Set Theory, Relations, Functions: Preliminaries: Part IX: (tutorial problems)

Reference: Introductory Real Analysis, Kolmogorov and Fomin, Dover Publications.

Problem 1:

Prove that if A \bigcup B=A and A \bigcap B=A, then A=B.

Problem 2:

Show that in general (A-B)\bigcup B \neq A.

Problem 3:

Let A = \{ 2,4, \ldots, 2n, \ldots\} and B= \{ 3,6,\ldots, 3n, \ldots\}. Find A \bigcap B and A - B.

Problem 4:

Prove that (a) (A-B)\bigcap (C)=(A \bigcap C)-(B \bigcap C)

Prove that (b) A \Delta B = (A \bigcup B)-(A \bigcap B)

Problem 5:

Prove that \bigcup_{a}A_{\alpha}-\bigcup_{a}B_{\alpha}=\bigcup_{\alpha}(A_{\alpha}-B_{\alpha})

Problem 6:

Let A_{n} be the set of all positive integers divisible by n. Find the sets (i) \bigcup_{n=2}^{\infty}A_{n} (ii) \bigcap_{n=2}^{\infty}A_{n}.

Problem 7:

Find (i) \bigcup_{n=1}^{\infty}[n+\frac{1}{n}, n - \frac{1}{n}] (ii) \bigcap_{n=1}^{\infty}(a-\frac{1}{n},b+\frac{1}{n})

Problem 8:

Let A_{\alpha} be the set of points lying on the curve y=\frac{1}{x^{\alpha}} where (0<x<\infty). What is \bigcap_{\alpha \geq 1}A_{\alpha}?

Problem 9:

Let y=f(x) = <x> for all real x, where <x> is the fractional part of x. Prove that every closed interval of length 1 has the same image under f. What is the image? Is f one-to-one? What is the pre-image of the interval \frac{1}{4} \leq y \leq \frac{3}{4}? Partition the real line into classes of points with the same image.

Problem 10:

Given a set M, let \mathscr{R} be the set of all ordered pairs on the form (a,a) with a \in M, and let aRb if and only if (a,b) \in \mathscr{R}. Interpret the relation R.

Problem 11:

Give an example of a binary relation which is:

  • Reflexive and symmetric, but not transitive.
  • Reflexive, but neither symmetric nor transitive.
  • Symmetric, but neither reflexive nor transitive.
  • Transitive, but neither reflexive nor symmetric.

We will continue later, 🙂 🙂 🙂

PS: The above problem set, in my opinion, will be very useful to candidates appearing for the Chennai Mathematical Institute Entrance Exam also.

Nalin Pithwa

 

 

A quadratic and trigonometry combo question: RMO and IITJEE maths coaching

Question:

Given that \tan {A} and \tan {B} are the roots of the quadratic equation x^{2}+px+q=0, find the value of

\sin^{2}{(A+B)}+ p \sin{(A+B)}\cos{(A+B)} + q\cos^{2}{(A+B)}

Solution:

Let \alpha=\tan{A} and \beta=\tan{B} be the two roots of the given quadratic equation: x^{2}+px+q=0

By Viete’s relations between roots and coefficients:

\alpha+\beta=\tan{A}+\tan{B}=-p and \alpha \beta = \tan{A}\tan{B}=q but we also know that \tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\frac{-p}{1-q}=\frac{p}{q-1}

Now, let us call E=\sin^{2}{(A+B)}+p\sin{(A+B)\cos{(A+B)}}+\cos^{2}{(A+B)} which in turn is same as

\cos^{2}{(A+B)}(\tan^{2}{(A+B)}+p\tan{(A+B)}+q)

We have already determined \tan{(A+B)} in terms of p and q above.

Now, again note that \sin^{2}{\theta}+\cos^{2}{\theta}=1 which in turn gives us that \tan^{2}{\theta}+1=\sec^{2}{\theta} so we get:

\sec^{2}{(A+B)}=1+\tan^{2}{(A+B)}=1+\frac{p^{2}}{(q-1)^{2}}=\frac{p^{2}+(q-1)^{2}}{(q-1)^{2}} so that

\cos^{2}{(A+B)}=\frac{1}{\sec^{2}{(A+B)}}=\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}}

Hence, the given expression E becomes:

(\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}})(\frac{p^{2}}{(q-1)^{2}}+\frac{p^{2}}{q-1}+q), which is the desired solution.

🙂 🙂 🙂

Nalin Pithwa.

|log{xx_{1}}| + |log{xx_{2}}| + …+ |log{xx_{n}}| + |log{x/x_{1}}| + |log{x/x_{2}}| + …+|log{x/x_{n}}|= |log{x_{1}}+ log{x_{2}}+ ….+log{x_{n}}|

Solve the following :

Find all positive real numbers x, x_{1}, x_{2}, \ldots, x_{n} such that

|\log{xx_{1}}|+|\log{xx_{2}}| + \ldots + |\log{xx_{n}}| + |\log{\frac{x}{x_{1}}}| + |\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}|= |\log{x_{1}}+ \log{x_{2}}+\log{x_{3}}+ \ldots + \log{x_{n}}|
…let us say this is given equality A

Solution:

Use the following inequality: |a-b| \leq |a| + |b| with equality iff ab \leq 0

So, we observe that : |\log{xx_{1}}|+|\log{\frac{x}{x_{1}}}| \geq |\log{xx_{1}}-\log{\frac{x}{x_{1}}}| = |\log{x_{1}^{2}}|=2 |\log{x_{1}}|,

Hence, LHS of the given equality is greater than or equal to:

2(|\log{x_{1}}|+|\log{x_{2}}|+|\log{x_{3}}|+ \ldots + |\log{x_{n}}|)

Now, let us consider the RHS of the given equality A:

we have to use the following standard result:

|\pm a_{} \pm a_{2} \pm a_{3} \ldots \pm a_{n}| \leq |a_{1}|+|a_{2}| + \ldots + |a_{n}|

So, applying the above to RHS of A:

|\log{x_{1}}+\log{x_{2}}+\ldots + \log{x_{n}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|.

But, RHS is equal to LHS as given in A:

That is, |\log{xx_{1}}|+|\log{xx_{2}}|+ \ldots + |\log{xx_{n}}| +|\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}|+ \ldots + |\log{\frac{x}{x_{n}}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|

Now, just a few steps before we proved that LHS is also greater than or equal to : That is,

|\log{xx_{1}}|+|\log{xx_{2}}|+\ldots + |\log{xx_{n}}|+ |\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}| \geq 2(|\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|)

The above two inequalities are like the following: x \leq y and x \geq 2y; so what is the conclusion? The first inequality means x2y or x=2y; clearly it means the only valid solution is x=2y.

Using the above brief result, we have here:

|\log{x_{1}}|+|\log{x_{2}}|+ \ldots +|\log{x_{n}}| =2(|\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|)

Hence, we get |\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|=0, which in turn means that (by applying the definition of absolute value):

|\log{x_{1}}|=|\log{x_{2}}|= \ldots =|\log{x_{n}}|, which implies that x_{1}=x_{2}= \ldots  x_{n}=1.

Substituting these values in the given logarithmic absolute value equation, we get:

n \times |\log{x}|+ n \times |\log{x}|=0, that is 2n \times |\log{x}|=0, and as n \neq 0, this implies that |\log{x}|=0 which in turn means x=1 also.

You and your research or you and your studies for competitive math exams

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

Axiomatic Method : A little explanation

I) Take an English-into-English dictionary (any other language will also do). Start with any word and note down any word occurring in its definition, as given in the dictionary. Take this new word and note down any word appearing in it until a vicious circle results. Prove that a vicious circle is unavoidable no matter which word one starts with , (Caution: the vicious circle may not always involve the original word).

For example, in geometry the word “point” is undefined. For example, in set theory, when we write or say : a \in A ; the element “a” ‘belongs to’ “set A” —- the word “belong to” is not defined.

So, in all branches of math or physics especially, there are such “atomic” or “undefined” terms that one starts with.

After such terms come the “axioms” — statements which are assumed to be true; that is, statements whose proof is not sought.

The following are the axioms based on which equations are solved in algebra:

  1. If to equals we add equals, we get equals.
  2. If from equals we take equals, the remainders are equal.
  3. If equals are multiplied by equals, the products are equal.
  4. If equals are divided by equals (not zero), the quotients are equal.

More later,

Nalin Pithwa.

Check your mathematical induction concepts

Discuss the following “proof” of the (false) theorem:

If n is any positive integer and S is a set containing exactly n real numbers, then all the numbers in S are equal:

PROOF BY INDUCTION:

Step 1:

If n=1, the result is evident.

Step 2: By the induction hypothesis the result is true when n=k; we must prove that it is correct when n=k+1. Let S be any set containing exactly k+1 real numbers and denote these real numbers by a_{1}, a_{2}, a_{3}, \ldots, a_{k}, a_{k+1}. If we omit a_{k+1} from this list, we obtain exactly k numbers a_{1}, a_{2}, \ldots, a_{k}; by induction hypothesis these numbers are all equal:

a_{1}=a_{2}= \ldots = a_{k}.

If we omit a_{1} from the list of numbers in S, we again obtain exactly k numbers a_{2}, \ldots, a_{k}, a_{k+1}; by the induction hypothesis these numbers are all equal:

a_{2}=a_{3}=\ldots = a_{k}=a_{k+1}.

It follows easily that all k+1 numbers in S are equal.

*************************************************************************************

Comments, observations are welcome 🙂

Regards,

Nalin Pithwa

Tutorial on Basic Set Theory and Functions: for PRMO, RMO and IITJEE Mains maths

I) Prove that every function can be represented as a sum of an even function and an odd function.

II)Let A, B, C be subsets of a set S. Prove the following statements and illustrate them with Venn Diagrams:

2a) The famous DeMorgan’s laws in their basic forms: A^{'} \bigcup B^{'} = (A \bigcap B)^{'} and A^{'} \bigcap B^{'} = (A \bigcup B)^{'}. Assume that both sets A and B are subsets of Set S. In words, the first is: union of complements is the complement of intersection; the second is: intersection of two complements is the complement of the union of the two sets.

Sample Solution:

Let us say that we need to prove: A^{'}\bigcap B^{'}=(A \bigcup B)^{'}.

Proof: It must be shown that the two sets have the same elements; in other words, that each element of the set on LHS is an element of the set on RHS and vice-versa.

If x \in A^{'} \bigcap B^{'}, then x \in A^{'} and x \in B^{'}. This means that x \in S, and x \notin A and x \notin B. Since x \notin A and x \notin B, hence x \notin A \bigcup B. Hence, x \in (A \bigcup B)^{'}.

Conversely, if x \in (A \bigcup B)^{'}, then x \in S  and x \notin A \bigcup B. Therefore, x \notin A and x \notin B. Thus, x \in A^{'} and x \in Y^{'}, so that x \in A^{'} \bigcap B^{'}. QED.

2b) A \bigcap (B \bigcup C) = (A \bigcap B)\bigcup (A \bigcap C).

2c) A \bigcup (B \bigcap C) = (A \bigcup B) \bigcap (A \bigcup C)

III) Prove that if I and S are sets and if for each i \in I, we have X_{i} \subset S, then (\bigcap_{i \in I} X_{i})^{'} = \bigcup_{i \in I}(X_{i})^{'}.

Sample Solution: 

It must be shown that each element of the set on the LHS is an element of the set on RHS, and vice-versa.

If x \in (\bigcap_{i \in I} X_{i})^{'}, then x \in S and x \notin \bigcap_{i \in I} X_{i}. Therefore, x \notin X_{i}, for at least one j \in I. Thus, x \in (X_{i})^{'}, so that x \in \bigcup_{i \in I}(X_{i})^{'}.

Conversely, if x \in \bigcup_{i \in I}(X_{i})^{i}, then for some j \in I, we have x \in (X_{i})^{'}. Thus, x \in S and x \notin X_{i}. Since x \notin X_{i}, we have x \notin \bigcap_{i \in I}X_{i}. Therefore, x \in \bigcap_{i \in I}(X_{i})^{'}. QED.

IV) If A, B and C are sets, show that :

4i) (A-B)\bigcap C = (A \bigcap C)-B

4ii) (A \bigcup B) - (A \bigcap B)=(A-B) \bigcup (B-A)

4iii) A-(B-C)=(A-B)\bigcup (A \bigcap B \bigcap C)

4iv) (A-B) \times C = (A \times C) - (B \times C)

V) Let I be a nonempty set and for each i \in I let X_{i} be a set. Prove that

5a) for any set B, we have : B \bigcap \bigcup_{i \in I} X_{i} = \bigcup_{i \in I}(B \bigcap X_{i})

5b) if each X_{i} is a subset of a given set S, then (\bigcup_{i \in I}X_{i})^{'}=\bigcap_{i \in I}(X_{i})^{'}

VI) Prove that if f: X \rightarrow Y, g: Y \rightarrow Z, and Z \rightarrow W are functions, then : h \circ (g \circ f) = (h \circ g) \circ f

VII) Let f: X \rightarrow Y be a function, let A and B be subsets of X, and let C and D be subsets of Y. Prove that:

7i) f(A \bigcup B) = f(A) \bigcup f(B); in words, image of union of two sets is the union of two images;

7ii) f(A \bigcap B) \subset f(A) \bigcap f(B); in words, image of intersection of two sets is a subset of the intersection of the two images;

7iii) f^{-1}(C \bigcup D) = f^{-1}(C) \bigcup f^{-1}(D); in words, the inverse image of the union of two sets is the union of the images of the two sets.

7iv) f^{-1}(C \bigcap D)=f^{-1}(C) \bigcap f^{-1}(D); in words, the inverse image of intersection of two sets is intersection of the two inverse images.

7v) f^{-1}(f(A)) \supset A; in words, the inverse of the image of a set contains the set itself.

7vi) f(f^{-1}(C)) \subset C; in words, the image of an inverse image of a set is a subset of that set.

For questions 8 and 9, we can assume that the function f is f: X \rightarrow Y and a set A lies in domain X and a set C lies in co-domain Y.

8) Prove that a function f is 1-1 if and only if f^{-1}(f(A))=A for all A \subset X; in words, a function sends different inputs to different outputs iff a set in its domain is the same as the inverse of the image of that set itself.

9) Prove that a function f is onto if and only if f(f^{-1}(C))=C for all C \subset Y; in words, the image of a domain is equal to whole co-domain (which is same as range) iff a set in its domain is the same as the image of the inverse image of that set.

Cheers,

Nalin Pithwa

Check your talent: are you ready for math or mathematical sciences or engineering

At the outset, let me put a little sweetener also: All I want to do is draw attention to the importance of symbolic manipulation. If you can solve this tutorial easily or with only a little bit of help, I would strongly feel that you can make a good career in math or applied math or mathematical sciences or engineering.

On the other hand, this tutorial can be useful as a “miscellaneous or logical type of problems” for the ensuing RMO 2019.

I) Let S be a set having an operation * which assigns an element a*b of S for any a,b \in S. Let us assume that the following two rules hold:

i) If a, b are any objects in S, then a*b=a

ii) If a, b are any objects in S, then a*b=b*a

Show that S can have at most one object.

II) Let S be the set of all integers. For a, b in S define * by a*b=a-b. Verify the following:

a) a*b \neq b*a unless a=b.

b) (a*b)*c \neq a*{b*c} in general. Under what conditions on a, b, c is a*(b*c)=(a*b)*c?

c) The integer 0 has the property that a*0=a for every a in S.

d) For a in S, a*a=0

III) Let S consist of two objects \square and \triangle. We define the operation * on S by subjecting \square and \triangle to the following condittions:

i) \square * \triangle=\triangle = \triangle * \square

ii) \square * \square = \square

iii) \triangle * \triangle = \square

Verify by explicit calculation that if a, b, c are any elements of S (that is, a, b and c can be any of \square or \triangle) then:

i) a*b \in S

ii) (a*b)*c = a*(b*c)

iii) a*b=b*a

iv) There is a particular a in S such that a*b=b*a=b for all b in S

v) Given b \in S, then b*b=a, where a is the particular element in (iv) above.

This will be your own self-appraisal !!

Regards,

Nalin Pithwa

Some random problems in algebra (part b) for RMO and INMO training

1) Solve in real numbers the system of equations:

y^{2}+u^{2}+v^{2}+w^{2}=4x-1

x^{2}+u^{2}+v^{2}+w^{2}=4y-1

x^{2}+y^{2}+v^{2}+w^{2}=4u-1

x^{2}+y^{2}+u^{2}+w^{2}=4v-1

x^{2}+y^{2}+u^{2}+v^{2}=4w-1

Hints: do you see some quadratics ? Can we reduce the number of variables? …Try such thinking on your own…

2) Let a_{1}, a_{2}, a_{3}, a_{4}, a_{5} be real numbers such that a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=0 and \max_{1 \leq i <j \leq 5} |a_{i}-a_{j}| \leq 1. Prove that a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2} \leq 10.

3) Let a, b, c be positive real numbers. Prove that

\frac{1}{2a} + \frac{1}{2b} + \frac{1}{2d} \geq \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}

More later

Nalin Pithwa.