iron Curtain, so that he became vital link between the East and the West.

In 1938, with Europe on the brink of war, Erdos fled to the United States and embarked on his mathematical journeys. This book is the story of those adventures. Because they took Erdos everywhere mathematics is done, this is also the story of the world of mathematics, a world virtually unknown to outsiders.. Today perhaps the only mathematician most people can name is Theodore Kacznyski. The names of Karl Friedrich Gauss, Bernhard Riemann, Georg Cantor and Leonhard Euler, who are to Mathematics what Shakespeare is to literature and Mozart to music, are virtually unknown outside of the worlds of math and science.

For all frequent flier miles Erdos collected, his true voyages were journeys of the mind. Erdos carefully constructed his life to allow himself as much time as possible for those inward journeys, so a true biography of Erdos should spend almost as much time in the Platonic realm of mathematics as in the real world. For a layman this may seem to be a forbidding prospect. Fortunately, many of the ideas that fascinated Erdos can be easily grasped by anyone with a modest recollection of high school mathematics. The proofs and conjectures that made Erdos famous are, of course, far more difficult to follow, but that should not be of much concern to the reader. As Ralph Boas wrote, “Only professional mathematicians learn anything from proofs. Other people learn from explanations.” Just as it is not necessary to understand how Glenn Gould fingers a difficult passage to be dazzled by his performance of thee “Goldberg Variations,” one does not have to understand the details of Erdos’s elegant proofs to appreciate the beauty of mathematics. And, it is the nature of Erdos’s work that while his proofs are difficult, the questions he asks can be quite easy to understand. Erdos often offered money for the solution to problems he proposed. Some of those problems are enough for readers of this book to understand — and, perhaps, even solve. Those who decide to try should be warned that, as Erdos has pointed out, when the number of hours it takes to solve one of his problems is taken into account, the cash prizes rarely exceed minimum wage. The true prize is to share in the joy that Erdos knew so well, joy in understanding a page of the eternal book of mathematics.

— shared by Nalin Pithwa (to motivate his students and readers.)

Nalin Pithwa.

Architecture and Math: Fibonacci and Golden ratio

https://blogs.ams.org/blogonmathblogs/2018/01/22/on-seashells-spirals-and-solids/

On seashells, spirals and solids.

Just sharing for motivation towards math especially…

Nalin Pithwa.

The infinite hotel paradox : due Jeff Dekofsky

This hardcore stuff about “infinity” , quite nicely, explained was pointed out to me by my ISC XII student, Mr. Utkarsh Malhotra! 🙂

Genius of Srinivasa Ramanujan

1. In December 1914, Ramanujan was asked by his friend P.C. Mahalanobis to solve a puzzle that appeared in Strand magazine as “Puzzles at a Village Inn”. The puzzle stated that n houses on one side of the street are numbered sequentially starting from 1. The sum of the house numbers on the left of a particular house having the number m, equals that of the houses on the right of this particular house. It is given that n lies between 50 and 500 and one has to determine the values of m and n. Ramanujan immediately rattled out a continued fraction generating all possible values of m without having any restriction on the values of n. List the first five values of m and n.
2. Ramanujan had posed the following problem in a journal: $\sqrt{1+2\sqrt{1+3\sqrt{\ldots}}}=x$, find x. Without receiving an answer from the readers, after three months he gave answer as 3. This he could say because he had an earlier general result stating $1+x=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\ldots}}}}$ is true for all x. Prove this result, then $x=2$ will give the answer to Ramanujan’s problem.

Try try until you succeed!!

Nalin Pithwa.

Limits that arise frequently

We continue our presentation of basic stuff from Calculus and Analytic Geometry, G B Thomas and Finney, Ninth Edition. My express purpose in presenting these few proofs is to emphasize that Calculus, is not just a recipe of calculation techniques. Or, even, a bit further, math is not just about calculation. I have a feeling that such thinking nurtured/developed at a young age, (while preparing for IITJEE Math, for example) makes one razor sharp.

We verify a few famous limits.

Formula 1:

If $|x|<1$, $\lim_{n \rightarrow \infty}x^{n}=0$

We need to show that to each $\in >0$ there corresponds an integer N so large that $|x^{n}|<\in$ for all n greater than N. Since $\in^{1/n}\rightarrow 1$, while $|x|<1$. there exists an integer N for which $\in^{1/n}>|x|$. In other words,

$|x^{N}|=|x|^{N}<\in$. Call this (I).

This is the integer we seek because, if $|x|<1$, then

$|x^{n}|<|x^{N}|$ for all $n>N$. Call this (II).

Combining I and II produces $|x^{n}|<\in$ for all $n>N$, concluding the proof.

Formula II:

For any number x, $\lim_{n \rightarrow \infty}(1+\frac{x}{n})^{n}=e^{x}$.

Let $a_{n}=(1+\frac{x}{n})^{n}$. Then, $\ln {a_{n}}=\ln{(1+\frac{x}{n})^{n}}=n\ln{(1+\frac{x}{n})}\rightarrow x$,

as we can see by the following application of l’Hopital’s rule, in which we differentiate with respect to n:

$\lim_{n \rightarrow \infty}n\ln{(1+\frac{x}{n})}=\lim_{n \rightarrow \infty}\frac{\ln{(1+x/n)}}{1/n}$, which in turn equals

$\lim_{n \rightarrow \infty}\frac{(\frac{1}{1+x/n}).(-\frac{x}{n^{2}})}{-1/n^{2}}=\lim_{n \rightarrow \infty}\frac{x}{1+x/n}=x$.

Now, let us apply the following theorem with $f(x)=e^{x}$ to the above:

(a theorem for calculating limits of sequences) the continuous function theorem for sequences:

Let $a_{n}$ be a sequence of real numbers. If $\{a_{n}\}$ be a sequence of real numbers. If $a_{n} \rightarrow L$ and if f is a function that is continu0us at L and defined at all $a_{n}$, then $f(a_{n}) \rightarrow f(L)$.

So, in this particular proof, we get the following:

$(1+\frac{x}{n})^{n}=a_{n}=e^{\ln{a_{n}}}\rightarrow e^{x}$.

Formula 3:

For any number x, $\lim_{n \rightarrow \infty}\frac{x^{n}}{n!}=0$

Since $-\frac{|x|^{n}}{n!} \leq \frac{x^{n}}{n!} \leq \frac{|x|^{n}}{n!}$,

all we need to show is that $\frac{|x|^{n}}{n!} \rightarrow 0$. We can then apply the Sandwich Theorem for Sequences (Let $\{a_{n}\}$, $\{b_{n}\}$ and $\{c_{n}\}$ be sequences of real numbers. if $a_{n}\leq b_{n}\leq c_{n}$ holds for all n beyond some index N, and if $\lim_{n\rightarrow \infty}a_{n}=\lim_{n\rightarrow \infty}c_{n}=L$,, then $\lim_{n\rightarrow \infty}b_{n}=L$ also) to  conclude that $\frac{x^{n}}{n!} \rightarrow 0$.

The first step in showing that $|x|^{n}/n! \rightarrow 0$ is to choose an integer $M>|x|$, so that $(|x|/M)<1$. Now, let us the rule (formula 1, mentioned above), so we conclude that:$(|x|/M)^{n}\rightarrow 0$. We then restrict our attention to values of $n>M$. For these values of n, we can write:

$\frac{|x|^{n}}{n!}=\frac{|x|^{n}}{1.2 \ldots M.(M+1)(M+2)\ldots n}$, where there are $(n-M)$ factors in the expression $(M+1)(M+2)\ldots n$, and

the RHS in the above expression is $\leq \frac{|x|^{n}}{M!M^{n-M}}=\frac{|x|^{n}M^{M}}{M!M^{n}}=\frac{M^{M}}{M!}(\frac{|x|}{M})^{n}$. Thus,

$0\leq \frac{|x|^{n}}{n!}\leq \frac{M^{M}}{M!}(\frac{|x|}{M})^{n}$. Now, the constant $\frac{M^{M}}{M!}$ does not change as n increases. Thus, the Sandwich theorem tells us that $\frac{|x|^{n}}{n!} \rightarrow 0$ because $(\frac{|x|}{M})^{n}\rightarrow 0$.

That’s all, folks !!

Aufwiedersehen,

Nalin Pithwa.

Cauchy’s Mean Value Theorem and the Stronger Form of l’Hopital’s Rule

Reference: Thomas, Finney, 9th edition, Calculus and Analytic Geometry.

Continuing our previous discussion of “theoretical” calculus or “rigorous” calculus, I am reproducing below the proof of the finite limit case of the stronger form of l’Hopital’s Rule :

L’Hopital’s Rule (Stronger Form):

Suppose that

$f(x_{0})=g(x_{0})=0$

and that the functions f and g are both differentiable on an open interval $(a,b)$ that contains the point $x_{0}$. Suppose also that $g^{'} \neq 0$ at every point in $(a,b)$ except possibly at $x_{0}$. Then,

$\lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)}=\lim_{x \rightarrow x_{0}}\frac{f^{x}}{g^{x}}$ ….call this equation I,

provided the limit on the right exists.

The proof of the stronger form of l’Hopital’s Rule is based on Cauchy’s Mean Value Theorem, a mean value theorem that involves two functions instead of one. We prove Cauchy’s theorem first and then show how it leads to l’Hopital’s Rule.

Cauchy’s Mean Value Theorem:

Suppose that the functions f and g are continuous on $[a,b]$ and differentiable throughout $(a,b)$ and suppose also that $g^{'} \neq 0$ throughout $(a,b)$. Then there exists a number c in $(a,b)$ at which

$\frac{f^{'}(c)}{g^{'}(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$…call this II.

The ordinary Mean Value Theorem is the case where $g(x)=x$.

Proof of Cauchy’s Mean Value Theorem:

We apply the Mean Value Theorem twice. First we use it to show that $g(a) \neq g(b)$. For if $g(b)$ did equal to $g(a)$, then the Mean Value Theorem would give:

$g^{'}(c)=\frac{g(b)-g(a)}{b-a}=0$ for some c between a and b. This cannot happen because $g^{'}(x) \neq 0$ in $(a,b)$.

We next apply the Mean Value Theorem to the function:

$F(x) = f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}[g(x)-g(a)]$.

This function is continuous and differentiable where f and g are, and $F(b) = F(a)=0$. Therefore, there is a number c between a and b for which $F^{'}(c)=0$. In terms of f and g, this says:

$F^{'}(c) = f^{'}(c)-\frac{f(b)-f(a)}{g(b)-g(a)}[g^{'}(c)]=0$, or

$\frac{f^{'}(c)}{g^{'}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$, which is II above. QED.

Proof of the Stronger Form of l’Hopital’s Rule:

We first prove I for the case $x \rightarrow x_{o}^{+}$. The method needs no  change to apply to $x \rightarrow x_{0}^{-}$, and the combination of those two cases establishes the result.

Suppose that x lies to the right of $x_{o}$. Then, $g^{'}(x) \neq 0$ and we can apply the Cauchy’s Mean Value Theorem to the closed interval from $x_{0}$ to x. This produces a number c between $x_{0}$ and x such that $\frac{f^{'}(c)}{g^{'}(c)}=\frac{f(x)-f(x_{0})}{g(x)-g(x_{0})}$.

But, $f(x_{0})=g(x_{0})=0$ so that $\frac{f^{'}(c)}{g^{'}(c)}=\frac{f(x)}{g(x)}$.

As x approaches $x_{0}$, c approaches $x_{0}$ because it lies between x and $x_{0}$. Therefore, $\lim_{x \rightarrow x_{0}^{+}}\frac{f(x)}{g(x)}=\lim_{x \rightarrow x_{0}^{+}}\frac{f^{'}(c)}{g^{'}(c)}=\lim_{x \rightarrow x_{0}^{+}}\frac{f^{'}(x)}{g^{'}(x)}$.

This establishes l’Hopital’s Rule for the case where x approaches $x_{0}$ from above. The case where x approaches $x_{0}$ from below is proved by applying Cauchy’s Mean Value Theorem to the closed interval $[x,x_{0}]$, where $x< x_{0}$QED.

The Sandwich Theorem or Squeeze Play Theorem

It helps to think about the core concepts of Calculus from a young age, if you want to develop your expertise or talents further in math, pure or applied, engineering or mathematical sciences. At a tangible level, it helps you attack more or many questions of the IIT JEE Advanced Mathematics. Let us see if you like the following proof, or can absorb/digest it:

Reference: Calculus and Analytic Geometry by Thomas and Finney, 9th edition.

The Sandwich Theorem:

Suppose that $g(x) \leq f(x) \leq h(x)$ for all x in some open interval containing c, except possibly at $x=c$ itself. Suppose also that $\lim_{x \rightarrow c}g(x)= \lim_{x \rightarrow c}h(x)=L$. Then, $\lim_{x \rightarrow c}f(x)=c$.

Proof for Right Hand Limits:

Suppose $\lim_{x \rightarrow c^{+}}g(x)=\lim_{x \rightarrow c^{+}}h(x)=L$. Then, for any $\in >0$, there exists a $\delta >0$ such that for all x, the inequality $c implies $L-\in and $L-\in ….call this (I)

These inequalities combine with the inequality $g(x) \leq f(x) \leq h(x)$ to give

$L-\in

$L-\in

$-\in ….call this (II)

Therefore, for all x, the inequality $c implies $|f(x)-L|<\in$. …call this (III)

Proof for LeftHand Limits:

Suppose $\lim_{x \rightarrow c^{-}} g(x)=\lim_{x \rightarrow c^{-}}=L$. Then, for $\in >0$ there exists a $\delta >0$ such that for all x, the inequality $c-\delta implies $L-\in and $L-\in …call this (IV).

We conclude as before that for all x, $c-\delta implies $|f(x)-L|<\in$.

Proof for Two sided Limits:

If $\lim_{x \rightarrow c}g(x) = \lim_{x \rightarrow c}h(x)=L$, then $g(x)$ and $h(x)$ both approach L as $x \rightarrow c^{+}$ and as $x \rightarrow c^{-}$ so $\lim_{x \rightarrow c^{+}}f(x)=L$ and $\lim_{x \rightarrow c^{-}}f(x)=L$. Hence, $\lim_{x \rightarrow c}f(x)=L$. QED.

Let me know your feedback on such stuff,

Nalin Pithwa