Author Archives: Nalin Pithwa

I am a DSP Engineer and a mathematician working in closely related areas of DSP, Digital Control, Digital Comm and Error Control Coding. I have a passion for both Pure and Applied Mathematics.

Various proofs of important algebraic identity: a^{3}+b^{3}+c^{3}-3abc

We know the following factorization: a^{3}+b^{3}+c^{3}-3abc= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

Proof 1:

Let a, b, c be roots of a polynomial P(X) Then by fundamental theorem of algebra

P(X)=(X-a)(X-b)(X-c) = X^{3}-(a+b+c)X^{2}+(ab+bc+ca)X-abc.

Now, once again basic algebra says that as a, b, c are roots/solutions of the above:

P(a)=a^{3}-(a+b+c)a^{2}+(ab+bc+ca)a-abc=0

P(b)=b^{3}-(a+b+c)b^{2}+(ab+bc+ca)b-abc=0

P(c)=c^{3}-(a+b+c)c^{2}+(ab+bc+ca)c-abc=0$

Adding all the above:

0= a^{3}+b^{3}+c^{3}-(a+b+c)(a^{2}+b^{2}+c^{2})+(ab+bc+ca)(a+b+c)-3abc

So, we get a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

Also, the above formula can be written as a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(\frac{1}{2})((a-b)^{2}+(b-c)^{2}+(c-a)^{2})

Proof 2:

Consider the following determinant D: \left| \begin{array}{ccc} a & b & c\\c & a & b\\ b & c & a \end{array} \right|

On adding all three columns to the first column: we know that the value of the determinant is unchanged: So we get the following:

D = \left| \begin{array}{ccc} a+b+c & b & c \\a+b+c & a & b\\a+b+c & c & a \end{array} \right|. Note that columns 2 and 3 of the three by three determinant do not change.

On expanding the original determinant D, we get

D = a(a^{2}-bc)-b(ac-b^{2})+c(c^{2}-ab)

D= a^{3}-abc-bac+b^{3}+c^{3}-cab

D= a^{3}+b^{3}+c^{3}-3abc

Whereas we get from the other transformed but equal D:

D =(a+b+c) \left| \begin{array}{ccc} 1 & b & c \\ 1 & a & b \\ 1 & c & a \end{array}\right|

D=(a+b+c)((a^{2}-bc)-b(a-b)+c(c-a))

So, that we again get a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

Proof 3:

Now, let us consider E=a^{2}+b^{2}+c^{2}-ab-bc-ca as a quadratic in a with b and c as parameters.

That is, E= a^{2} -(b+c)a + b^{2}+c^{2}-bc

Then, the discriminant is given by

\triangle = (b+c)^{2}-4 \times 1 \times (b^{2}+c^{2}-bc), which in turn equals:

\triangle = (b^{2}+c^{2}+2bc)-4(b^{2}+c^{2}-bc) = -3b^{2}-3c^{2}+6bc=-3(b-c)^{2}

a_{1}, a_{2} = \frac{(b+c) \pm i\sqrt{3}(b-c)}{2}

a_{1}= b(\frac{1+i\sqrt{3}}{2})=c(\frac{1-i\sqrt{3}}{2}) = -b\omega - c\omega^{2}

a_{2}=b(\frac{1-i\sqrt{3}}{2})+c(\frac{1-i\sqrt{3}}{2})=-b\omega^{2}-c\omega

Hence, the factorization of the above quadratic in a is given as:

a^{2}+b^{2}+c^{2}-ab-bc-ca = (a+b\omega+c\omega^{2})(a+b\omega^{2}+c\omega)

So, the other non-trivial factorization of the above famous algebraic identity is:

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a+b\omega+ c\omega^{2})(a+b\omega^{2}+c\omega)

where 1, \omega, \omega^{2} are cube roots of unity.

Proof 4: 

Factorize the expression a^{3}+b^{3}+c^{3}-3abc by (a+b+c).

Solution 4: 

We can carry out the above polynomial division by considering the dividend to be a polynomial in a single variable, say a, (and assuming b and c are just parameters; so visualize them as arbitrary but fixed constants); further arrange the dividend in descending powers of a; so also arrange the divisor in descending powers of a (well, of course, it is just linear in a; and assume b and c are parameters also in dividend).

Proof 5:

Prove that the eliminant of

ax+cy+bz=0

cx+by+az=0

bx+ay+cz=0

is a^{3}+b^{3}+c^{3}-3abc=0

Proof 5:

By Cramer’s rule, the eliminant is given by determinant \left| \begin{array}{ccc}a & c & b \\ c & b & a\\b & a & c \end{array}\right|=0.

On expansion using the first row:

a(bc-a^{2})-c(c^{2}-ab)+b(ac-b^{2})=0

a^{3}+b^{3}+c^{3}-3abc=0 upon multiplying both the sides of the above equation by (-1). Of course, we have only been able to generate the basic algebraic expression but we have done so by encountering a system of three linear equations in x, y, z. (we could append any of the above factorization methods to this further!! :-)))

Cheers,

Nalin Pithwa

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

 

 

 

 

 

 

 

 

Set Theory, Relations, Functions Preliminaries: I

In these days of conflict between ancient and modern studies there must surely be something to be said of a study which did not begin with Pythagoras and will not end with Einstein. — G H Hardy (On Set Theory)

In every day life, we generally talk about group or collection of objects. Surely, you must have used the words such as team, bouquet, bunch, flock, family for collection of different objects.

It is very important to determine whether a given object belongs to a given collection or not. Consider the following conditions:

i) Successful persons in your city.

ii) Happy people in your town.

iii) Clever students in your class.

iv) Days in a week.

v) First five natural numbers.

Perhaps, you have already studied in earlier grade(s) —- can you state which of the above mentioned collections are sets? Why? Check whether your answers are as follows:

First three collections are not examples of sets but last two collections represent sets. This is because in first three collections, we are not sure of the objects. The terms ‘successful persons’, ‘happy people’, ‘clever students’ are all relative terms. Here, the objects are not well-defined. In the last two collections, we can determine the objects clearly (meaning, uniquely, or without ambiguity). Thus, we can say that the objects are well-defined.

So what can be the definition of a set ? Here it goes:

A collection of well-defined objects is called a set. (If we continue to “think deep” about this definition, we are led to the famous paradox, which Bertrand Russell had discovered: Let C be a collection of all sets such which are not elements of themselves. If C is allowed to be a set, a contradiction arises when one inquires whether or not C is an element of itself. Now plainly, there is something suspicious about the idea of a set being an element of itself, and we shall take this as evidence that the qualification “well-defined” needs to be taken seriously. Bertrand Russell re-stated this famous paradox in a very interesting way: In the town of Seville lives a barber who shaves everyone who does not shave himself. Does the barber shave himself?…)

The objects in a set are called elements or members of that set.

We denote sets by capital letters : A, B, C etc. The elements of a set are represented by small letters : a, b, c, d, e, f ….etc. If x is an element of a set A, we write x \in A. And, we read it as “x belongs to A.” If x is not an element of a set A, we write x \not\in A, and read as ‘x does not belong to A.’e.g., 1 is a “whole” number but not a “natural” number.

Hence, 0 \in W, where W is the set of whole numbers and 0 \not\in N, where N is a set of natural numbers.

There are two methods of representing a set:

a) Roster or Tabular Method or List Method (b) Set-Builder or Ruler Method

a) Roster or Tabular or List Method:

Let A be the set of all prime numbers less than 20. Can you enumerate all the elements of the set A? Are they as follows?

A=\{ 2,3,5,7,11,15,17,19\}

Can you describe the roster method? We can describe it as follows:

In the Roster method, we list all the elements of the set within braces \{, \} and separate the elements by commas.

In the following examples, state the sets using Roster method:

i) B is the set of all days in a week

ii) C is the set of all consonants in English alphabets.

iii) D is the set of first ten natural numbers.

2) Set-Builder Method:

Let P be the set of first five multiples of 10. Using Roster Method, you must have written the set as follows:

P = \{ 10, 20, 30, 40, 50\}

Question: What is the common property possessed by all the elements of the set P?

Answer: All the elements are multiples of 10.

Question: How many such elements are in the set?

Answer: There are 5 elements in the set.

Thus, the set P can be described using this common property. In such a case, we say that set-builder method is used to describe the set. So, to summarize:

In the set-builder method, we describe the elements of the set by specifying the property which determines the elements of the set uniquely.

Thus, we can write : P = \{ x: x =10n, n \in N, n \leq 5\}

In the following examples, state the sets using set-builder method:

i) Y is the set of all months of a year

ii) M is the set of all natural numbers

iii) B is the set of perfect squares of natural numbers.

Also, if elements of a set are repeated, they are written once only; while listing the elements of a set, the order in which the elements are listed is immaterial. (but this situation changes when we consider sets from the view-point of permutations and combinations. Just be alert in set-theoretic questions.)

Subset: A set A is said to be a subset of a set B if each element of set A is an element of set B. Symbolically, A \subseteq B.

Superset: If A \subset B, then B is called the superset of set A. Symbolically: B \supset A

Proper Subset: A non empty set A is said to be a proper subset of the set B, if and only if all elements of set A are in set B, and at least one element of B is not in A. That is, if A \subseteq B, but A \neq B then A is called a proper subset of B and we write A \subset B.

Note: the notations of subset and proper subset differ from author to author, text to text or mathematician to mathematician. These notations are not universal conventions in math.

Intervals: 

  1. Open Interval : given a < b, a, b \in R, we say a<x<b is an open interval in \Re^{1}.
  2. Closed Interval : given a \leq x \leq b = [a,b]
  3. Half-open, half-closed: a <x \leq b = (a,b], or a \leq x <b=[a,b)
  4. The set of all real numbers greater than or equal to a : x \geq a =[a, \infty)
  5. The set of all real numbers less than or equal to a is (-\infty, a] = x \leq a

Types of Sets:

  1. Empty Set: A set containing no element is called the empty set or the null set and is denoted by the symbol \phi or \{ \} or void set. e.g., A= \{ x: x \in N, 1<x<2\}
  2. Singleton Set: A set containing only one element is called a singleton set. Example : (i) Let A be a set of all integers which are neither positive nor negative. Then, A = \{ 0\} and example (ii) Let B be a set of capital of India. Then B= \{ Delhi\}

We will define the following sets later (after we giving a working definition of a function): finite set, countable set, infinite set, uncountable set.

3. Equal sets: Two sets are said to be equal if they contain the same elements, that is, if A \subseteq B and B \subseteq A. For example: Let X be the set of letters in the word ‘ABBA’ and Y be the set of letters in the word ‘BABA’. Then, X= \{ A,B\} and Y= \{ B,A\}. Thus, the sets X=Y are equal sets and we denote it by X=Y.

How to prove that two sets are equal?

Let us say we are given the task to prove that A=B, where A and B are non-empty sets. The following are the steps of the proof : (i) TPT: A \subset B, that is, choose any arbitrary element x \in A and show that also x \in B holds true. (ii) TPT: B \subset A, that is, choose any arbitrary element y \in B, and show that also y \in A. (Note: after we learn types of functions, we will see that a fundamental way to prove two sets (finite) are equal is to show/find a bijection between the two sets).

PS: Note that two sets are equal if and only if they contain the same number of elements, and the same elements. (irrespective of order of elements; once again, the order condition is changed for permutation sets; just be alert what type of set theoretic question you are dealing with and if order is important in that set. At least, for our introduction here, order of elements of a set is not important).

PS: Digress: How to prove that in general, x=y? The standard way is similar to above approach: (i) TPT: x < y (ii) TPT: y < x. Both (i) and (ii) together imply that x=y.

4. Equivalent sets: Two finite sets A and B are said to be equivalent if n(A)=n(B). Equal sets are always equivalent but equivalent sets need not be equal. For example, let A= \{ 1,2,3 \} and B = \{ 4,5,6\}. Then, n(A) = n(B), so A and B are equivalent. Clearly, A \neq B. Thus, A and B are equivalent but not equal.

5. Universal Set: If in a particular discussion all sets under consideration are subsets of a set, say U, then U is called the universal set for that discussion. You know that the set of natural numbers the set of integers are subsets of set of real numbers R. Thus, for this discussion is a universal set. In general, universal set is denoted by or X.

6. Venn Diagram: The pictorial representation of a set is called Venn diagram. Generally, a closed geometrical figures are used to represent the set, like a circle, triangle or a rectangle which are known as Venn diagrams and are named after the English logician John Venn.

In Venn diagram the elements of the sets are shown in their respective figures.

Now, we have these “abstract toys or abstract building-blocks”, how can we get new such “abstract buildings” using these “abstract building blocks”. What I mean is that we know that if we are a set of numbers like 1,2,3, …, we know how to get “new numbers” out of these by “adding”, subtracting”, “multiplying” or “dividing” the given “building blocks like 1, 2…”. So, also what we want to do now is “operations on sets” so that we create new, more interesting or perhaps, more “useful” sets out of given sets. We define the following operations on sets:

  1. Complement of a set: If A is a subset of the universal set U then the set of all elements in U which are not in A is called the complement of the set A and is denoted by A^{'} or A^{c} or \overline{A} Some properties of complements: (i) {A^{'}}^{'}=A (ii) \phi^{'}=U, where U is universal set (iii) U^{'}= \phi
  2. Union of Sets: If A and B are two sets then union of set A and set B is the set of all elements which are in set A or set B or both set A and set B. (this is the INCLUSIVE OR in digital logic) and the symbol is : $latex A \bigcup B
  3. Intersection of sets: If A and B are two sets, then the intersection of set A and set B is the set of all elements which are both in A and B. The symbol is A \bigcap B.
  4. Disjoint Sets: Let there be two sets A and B such that A \bigcap B=\phi. We say that the sets A and B are disjoint, meaning that they do not have any elements in common. It is possible that there are more than two sets A_{1}, A_{2}, \ldots A_{n} such that when we take any two distinct sets A_{i} and A_{j} (so that i \neq j, then A_{i}\bigcap A_{j}= \phi. We call such sets pairwise mutually disjoint. Also, in case if such a collection of sets also has the property that \bigcup_{i=1}^{i=n}A_{i}=U, where U is the Universal Set in the given context, We then say that this collection of sets forms a partition of the Universal Set.
  5. Difference of Sets: Let us say that given a universal set U and two other sets A and B, B-A denotes the set of elements in B which are not in A; if you notice, this is almost same as A^{'}=U-A.
  6. Symmetric Difference of Sets: Suppose again that we are two given sets A and B, and a Universal Set U, by symmetric difference of A and B, we mean (A-B)\bigcup (B-A). The symbol is A \triangle B. Try to visualize this (and describe it) using a Venn Diagram. You will like it very much. Remark : The designation “symmetric difference” for the set A \triangle B is not too apt, since A \triangle B has much in common with the sum A \bigcup B. In fact, in A \bigcup B the statements “x belongs to A” and “x belongs to B” are joined by the conjunction “or” used in the “either …or …or both…” sense, while in A \triangle B the same two statements are joined by “or” used in the ordinary “either…or….” sense (as in “to be or not to be”). In other words, x belongs to A \bigcup B if and only if x belongs to either A or B or both, while x belongs to A \triangle B if and only if x belongs to either A or B but not both. The set A \triangle B can be regarded as a kind of a “modulo-two-sum” of the sets A and B, that is, a sum of the sets A and B in which elements are dropped if they are counted twice (once in A and once in B).

Let us now present some (easily provable/verifiable) properties of sets:

  1. A \bigcup B = B \bigcup A (union of sets is commutative)
  2. (A \bigcup B) \bigcup C = A \bigcup (B \bigcup C) (union of sets is associative)
  3. A \bigcup \phi=A
  4. A \bigcup A = A
  5. A \bigcup A^{'}=U where U is universal set
  6. If A \subseteq B, then A \bigcup B=B
  7. U \bigcup A=U
  8. A \subseteq (A \bigcup B) and also B \subseteq (A \bigcup B)

Similarly, some easily verifiable properties of set intersection are:

  1. A \bigcap B = B \bigcap A (set intersection is commutative)
  2. (A \bigcap B) \bigcap C = A \bigcap (B \bigcap C) (set intersection is associative)
  3. A \bigcap \phi = \phi \bigcap A= \phi (this matches intuition: there is nothing common in between a non empty set and an empty set :-))
  4. A \bigcap A =A (Idempotent law): this definition carries over to square matrices: if a square matrix is such that A^{2}=A, then A is called an Idempotent matrix.
  5. A \bigcap A^{'}=\phi (this matches intuition: there is nothing in common between a set and another set which does not contain any element of it (the former set))
  6. If A \subseteq B, then A \bigcap B =A
  7. U \bigcap A=A, where U is universal set
  8. (A \bigcap B) \subseteq A and (A \bigcap B) \subseteq B
  9. i: A \bigcap (B \bigcap )C = (A \bigcap B)\bigcup (A \bigcap C) (intersection distributes over union) ; (9ii) A \bigcup (B \bigcap C)=(A \bigcup B) \bigcap (A \bigcup C) (union distributes over intersection). These are the two famous distributive laws.

The famous De Morgan’s Laws for two sets are as follows: (it can be easily verified by Venn Diagram):

For any two sets A and B, the following holds:

i) (A \bigcup B)^{'}=A^{'}\bigcap B^{'}. In words, it can be captured beautifully: the complement of union is intersection of complements.

ii) (A \bigcap B)^{'}=A^{'} \bigcup B^{'}. In words, it can be captured beautifully: the complement of intersection is union of complements.

Cardinality of a set: (Finite Set) : (Again, we will define the term ‘finite set’ rigorously later) The cardinality of a set is the number of distinct elements contained in a finite set A and we will denote it as n(A).

Inclusion Exclusion Principle:

For two sets A and B, given a universal set U: n(A \bigcup B) = n(A) + n(B) - n(A \bigcap B).

For three sets A, B and C, given a universal set U: n(A \bigcup B \bigcup C)=n(A) + n(B) + n(C) -n(A \bigcap B) -n(B \bigcap C) -n(C \bigcup A) + n(A \bigcap B \bigcap C).

Homework Quiz: Verify the above using Venn Diagrams. 

Power Set of a Set:

Let us consider a set A (given a Universal Set U). Then, the power set of A is the set consisting of all possible subsets of set A. (Note that an empty is also a subset of A and that set A is a subset of A itself). It can be easily seen (using basic definition of combinations) that if n(A)=p, then n(power set A) = 2^{p}. Symbol: P(A).

Homework Tutorial I:

  1. Describe the following sets in Roster form: (i) \{ x: x \hspace{0.1in} is \hspace{0.1in} a \hspace{0.1in} letter \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} word \hspace{0.1in}  PULCHRITUDE\} (II) \{ x: x \hspace{0.1in } is \hspace{0.1in} an \hspace{0.1in} integer \hspace{0.1in} with \hspace{0.1in} \frac{-1}{2} < x < \frac{1}{2} \} (iii) \{x: x=2n, n \in N\}
  2. Describe the following sets in Set Builder form: (i) \{ 0\} (ii) \{ 0, \pm 1, \pm 2, \pm 3\} (iii) \{ \}
  3. If A= \{ x: 6x^{2}+x-15=0\} and B= \{ x: 2x^{2}-5x-3=0\}, and x: 2x^{2}-x-3=0, then find (i) A \bigcup B \bigcup C (ii) A \bigcap B \bigcap C
  4. If A, B, C are the sets of the letters in the words, ‘college’, ‘marriage’, and ‘luggage’ respectively, then verify that \{ A-(B \bigcup C)\}= \{ (A-B) \bigcap (A-C)\}
  5. If A= \{ 1,2,3,4\}, B= \{ 3,4,5, 6\}, C= \{ 4,5,6,7,8\} and universal set X= \{ 1,2,3,4,5,6,7,8,9,10\}, then verify the following:

5i) A\bigcup (B \bigcap C) = (A\bigcup B) \bigcap (A \bigcup C)

5ii) A \bigcap (B \bigcup C)= (A \bigcap B) \bigcup (A \bigcap C)

5iii) A= (A \bigcap B)\bigcup (A \bigcap B^{'})

5iv) B=(A \bigcap B)\bigcup (A^{'} \bigcap B)

5v) n(A \bigcup B)= n(A)+n(B)-n(A \bigcap B)

6. If A and B are subsets of the universal set is X, n(X)=50, n(A)=35, n(B)=20, n(A^{'} \bigcap B^{'})=5, find (i) n(A \bigcup B) (ii) n(A \bigcap B) (iii) n(A^{'} \bigcap B) (iv) n(A \bigcap B^{'})

7. In a class of 200 students who appeared certain examinations, 35 students failed in MHTCET, 40 in AIEEE, and 40 in IITJEE entrance, 20 failed in MHTCET and AIEEE, 17 in AIEEE and IITJEE entrance, 15 in MHTCET and IITJEE entrance exam and 5 failed in all three examinations. Find how many students (a) did not flunk in any examination (b) failed in AIEEE or IITJEE entrance.

8. From amongst 2000 literate and illiterate individuals of a town, 70 percent read Marathi newspaper, 50 percent read English newspapers, and 32.5 percent read both Marathi and English newspapers. Find the number of individuals who read

8i) at least one of the newspapers

8ii) neither Marathi and English newspaper

8iii) only one of the newspapers

9) In a hostel, 25 students take tea, 20 students take coffee, 15 students take milk, 10 students take both tea and coffee, 8 students take both milk and coffee. None of them take the tea and milk both and everyone takes at least one beverage, find the number of students in the hostel.

10) There are 260 persons with a skin disorder. If 150 had been exposed to chemical A, 74 to chemical B, and 36 to both chemicals A and B, find the number of persons exposed to  (a) Chemical A but not Chemical B (b) Chemical B but not Chemical A (c) Chemical A or Chemical B.

11) If A = \{ 1,2,3\} write down the power set of A.

12) Write the following intervals in Set Builder Form: (a) (-3,0) (b) [6,12] (c) (6,12] (d) [-23,5)

13) Using Venn Diagrams, represent (a) (A \bigcup B)^{'} (b) A^{'} \bigcup B^{'} (c) A^{'} \bigcap B (d) A \bigcap B^{'}

Regards,

Nalin Pithwa.

IITJEE Foundation Maths: Variation

DEFINITION:

One quantity A is said to vary directly as another B, when the two quantities depend upon each other in such a manner that if B is changed, A is changed in the same ratio.

NOTE: The word directly is often omitted, and A is said to vary as B.

For instance: if a train a moving at a uniform rate travels 40 miles in 60 minutes, it will travel 20 miles in 30 minutes, 80 miles in 120 minutes, and so on; the distance in each case being increased or diminished in the same ratio as the time. This is expressed by saying that when the velocity is uniform the distance is proportional to the time, or the distance varies as the time.

NOTATION: The symbol \alpha is used to denote variation; so that, A \alpha B is read as “A varies as B.”

Theorem I: If A varies as B, then A is equal to B multiplied by some constant quantity.

Note: If any pair of corresponding values of A and B are known, the constant m can be determined. For instance, if A=B, when B=12, we have 3=m \times 12; and m=\frac{1}{4}, and A=\frac{1}{4}B

DEFINITION: One quantity A is said to vary inversely as another B, when A varies directly as the reciprocal of B.

The following is an illustration of inverse variation: If 6 men do a certain work in 8 hours, 12 men would do the same work in 4 hours; 2 men in 24 hours; and so on. Thus, it appears that when the number of men is increased, the same is proportionately decreased; and vice-versa.

Example 1: The cube root of x varies inversely as the square of y; if x=8, when y=3; find x when y=\frac{3}{2}.

Solution 1: By supposition, \sqrt[3]{x}=\frac{m}{y^{2}}, where m is constant. Putting x=8, y=3, we have 2=\frac{m}{9}, so m=18, and \sqrt[3]{x}=\frac{18}{y^{2}}; hence, by putting y=\frac{3}{2}, we obtain x=512.

Example 2: The square of the time of a planet’s revolution varies as the cube of its distance from the Sun; find the time of Venus’s revolution, assuming the distances of the Earth and Venus from the Sun to be 91\frac{1}{4} and 66 millions of miles respectively.

Let P be the periodic time measured in days, D the distance in millions of miles; we have

P^{2} \alpha D^{3}, or P^{2}=k \times D^{3}, where k is some constant.

For the Earth, 365 \times 365 = k \times 91 \frac{1}{4} \times 91 \frac{1}{4} \times 91 \times \frac{1}{4}

hence, k=\frac{4 \times 4 \times 4}{3.5}

so that P^{2}=\frac{4 \times 4 \times 4}{365}D^{3}

For Venus, P^{2}=\frac{4 \times 4 \times 4}{365} \times 66 \times 66 \times 66

hence, P= A \times 66 \times \sqrt{\frac{264}{365}} = 264 \times \sqrt{0.7233}, approximately.

P^{2}=264 \times 0.85=224.4.

Hence, the time of revolution is nearly 224\frac{1}{2}.

DEFINITION: One quantity is said to vary jointly as a number of others, when it varies directly as their product.

Thus, A varies jointly as B and C, when A=m \times BC. For instance, the interest on a sum of money varies jointly as the principal, the time, and the rate per cent.

DEFINITION: 

A is said to vary directly as B and inversely as C, when A varies as \frac{B}{C}.

Theorem:

If A varies as B, when C is constant, and A varies as C when B is constant, then A will vary as BC when both B and C vary.

Proof:

Homework

The following are some illustrations of the theorems stated above:

The amount of work done by a given number of men varies directly as the number of days they work, and the amount of work done in a given time varies directly as the number of men; therefore, when the number of days and the number of men are both variable, the amount of work will vary as the product of the number of men and the number of days.

Again, in plane geometry, the area of a triangle varies directly as its base when the height is constant, and directly as the height when the base is constant; and when both the height and base are variable, the area varies as the product of the numbers representing the height and the base.

Example:

The volume of a right circular cone varies as the square of the radius of the base when the height is constant, and as the height when the base is constant. If the radius of base is 7cm, and the height is 15 cm, the volume is 770 cc, find the height of a cone whose volume is 132 cubic cm, and which stands on a base whose radius is 3cm.

Solution:9 

Let h and r denote respectively the height and radius of the base measured in cm.; also let V be the volume in cubic cm.

Then. V=m \times r^{2} \times h, where m is constant.

By assumption, 770=m \times 7^{2} \times 15

hence, m = \frac{22}{21} so that V=\frac{22}{21}r^{2}h.

By substituting V=132, r=3, we get the following:

132=\frac{22}{21} \times 9 \times h

so that h=14; and, therefore the height is 14 cm.

Note:

A quantity A can vary jointly as (a product of) more than two variables also as is most often the case in real engineering. Further, the variations may be either direct or inverse. The principle is interesting because of its frequent occurence in physical sciences or engineering. For example, Boyle’s law in chemistry: It is found by experiment that the pressure (P) of a gas varies as the “absolute temperature” (T) (in Kelvin) when its volume (V) is constant and that the pressure varies inversely as the volume when the temperature is constant; that is,

P \alpha T, when V is constant and P \alpha \frac{1}{V} when T is constant.

From these results we should expect that, when both t and v are variable, we should have the formula:

P \alpha \frac{T}{V}, or PV=kT, where k is a constant (based on laws of chemistry). And, by actual experiment this is found to be true.

Example:

The duration of a railway journey varies ditectly as the distance and inversely as the velocity; the velocity varies directly as the square root of the quantity of coal used per kilometer (don’t worry the days of steam engine/coal engine and resulting environmental degradation are over; but this is only a simple engineering application), and inversely as the number of carriages in the train. In a journey of 50 kilometers, in half an hour with 18 carriages 100 kg of coal is required; how much coal will be consumed in a journey of 42 kilometers, in 28 minutes with 16 carriages?

Solution:

Let t be the time expressed in hours; let d be the distance in kilometers; let v be the velocity in kmph; let q be the mass of coal (in kg) used per kilometers; and let c be the number of carriages.

We have t \alpha \frac{d}{v} and v \alpha \frac{\sqrt{q}}{c}, and hence, t \alpha \frac{cd}{\sqrt{q}}, or t=\frac{kcd}{\sqrt{q}}, where k is a constant.

Substituting the values given, we have (since q=2),

\frac{1}{2} = \frac{k \times 18 \times 50}{\sqrt{2}}

that is, k=\frac{1}{\sqrt{2} \times 18 \times 50}.

Hence, t=\frac{cd}{\sqrt{2} \times 18 \times 50\sqrt{q}}

Substituting now the values of t, c, d given in the second part of the question, we have

\frac{28}{60}=\frac{16 \times 42}{\sqrt{2} \times 18 \times 50 \times \sqrt{q} }

that is, \sqrt{q} = 4 \sqrt{2}, hence q=32.

Hence, the quantity of coal is 42 \times 32 = 1344 kg.

Tutorial problems on Variation:

  1. If x varies as y, and x=8, when y=15, find x when y=10
  2. If P varies as Q, and P=7 when Q=3, find P when Q=2\frac{1}{3}.
  3.  If the square of x varies as the cube of y, and x=3, when y=4, find the value of y when x=\frac{1}{\sqrt{3}}.
  4. A varies as B and C jointly; if A=2 when B=\frac{3}{5} and C=\frac{10}{27}, find C when A=54 and B=3.
  5. If A varies as C, and B varies as C, then A \pm B and \sqrt{AB} will each vary as C.
  6. If A varies as BC, then B varies inversely as \frac{C}{A}.
  7. P varies directly as Q and inversely as R; also P=\frac{2}{3} when Q=\frac{3}{7} and R=\frac{9}{14}; find Q when P=\sqrt{48} and R=\sqrt{75}.
  8. If x varies as y, prove that x^{2}+y^{2} varies as x^{2}-y^{2}.
  9. If y varies as the sum of two quantities, of which one varies directly as x and the other inversely as x; and if y=6 then x=4, and y=3\frac{1}{3} when x=3, find the equation between x and y.
  10. If y is equal to the sum of two quantities one of which varies as x directly, and the other as x^{2} inversely; and, if y=19 when x=2, or 3; find y in terms of x.
  11. If A varies directly as the square root of B and inversely as the cube of C, and if A=3, when B=256 and C=2, find B when A=24 and C = \frac{1}{2}
  12. Given that x+y varies as x+\frac{1}{x}, and that x-y varies as z- \frac{1}{z}, find the relation between x and z, provided that z=2 when x=3 and y=1.
  13. If A varies as B and C jointly, write B varies as D^{2}, and C varies inversely as A, show that A varies as D.
  14. If y varies as the sum of three quantities of which the first is a constant, the second varies as x, and the third as x^{2}; and, if y=0 when x=1, y=1, when x=2, and y=4 when x=3; find y when x=7.
  15. When a body falls down from rest the distance from the starting point varies as the square of the time it has been falling; if a body falls through 122.6 meters in 5 seconds, how far does it fall in 10 seconds? Also, how far does it fall in the tenth second?
  16. Given that the volume of a sphere varies as the cube of its radius, and that when the radius is 3.5 cm, the volume is 176.7 cubic cm; find the volume when the radius is 1.75 cm.
  17. The weight of a circular disc varies as the square of the radius when the thickness remains the same; it also varies as the thickness when the radius remains the same. Two discs have their thicknesses in the ratio of 9:8; find the ratio of their radii if the weight of the first is twice that of the second.
  18. At a certain regatta, the numbers of races on each day varied jointly as the number of days from the beginning and end of the regatta up to and including the day in question. On three successive days there were respectively 6, 5 and 3 races. Which days were these, and how long did the regatta last?
  19. The price of a diamond varies as the square of its weight (mass). Three rings of equal weight, each composed of a diamond set in gold, have values INR a, INR b, INR c, the diamonds in them weighing 3, 4, 5 carats respectively. Show that the value of a diamond of one carat is INR (\frac{a+c}{2}-b), the cost of workmanship being the same for each ring.
  20. Two persons are awarded pensions in proportion to the square root of the number of root of the number of years they have served. One has served 9 years longer than the other and receives a pension greater by INR 500. If the length of service of the first had exceeded that of the second by 4\frac{1}{4} years their pensions would have been in the proportion of 9:8. How long had they served and what were their respective pensions?
  21. The attraction of a planet on the satellites varies directly as the mass (M) of the planet, and inversely as the square of the distance (D); also the square of a satellite’s time of revolution varies directly as the distance and inversely as the force of attraction. If m_{1}, d_{1}, t_{1} and m_{2}, d_{2}, t_{2} are simultaneous values of M, D, T respectively, prove that \frac{m_{1}t_{1}^{2}}{m_{2}t_{2}^{2}} = \frac{d_{1}^{3}}{d_{2}^{3}}. Hence, find the time of revolution of that moon of Jupiter whose distance is to the distance of our Moon as 35:31, having given that the mass of Jupiter is 343 times that of the Earth, and that the Moon’s period is 27.32 days.
  22. The consumption of coal by a locomotive varies as the square of the velocity; when the speed is 32 kmph the consumption of coal per hour is 2 tonnes: if the price of coal is INR 10 per tonne, and the other expenses of the engine be INR 11.25 an hour, find the least cost of a journey of 100 km.

Cheers,

Nalin Pithwa

 

 

 

 

 

 

 

 

 

 

Two powerful wise quotes

  1. I am a pessimist by logic, but optimist by will-power. — Anon.
  2. The only thing greater than the power of the human mind is the courage of the human heart. — John Forbes Nash, Jr., Nobel Laureate mathematician (Economics Prize), Abel Laureate, victim of paranoid schizophrenia for 30 years.

 

Some fun – Math Late Show with David Letterman and Daniel Tammet

Stretching is a good exercise but…

Stretching is a good exercise, but stretching the mind through math is even better !!

Ha…ha…ha…LOL 🙂

Nalin Pithwa

PS: An object does not change topologically if it is “stretched”…Ha, ha, ha…LOL 🙂

Games, social behavior, chess, economics and maths

The following is some trivia but in fact, not so trivia, in this age of data science, data analytics, social media platforms, on-line gaming etc…If you decide to ponder over deep…you will become a giant mathematician or applied mathematician or of course, a computer science wunderkind..

The following is “picked out as it is” from a famous biography, (which regular readers of my blog will now know, perhaps, is a favorite mathematical biography for me)…A Beautiful Mind by Sylvia Nasar, biography of mathematical genius, John Forbes Nash, Jr, Nobel Laureate (Economics) and Abel Laureate:

“It was the great Hungarian-born polymath John von Neumann who first recognized that social behaviour could be analyzed as games. Von Neumann’s 1928 article on parlor games was the first successful attempt to derive logical and mathematical rules about rivalries. Just as William Blake saw the universe in a grain of sand, great scientists have often looked for clues in vast and complex problems in the small, familiar phenomena of daily life. Isaac Newton reached insights about the heavens by juggling wooden balls. Albert Einstein contemplated a boat paddling upriver. Von Neumann pondered the game of poker.

A seemingly trivial and playful pursuit like poker, von Neumann argued, might hold the key to more serious human affairs for two reasons. Both poker and economic competition require a certain type of reasoning, namely the rational calculation of advantage and disadvantage based on some internally consistent system of values (“more is better than less’). And, in both, the outcome for any individual actor depends not only on his own actions, but on the independent actions of others.

More than a century earlier, the French economist Antoine-Augustin Cournot had pointed out that problems of economic cnoice were greatly simplified when either none or a large number of other agents were present. Alone on his island, Robinson Crusoe does not have to worry about whose actions might affect him. Neither do Adam Smith’s butchers and bakers. They live in a world with so many others that their actions, in effect, cancel each other out. But when there is more than one agent but not so many that their influence may be safely ignored, strategic behavior raises a seemingly insoluble problem:”I think that he thinks that I think that he thingks,” and so forth…

So play games but think math ! 🙂

Nalin Pithwa