# |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.

# Every function can be written as a sum of an even and an odd function

Let $f(x)$ be any well-defined function.

We want to express it as a sum of an even function and an odd function.

Let us define two other functions as follows:

$F(x) = \frac{f(x)+f(-x)}{2}$ and $G(x)=\frac{f(x)-f(-x)}{2}$.

Claim I: F(x) is an even function.

Proof I; Since by definition $F(x)= \frac{f(x)+f(-x)}{2}$, so $F(-x) = \frac{f(-x) +f(-(-x))}{2}=\frac{f(-x)+f(x)}{2} \Longrightarrow F(x) = F(-x)$ so that F(x) is indeed an even function.

Claim 2: G(x) is an odd function.

Proof 2: Since by definition $G(x) = \frac{f(x)-f(-x)}{2}$, so $G(-x) = \frac{f(-x)-f(-(-x))}{2} = \frac{f(-x)-f(x)}{2} = -\frac{f(x)-f(-x)}{2} = -G(-x) \Longrightarrow G(x) = -G(-x)$ so that G(x) is indeed an odd function.

Claim 3: $f(x)= F(x) + G(x)$

Proof 3: $F(x)+ G(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2} = \frac{f(x)}{2} + \frac{f(-x)}{2} + \frac{f(x)}{2} - \frac{f(-x)}{2} = f(x)$ indeed.

# Number theory : A set of friendly examples

Even and odd numbers

Two whole numbers are added together. If their sum is odd, which statements below are
always true? Which are always false? Which are sometimes true and sometimes false?
1 Their quotient is not a whole number.
2 Their product is even.
3 Their difference is even.
4 Their product is more than their sum.
5 If 1 is added to one of the numbers and the product is found, it will be even.
The Collatz conjecture

If it is odd, multiply it by 3 and then add 1.
If it is even, divide it by 2. Then repeat this process on the number just obtained. Keep repeating the procedure.

For example, if you start with 58, the resulting chain of numbers is
58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, …

The Collatz conjecture, made by Lothar Collatz in 1937, claims that, if you repeat this
process over and over, starting with any whole number greater than zero, eventually you will finish up with the sequence … 4, 2, 1, 4, 2, 1.

A conjecture is a statement that is thought to be true but has not been proved mathematically to be true for all cases. Although the Collatz conjecture has been shown to work − often very quickly − for many whole numbers, there are some quite small numbers that take a very long time to come down to … 4, 2, 1, 4, 2, 1.Apply this process to all the whole numbers greater than zero and less than or equal to 30.
For each one, find:
• how many steps it takes to reach 1 the frst time
• the largest number in the sequence. (For the sequence above, 58 takes 19 steps and reaches a maximum of 88.)
Look for shortcuts and work with a partner if you like.

Long division
Here is a way to check how good your long division skills are. If you are able to follow it
through and get to the end without making a mistake, you can consider yourself a qualiꏨed long division champion.
• Start with any two-digit number (for example, 58). Write it three times so that a six-digit  number is formed (585 858).

• Divide this number by 21. There should not be any remainder. If there is, try and find out where you made your mistake and fix it.
• Now divide this new four- or possibly five-digit number by 37. Once again, there should be no remainder.
• Finally, divide this number − which should by now have only three or four digits − by 13.
You will know if you got it right by looking at the number you are left with.
Explain why this exercise works.
(Doing any of this exercise on a calculator is still interesting but is de뀠nitely wimping out!)

Totient numbers

1 A totient number is the number of fractions between 0 and 1 (not including 0 or 1) for
a given denominator that cannot be reduced to a simpler equivalent fraction. The totient
number of 2 is 1, since we have $\frac{1}{2}$; of 3 it is 2, since we have $\frac{1}{3}$ and $\frac{2}{3}$; and of 4 it is also 2, since we have $\frac{1}{4}$
and $\frac{3}{4}$ ($\frac{2}{4}$ can be reduced to $\frac{1}{2}$). The totient number of 5 is 4, since we have $\frac{1}{5}$, $\frac{2}{5}$, $\frac{3}{5}$, $\frac{4}{5}$; and of 6 it is 2, since we have $\frac{1}{5}$ and
$\frac{5}{6}$. Find the totient numbers forall denominators up to 12.

2 For any denominator n, there are n fractions between 0 and 1 (including 0 but not 1). Of
these fractions, some will be counted towards the totient number of n, but others will
cancel down and count towards the totient number of one of the factors of n. Using this
information and the totient numbers from the previous question, calculate the totient
numbers for 15, 18, 20 and 24.

3 The totient number is related to the prime factors of the original number, since these will
determine which fractions can be cancelled. Using this information, calculate the totient numbers of 72, 81, 98 and 100.

$\bf{Last \hspace{0.1in}Digits \hspace{0.1in} of \hspace{0.1in}powers}$
$\bf{Square \hspace{0.1in}Numbers}$

Without using a calculator, can you say which of this set of numbers could not be square numbers?

8116801, 251301659, 3186842, 20720704.

You can just by checking the last digit (units digit) of each number.

Do a bit of experimentation with a calculator and find the four digits that square numbers end in. (This eliminates the third number in this set).

Now check out the pairs of digits that your odd square numbers end with. What digits are possible in the tens position of an odd square number? (This number eliminates the second number in the set).

Complete these sentences with what you have discovered:

* In a square number, the last digit (units digit) can only be _____, _______, _______, _______, _______ or _______.
* The second last digit (tens place) of an odd square number is always _______.

$\bf{Cube \hspace {0.1in} Numbers}$

Cube numbers behave rather differently.

A bit of experimentation will show that cube numbers can end in ANY digit (units place). This digit depends on the last digit (units place) of the original number being cubed.

Complete this table:
$\left| \begin{array}{cc} \mbox {if a number ends in} & \mbox{its cube will end in}\\ 0 & \\ 1 & \\ 2 & \\ 3 & \\ 4 & \\ 5 & \\ 6 & \\ 7 & \\ 8 & \\ 9 & \end{array}\right|$

$\bf{Fourth \hspace{0.1in}Powers}$

Fourth powers are in fact just square numbers that have been squared. For example, $7^{4}=7^{2} \times 7^{2}= 49^{2}=2401$.

Since $4^{2}=16$ and $9^{2}=81$, the last digit of a fourth power can only be 0, 1, 5 or 6.

$\bf{Fifth \hspace{0.1in}powers}$

Fifth powers have a magic of their own. Do a bit of experimentation to find out what it is. In p, particular, I suggest you create tables of second powers, third powers, fourth powers and fifth powers of all numbers from zero to 20. Check, compare…actually, it is fun to “compare rate of growth of powers with increasing integers”…this idea involves rudimentary ideas of calculus…

$\bf{Obstinate \hspace{0.1in} numbers}$

An odd number can usually be written as a sum of a prime number and another number, which is a power of two. This is true for all odd numbers greater than one but less than 100.

For example, if we choose 23, we can say that it is equal to $23=19 + 2^{2}$. There is one more way to represent 23: it is $7 + 2^{4}$. So, there are two ways to represent 23 as a sum of a prime number and a power of two. But, $21+2^{1}$ and $15+2^{3}$ do not work as both 21 and 15 are not prime numbers.

Some odd numbers like this can be expressed in many ways.

Try to find various representations as sum(s) of prime number and a power(s) of two of the following integers: 45, 29, 59, 95.

If you are adventurous or courageous, try to find such representations of all odd numbers lying from 1 to 100. You need a lot of patience and stamina and grit…but you will develop an “intuitive feel or tactile feel for numbers”…that’s the way math begins…

There are some odd numbers which cannot be expressed as a sum of a prime number and a power of two. Such numbers are called $\bf{obstinate \hspace{0.1in} numbers}$.

An example of an obstinate number is $\bf{251}$ as the working below shows:

$251-2^{1}=249=3 \times 83$

$251-2^{2}=247=13 \times 19$

$251-2^{3}=243=3 \times 81$

$251-2^{4}=235= 5 \times 47$

$251-2^{5}=219= 3 \times 73$

$251-2^{6}= 187= 11 \times 17$

$251-2^{7}=123=3 \times 41$

The next power of 2 is $2^{8}=256$, which is clearly greater than 251. Hence, 251 is an obstinate number.

In fact, 251 is the third obstinate number. The first two lie between 100 and 150. Find these two odd numbers keeping track of how you eliminated the other twenty three odd numbers between 100 and 150.

$\it{Remember \hspace{0.1in} to \hspace{0.1in} be \hspace{0.1in} systematic \hspace{0.1in}}$.

Making a list of the powers of two up to $2^{8}$ might be a good place to start with. Look for short cuts and patterns as you proceed further.

Have fun with numbers !!

Regards,
Nalin Pithwa.

# Cyclic expressions, fractions: Pre RMO, PRMO, IITJEE foundation 2019

In order to solve the following tutorial sheet, it helps to solve/understand and then apply the following beautiful cyclic relations or identities:

(Note if these look new to you, then you need to check the truth of all them; if all are v v familiar to you, just go ahead and crack the tutorial sheet below):

Core Identities in Cyclic Expressions:
1) $(b-c)+(c-a)+(a-b)=0$
2) $a(b-c)+b(c-a)+c(a-b)=0$
3) $a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)=-(a-b)(b-c)(c-a)$
4) $bc(b-c)+ca(c-a)+ab(a-b)=-(a-b)(b-c)(c-a)$
5) $a(b^{2}-c^{2})+b(c^{2}-a^{2})+c(a^{2}-b^{2})=(a-b)(b-c)(c-a)$

Solve or simplify the following:

1) $\frac{a}{(a-b)(a-c)} + \frac{b}{(b-c)(b-a)} + \frac{c}{(c-a)(c-b)}$
2) $\frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)}$
3) $\frac{a^{2}}{(a-b)(a-c)} + \frac{b^{2}}{(b-c)(b-a)} + \frac{c^{2}}{(c-a)(c-b)}$
4) $\frac{a^{3}}{(a-b)(a-c)} + \frac{b^{3}}{(b-c)(b-a)} + \frac{c^{3}}{(c-a)(c-b)}$
5) $\frac{a(b+c)}{(a-b)(c-a)} + \frac{b(a+c)}{(a-b)(b-c)} + \frac{a(a+b)}{(c-a)(b-c)}$
6) $\frac{1}{a(a-b)(a-c)} + \frac{1}{b(b-c)(b-a)} + \frac{1}{c(c-a)(c-b)}$
7) $\frac{bc}{a(a^{2}-b^{2})(a^{2}-c^{2})} + \frac{ca}{b(b^{2}-c^{2})(b^{2}-a^{2})} + \frac{ab}{c(c^{2}-a^{2})(c^{2}-b^{2})}$
8) $\frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$
9) $\frac{bc(a+d)}{(a-b)(a-c)} + \frac{ca(b+d)}{(b-c)(b-a)} + \frac{ab(c+d)}{(c-a)(c-b)}$
10) $\frac{1}{(a-b)(a-c)(x-a)} + \frac{1}{(b-c)(b-a)(x-b)} + \frac{1}{(c-a)(c-b)(x-c)}$
11) $\frac{a^{2}}{(a-b)(a-c)(x+a)} + \frac{b^{2}}{(b-c)(b-a)(x+b)} + \frac{c^{2}}{(c-a)(c-b)(x+c)}$
12) $a^{2}\frac{(a+b)(a+c)}{(a-b)(a-c)} + b^{2}\frac{(b+c)(b+a)}{(b-c)(b-a)} + c^{2}\frac{(c+a)(c+b)}{(c-a)(c-b)}$
13) $\frac{a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)}{(b-c)^{3}+(c-a)^{3}+(a-b)^{3}}$
14) $\frac{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)+2(c-a)(a-b)(b-c)}{(b-c)^{3}+(c-a)^{3}+(a-b)^{3}}$
15) $\frac{a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)}{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)}$
16) $\frac{a^{2}(b-c)^{3}+b^{2}(c-a)^{3}+c^{2}(a-b)^{3}}{(a-b)(b-c)(c-a)}$
17) $\frac{\frac{b-c}{a} + \frac{c-a}{b} + \frac{a-b}{c}}{\frac{1}{a}(\frac{1}{b^{2}}-\frac{1}{c^{2}})+\frac{1}{b}(\frac{1}{c^{2}}-\frac{1}{a^{2}})+\frac{1}{c}(\frac{1}{a^{2}}-\frac{1}{b^{2}})}$^
18) $\frac{a^{2}(\frac{1}{a^{2}}-\frac{1}{b^{2}})+b^{2}(\frac{1}{a^{2}}-\frac{1}{c^{2}})+c^{2}(\frac{1}{b^{2}}-\frac{1}{a^{2}})}{\frac{1}{bc}(\frac{1}{c}-\frac{1}{b})+\frac{1}{ca}(\frac{1}{a}-\frac{1}{c})+\frac{1}{ab}(\frac{1}{b}-\frac{1}{c})}$
19) $\frac{a}{(a-b)(a-c)(x-a)} + \frac{b}{(b-c)(b-a)(x-b)} + \frac{c}{(c-a)(c-b)(x-c)}$

More later,
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