Monthly Archives: September 2019

The personality of Leonhard Euler

The portrait of Euler that emerges from his publications and letters is that of a genial man of simple tastes and conventional religious faith. He was even wealthy, at least in the second half of his life, but ostentation was not part of his lifestyle. His memory was prodigious, and contemporary accounts have emphasized this. He would delight relatives, friends, and acquaintances with a literal recitation of any song from Virgil’s Aenesis, and he would remember minutes of Academy meetings years after they were held. He was not given to envy, and when someone made an advance on his work his happiness was genuine. For example, when he learnt of Lagrange’s improvements on his work on elliptic integrals, he wrote to him that his admiration knew no bounds and then proceeded to improve upon Lagrange!

But, what is most characteristic of his work is its clarity and openness. He never tries to hide the difficulties from the reader. This is in stark contrast to Newton, who was prone to hide his methods in obscure anagrams, and even from his successor, Gauss, who very often erased his steps to present a monolithic proof that was seldom illuminating. In Euler’s writings there are no comments on how profound his results are, and in his papers one can follow his ideas step by step with the greatest of ease. Nor was he chary of giving credit to others; his willingness to share his summation formula with Maclaurin, his proper citations to Fuguano when he started his work on algebraic integrals, his open admiration for Lagrange when the latter improved on his work in calculus of variations are all instances of his serene outlook. One can only contrast this with Gauss’s reaction to Bolyai’s discovery of non-Euclidean postulates. Euler was secure in his knowledge of what he had achieved but never insisted that he should be the only one on top of the mountain.

Perhaps, the most astounding aspect of his scientific opus is its universality. He worked on everything that had any bearing on mathematics. For instance, his early training under Johann Bernoulli did not include number theory; nevertheless, within a couple of years after reaching St. Petersburg he was deeply immersed in it, recreating the entire corpus of Fermat’s work in that area and then moving well beyond him. His founding of graph theory as a separate discipline, his excursions in what we call combinatorial topology, his intuition that suggested to him the idea of exploring multizeta values are all examples of a mind that did not have any artificial boundaries. He had no preferences about which branch of mathematics was dear to him. To him, they were all filled with splendour, or Herrlichkeit, to use his own favourite word.

Hilbert and Poincare were perhaps last of the universalists of modern era. Already von Neumann had remarked that it would be difficult even to have a general understanding of more than a third of the mathematicians of his time. With the explosive growth of mathematics in the twentieth century we may never see again the great universalists. But who is to say what is and is not possible for the human mind?

It is impossible to read Euler and not fall under his spell. He is to mathematics what Shakespeare is to literature and Mozart to music: universal and sui generis.

Reference:

Euler Through Time: A New Look at Old Themes by V S Varadarajan:

Hindustan Book Agency;

http://www.hindbook.com/index.php/euler-through-time-a-new-look-at-old-themes;

Amazon India link:

https://www.amazon.in/Euler-Through-Time-Look-Themes/dp/9380250592/ref=sr_1_1?keywords=Euler+Through+Time&qid=1568316624&s=books&sr=1-1

 

 

Rules for Inequalities

If a, b and c are real numbers, then

  1. a < b \Longrightarrow a + c< b + c
  2. a < b \Longrightarrow a - c < b - c
  3. a < b \hspace{0.1in} and \hspace{0.1in}c > 0 \Longrightarrow ac < bc
  4. a < b \hspace{0.1in} and \hspace{0.1in}c < 0 \Longrightarrow bc < ac special case: a < b \Longrightarrow -b < -a
  5. a > 0 \Longrightarrow \frac{1}{a} > 0
  6. If a and b are both positive or both negative, then a < b \Longrightarrow \frac{1}{b} < \frac{1}{a}.

Remarks:

Notice the rules for multiplying an inequality by a number: Multiplying by a positive number preserves the inequality; multiplying by a negative number reverses the inequality. Also, reciprocation reverses the inequality for numbers of the same sign.

Regards,

Nalin Pithwa.

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<y. 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