Category Archives: ISI Kolkatta Entrance Exam

Set Theory Primer : Some basic thinking and problem solving

Reference: AMS, Student Mathematical Library: Basic Set Theory by A. Shen, et al. Chapter 1. Section 1.

Problem 1:

Consider the oldest mathematician amongst chess players and the oldest chess player amongst mathematicians. Could they be two different people?

Problem 2:

The same question for the best mathematician amongst chess players and the best chess player amongst mathematicians.

Problem 3:

One tenth of mathematicians are chess players, and one sixth of chess players are mathematicians. Which group (mathematicians or chess players) is bigger? What is the ratio of sizes of these two groups?

Problem 4:

Do there exist sets A, B and C such that A \bigcap B \neq \phi, A \bigcap C = \phi and (A\bigcap B)-C = \phi ?

Problem 5:

Which of the following formulas are true for arbitrary sets A, B and C:

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

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

iii) (A \bigcup B) - C = (A-C)\bigcup B

iv) (A \bigcap B) - C = (A - C) \bigcap B

v) A - (B \bigcup C) = (A-B) \bigcap (A-C)

vi) A - (B \bigcap C) = (A-B) \bigcup (A-C)

Problem 6:

Give formal proofs of all valid formulas from the preceding problem. (Your proof should go like this : “We have to prove that the left hand side equals the right hand side. Let x be any element of the left hand side set. Then, ….Therefore, x belongs to the right hand side set. On the other hand, let…”)

Please give counterexamples to the formulas which are not true.

Problem 7:

Prove that the symmetric difference is associative:

A \triangle (B \triangle C) = (A \triangle B) \triangle C for any sets A, B and C. Hint: Addition modulo two is associative.

Problem 8:

Prove that:

(A_{1}\bigcap A_{2}\bigcap \ldots A_{n}) \triangle (B_{1} \bigcap B_{2} \bigcap \ldots B_{n}) = (A_{1} \triangle B_{1}) \bigcup (A_{2} \triangle B_{2}) \ldots (A_{n} \triangle B_{n}) for arbitrary sets A_{1}, A_{2}, \ldots, A_{n}, B_{1}, B_{2}, \ldots, B_{n}.

Problem 9:

Consider an inequality whose left hand side and right hand side contain set variables and operations \bigcap, \bigcup and -. Prove that if this equality is false for some sets, then it is false for some sets that contain at most one element.

Problem 10:

How many different expressions can be formed from set variables A and B by using union, intersection and set difference? (Variables and operations can be used more than once. Two expressions are considered identical if they assume the same value for each set of values of the variables involved.) Solve the same problem for three sets and for n sets. (Answer: In the general case, 2^{2^{n}-1})

Problem 11:

Solve the same problem if only \bigcup and \bigcap are allowed. For n=2 and n-3, this problem is easy to solve; however, no general formula for any n is known. This problem is also called “counting monotone Boolean functions in n variables”.)

Problem 12:

How many subsets does an n-element subset have?

Problem 13:

Assume that A consists of n elements and B \subset A consists of k elements. Find the number of different sets C such that B \subset C \subset A.

Problem 14:

A set U contains 2n elements. We select k subsets of A in such a way that none of them is a subset of another one. What is the maximum possible value of k? (Hint: Maximal k is achieved when all subsets have n elements. Indeed, imagine the following process: We start with an empty set and add random elements one by one until we get U. At most one selected set can appear in this process. On the other hand, the expected number of selected sets that appear during this process can be computed using the linearity of expectation. Take into account that the probability to come across some set Z \subset U is minimal when Z contains n elements, since all the sets of a given size are equiprobable.)

Your comments/solutions are welcome.


Nalin Pithwa.

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Binomial Theorem Tutorial problems I: IITJEE mains practice

I. Expand up to 5 terms the following expressions:

  1. (1+x)^{\frac{1}{2}}
  2. (1+x)^{\frac{7}{2}}
  3. (1-x)^{\frac{2}{5}}
  4. (1+x^{2})^{-2}
  5. (1-3x)^{\frac{1}{3}}
  6. (1-3x)^{\frac{-1}{2}}
  7. (1+2x)^{-\frac{1}{2}}
  8. (1+\frac{x}{3})^{-2}
  9. (1+\frac{2x}{3})^{\frac{3}{2}}
  10. (1+\frac{1}{2}a)^{-4}
  11. (2+x)^{-2}
  12. (9+2x)^{\frac{1}{2}}
  13. (8+12a)^{\frac{3}{2}}
  14. (9-6x)^{-\frac{3}{2}}
  15. (4a-8x)^{-\frac{1}{2}}

II. Write down and simplify:

  1. The 8th term of (1+2x)^{-\frac{1}{2}}
  2. The 11th term of (1-2x^{3})^{\frac{11}{2}}
  3. The 16th term of (1+3a^{2})^{\frac{16}{3}}
  4. The 6th term of (3a-2b)^{-1}
  5. The (r+1)^{th} term of (1-x)^{-2}
  6. The (r+1)^{th} term of (1-x)^{-4}
  7. The (r+1)^{th} term of (1+x)^{\frac{1}{2}}
  8. The (r+1)^{th} term of (1+x)^{\frac{11}{3}}
  9. The 14th term of (2^{10}-2^{7}x)^{\frac{13}{2}}
  10. The 7th term of (3^{8}+6^{4}x)^{\frac{11}{4}}


Nalin Pithwa

best explanation of epsilon delta definition

Refer any edition of (i) Calculus and Analytic Geometry by Thomas and Finney (ii) recent editions which go by the title “Thomas’ Calculus”. If you need, you will have to go through the previous stuff (given in the text) on “preliminaries” and/or functions also. For Sets, Functions and Relations, I have also presented a long series of articles on this blog.


Theory of Quadratic Equations: Part III: Tutorial practice problems: IITJEE Mains and preRMO

Problem 1:

Find the condition that a quadratic function of x and y may be resolved into two linear factors. For instance, a general form of such a function would be : ax^{2}+2hxy+by^{2}+2gx+2fy+c.

Problem 2:

Find the condition that the equations ax^{2}+bx+c=0 and a^{'}x^{2}+b^{'}x+c^{'}=0 may have a common root.

Using the above result, find the condition that the two quadratic functions ax^{2}+bxy+cy^{2} and a^{'}x^{2}+b^{'}xy+c^{'}y^{2} may have a common linear factor.

Problem 3:

For what values of m will the expression y^{2}+2xy+2x+my-3 be capable of resolution into two rational factors?

Problem 4:

Find the values of m which will make 2x^{2}+mxy+3y^{2}-5y-2 equivalent to the product of two linear factors.

Problem 5:

Show that the expression A(x^{2}-y^{2})-xy(B-C) always admits of two real linear factors.

Problem 6:

If the equations x^{2}+px+q=0 and x^{2}+p^{'}x+q^{'}=0 have a common root, show that it must be equal to \frac{pq^{'}-p^{'}q}{q-q^{'}} or \frac{q-q^{'}}{p^{'}-p}.

Problem 7:

Find the condition that the expression lx^{2}+mxy+ny^{2} and l^{'}x^{2}+m^{'}xy+n^{'}y^{2} may have a common linear factor.

Problem 8:

If the expression 3x^{2}+2Pxy+2y^{2}+2ax-4y+1 can be resolved into linear factors, prove that P must be be one of the roots of the equation P^{2}+4aP+2a^{2}+6=0.

Problem 9:

Find the condition that the expressions ax^{2}+2hxy+by^{2} and a^{'}x^{2}+2h^{'}xy+b^{'}y^{2} may be respectively divisible by factors of the form y-mx and my+x.

Problem 10:

Prove that the equation x^{2}-3xy+2y^{2}-2x-3y-35=0 for every real value of x, there is a real value of y, and for every real value of y, there is a real value of x.

Problem 11:

If x and y are two real quantities connected by the equation 9x^{2}+2xy+y^{2}-92x-20y+244=0, then will x lie between 3 and 6, and y between 1 and 10.

Problem 11:

If (ax^{2}+bx+c)y+a^{'}x^{2}+b^{'}x+c^{'}=0, find the condition that x may be a rational function of y.

More later,


Nalin Pithwa.

Theory of Quadratic Equations: part II: tutorial problems: IITJEE Mains, preRMO

Problem 1:

If x is a real number, prove that the rational function \frac{x^{2}+2x-11}{2(x-3)} can have all numerical values except such as lie between 2 and 6. In other words, find the range of this rational function. (the domain of this rational function is all real numbers except x=3 quite obviously.

Problem 2:

For all real values of x, prove that the quadratic function y=f(x)=ax^{2}+bx+c has the same sign as a, except when the roots of the quadratic equation ax^{2}+bx+c=0 are real and unequal, and x has a value lying between them. This is a very useful famous classic result. 


a) From your proof, you can conclude the following also: The expression ax^{2}+bx+c will always have the same sign, whatever real value x may have, provided that b^{2}-4ac is negative or zero; and if this condition is satisfied, the expression is positive, or negative accordingly as a is positive or negative.

b) From your proof, and using the above conclusion, you can also conclude the following: Conversely, in order that the expression ax^{2}+bx+c may be always positive, b^{2}-4ac must be negative or zero; and, a must be positive; and, in order that ax^{2}+bx+c may be always negative, b^{2}-4ac must be negative or zero, and a must be negative.

Further Remarks:

Please note that the function y=f(x)=ax^{2}+bx+c, where a, b, c \in \Re and a \neq 0 is a parabola. The roots of this y=f(x)=0 are the points where the parabola cuts the y axis. Can you find the vertex of this parabola? Compare the graph of the elementary parabola y=x^{2}, with the graph of y=ax^{2} where a \neq 0 and further with the graph of the general parabola y=ax^{2}+bx+c. Note you will just have to convert the expression ax^{2}+bx+c to a perfect square form.

Problem 3:

Find the limits between which a must lie in order that the rational function \frac{ax^{2}-7x+5}{5x^{2}-7x+a} may be real, if x is real.

Problem 4:

Determine the limits between which n must lie in order that the equation 2ax(ax+nc)+(n^{2}-2)c^{2}=0 may have real roots.

Problem 5:

If x be real, prove that \frac{x}{x^{2}-5x+9} must lie between 1 and -\frac{1}{11}.

Problem 6:

Prove that the range of the rational function y=f(x)=\frac{x^{2}-x+1}{x^{2}+x+1} lies between 3 and \frac{1}{3} for all real values of x.

Problem 7:

If x \in \Re, Prove that the rational function y=f(x)=\frac{x^{2}+34x-71}{x^{2}+2x-7} can have no value between 5 and 9. In other words, prove that the range of the function is (x <5)\bigcup(x>9).

Problem 8:

Find the equation whose roots are \frac{\sqrt{a}}{\sqrt{a} \pm \sqrt(a-b)}.

Problem 9:

If \alpha, \beta are roots of the quadratic equation x^{2}-px+q=0, find the value of (a) \alpha^{2}(\alpha^{2}\beta^{-1}-\beta)+\beta^{2}(\beta^{2}\alpha^{-1}-\alpha) (b) (\alpha-p)^{-4}+(\beta-p)^{-4}.

Problem 10:

If the roots of lx^{2}+mx+n=0 be in the ratio p:q, prove that \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0

Problem 11:

If x be real, the expression \frac{(x+m)^{2}-4mn}{2(x-n)} admits of all values except such as those that lie between 2n and 2m.

Problem 12:

If the roots of the equation ax^{2}+2bx+c=0 are \alpha and \beta, and those of the equation Ax^{2}+2Bx+C=0 be \alpha+\delta and \beta+\delta, prove that \frac{b^{2}-ac}{a^{2}} = \frac{B^{2}-AC}{A^{2}}.

Problem 13:

Prove that the rational function y=f(x)=\frac{px^{2}+3x-4}{p+3x-4x^{2}} will be capable of all values when x is real, provided that p has any real value between 1 and 7. That is, under the conditions on p, we have to show that the given rational function has as its range the full real numbers. (Of course, the domain is real except those values of x for which the denominator is zero).

Problem 14:

Find the greatest value of \frac{x+2}{2x^{2}+3x+6} for any real value of x. (Remarks: this is maxima-minima problem which can be solved with algebra only, calculus is not needed). 

Problem 15:

Show that if x is real, the expression (x^{2}-bc)(2x-b-c)^{-1} has no real value between b and a.

Problem 16:

If the roots of ax^{2}+bx+c=0 be possible (real) and different, then the roots of (a+c)(ax^{2}+2bx+c)=2(ac-b^{2})(x^{2}+1) will not be real, and vice-versa. Prove this.

Problem 17:

Prove that the rational function y=f(x)=\frac{(ax-b)(dx-c)}{(bx-a)(cx-a)} will be capable of all real values when x is real, if a^{2}-b^{2} and c^{2}-a^{2} have the same sign.


Nalin Pithwa

Theory of Quadratic Equations: Tutorial problems : Part I: IITJEE Mains, preRMO

I) Form the equations whose roots are:

a) -\frac{4}{5}, \frac{3}{7} (b) \frac{m}{n}, -\frac{n}{m} (c) \frac{p-q}{p+q}, -\frac{p+q}{p-q} (d) 7 \pm 2\sqrt{5} (e) -p \pm 2\sqrt{2q} (f) -3 \pm 5i (g) -a \pm ib (h) \pm i(a-b) (i) -3, \frac{2}{3}, \frac{1}{2} (j) \frac{a}{2},0, -\frac{2}{a} (k) 2 \pm \sqrt{3}, 4

II) Prove that the roots of the following equations are real:

i) x^{2}-2ax+a^{2}-b^{2}-c^{2}=0

ii) (a-b+c)x^{2}+4(a-b)x+(a-b-c)=0

III) If the equation x^{2}-15-m(2x-8)=0 has equal roots, find the values of m.

IV) For what values of m will the equation x^{2}-2x(1+3m)+7(3+2m)=0 have equal roots?

V) For what value of m will the equation \frac{x^{2}-bx}{ax-c} = \frac{m-1}{m+1} have roots equal in magnitude but opposite in sign?

VI) Prove that the roots of the following equations are rational:

(i) (a+c-b)x^{2}+2ax+(b+c-a)=0

(ii) abc^{2}x^{2}+3a^{2}cx+b^{2}ax-6a^{2}-ab+2b^{2}=0

VII) If \alpha, \beta are the roots of the equation ax^{2}+bx+c=0, find the values of

(i) \frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}

(ii) \alpha^{4}\beta^{7}+\alpha^{7}\beta^{4}

(iii) (\frac{\alpha}{\beta}-\frac{\beta}{\alpha})^{2}

VIII) Find the value of:

(a) x^{3}+x^{2}-x+22 when x=1+2i

(b) x^{3}-3x^{2}-8x+16 when x=3+i

(c) x^{3}-ax^{2}+2a^{2}x+4a^{3} when \frac{x}{a}=1-\sqrt{-3}

IX) If \alpha and \beta are the roots of x^{2}+px+q=0 form the equation whose roots are (\alpha-\beta)^{2} and (\alpha+\beta)^{2}/

X) Prove that the roots of (x-a)(x-b)=k^{2} are always real.

XI) If \alpha_{1}, \alpha_{2} are the roots of ax^{2}+bx+c=0, find the value of (i) (ax_{1}+b)^{-2}+(ax_{2}+b)^{-2} (ii) (ax_{1}+b)^{-3}+(ax_{2}+b)^{-3}

XII) Find the condition that one root of ax^{2}+bx+c=0 shall be n times the other.

XIII) If \alpha, \beta are the roots of ax^{2}+bx+c=0 form the equation whose roots are \alpha^{2}+\beta^{2} and \alpha^{-2}+\beta^{-2}.

XIV) Form the equation whose roots are the squares of the sum and of the differences of the roots of 2x^{2}+2(m+n)x+m^{2}+n^{2}=0.

XV) Discuss the signs of the roots of the equation px^{2}+qx+r=0

XVI) If a, b and c are odd integers, prove that the roots of the equation ax^{2}+bx+c=0 cannot be rational numbers.

XVII) Given that the equation x^{4}+px^{3}+qx^{2}+rx+s=0 has four real positive roots, prove that (a) pr-16s \geq 0 (b) q^{2}-36s \geq 0, where equality holds, in each case, if and only if the roots are equal.

XVIII) Let p(x)=x^{2}+ax+b be a quadratic polynomial in which a and b are integers. Given any integer n, show that there is an integer M such that p(n)p(n+1)=p(M).


Nalin Pithwa.

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}.


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.


Nalin Pithwa.

Set Theory, Relations, Functions Preliminaries: Part III


Functions as a special kind of relation:

Let us first consider an example where set A is a set of men, and B is a set of positive real numbers. Let us say f is a relation from A to B given by : f = \{ (x,y) : x \in A, y \hspace{0.1in} is \hspace{0.1in} the \hspace{0.1in} weight \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} person \hspace{0.1in} x \}

Hence, f “relates” every man in set A to his weight in set B. That is,

i) Every man has some weight associated with him in set B. (ii) That weight is unique. That is, a person cannot have more than one weight (at a given time, of course) !! 🙂 This, of course, does not mean that two different persons, say P and Q may not have the same weight 100 kg ( the same element of set B). The only thing it means is that any one person, say P will have one and only one weight (100kg) at the time instant of measurement and not more than one weights (which would be crazy) at a time instant it is measured !!

Definition I (a function defined as a relation):

A function f from a set A (called domain) to a set B (called codomain) is a relation that associates or “pairs up” every element of domain A with a unique element of codomain B. (Note that whereas a relation from a set A to a set B is just a subset of the cartesian product A \times B).

Some remarks: The above definition is also motivated by an example of a function as a relation. On the other hand, another definition of a function can be motivated as follows:

We know that the boiling point of water depends on the height of water above sea level. We also know that the simple interest on a deposit in a bank depends on the duration of deposit held in the bank. In these and several such examples, one quantity, say y, depends on another quantity “x”.

Symbol: f: A \longrightarrow B; if x \in A, y \in B, then we also denote: f: x \longmapsto y; we also write y=f(x), read as “y is f of x”.

Here, y is called image of x under f and x is called the preimage of y under f.

Definition: Range: The set of all images in B is called the range of f. That is, Range = \{ f(x): x \in A\}

Note: (i) Every function is a relation but every relation need not be a function. (Homework quiz: find illustrative examples for the same) (ii) If the domain and codomain are not specified, they are assumed to be the set of real numbers.

In calculus, we often want to refer to a generic function without having any particular formula in mind. Leonhard Euler invented a symbolic way to say “y is a function of x” by writing

y = f(x) (“y equals f of x”)

In this equation, the symbol f represents the function. The letter x, called the independent variable, represents an input value from the domain of f, and y, the dependent variable, represents the corresponding output value f(x) in the range of f. Here is the formal definition of function: (definition 2):

function from a set D to a set \Re is a rule that assigns a unique element f(x) in \Re to each element x in D.

In this definition, D=D(f) (read “D of f”) is the domain of the function f and \Re is the range (or codomain containing the range of f).

Think of a function f as a kind of machine that produces an output value f(x) in its range whenever we feed it an input value x from its domain. In our scope, we will usually define functions in one of two ways:

a) by giving a formula such as y=x^{2} that uses a dependent variable y to denote the value of the function, or

b) by giving a formula such as f(x)=x^{2} that defines a function symbol f to name the function.

NOTE: there can be well-defined functions which do not have any formula at all; for example, let f(x) = 0 when x \in Q and f(x)=1, when x \in Q^{'}.

Strictly speaking, we should call the function f and not f(x) as the latter denotes the value of the function at the point x. However, as is common usage, we will often refer to the function as f(x) in order to name the variable on which f depends.

It is sometimes convenient to use a single letter to denote both a function and the dependent variable. For instance, we might say that the area A of a circle of radius r is given by the function : A(r)=\pi r^{2}.


As we said earlier, most of the functions in our scope will be real-valued function of a real variable, functions whose domains and ranges are sets of real numbers. We evaluate such functions by susbtituting particular values from the domain into the function’s defining rule to calculate the corresponding values in the range.

Example 1:

The volume V of a ball (solid sphere) r is given by the function: V(r)=\frac{4}{3}\pi  r^{3}.

The volume of a ball of radius 3 meters is : V(3)=\frac{4}{3}\pi (3)^{3}=36 \pi m^{3}.

Example 2:

Suppose that the function F is defined for all real numbers t by the formula: F(t)=2(t-1)+3.

Evaluate F at the output values 0, 2, x+2, and F(2).

Solution 2:

In each case, we substitute the given input value for t into the formula for F:





The Domain Convention

When we define a function y=f(x) with a formula and the domain is not stated explicitly, the domain is assumed to be the largest set of x-values for which the formula gives real x-values. This is the function’s so-called natural domain. If we want the domain to be restricted in some way, we must say so.

The domain of the function y=x^{2} is understood to be the entire set of real numbers. The formula gives a real value y-value for every real number x. If we want to restrict the domain to values of x greater than or equal to 2, we must write ” y=x^{2}” for x \geq 2.

Changing the domain to which we apply a formula usually changes the range as well. The range of y=x^{2} is [0, \infty). The  range of y=x^{2} where x \geq 2 is the set of all numbers obtained by squaring numbers greater than or equal to 2. In symbols, the range is \{ x^{2}: x \geq 2\} or \{ y: y \geq 4\} or [4,\infty)

Example 3:

Function : y = \sqrt{1-x^{2}}; domain [-1,1]; Range (y) is [0,1]

Function: y=\frac{1}{x}; domain (-\infty,0) \bigcup (0,\infty); Range (y) is (-\infty,0)\bigcup (0,\infty)

Function: y=\sqrt{x}; domain (0,\infty) and range (y) is (0,\infty)

Function y = \sqrt{4-x}, domain (-\infty,,4], and range (y) is [0, \infty)

Graphs of functions:

The graph of a function f is the graph of the equation y=f(x). It consists of the points in the Cartesian plane whose co-ordinates (x,y) are input-output pairs for f.

Not every curve you draw is the graph of a function. A function f can have only one value f(x) for each x in its domain so no vertical line can intersect the graph of a function more than once. Thus, a circle cannot be the graph of a function since some vertical line intersect the circle twice. If a is in the domain of a function f, then the vertical line x=a will intersect the graph of f in the single point (a, f(a)).

Example 4: Graph the function y=x^{2} over the interval [-2.2]. (homework).Thinking further: so plotting the above graph requires a table of x and y values; but how do we connect the points ? Should we connect two points by a straight line, smooth line, zig-zag line ??? How do we know for sure what the graph looks like between the points we plot? The answer lies in calculus, as we will see in later chapter. There will be a marvelous mathematical tool called the derivative to find a curve’s shape between plotted points. Meanwhile, we will have to settle for plotting points and connecting them as best as we can. 

PS: (1) you can use GeoGebra, a beautiful freeware for plotting various graphs, and more stuff (2) If you wish, you can use a TI-graphing calculator. This is a nice investment for many other things like number theory also. See for example,

Meanwhile, you need to be extremely familiar with graphs of following functions; plot and check on your own:

y=x^{3}, y=x^{2/3}, y=\sqrt{x}, y=\sqrt[3]{x}, y=\frac{1}{x}, y=\frac{1}{x^{2}}, y=mx, where m \in Z, y=x^{3/2}

Sums, Differences, Products and Quotients

Like numbers, functions can be added, subtracted, multiplied and divided (except where the the denominator is zero) to produce new functions. If f and g are functions, then for every x that belongs to the domains of BOTH f and g, we define functions: f+g, f-g, fg by the formulas:




At any point D(f) \bigcap D(g) at which g(x) \neq 0, we can also define the function f/g by the formula:

(\frac{f}{g})(x)=\frac{f(x)}{g(x)}, where g(x) \neq 0

Functions can also be multiplied by constants. If c is a real number, then the function cf is defined for all x in the domain of f by (cf)(x)=cf(x)

Example 5:

Function f, formula y=\sqrt{x}, domain [0,\infty)

Function g, formula g(x)=\sqrt{(1-x)}, domain (-\infty, 1]

Function 3g, formula 3g(x)=3\sqrt{(1-x)}, domain (-\infty, 1]

Function f+g, formula (f+g)(x)=\sqrt{x}+\sqrt{(1-x)}, domain [0,1]=D(f) \bigcap D(g)

Function f-g, formula (f-g)(x)=\sqrt{x}-\sqrt{(1-x)}, domain [0.1]

Function g-f, formula (g-f)(x)=\sqrt{(1-x)}-\sqrt{x}, domain [0,1]

Function f . g, formula (f . g)(x)=f(x)g(x) = \sqrt{x(1-x)}, domain [0,1]

Function \frac{f}{g}, formula \frac{f}{g}(x)=\frac{f(x)}{g(x)}=\sqrt{\frac{x}{1-x}}, domain is [0,1)

Function \frac{g}{f}(x) = \frac{g(x)}{f(x)}=\sqrt{\frac{1-x}{x}}, domain (0,1]

Composite Functions:

Composition is another method for combining functions.


If f and g are functions, the composite function f \circ g (f “circle” g) is defined by (f \circ g)(x)=f(g(x)). The domain of f \circ g consists of the numbers x in the domain of g for which g(x) lies in the domain of f.

The definition says that two functions can be composed when the image of the first lies in the domain of the second. To (f \circ g)(x) we first find g(x) and second find f(g(x)).

Clearly, in general, (f \circ g)(x) \neq (g \circ f)(x). That is, composition of functions is not commutative.

Example 6:

If f(x)=\sqrt{x} and g(x)=x+1, find (a) (f \circ g)(x) (b) (g \circ f)(x) (c) (f \circ f)(x) (d) (g \circ g)(x)

Solution 6:

a) (f \circ g)(x) = f(g(x))=\sqrt{g(x)}=\sqrt{x+1}, domain is [-1, \infty)

b) (g \circ f)(x)=g(f(x))=f(x)+1=\sqrt{x}+1, domain is [0, \infty)

c) (f \circ f)(x)=f(f(x))=\sqrt{f(x)}=\sqrt{\sqrt{x}}=x^{\frac{1}{4}}, domain is [0, \infty)

d) (g \circ g)(x)=g(g(x))=g(x)+1=(x+1)+1=x+2, domain is \Re or (-\infty, \infty)

Even functions and odd functions:

A function f(x) is said to be even if f(x)=f(-x). That is, the function possesses symmetry about the y-axis. Example, y=f(x)=x^{2}.

A function f(x) is said to be odd if f(x)=-f(-x). That is, the function possesses symmetry about the origin. Example y=f(x)=x^{3}.

Any function can be expressed as a sum of an even function and an odd function.

A function could be neither even nor odd.

Note that a function like y^{2}=x possesses symmetry about the x-axis !!

Piecewise Defined Functions:

Sometimes a function uses different formulas or formulae over different parts of its domain. One such example is the absolute value function:

y=f(x) = |x|=x, when x \geq 0 and y=-x, when x<0.

Example 7:

The function f(x)=-x, when x<0, y=f(x)=x^{2}, when 0 \leq x \leq 1, and f(x)=1, when x>1.

Example 8:

The greatest integer function:

The function whose value at any number x is the greatest integer less than or equal to x is called the greatest integer function or the integer floor function. It is denoted by \lfloor x \rfloor.

Observe that \lfloor 2.4 \rfloor =2; \lfloor 1.4 \rfloor =1; \lfloor 0 \rfloor =0; \lfloor -1.2 \rfloor =-2; \lfloor 2 \rfloor =2; \lfloor 0.2 \rfloor =0\lfloor -0.3 \rfloor =-1; \lfloor -2 \rfloor =-2.

Example 9:

The least integer function:

The function whose value at any number x is the smallest integer greater than or equal to x is called the least integer function or the integer ceiling function. It is denoted by \lceil x \rceil. For positive values of x, this function might represent, for example, the cost of parking x hours in a parking lot which charges USD 1 for each hour or part of an hour.


Nalin Pithwa




Set Theory, Relations, Functions Preliminaries: II


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;


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









You and your research ( You and your studies) : By Richard Hamming, AT and T, Bell Labs mathematician;

Although the title is grand (and quite aptly so)…the reality is that it can be applied to serious studies for IITJEE entrance, CMI entrance, highly competitive math olympiads, and also competitive coding contests…in fact, to various aspects of student life and various professional lifes…

Please read the whole article…apply it wholly or partially…modified or unmodified to your studies/research/profession…these are broad principles of success…