# Meaning of function or mapping or transformation in brief words

DEFINITION:

Given two sets X and Y, a transformation (also called a function or mapping) $f: X \rightarrow Y$ of X into Y is a triple $(X,Y,G)$ where G itself is a collection of ordered pairs $(x,y)$, the first element of each pair being an element of X, and the second an element of Y, with the condition that each element of X appears as the first element of exactly one pair of G.

If each element of Y appears as the second element of some pair in G, then the transformation is said to be onto.

If each element of Y which appears at all, appears as the second element of exactly one pair in G, then f is said to be one-to-one. Note that a transformation can be onto without being one-to-one and conversely.

As an aid in understanding the above definition, consider the equation $y=x^{2}$ where x is a real number. We may take X to be the set of all real numbers and then the collection G is the set of pairs of $(x,x^{2})$. Taking Y to be just the set of nonnegative reals will cause f to be onto. But if Y is all real numbers, or all reals greater than -7, or any other set containing the nonnegative reals as a proper subset, the transformation is not onto. With each new choice of Y, we change the triple and hence the transformation.

Continuing with the same example, we could assume that X is the set of nonnegative reals also. Then the transformation is one-to-one, as is easily seen. Depending upon the the choice of Y, the transformation may or may not be onto, of course. Thus, we see that we have stated explicitly the conditions usually left implicit in defining a function in analysis. The reader will find that the seemingly pedantic distinctions made here are really quite necessary.

If $f: X \rightarrow Y$ is a transformation of X into Y, and x is an element of the set X, then we let $f(x)$ denote the second element of the pair in G whose first element is x. That is, $f(x)$ is the “functional value” in Y of the point x. Similarly, if Z is a subset of X, then $f(Z)$ denotes that subset of Y composed of all points $f(z)$, where z is a point in Z. If y is a point of Y, then by $f^{-1}(y)$ is meant the set of all points in X for which $f(x)=y$; and if W is a subset of Y, then $f^{-1}(W)$ is the set theoretic union of the sets $f^{-1}(w)$, where w is in W. Note that $f^{-1}$ can be used as a symbol to denote the triple $(X,Y,G^{'})$ wjere $G^{'}$ consists of all pairs $(y,x)$ that are reversals of pairs in G. But $f^{-1}$ is a transformation only if f is both one-to-one and onto. If A is a subset of X and if $f: X \rightarrow Y$, then f may be restricted to A to yield a transformation denoted by $f|A: A \rightarrow Y$, and called the restriction of f to A.

Cheers,

Nalin Pithwa

# Notes I: Sets and Functions: Solutions

Problem 1:

If $\{ A_{i}\}$ and $\{ B_{j}\}$ are two classes of sets such that $\{ A_{i}\} \subseteq \{ B_{j}\}$, prove that $\bigcup_{i}A_{i} \subseteq \bigcup_{i}B_{i}$ and $\bigcap_{j}B_{j} \subseteq \bigcap_{i}A_{i}$.

The first part is clear as the hypothesis says the set of sets $\{ A_{i}\}$ is a subset of the set of sets $\{ B_{j}\}$. So, quite clearly $\bigcup_{i}A_{i} \subseteq \bigcup_{j}A_{j}$. For the second part, we can use the generalized De Morgan’s laws. (Note: if you try to derive it straight, you will prove exactly what is opposite of the required result :-)). For the second part note that $X \subseteq Y \Longleftrightarrow Y^{'} \bigcap X^{'}$. Using the result in the first part, that is, consider $\bigcup_{i}A_{i} \subseteq \bigcup_{j}A_{j}$ and now take the complement of both sides so we get: $\bigcup_{i}A_{i} \subseteq \bigcup_{j}B_{j} \Longleftrightarrow (\bigcup_{j}B_{j})^{'} \subseteq (\bigcup_{i}A_{i})^{'}$, that is, $\bigcap_{j}B_{j} \subseteq \bigcap_{i}A_{i}$. QED.

Problem 2:

The difference between two sets A and B, denoted by $A-B$, is the set of all elements in A and not in B, thus $A-B = A \bigcap B^{'}$. Show the following:

(a)$A-B = A - (A \bigcap B) = (A \bigcup B) -B$.

Answer (a): $A-B$ is the set of elements $x \in A$, but not in B. Hence, $A-B = A-(A \bigcap B)$, simply by definition. Again, if we consider elements in A or B or both, and then take away elements of B, we are just left with elements of A only, but not of B. That is, $A-B = A-(A \bigcap B) = (A \bigcup B) - B$. QED. ps: A venn diagram may help visualize v well.

(b) TPT: $(A-B) - C = A - (B \bigcup C)$ (PS: a Venn diagram can help visualize and perhaps, guide the writing of the proof also).

Let $x \in A-B$, but $x \notin C$. Then, $x \in A$, but $x \notin B$, or $x \notin C$, or not in both B and C. That is, $x \in A-(B \bigcup C)$. By reversing the argument, we get the other subset relationship. Hence, $(A-B)-C = A- (B \bigcap C)$. QED.

(c) TPT: $A - (B-C) = (A-B) \bigcup (A \bigcap C)$.

Answer (c): Let $x \in A-(B-C)$. That is, $x \in A$ and $x \notin (B-C)$. That is, $x \in A, x \notin B, x \in C$. That is, $x \in A, x \notin B$, or x could be both in A and C. That is, $x \in (A-B) \bigcup (A \bigcap C)$. Simply reversing the arguments, we get the reverse subset relation. Hence, the two sets are equal. Hence, $A-(B-C) = (A-B) \bigcup (A \bigcap C)$. QED.

(d) TPT: $(A \bigcup B) - C = (A-C) \bigcup (B-C)$.

answer (d): let $x \in A\bigcup B$, but $x \notin C$. Then, $x \in A$, or $x \in B$, or x is an element of both A and B, but x is not an element of C. Which clearly means, x is an element of A but not C, OR x is an element of B but not C. That is, $x \in (A-C) \bigcup (B-C)$ (upon reversing the arguments, we get the other subset relations. Hence, the two sets are equal). QED.

problem (e): TPT: $A-(B \bigcup C) = (A-B) \bigcap (A-C)$ (once again, note that for set theoretic relations with up to three sets, venn diagrams are helpful to visualize and guide the proofs..)

answer (c): let $x \in A$, but not in $B \bigcup C$. That is, x is an element of A, but not of B, not of C, or not both of B and C. That is, x is an element of A not of B, and x is an element of A not of C. That is, $x \in (A-B) \bigcap (A-C)$. Reversing the arguments, we get the other subset relationship. Hence, the two sets are equal. Hence, $A - (B \bigcup C) = (A-B) \bigcap (B-C)$. QED.

Problem 3:

The symmetric difference of two sets A and B, denoted by $A \triangle B$, is defined by $A \triangle B = (A-B) \bigcup (B-A)$; it is the union of the difference of two sets in opposite orders. Prove the following:

3(a) Symmetric difference is associative: $A \triangle (B \triangle C) = (A \triangle B) \triangle C$.

3(b): TPT: $A \triangle \phi = A$; and $A \triangle A = \phi$. (this is some sort of ‘existence of additive inverse of ‘in a symmetric relation’).

Answer 3b: both results are obvious from definition.

3(c): TPT: Symmetric difference is commutative operation: $A \triangle B = B\triangle A$.

answer 3c: By definition, LHS is $(A-B)\bigcup (B-A) = (B-A) \bigcup (A-B)$, which is RHS. Hence, done.

3(d): TPT: Some sort of distributive law: $A \bigcap (B \triangle C) = (A \bigcap B) \triangle (A \bigcap C)$.

answer 3d: Let $x \in A$, and also simultaneously $x \in (B \triangle C)$. That is, $x \in A$, and $x \notin (B \bigcap C)$. That is, $x \in A, x \notin (A \bigcap B)\bigcap (B \bigcap C)$. That is, $x \in RHS$. Reversing the arguments, we get the other subset relation. Hence, the two sets are equal. Hence, we can say “intersection distributes over symmetric difference.” QED.

Problem 4:

A ring of sets is a non-empty class A of sets such that if A and B are in A then $A \triangle B$ is in A and $A \bigcap B$ is also in A. Show that A must contain the empty set, $A \bigcup B$ and $A-B$. Also, show that if a non empty class of sets contains the union and difference of any pair of the sets, then it is a ring of sets. Also, prove that a Boolean algebra of sets is a ring of sets.

Answer 4(i): TPT: A must also contain the empty set, the set $A \bigcup B$ and the set $A-B$. As $A \triangle B$ is in A, so also $A \triangle A = \phi \in A$; now, it is already given that the ring of sets is non-empty, so it contains a non empty universal set also, call it U. We know from problem 2 in this blog that $A - B = A \bigcap B^{'}$. As U exists, so does $B^{'}$ by definition and so $A-B$ is in A. Also, as $A-B$ exists, $A-B^{'}$ exists in A, but $A-B^{'}=A\bigcap B$ also exists in A. QED.

Answer 4(ii): TPT: If a non empty class of sets contains the union and difference of any pair of the sets, then it is a ring of sets.

Let A be any non-empty class of sets. Clearly, it therefore has a non empty universal set. As $\phi \in U$, so also $\phi$ is contained in A. Hence, the complement of a set A is also in A. Now, $A \bigcup B \in A$, so $U - (A \bigcup B)$ is in A, that is, by De Morgan law, $B^{'} \bigcap A^{'}$ is in A. That is, again, $A \bigcap B$ is in A. Now, since $A-B$ is in A, hence, $B-A$ is in A, and since $A \bigcup B$ is in A, then $(A-B) \bigcup (B-A)$ is in A also. Hence, A is a ring of sets.

4(iii) TPT: Show that a boolean algebra of sets is a ring of sets.

Answer 4(iii); Firstly, as a boolean algebra of sets is non-empty, and it also contains the empty set and the Universal set. As A is in A implies that $A^{'}$ is in A, so $A \bigcup B^{'}$ is also in A (by definition of Boolean algebra), but this is precisely that $A-B$ is in A. Now, as $A \triangle B = (A-B) \bigcup (B-A)$ clearly is also in A. Hence, a Boolean algebra of sets is a ring of sets.

Problem 5:

Show that the class of all finite subsets (including the empty set) of an infinite set is a ring of sets but is not a Boolean algebra.

In problem 4 above, we showed that if a non empty class of sets contains the union and difference of any pair of the sets, then it is a ring of sets. The given class is a subclass of all an infinite set such that it contains finite subsets only; that is, again, we know that finite unions of finite subsets and finite intersections of finite subsets also lie in that subclass, but the complements might be infinite sets so that it need not always be a Boolean algebra of sets.

Problem 6:

Show that the class of all finite unions of closed-open intervals on the real line is a ring of sets but is not a Boolean algebra of sets.

Same reasoning as above tells us that it is indeed a ring of sets; but consider for example the closed-open interval $[a,b)$, the complement of this is not in that class as it is an infinite interval so that it is not a Boolean algebra of sets.

Problem 7:

Assuming that the Universal set U is non empty, show that Boolean algebras of sets can be described as rings of sets which contain U.

As $U \neq \phi$, by definition of Boolean algebras, $U^{'}$ is in this class of sets, so that the empty set is also a part of this class of sets. Again, by definition of Boolean algebras, the union of two sets of this class is also in this set so also the intersection of any two sets of this class is also in this class. So, for any two sets A and B, $A-B$ is also in this set (as complement of B exists), and so also, $A \triangle B = (A-B) \bigcup (B-A)$ is also in this set. In other words, this is also a ring of sets.

Regards,

Nalin Pithwa

# Set theory: more basic problems to solve and clear and apply

I am producing the list of questions first so that the motivated reader can first read and try them …and can compare with my answers by scrolling down only much later; here we go:

1. If $\{ A_{i}\}$ and $\{ B_{j}\}$ are two classes of sets such that $\{ A_{i}\} \subseteq \{ B_{j}\}$, then prove that $\bigcup_{i}A_{i} \subseteq \{ B_{j}\}$ and $\bigcap_{j} B_{j} \subseteq \bigcap_{i}A_{i}$.
2. The difference between two sets A and B, denoted by $A-B$, is the set of elements in A and not in B; thus, $A - B = A \bigcap B^{'}$. Prove the following simple properties: (a) $A-B = A-(A \bigcap B) = (A \bigcup B)-B$; (b) $(A-B)-C = A-(B \bigcup C)$; (c) $A - (B-C) = (A-B) \bigcup (A \bigcap C)$; (d) $(A \bigcup B) - C = (A-C) \bigcup (B-C)$; (e) $A - (B \bigcup C) = (A-B) \bigcap (A-C)$
3. The symmetric difference of two sets A and B, denoted by $A \triangle B$, is defined by $A \triangle B = (A-B) \bigcup (B-A)$; it is thus the union of their differences in opposite orders. Prove the following : (a) Symmetric difference is associative : $A \triangle (B \triangle C) = (A \triangle B) \triangle C$ (b) $A \triangle \phi=A$ (c) $A \triangle A = \phi$ (c) Symmetric difference is commutative: $A \triangle B = B \triangle A$ (d) Some sort of distributive rule also holds: $A \bigcap (B \triangle C) = (A \bigcap B) \triangle (A \bigcap C)$
4. A ring of sets is a non-empty class A of sets such that if A and B are in A, then $A \triangle B$ and $A \bigcap B$ are also in A. Show that A must also contain the empty set, $A \bigcup B$, and $A-B$. Show that if a non-empty class of sets contains the union and difference of sets any pair of its sets, then it is a ring of sets. Prove that a Boolean algebra of sets is a ring of sets.
5. Show that the class of all finite subsets (including the empty set) of an infinite set is a ring of sets but is not a Boolean algebra of sets.
6. Show that the class of all finite unions of closed-open intervals on the real line is a ring of sets but is not a Boolean algebra of sets.
7. Assuming that the universal set U is non-empty, show that Boolean algebras of sets can be described as a ring of sets which contain U.

I will put up my solutions as soon as I can.

Regards,

Nalin Pithwa.

# Symmetric difference is associative

We want to show that : $A \triangle (B \triangle C) = (A \triangle B) \triangle C$

Part I: TPT: $LHS \subset RHS$

Proof of Part I: let $x \in LHS$

Then, $x \in A \triangle (B \triangle C)$. By definition of symmetric difference,

$x \in \{ A - (B \triangle C)\} \bigcup \{(B \triangle C) -A \}$

Hence, $x \in A$, $x \notin (B \triangle C)$, OR $x \in B \triangle C$, $x \in A$

That is, $x \notin A \bigcap (B \triangle C)$.

Hence, $x \notin A$, and $x \notin B \triangle C$

Hence, $x \notin A$ and $x \in B \bigcap C$.

Hence, $x \in (B \bigcap C)-A$.

Hence, $x \in B$, $x \in C$, but $x \notin A$.

Therefore, $x \in B$, but $x \notin A \bigcap C$. —- Call this I.

Consider $y \in (A \triangle B) \triangle C$.

Therefore, $y \notin (A \triangle B) \bigcap C$.

Therefore, $y \notin C$, and $y \notin A \triangle B$

Therefore, $y \notin C$, and $y \in A \bigcap B$.

Therefore, $y \in (A \bigcap B)-C$.

Hence, $y \in A$, $y \in B$, but $y \notin C$.

That is, $y \in B$, $y \notin A \bigcap C$. Call this II.

From I and II, $LHS \subset RHS$.

Part II: TPT: $RHS \subset LHS$.

Quite simply, reversing the above steps we prove part II.

QED.

Cheers,

Nalin Pithwa

# 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