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

Pre-RMO training; a statement and its converse; logic and plane geometry

I hope the following explanation is illuminating to my readers/students:

How to prove that two lines are parallel ? (Note that we talk of parallel lines only when they lie in the same plane; on the other hand: consider the following scenario — your study table and the floor on which it stands. Let us say you draw a straight line AB on your study table and another line PQ on the floor on which the study table is standing; then, even though lines AB and PQ never meet, we do not say that they are parallel because they lie in different planes. Such lines are called skew lines. They are dealt with in solid geometry or 3D geometry or vector spaces).

Coming back to the question — when can we say that two lines are parallel?

Answer:

Suppose that a transversal crosses two other lines.

1) If the corresponding angles are equal, then the lines are parallel.
2) If the alternate angles are equal, then the lines are parallel.
3) If the co-interior angles are supplementary, then the lines are parallel.

A STATEMENT AND ITS CONVERSE

Let us first consider the following statements:

A transversal is a line that crosses two other lines. If the lines crossed by a transversal are parallel, then the corresponding angles are equal; if the lines crossed by a transversal are parallel, then the alternate interior angles are equal; if the lines crossed by a transversal are parallel, then the co-interior angles are supplementary.

The statements given below are the converses of the statement given in the above paragraph; meaning that they are formed from the former statements by reversing the logic. For example:

STATEMENT: If the lines are parallel then the corresponding angles are equal.

CONVERSE: If the corresponding angles are equal, then the lines are parallel.

Pairs such as these, a statement and its converse, occur routinely through out mathematics, and are particularly prominent in geometry. In this case, both the statement and its converse are true. It is important to realize that a statement and its converse are, in general, quite different. NEVER ASSUME THAT BECAUSE A STATEMENT IS TRUE, SO ITS CONVERSE IS ALSO TRUE. For example, consider the following:

STATEMENT: If a number is a multiple of 4, then it is even.
CONVERSE: If a number is even, then it is a multiple of 4.

The first statement is clearly true. But, let us consider the number 18. It is even. But 18 is not a multiple of 4. So, the converse is not true always.

\it Here \hspace{0.1in} is \hspace{0.1in}an \hspace {0.1in}example \hspace{0.1in}from \hspace{0.1in}surfing

STATEMENT: If you catch a wave, then you will be happy.
CONVERSE: If you are happy, then you will catch a wave.

Many people would agree with the first statement, but everyone knows that its converse is plain silly — you need skill to catch waves.

Thus, the truth of a statement has little to do with its converse. Separate justifications (proofs) are required for the converse and its statements.

Regards,
Nalin Pithwa.

Reference: (I found the above beautiful, simple, lucid explanation in the following text): ICE-EM, year 7, book 1; The University of Melbourne, Australian Curriculum, Garth Gaudry et al.

Prof. Tim Gowers’ on recognising countable sets

https://gowers.wordpress.com/2008/07/30/recognising-countable-sets/

Thanks Dr. Gowers’. These are invaluable insights into basics. Thanks for giving so much of your time.

Prof. Tim Gowers’ on functions, domains, etc.

https://gowers.wordpress.com/2011/10/13/domains-codomains-ranges-images-preimages-inverse-images/

Thanks a lot Prof. Gowers! Math should be sans ambiguities as far as possible…!

I hope my students and readers can appreciate the details in this blog article of Prof. Gowers.

Regards,
Nalin Pithwa