Category Archives: Basic Set Theory and Logic

Set theory, relations, functions: preliminaries: Part V

Types of functions: (please plot as many functions as possible from the list below; as suggested in an earlier blog, please use a TI graphing calculator or GeoGebra freeware graphing software): 

  1. Constant function: A function f:\Re \longrightarrow \Re given by f(x)=k, where k \in \Re is a constant. It is a horizontal line on the XY-plane.
  2. Identity function: A function f: \Re \longrightarrow \Re given by f(x)=x. It maps a real value x back to itself. It is a straight line passing through origin at an angle 45 degrees to the positive X axis.
  3. One-one or injective function: If different inputs give rise to different outputs, the function is said to be injective or one-one. That is, if f: A \longrightarrow B, where set A is domain and set B is co-domain, if further, x_{1}, x_{2} \in A such that x_{1} \neq x_{2}, then it follows that f(x_{1}) \neq f(x_{2}). Sometimes, to prove that a function is injective, we can prove the conrapositive statement of the definition also; that is, y_{1}=y_{2} where y_{1}, y_{2} \in codomain \hspace{0.1in} range, then it follows that x_{1}=x_{2}. It might be easier to prove the contrapositive. It would be illuminating to construct your own pictorial examples of such a function. 
  4. Onto or surjective: If a function is given by f: X \longrightarrow Y such that f(X)=Y, that is, the images of all the elements of the domain is full of set Y. In other words, in such a case, the range is equal to co-domain. it would be illuminating to construct your own pictorial examples of  such a function.
  5. Bijective function or one-one onto correspondence: A function which is both one-one and onto is called a bijective function. (It is both injective and surjective). Only a bijective function will have a well-defined inverse function. Think why! This is the reason why inverse circular functions (that is, inverse trigonometric functions have their domains restricted to so-called principal values). 
  6. Polynomial function: A function of the form f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\ldots + a_{n}x^{n}, where n is zero or positive integer only and a_{i} \in \Re is called a polynomial with real coefficients. Example. f(x)=ax^{2}=bx+c, where a \neq 0, a, b, c \in \Re is called a quadratic function in x. (this is a general parabola).
  7. Rational function: The function of the type \frac{f(x)}{g(x)}, where g(x) \neq 0, where f(x) and g(x) are polynomial functions of x, defined in a domain, is called a rational function. Such a function can have asymptotes, a term we define later. Example, y=f(x)=\frac{1}{x}, which is a hyperbola with asymptotes X and Y axes.
  8. Absolute value function: Let f: \Re \longrightarrow \Re be given by f(x)=|x|=x when x \geq 0 and f(x)=-x, when x<0 for any x \in \Re. Note that |x|=\sqrt{x^{2}} since the radical sign indicates positive root of a quantity by convention.
  9. Signum function: Let f: \Re \longrightarrow \Re where f(x)=1, when x>0 and f(x)=0 when x=0 and f(x)=-1 when x<0. Such a function is called the signum function. (If you can, discuss the continuity and differentiability of the signum function). Clearly, the domain of this function  is full \Re whereas the range is \{ -1,0,1\}.
  10. In part III of the blog series, we have already defined the floor function and the ceiling function. Further properties of these functions are summarized (and some with proofs in the following wikipedia links): (once again, if you can, discuss the continuity and differentiablity of the floor and ceiling functions): https://en.wikipedia.org/wiki/Floor_and_ceiling_functions
  11. Exponential function: A function f: \Re \longrightarrow \Re^{+} given by f(x)=a^{x} where a>0 is called an exponential function. An exponential function is bijective and its inverse is the natural logarithmic function. (the logarithmic function is difficult to define, though; we will consider the details later). PS: Quiz: Which function has a faster growth rate — exponential or a power function ? Consider various parameters.
  12. Logarithmic function: Let a be a positive real number with a \neq 1. If a^{y}=x, where x \in \Re, then y is called the logarithm of x with base a and we write it as y=\ln{x}. (By the way, the logarithmic function is used in the very much loved mp3 music :-))

Regards,

Nalin Pithwa

Set Theory, Relations and Functions: Preliminaries: IV:

Problem Set based on previous three parts:

I) Solve the inequalities in the following exercises expressing the solution sets as intervals or unions of intervals. Also, graph each solution set on the real line:

a) |x| <2 (b) |x| \leq 2 (c) |t-1| \leq 3 (d) |t+2|<1 (e) |3y-7|<4(f) |2y+5|<1 (g) |\frac{z}{5}-1| \leq 1 (h) | \frac{3}{2}z-1| \leq 2 (i) |3-\frac{1}{x}|<\frac{1}{2} (j) |\frac{2}{x}-4|<3 (k) |2x| \geq 4 (l) |x+3| \geq \frac{1}{2} (m) |1-x| >1 (n) |2-3x| > 5 (o) |\frac{x+1}{2}| \geq 1 (p) |\frac{3x}{5}-1|>\frac{2}{5}

II) Quadratic Inequalities:

Solve the inequalities in the following exercises. Express the solution sets as intervals or unions of intervals and graph them. Use the result \sqrt{a^{2}}=|a| as appropriate.

(a) x^{2}<2 (b) 4 \leq x^{2} (c) 4<x^{2}<9 (d) \frac{1}{9} < x^{2} < \frac{1}{4} (e) (x-1)^{2}<4 (f) (x+3)^{2}<2 (g) x^{2}-x<0 (h) x^{2}-x-2 \geq 0

III) Theory and Examples:

i) Do not fall into the trap |-a|=a. For what real numbers a is the equation true? For what real numbers is it false?

ii) Solve the equation: |x-1|=1-x.

iii) A proof of the triangle inequality: 

Give the reason justifying each of the marked steps in the following proof of the triangle inequality:

|a+b|^{2}=(a+b)^{2}…..why ?

=a^{2}+2ab++b^{2}

\leq a^{2}+2|a||b|+b^{2}….why ?

\leq |a|^{2}+2|a||b|+|b|^{2}….why?

=(|a|+|b|)^{2}….why ?

iv) Prove that |ab|=|a||b| for any numbers a and b.

v) If |x| \leq 3 and x>-\frac{1}{2}, what can you say about x?

vi) Graph the inequality: |x|+|y| \leq 1

Questions related to functions:

I) Find the domain and range of each function:

a) f(x)=1-\sqrt{x} (b) F(t)=\frac{1}{1+\sqrt{t}} (c) g(t)=\frac{1}{\sqrt{4-t^{2}}}

II) Finding formulas for functions:

a) Express the area and perimeter of an equilateral triangle as a function of the triangle’ s side with length s.

b) Express the side length of a square as a function of the cube’s diagonal length d. Then, express the surface area  and volume of the cube as a function of the diagonal length.

c) A point P in the first quadrant lies on the graph of the function f(x)=\sqrt{x}. Express the coordinates of P as functions of the slope of the line joining P to the origin.

III) Functions and graphs:

Graph the functions in the questions below. What symmetries, if any, do the graphs have?

a) y=-x^{3} (b) y=-\frac{1}{x^{2}} (c) y=-\frac{1}{x} (d) y=\frac{1}{|x|} (e) y = \sqrt{|x|} (f) y=\sqrt{-x} (g) y=\frac{x^{3}}{8} (h) y=-4\sqrt{x} (i) y=-x^{\frac{3}{2}} (j) y=(-x)^{\frac{3}{2}} (k) y=(-x)^{\frac{2}{3}} (l) y=-x^{\frac{2}{3}}

IV) Graph the following equations ad explain why they are not graphs of functions of x. (a) |y|=x (b) y^{2}=x^{2}

V) Graph the following equations and explain why they are not graphs of functions of x: (a) |x|+|y|=1 (b) |x+y|=1

VI) Even and odd functions:

In the following questions, say whether the function is even, odd or neither.

a) f(x)=3 (b) f(x=x^{-5} (c) f(x)=x^{2}+1 (d) f(x)=x^{2}+x (e) g(x)=x^{4}+3x^{2}-1 (f) g(x)=\frac{1}{x^{2}-1} (g) g(x)=\frac{x}{x^{2}-1} (h) h(t)=\frac{1}{t-1} (i) h(t)=|t^{3}| (j) h(t)=2t+1 (k) h(t)=2|t|+1

Sums, Differences, Products and Quotients:

In the two questions below, find the domains and ranges of f, g, f+g, and f-g.

i) f(x)=x, g(x)=\sqrt{x-1} (ii) f(x)=\sqrt{x+1}, g(x)=\sqrt{x-1}

In the two questions below, find the domains and ranges of f, g, \frac{f}{g} and \frac{g}{f}

i) f(x)=2, g(x)=x^{2}+1

ii) f(x)=1, g(x)=1+\sqrt{x}

Composites of functions:

  1. If f(x)=x+5, and g(x)=x^{2}-5, find the following: (a) f(g(0)) (b) g(f(0)) (c) f(g(x)) (d) g(f(x)) (e) f(f(-5)) (f) g(g(2)) (g) f(f(x)) (h) g(g(x))
  2. If f(x)=x-1 and g(x)=\frac{1}{x+1}, find the following: (a) f(g(\frac{1}{2})) (b) g(f(\frac{1}{2})) (c) f(g(x)) (d) g(f(x)) (e) f(f(2)) (f) g(g(2)) (g) f(f(x)) (h) g(g(x))
  3. If u(x)=4x-5, v(x)=x^{2}, and f(x)=\frac{1}{x}, find formulas or formulae for the following: (a) u(v(f(x))) (b) u(f(v(x))) (c) v(u(f(x))) (d) v(f(u(x))) (e) f(u(v(x))) (f) f(v(u(x)))
  4. If f(x)=\sqrt{x}, g(x)=\frac{x}{4}, and h(x)=4x-8, find formulas or formulae for the following: (a) h(g(f(x))) (b) h(f(g(x))) (c) g(h(f(x))) (d) g(f(h(x))) (e) f(g(h(x))) (f) f(h(g(x)))

Let f(x)=x-5, g(x)=\sqrt{x}, h(x)=x^{3}, and f(x)=2x. Express each of the functions in the questions below as a composite involving one or more of f, g, h and j:

a) y=\sqrt{x}-3 (b) y=2\sqrt{x} (c) y=x^{\frac{1}{4}} (d) y=4x (e) y=\sqrt{(x-3)^{3}} (f) y=(2x-6)^{3} (g) y=2x-3 (h) y=x^{\frac{3}{2}} (i) y=x^{9} (k) y=x-6 (l) y=2\sqrt{x-3} (m) \sqrt{x^{3}-3}

Questions:

a) g(x)=x-7, f(x)=\sqrt{x}, find (f \circ g)(x)

b) g(x)=x+2, f(x)=3x, find (f \circ g)(x)

c) f(x)=\sqrt{x-5}, (f \circ g)(x)=\sqrt{x^{2}-5}, find g(x).

d) f(x)=\frac{x}{x-1}, g(x)=\frac{x}{x-1}, find (f \circ g)(x)

e) f(x)=1+\frac{1}{x}, (f \circ g)(x)=x, find g(x).

f) g(x)=\frac{1}{x}, (f \circ g)(x)=x, find f(x).

Reference: Calculus and Analytic Geometry, G B Thomas. 

NB: I have used an old edition (printed version) to prepare the above. The latest edition may be found at Amazon India link:

https://www.amazon.in/Thomas-Calculus-George-B/dp/9353060419/ref=sr_1_1?crid=1XDE2XDSY5LCP&keywords=gb+thomas+calculus&qid=1570492794&s=books&sprefix=G+B+Th%2Caps%2C255&sr=1-1

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.

Permutations and Combinations: A primer only

The infinite hotel paradox : due Jeff Dekofsky

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

 

Logicalympics — 100 meters!!!

Just as you go to the gym daily and increase your physical stamina, so also, you should go to the “mental gym” of solving hard math or logical puzzles daily to increase your mental stamina. You should start with a laser-like focus (or, concentrate like Shiva’s third eye, as is famous in Hindu mythology/scriptures!!) for 15-30 min daily and sustain that pace for a month at least. Give yourself a chance. Start with the following:

The logicalympics take place every year in a very quiet setting so that the competitors can concentrate on their events — not so much the events themselves, but the results. At the logicalympics every event ends in a tie so that no one goes home disappointed 🙂 There were five entries in the room, so they held five races in order that each competitor could win, and so that each competitor could also take his/her turn in 2nd, 3rd, 4th, and 5th place. The final results showed that each competitor had duly taken taken their turn in finishing in each of the five positions. Given the following information, what were the results of each of the five races?

The five competitors were A, B, C, D and E. C didn’t win the fourth race. In the first race A finished before C who in turn finished after B. A finished in a better position in the fourth race than in the second race. E didn’t win the second race. E finished two places  behind C in the first race. D lost the fourth race. A finished ahead of B in the fourth race, but B finished before A and C in the third race. A had already finished before C in the second race who in turn finished after B again. B was not first in the first race and D was not last. D finished in a better position in the second race than in the first race and finished before B. A wasn’t second in the second race and also finished before B.

So, is your brain racing now to finish this puzzle?

Cheers,

Nalin Pithwa.

PS: Many of the puzzles on my blog(s) are from famous literature/books/sources, but I would not like to reveal them as I feel that students gain the most when they really try these questions on their own rather than quickly give up and ask for help or look up solutions. Students have finally to stand on their own feet! (I do not claim creating these questions or puzzles; I am only a math tutor and sometimes, a tutor on the web.) I feel that even a “wrong” attempt is a “partial” attempt; if u can see where your own reasoning has failed, that is also partial success!

Pick’s theorem to pick your brains!!

Pick’s theorem:

Consider a square lattice of unit side. A simple polygon (with non-intersecting sides) of any shape is drawn with its vertices at the lattice points. The area of the polygon can be simply obtained as (B/2)+I-1 square units, where B is number of lattice points on the boundary, I is number of lattice points in the interior of the polygon. Prove this theorem!

Do you like this challenge?

Nalin Pithwa.

Genius of Srinivasa Ramanujan

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

Try try until you succeed!!

Nalin Pithwa.

Men, monkey and coconuts problem

Five men and a monkey collected a heap of coconuts in a desert island. The men went to sleep in the night. One of them got up in the middle of the night and divided the heap into five equal parts and found an extra coconut, which he gave to the monkey. He hid his share and made one heap combining the rest of the coconuts. Then, the second person got up and divided this new heap into five equal parts and again found an extra coconut, which he gave to the monkey. He also hid his share and made a new heap of the remaining coconuts. This process continued until the last person did exactly the same thing as the others. What must be the minimum number of coconuts in the original heap if

(a) the next morning when they divided the last remaining heap into five equal parts, there was no coconut left for the monkey.

(b) the next morning they divided the last remaining heap into five equal parts, they again found an extra one, which they gave to the monkey.

Cheers,

Nalin Pithwa.

PS: One good way to increase concentration, motivation, intellectual stamina for solving such demanding puzzles or math problems for the IITJEE or RMO/INMO or even Mensa challenges is to give yourself a “reward” after you actually solve that problem. So, you may go ahead and have chilled coconut juice after you solve this puzzle. Anyway, these are the hot summer days in India.