# Miscellaneous Questions: part I: solution to chess problem by my student RI

Some blogs away I had posted several interesting, non-trivial, yet do-able-with-some-effort problems for preRMO and RMO.

A student of mine, RI has submitted the following beautiful solution to the chess problem. I am reproducing the question for convenience of the readers:

Question:

The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge on vertex. Thus, a square can have 8, 5 or 3 neighbours depending on its position.) Show that all the sixty four entries are in fact equal.

Let us denote the set of all integers on the chess board by S (assume they are distinct). [Now, we can use the Well-ordering principle: every non-empty set of non-negative integers contains a least element. That is, every non-empty set S of non-negative integers contains an element a in S such that $a \leq b$ for all elements b of S}. So, also let “a” be the least element of set S here. As it is the average of the neighbouring elements, it can’t be less than each of them. But it can’t be greater than all of them also. So, all the elements of S are equal.

QED.

Three cheers for RI 🙂 🙂 🙂

Regards,

Nalin Pithwa

# Miscellaneous questions: part II: solutions to tutorial practice for preRMO and RMO

Refer the blog questions a few days before:

Question 1:

Let $a_{1}, a_{2}, \ldots, a_{10}$ be ten real numbers such that each is greater than 1 and less than 55. Prove that there are three among the given numbers which form the lengths of the sides of a triangle.

Without loss of generality, we may take $1…..call this relation (i).

Let, if possible, no three of the given numbers be the lengths of the sides of a triangle. (That is, no three satisfy the triangle inequality. Note that when we say three numbers a, b and c satisfy the triangle inequality —- it means all the following three inequalities have to hold simultaneously: $a+b>c$, $a+c>b$ and $b+c>a$). We will consider triplets $a_{i}, a_{i+1}, a_{i+2}$ and $1 \leq i \leq 8$. As these numbers do not form the lengths of the sides of a triangle, the sum of the smallest two numbers should not exceed the largest number, that is, $a_{i}+a_{i+1} \leq a_{i+2}$. Hence, we get the following set of inequalities:

$i=1$ gives $a_{1}+a_{2} \leq a_{3}$ giving $2 < a_{3}$.

$i=2$ gives $a_{2}+a_{3} \leq a_{4}$ giving $3 < a_{4}$

$i=3$ gives $a_{3}+a_{4} \leq a_{5}$ giving $5 < a_{5}$

$i=4$ gives $a_{4}+a_{5} \leq a_{6}$ giving $8 < a_{6}$

$i=5$ gives $a_{5}+a_{6} \leq a_{7}$ giving $13 < a_{7}$

$i=6$ gives $a_{6}+a_{7} \leq a_{8}$ giving $21 < a_{8}$

$i=7$ gives $a_{7}+a_{8} \leq a_{9}$ giving $34 < a_{9}$

$i=8$ gives $a_{8}+a_{9} \leq a_{10}$ giving $55

contradicting the basic hypothesis. Hence, there exists three numbers among the given numbers which form the lengths of the sides of a triangle.

Question 2:

In a collection of 1234 persons, any two persons are mutual friends or enemies. Each person has at most 3 enemies. Prove that it is possible to divide the collection into two parts such that each person has at most 1 enemy in his sub-collection.

Let C denote the collection of given 1234 persons. Let $\{ C_{1}, C_{2}\}$ be a partition of C. Let $e(C_{1})$ denote the total number of enemy pairs in $C_{1}$. Let $e(C_{2})$ denote the total number of enemy pairs in $C_{2}$.

Let $e(C_{1}, C_{2})= e(C_{1})+e(C_{2})$ denote the total number of enemy pairs corresponding to the partition $\{ C_{1}, C_{2}\}$ of C. Note $e(C_{1}, C_{2})$ is an integer greater than or equal to zero. Hence, by Well-Ordering Principle, there exists a partition having the least value of $e(C_{1}, C_{2})$.

Claim: This is “the” required partition.

Proof: If not, without loss of generality, suppose there is a person P in $C_{1}$ having at least 2 enemies in $C_{1}$. Construct a new partition $\{D_{1}, D_{2}\}$ of C as follows: $D_{1}=C_{1}-\{ P \}$ and $D_{2}=C_{2}- \{P\}$. Now, $e(D_{1}, D_{2})=e(D_{1})+e(D_{2}) \leq \{ e(C_{1})-2\} + \{ e(C_{2})+1\}=e(C_{1}, C_{2})-1$. Hence, $e(D_{1}, D_{2}) contradicting the minimality of $e(C_{1}, C_{2})$. QED.

Problem 3:

A barrel contains 2n balls, numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw.

What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, if any is repeated?

The total number of possible outcomes is $N=2n \times 2n \times 2n=8n^{3}$. To find the total number of favourable outcomes we proceed as follows:

Let a be any odd integer such that $1 \leq a \leq 2n-1$ and let us count the sequences having a as least element.

(i) There is only one sequence $(a,a,a)$ with a repeated thrice.

(ii) There are $2n-a$ sequences of the form $(a,a,b)$ with $a. For each such sequence there are three distinct permutations possible. Hence, there are in all $3(2n-a)$ sequences with a repeated twice.

iii) When $n>1$, for values of a satisfying $1 \leq a \leq (2n-3)$, sequences of the form $(a,b,c,)$ with $a are possible and the number of such sequences is $r=1+2+3+\ldots+(2n+a-1)=\frac{1}{2}(2n-a)(2n-a-1)$. For each such sequence, there are six distinct permutations possible. Hence, there are $6r=3(2n-a)(2n-a-1)$ sequences in this case.

Hence, for odd values of a between 1 and $2n-1$, the total counts of possibilities $S_{1}$, $S_{2}$, $S_{3}$ in the above cases are respectively.

$S_{1}=1+1+1+\ldots+1=n$

$S_{2}=3(1+3+5+\ldots+(2n-1))=3n^{2}$

$3(2 \times 3 + 4 \times 5 + \ldots+ (2n-2)(2n-1))=n(n-1)(4n+1)$.

Hence, the total number A of favourable outcomes is $A=S_{1}+S_{2}+S_{3}=n+3n^{2}+n(n-1)(4n+1)=4n^{3}$. Hence, the required probability is $\frac{A}{N} = \frac{4n^{3}}{8n^{3}} = \frac{1}{2}$. QED>

Cheers,

Nalin Pithwa

# Miscellaneous questions: part I : solutions: tutorial practice preRMO and RMO

The following questions were presented in an earlier blog (the questions are reproduced here) along with solutions. Please compare your attempts/partial attempts too are to be compared…that is the way to learn:

Problem 1:

The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the integers in the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8,5 or 3 neighbours depending on its position.) Show that all the sixty four squares are in fact equal.

Solution 1:

Consider the smallest value among the 64 entries on the board. Since it is the average of the surrounding numbers, all those numbers must be equal to this number as it is the smallest. This gives some more squares with the smallest value. Continue in this way till all the squares are covered.

Problem 2:

Let T be the set of all triples $(a,b,c)$ of integers such that $1 \leq a < b < c \leq 6$. For each triple $(a,b,c)$ in T, take the product abc. Add all these products corresponding to all triples in T. Prove that the sum is divisible by 7.

Solution 2:

For every triple $(a,b,c)$ in T, the triple $(7-c,7-b,7-a)$ is in T and these two are distinct as $7 \neq 2b$. Pairing off $(a,b,c)$ with $(7-c,7-b,7-a)$ for each $(a,b,c) \in T$, 7 divides $abc-(7-a)(7-b)(7-c)$.

Problem 3:

In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one is all the three. In a certain math examination 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists who are swimmers.

Solution 3:

Let S denote the set of all 25 students in the class, X the set of swimmers in S, Y the set of weight lifters in S, and Z the set of all cyclists. Since students in $X\bigcup Y \bigcup Z$ all get grades B and C, and six students get grades D or E, the number of students in $X\bigcup Y \bigcup Z \leq 25-6=19$. Now assign one point to each of the 17 cyclists, 13 swimmers and 8 weight lifters. Thus, a total of 38 points would be assigned among the students in $X \bigcup Y \bigcup Z$. Note that no student can have more than 2 points as no one is all three (swimmer, cyclist and weight lifter). Then, we should have $X \bigcup Y \bigcup Z \geq 19$ as otherwise 38 points cannot be accounted for. (For example, if there were only 18 students in $X \bigcup Y \bigcup Z$ the maximum number of points that could be assigned to them would be 36.) Therefore, $X \bigcup Y \bigcup Z=19$ and each student in $X \bigcup Y \bigcup Z$ is in exactly 2 of the sets X, Y and Z. Hence, the number of students getting grade $A=25-19-6=0$, that is, no student gets A grade. Since there are $19-8=11$ students who are not weight lifters all these 11 students must be both swimmers and cyclists. (Similarly, there are 2 who are both swimmers and weight lifters and 6 who are both cyclists and weight lifters.)

Problem 4:

Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:

A: I see three black and one white cap.

B: I see four white caps.

C: I see one black and three white caps.

D: I see four black caps.

Solution 4:

Suppose E is wearing a white cap. Then, D is lying and hence must be wearing a white cap. Since D and E both have white caps, A is lying and hence, he must be wearing white cap. If C is speaking truth, then C must be having a black cap and B must be wearing a black cap as observed by C. But then B must observe a cap on C. Hence, B must be lying. This implies that B is wearing a white cap which is a contradiction to C’s statement.

On the other hand, if C is lying, then C must be wearing a white cap. Thus, A, C, D and E are wearing white caps which makes B’s statement true. But, then B must be wearing a black cap and this makes C statement correct.

Thus, E must be wearing a black cap. This implies that B is lying and hence, must be having a white cap. But then D is lying and hence, must be having a white cap since B and D have white caps. A is not saying the truth. Hence, A must also be wearing a white cap. These together imply that C is truthful. Hence, C must be wearing a black cap. Thus, we have the following: A: white cap; B: white cap; C:black cap; D:white cap; E: black cap.

Problem 5:

Let f be a bijective function from the set $A=\{ 1,2,3,\ldots,n\}$ to itself. Show that there is a positive integer $M>1$ such that $f^{M}(i)=f(i)$ for each $i \in A$. Here $f^{M}$ denotes the composite function $f \circ f \circ \ldots \circ f$ repeated M times.

Solution 5:

Let us recall the following properties of a bijective function:

a) If $f:A \rightarrow A$ is a bijective function, then there is a unique bijective function $g: A \rightarrow A$ such that $f \circ g = g \circ f=I_{A}$ the identity function on A. The function g is called the inverse of f and is denoted by $f^{-1}$. Thus, $f \circ f^{-1}=I_{A}=f^{-1}\circ f$

b) $f \circ I_{A} = f = I_{A} \circ f$

c) If f and g are bijections from A to A, then so are $g \circ f$ and $f \circ g$.

d) If f, g, h are bijective functions from A to A and $f \circ g = f \circ h$, then $g=h$.

Apply $f^{-1}$ at left to both sides to obtain $g=h$.

Coming to the problem at hand, since A has n elements, we see that the there are only finitely many (in fact, n!) bijective functions from A to A as each bijective function f gives a permutation of $\{ 1,2,3,\ldots, n\}$ by taking $\{ f(1),f(2), \ldots, f(n)\}$. Since f is a bijective function from A to A, so is each of the function in the sequence:

$f^{2}, f^{3}, \ldots, f^{n}, \ldots$

All these cannot be distinct, since there are only finitely many bijective functions from A to A. Hence, for some two distinct positive integers m and n, $m > n$, say, we must have $f^{m}=f^{n}$

If $n=1$, we take $M=m$, to obtain the result. If $n>1$, multiply both sides by $(f^{-1})^{n-1}$ to get $f^{m-n+1}=f$. We take $M=m-n+1$ to get the relation $f^{M}=f$ with $M>1$. Note that this means $f^{M}(i)=f(i)$ for all $i \in A$. QED.

Problem 6:

Show that there exists a convex hexagon in the plane such that :

a) all its interior angles are equal

b) its sides are 1,2,3,4,5,6 in some order.

Solution 6:

Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$.

Since the sum of all angles of a hexagon is equal to $(6-2) \times 180=720 \deg$, it follows that each interior angle must be equal to $720/6=120\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have:

$\overline{AB}=(1,0)$

$\overline{BC}=(a\cos 60\deg, a\sin 60\deg)$

$\overline{CD}=(b\cos{120\deg},b\sin{120\deg})$

$\overline{DE}=(c\cos{180\deg},c\sin{180\deg})=(-c,0)$

$\overline{EF}=(d\cos{240\deg},d\sin{240\deg})$

$\overline{FA}=(e\cos{300\deg},e\sin{300\deg})$

This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively.

Since the sum of all these six vectors is $\overline{0}$, it implies that $1+\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2}+\frac{e}{2}=0$

and $(a+b-d-e)\frac{\sqrt{3}}{2}=0$

That is, $a-b-2c-d+e+2=0$….call this I

and $a+b-d-e=0$….call this II

Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have

$(a,b)=(2,5), (a,e)=(3,4), c=6$….(i)

$(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii)

$(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii)

The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\frac{1}{2}$ and interchange these two sets.

Case (i):

Here $(a-b)-(d-e)=2c-2=10$. Since $a-b=\pm 3, d-e=\pm 1$, this is not possible.

Case (ii):

Here $(a-b)-(d-e)=2c-2=2$. This is satisfied by $(a,b,d,e)=(6,3,5,4)$

Case (iii):

Here $(a-b)-(d-e)=2c-2=6$

Case (iv):

This is satisfied by $(a,b,d,e)=(6,2,3,5)$.

Hence, we have (essentially) two different solutions: $(1,6,3,2,5,4)$ and $(1,6,2,4,3,5)$. It may be verified that I and II are both satisfied by these sets of values.

Aliter: Embed the hexagon in an appropriate equilateral triangle, whose sides consist of some sides of the hexagon.

Solutions to the remaining problems from that blog will have to be tried by the student.

Cheers,

Nalin Pithwa.

# Miscellaneous questions: part II: tutorial practice for preRMO and RMO

Problem 1:

Let $a_{1}, a_{2}, \ldots, a_{10}$ be ten real numbers such that each is greater than 1 and less than 55. Prove that there are three among the given numbers which form the lengths of the sides of a triangle.

Problem 2:

In a collection of 1234 persons, any two persons are mutual friends or enemies. Each person has at most 3 enemies. Prove that it is possible to divide this collection into two parts such that each person has at most 1 enemy in his subcollection.

Problem 3:

A barrel contains 2n balls numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw. What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, if any, is repeated?

That’s all, folks !!

You will need to churn a lot…!! In other words, learn to brood now…learn to think for a long time on a single hard problem …

Regards,
Nalin Pithwa

# Miscellaneous questions: Part I: tutorial practice for preRMO and RMO

Problem 1:

The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8,5 or 3 neighbours depending on its position). Show that all sixty four entries are in fact equal.

Problem 2:

Let T be the set of all triples (a,b,c) of integers such that $1 \leq a < b < c \leq 6$. For each triple (a,b,c) in T, take the product abc. Add all these products corresponding to all triples in I. Prove that the sum is divisible by 7.

Problem 3:

In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one in all the three. In a certain mathematics examination, 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists, who are swimmers.

Problem 4:

Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:

A: I see three black and one white cap.
B: I see four white caps.
C: I see one black and three white caps.
D: I see four black caps.

Problem 5:

Let f be a bijective (one-one and onto) function from the set $A=\{ 1,2,3,\ldots,n\}$ to itself. Show that there is a positive integer $M>1$ such that $f^{M}(i)=f(i)$ for each $i \in A$. Note that $f^{M}$ denotes the composite function $f \circ f \circ f \ldots \circ f$ repeated M times.

Problem 6:

Show that there exists a convex hexagon in the plane such that:
a) all its interior angles are equal
b) its sides are 1,2,3,4,5,6 in some order.

Problem 7:

There are ten objects with total weights 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed on the pans of a balance.

Problem 8:

In each of the eight corners of a cube, write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so writtein down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero?

Problem 9:

Given the seven element set $A = \{ a,b,c,d,e,f,g\}$ find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.

Try these !!

Regards,
Nalin Pithwa

# Set Theory, Relations, Functions: Preliminaries: Part VIII

SETS and FUNCTIONS:

Reference: Introductory Real Analysis by A. N. Kolmogorov and S V Fomin, Dover books.

Operations on sets:

Let A and B be any two sets. Then, by the sum or union of A and B, denoted by $A \bigcup B$, is meant the set consisting of all elements which belong to at least one of the sets A and B. More generally, by the sum or union of an arbitrary number (finite or infinite) of sets $A_{\alpha}$ (indexed by some parameter $\alpha$), we mean the set, denoted by $\bigcup_{\alpha}A_{\alpha}$ of all elements belonging to at least one of the sets $A_{\alpha}$.

By the intersection $A \bigcap B$ of two given sets A and B, we mean the set consisting of all elements which belong to both A and B. For example, the intersection of the set of all even numbers and the set of all integers divisible by 3 is the set of all integers divisible by 6. By the intersection of an arbitrary number (finite or infinite) of sets $A_{\alpha}$, we mean the set, denoted by $\bigcap_{\alpha}A_{\alpha}$ of all elements belonging to every one of the sets $A_{\alpha}$. Two sets A and B are said to be disjoint if $A \bigcap B=\phi$, that is, if they have no elements in common. More generally, let $\mathscr{F}$ be a family of sets such that $A \bigcap B=\phi$ for every pair of sets A, B in $\mathscr{F}$. Then, the sets in $\mathscr{F}$ are said to be pairwise disjoint.

It is an immediate consequence of the above definitions that the operations $\bigcup$ and $\bigcap$ are commutative and associative, that is,

$A \bigcup B=B \bigcup A$ and $(A \bigcup B) \bigcup C= A \bigcup (B \bigcup C)$; $A \bigcap B=B \bigcap A$; and, $(A \bigcap B)\bigcap C = A \bigcap (B \bigcap C)$.

Moreover, the operations $\bigcup$ and $\bigcap$ obey the following distributive laws:

$(A \bigcup B)\bigcap C = (A \bigcap C) \bigcup (B \bigcap C)$….call this I

$(A \bigcap B)\bigcup C = (A \bigcup C) \bigcap (B \bigcup C)$….call this II.

For example, suppose that $x \in (A \bigcup B)\bigcap C$, so that x belongs to the left-hand side of I. Then, x belongs to both C and $A \bigcup B$, that is, x belongs to both C and at least one of the sets A and B. But then x belongs to at least one of the sets $A \bigcap C$ and $B \bigcap C$, that is, $x \in (A \bigcap C)\bigcup (B \bigcap C)$, so that x belongs to the right hand side of I. Conversely, suppose that $x \in (A \bigcap C)\bigcup (B \bigcap C)$. Then, x belongs to at least one of the two sets $A \bigcap C$ and $B \bigcap C$. It follows that x belongs to both C and at least one of the two sets A and B, that is, $x \in C$ and $x \in A \bigcup B$, or equivalently $x \in (A \bigcup B)\bigcap C$. This proves I, and II is proved similarly.

By the difference of two sets, $A - B$ between two sets A and B (in that order), we mean the set of all elements of A which do not belong to B. Note that it is not assumed that $A \supset B$. It is sometimes convenient (e.g., in measure theory) to consider the symmetric difference of two sets A and B, denoted by $A \delta B$ and defined as the union of the two differences $A-B$ and $B-A$: $A \delta B=(A-B)\bigcup (B-A)$.

We will often be concerned later with various sets which are all subsets of some underlying basic set R, for example, various sets of points on the real line. In this case, given a set A, the difference $R-A$ is called the complement of A, denoted by $A^{'}$.

An important role is played in set theory and its applications by the following duality principle:

$R - \bigcup_{\alpha}A_{\alpha}=\bigcap_{\alpha}(R-A_{\alpha})$…call this III

$R - \bigcap_{\alpha}A_{\alpha}=\bigcup_{\alpha}(R-A_{\alpha})$…call this IV.

In words, the complement of a union equal the intersection of the complements; and, the complement of an intersection equals the union of the complements. According to the duality principle, any theorem involving a family of subsets of a fixed set R can be converted automatically into another, “dual” theorem by replacing all subsets by their complements, all unions by intersections and all intersections by unions. To prove III, suppose that $x \in R - \bigcup_{\alpha}A_{\alpha}$….call this V.

Then, x does not belong to the union $\bigcup_{\alpha}A_{\alpha}$. …call this VI.

That is, x does not belong to any of the sets $A_{\alpha}$. It follows that x belongs to each of the complements $R - \bigcup_{\alpha}A_{\alpha}$, and hence, $x \bigcap_{\alpha}(R-A_{\alpha})$….call this VII.

Conversely, suppose that VII holds, so that x belongs to every set $R - A_{\alpha}$. Then, x does not belong to any of the sets $A_{\alpha}$, that is, x does not belong to the union VI, or equivalently V holds true. This proves 3.

Homework: Prove IV similarly.

Remark:

The designation “symmetric difference” for the set $A \Delta B$ is not too apt, since $A \Delta B$ has much in common with the sum $A \bigcup B$. In fact, in $A \bigcup B$ the two 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 $A \Delta B$ 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 $A \Delta B$ if and only if x belongs to either A or B but not both. The set $A \Delta B$ can be regarded as a kind of “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).

Functions and mappings. Images and preimages:

(Of course, we have dealt with this topic in quite detail so far in the earlier blog series; but as presented by Kolmgorov and Fomin here, the flavour is different; at any rate, it helps to revise. Revision refines the intellect.) 🙂 🙂 🙂

A rule associating a unique real number $y=f(x)$ with each element of a set real numbers X is said to define a (real) function f on X. The set X is called the domain (of definition) of f, and the set Y of all numbers $f(x)$ such that $x \in X$ is called the range of f.

More generally, let M and N be two arbitrary sets. Then a rule associating a unique element $b=f(a) \in N$ with each element $a \in M$ is again said to define a function f on M (or a function f with domain M). In this more general context, f is usually called a mapping of f M into N. By the same token, f is said to map M into N(and a into b).

If a is an element of M, the corresponding element $b=f(a)$ is called the image of a (under the mapping f). Every element of M with a given element $b \in N$ as its image is called a preimage of b. Note that in general b may have several pre-images. Moreover, N may contain elements with no pre-images at all. If b has a unique pre-image, we denote this pre-image by $f^{-1}(b)$.

If A is a subset of M, the set of all elements $f(a) \in N$ such that $a \in A$ is called the image of A, denoted by $f(A)$. The set of all elements of M whose images belong to a given set $B \subset N$ is called the preimage of B, denoted by $f^{-1}(B)$. If no element of B has a preimage, then $f^{-1}(B)=\phi$. A function f is said to map M into N if $f(M)=N$, as is always the case, and onto N if $f(M)=N$. Thus, every “onto” mapping is an “into” mapping, but converse is not true.

Suppose f maps M onto N. Then, f is said to be one-to-one if each element $b \in N$ has a unique preimage $f^{-1}(b)$. In this case, f is said to establish a one-to-one correspondence between M and N, and the mapping $f^{-1}$ associating $f^{-1}(b)$ is called the inverse of f.

Theorem I:

The preimage of the union of two sets is the union of the preimages of the sets $f^{-1}(A \bigcup B)=f^{-1}(A) \bigcup f^{-1}(B)$.

Proof of Theorem I:

If $x \in f^{-1}(A \bigcup B)$, then $f(x) \in A \bigcup B$ so that $f(x)$ belongs to at least one of the sets A and B. But, then x belongs to at least one of the sets $f^{-1}(A)$ and $f^{-1}(B)$, that is, $x \in f^{-1}(A) \bigcup f^{-1}(B)$.

Conversely, if $x \in f^{-1}(A) \bigcup f^{-1}(B)$, then x belongs to at least one of the sets $f^{-1}(A)$ and $f^{-1}(B)$. Therefore, $f(x)$ belongs to at least one of the sets A and B, that is, $f(x) \in A \bigcup B$. But, then $x \in f^{-1}(A \bigcup B)$.

QED.

Theorem 2:

The preimage of the intersection of two sets is the intersection of the preimages of the sets:

$f^{-1}(A \bigcap B)=f^{-1}(A) \bigcap f^{-1}(B)$.

Proof of Theorem 2:

If $x \in f^{-1}(A \bigcap B)$, then $f(x) \in A \bigcap B$, so that $f(x) \in A$ and (meaning, simultaneously) $f(x) \in B$. But, then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$, that is, $x \in f^{-1}(A) \bigcap f^{-1}(B)$.

Conversely, if $x \in f^{-1}(A) \bigcap f^{-1}(B)$, then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$. Therefore, $f(x) \in A$ and $f(x) \in B$, that is, $f(x) \in A \bigcap B$. But, then $x \in f^{-1}(A \bigcap B)$.

QED.

Theorem 3:

The image of the union of two sets equals the union of the images of the sets $f(A \bigcup B) = f(A) \bigcup f(B)$.

Proof of theorem 3:

If $y \in f(A \bigcup B)$, then $y=f(x)$ where x belongs to at least one of the sets A and B. Therefore, $y=f(x)$ belongs to at least one of the sets $f(A)$ and $f(B)$. That is, $x \in f(A) \bigcup f(B)$.

Conversely, if $y \in f(A) \bigcup f(B)$, then $y=f(x)$. where x belongs to at least one of the sets A and B, that is, $x \in A \bigcup B$ and hence, $y=f(x) \in f(A \bigcup B)$.

QED.

Remark I:

Surprisingly enough, the image of the intersection of two sets is not so “well-behaved”. The image of the intersection of two sets does not necessarily equal the intersection of the images of the sets. For example, suppose the mapping f projects the xy-plane onto the x-axis, carrying the point $(x,y)$ into the $(x,0)$. Then, the segments $0 \leq x \leq 1$ with $y=0$, and $0 \leq x \leq 1$ with $y=1$ do not intersect, although their images coincide.

Remark 2:

In the light of above remark: consider the following: If a function is not specified on elements, it is important in general to check that f is well-defined. That is, unambiguously defined. For example, if the set A is the union of two subsets $A_{1}$ and $A_{2}$, and we are considering a function $f: A \longrightarrow B$, then one can try to specify a function from set A to the set B $\{ 0,1\}$ by declaring that f is to map everything in $A_{1}$ to 0 and is to map everything in $A_{2}$ to 1. This unambiguously defines f unless $A_{1}$ and $A_{2}$ have elements in common in which case it is not clear whether these elements should map to 0 or to 1. Checking that this f is well-defined therefore amounts to checking that $A_{1}$ and $A_{2}$ have no intersection.

Remark 3:

Theorems 1-3 continue to hold for unions and intersections of an arbitrary number (finite or infinite) of sets $A_{\alpha}$:

$f^{-1}(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f^{-1}(A_{\alpha})$

$f^{-1}(\bigcap_{\alpha}A_{\alpha})=\bigcap_{\alpha}f^{-1}(A_{\alpha})$

$f(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f(A_{\alpha})$

Decomposition of a set into classes. Equivalence relations.

Decompositions of a given set into pairwise disjoint subsets play an important role in a great variety of problems. For example, the plane (regarded as a point set) can be decomposed into lines parallel to the x-axis, three dimensional space can be decomposed into concentric spheres, the inhabitants of a given city can be decomposed into different age groups, and so on. Any such representation of a given set M as the union of a family of pairwise disjoint subsets of M is called a decomposition or partition of M into classes.

A decomposition is usually made on the basis of a certain criterion, allowing us to assign the elements of M to one class or another. For example, the set of all triangles in the plane can be decomposed into classes of congruent triangles, or classes of triangles of equal area, the set of all functions of x can be decomposed into classes of functions all taking the same value at a given point x, and so on. Despite the great variety of such criteria, they are not completely arbitrary. For example, it is obviously impossible to partition all real numbers into classes by assigning the number b to the same class as number a if and only if $b>a$. In fact, if $b>a$, b must be assigned to same class as a but then a cannot be assigned to same class as b, since $a. Moreover, since a is not greater than itself, a cannot be assigned to the class containing itself!! As another example, it is impossible to partition the points of the plane into classes by assigning two points to the same class if and only if the distance between them is less than 1. In fact, if the distance between a and b is less than 1, and if the distance between b and c is less than 1, it does not follow that distance between a and c is less than 1. (Hint: Think of triangle inequality). Thus, by assigning a to the same class as b and b to the same class as c, we may well find that two points fall in the same class even though the distance between them is greater than 1!

These examples suggest conditions that which must be satisfied by any criterion it it is to be used as the basis for partitioning a given set into classes. Let M be a set, and let certain ordered pairs (a,b) of elements of M be called “labelled.” If $(a,b)$ is a labelled pair, we say that a is related to b by the binary relation R and we write it symbolically as aRb. For example, if a and b are real numbers, $aRb$ might mean $a, while if a and b are triangles, $aRb$ might mean that a and b have the same area. A relation between elements of M is called a relation on M, if there exists at least one labelled pair $(a,b)$ for every $a \in M$. A relation R on M is called an equivalence relation (on M) if it satisfies the following three conditions:

• Reflexivity: $aRa$ for every $a \in M$.
• Symmetry: If $aRb$, then $bRa$.
• Transititivity: If $aRb$ and $bRc$, then $aRc$.

🙂 🙂 🙂

Nalin Pithwa, more later