# Category: Study Notes

## SETS

In mathematics, Set is a well defined collection of distinct objects. The theory of Set as a mathematical discipline rose up with George Cantor, German mathematician, when he was working on some problems in Trigonometric series and series of real numbers, after he recognized the importance of some distinct collections and intervals.

Cantor defined the set as a ‘plurality conceived as a unity’ (many in one; in other words, mentally putting together a number of things and assigning them into one box).

Mathematically, a Set $S$ is ‘any collection’ of definite, distinguishable objects of our universe, conceived as a whole. The objects (or things) are called the elements or members of the set $S$ . Some sets which are often pronounced in real life are, words like ”bunch”, ”herd”, ”flock” etc. The set is a different entity from any of its members.

For example, a flock of birds (set) is not just only a single bird (member of the set). ‘Flock’ is just a mathematical concept with no material existence but ‘Bird’ or ‘birds’ are real.

## Representing sets

Sets are represented in two main ways:

## Euler’s (Prime to) Prime Generating Equation

The greatest number theorist in mathematical universe, Leonhard Euler had discovered some formulas and relations in number theory, which were based on practices and were correct to limited extent but still stun the mathematicians. The prime generating equation by Euler is a very specific binomial equation on prime numbers and yields more primes than any other relations out there in number theory. Euler told that the equation $f(x)=x^2+x+k$ yields many prime numbers with the values of x being input from x=0 to $x=k-2$ ; k being a prime.

Let’s see how many primes we can get by using different values of $k$ and $x$:

## Irrational Numbers and The Proofs of their Irrationality

“Irrational numbers are those real numbers which are not rational numbers!”

## Def.1: Rational Number

A rational number is a real number which can be expressed in the form of $\frac{a}{b}$ where $a$ and $b$ are both integers relatively prime to each other and $b$ being non-zero.
Following two statements are equivalent to the definition 1.
1. $x=\frac{a}{b}$ is rational if and only if $a$ and $b$ are integers relatively prime to each other and $b$ does not equal to zero.
2.   $x=\frac{a}{b}&space;\in&space;\mathbb{Q}&space;\iff&space;\mathrm{g.c.d.}&space;(a,b)&space;=1,&space;\&space;a&space;\in&space;\mathbb{Z},&space;\&space;b&space;\in&space;\mathbb{Z}&space;\setminus&space;\{0\}$.

## Def. 2: Relatively Prime Numbers

Two integers $a$ and $b$ are said to be relatively prime to each other if the greatest common divisor of $a$ and $b$ is $1$ .
For example: The pairs (2, 9); (4, 7) etc. are such that each element is relatively prime to other.

## Def. 3: Irrational Number

A real number, which does not fit well under the definition of rational numbers is termed as an irrational number.

A silly question: Let, in the definition of a rational numbers, $a=0$ and $b=8$ , then, as we know $\frac{0}{8}=0$ is a rational number, however $8$ can divide both integers $0$ and $8$ , i.e., $\mathrm{g.c.d.} (0,8) =8$ . (Why?) $\Box$

# Primary ways to prove the irrationality of a real number

It is all clear that any real, if not rational, is irrational. So, in order to prove a (real) number irrational, we need to show that it is not a rational number (i.e., not satisfying definition 1). Most popular method to prove irrationality in numbers, is the Proof by Contradiction, in which we first assume the given (irrational) number to be ‘almost’ rational and later we show that our assumption was untrue. There are many more ways to prove the irrational behavior of numbers but all those are more or less derived from the proof by contradiction.
Some methods which I’ll discuss here briefly are:

1. Pythagorean Approach
2. Using Euclidean Algorithm
3. Power series expansion of special numbers
4. Continued Fraction representation of irrational numbers.

## (1) Pythagorean Approach

This proof is due to Pythagoras and thus called Pythagorean Approach to irrationality. In this approach, we assume a number to be first. Later using the fundamental rules of arithmetic, we make sure whether or not our assumption was true. If our assumption was true, the number we took was rational, otherwise irrational.
For example:

Prove that the number $\sqrt{2}$ is irrational.$Proof: Suppose, to the contrary, that$ \sqrt{2}$is a rational number. Then as according to the definition 1, we can write$ \sqrt{2}=\frac{a}{b} \ldots (1)$where$ a$and$ b$are both integers with$ \mathrm{g.c.d.} (a,b) =1$and$ b \ne 0$. Squaring equation (1),$ 2=\frac{a^2}{b^2}$or,$ a^2=2b^2 \ldots (2)$From the equation (2), we can proceed our proof into two ways: Way I:$ \sqrt{2}$is a positive number, therefore we can assume$ a$and$ b$both to be positive. Since,$ a^2=2b^2$then$ a^2=2b \cdot b$or,$ b|a^2$(read as b divides a squared). Since,$ b$is positive integer,$ b \ge 1$. However,$ b=1$is impossible since corresponding$ a=\sqrt{2}$is not an integer. Thus,$ b > 1$and then according to fundamental theorem of arithmetic, there exists at least one prime$ p> 1$which divides$ b$. Mathematically,$ p|b$but as$ b|a^2$. It is clear that$ p|a^2$. This implies that$ p|a$. Since$ p|a$and$ p|b$, therefore$ \mathrm{g.c.d.}(a,b) \ge p$. So for given number,the greatest common divisor of$ a$and$ b$is not$ 1$, but another prime larger than$ 1$. Thus, it fails to satisfy the definition 1. Thus our claim that$ \sqrt{2}$is rational, is untrue. Therefore,$ \sqrt{2}$is an irrational number. Way II: As a deviation, we can proceed our proof from equation (2) by taking the fact into mind that$ \sqrt{2}$is positive. The number (natural number)$ a$can either be odd or even. Let$ a$be odd, i.e.,$ a=2k+1$where$ k\in \{0,1,2,3, \ldots \}$. Therefore$ a^2$would also be odd. Which contradicts (2), since$ 2b^2$is always even and that equals to$ a^2$. Therefore,$ a$must be an even number. Let$ a=2k$. Putting this into (2) we get,$ 4k^2=2b^2$or,$ b^2=2k^2$or,$ b=\sqrt{2} k \mathrm{g.c.d.}(a,b)=\sqrt{2} \ne 1$. Which is contradiction to our claim. Thus$ \sqrt{2}$is an irrational number. In similar ways, one can prove$ \sqrt{3}$,$ \sqrt{5}$,$ \sqrt{7}$etc. to be irrationals. ## (2) Using Euclidean Algorithm This is an interesting variation of Pythagorean proof. Let$ \sqrt{2}=\frac{a}{b}$with$ \mathrm{g.c.d.}(a,b)=1$, then according to Euclidean Algorithm, there must exist integers$ r$and$ s$, satisfying$ ar+bs=1$. or,$ \sqrt{2} \cdot 1 =\sqrt{2}(ar+bs)$or,$ \sqrt{2}=\sqrt{2}ar +\sqrt{2}bs$or,$ \sqrt{2}=(\sqrt{2}a)r+ (\sqrt{2}b)s$or,$ \sqrt{2}=2br +as$. (From$ \sqrt{2}=a/b$we put$ a=\sqrt{2} b$.) This representation of$ \sqrt{2}$leads us to conclude that$ \sqrt{2}$is an integer, which is completely false. Hence our claim that$ \sqrt{2}$can be written in form of$ \frac{a}{b}$is untrue. Thus,$ \sqrt{2}$is irrational. Similarly, we can use other numbers to prove so. ## (3) Power Series Expansion Some irrational numbers, like$ e$, can be proved to be irrational by expanding them and arranging the terms. Over all, it is another form of proof by contradiction but different from the Pythagorean Approach.$ e$can be defined by the following infinite series:$ e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!}+\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\ldots$. Suppose, to the contrary, that$ e$is rational, and$ e=\frac{a}{b}$(say) where$ a$and$ b$are positive integers. Then for any$ n>b$and also$ n>1$,$ N=n! \left({e-(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!})}\right)$is positive, since$ \left({e-(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!})}\right)$is positive. Or,$ N=n! \left({\dfrac{1}{(n+1)!}+\dfrac{1}{(n+2)!}+\dfrac{1}{(n+3)!}+\ldots}\right)$or,$ N=\left({\dfrac{n!}{(n+1)!}+\dfrac{n!}{(n+2)!}+\dfrac{n!}{(n+3)!}+\ldots}\right)$or,$ N=\left({\dfrac{1}{(n+1)}+\dfrac{1}{(n+1)(n+2)}+\dfrac{1}{(n+1)(n+2)(n+3)}+\ldots}\right) > 0$. It is clear that$ N$is less than$ \left({\dfrac{1}{(n+1)}+\dfrac{1}{(n+1)(n+2)}+\dfrac{1}{(n+2)(n+3)}+\ldots}\right)$. And$ \left({\dfrac{1}{(n+1)}+\dfrac{1}{(n+1)(n+2)}+\dfrac{1}{(n+2)(n+3)}+\ldots}\right) =\left({\dfrac{1}{(n+1)}+\dfrac{1}{(n+1)}-\dfrac{1}{(n+2)}+\dfrac{1}{(n+2)}-\dfrac{1}{(n+3)}+\ldots}\right) =\dfrac{2}{(n+1)}$. Thus,$ N < \dfrac{2}{(n+1)}<1$. So,$ N$being positive integer is less than$ 1$? It is impossible for any integer. Thus our claim is not true and hence$ e$is irrational. ## (4) Continued Fractions Any number, that can be expressed in form of an infinite continued fraction is always irrational. For example: 1)$ e$can be represented in form of infinite continued fractions, thus$ e$is irrational. 2) Similarly$ \pi$is irrational. 3)$ \sqrt{2}$is also irrational.$ \Box$## The Area of a Disk If you are aware of elementary facts of geometry, then you might know that the area of a disk with radius$ R$is$ \pi R^2$. The radius is actually the measure(length) of a line joining the center of disk and any point on the circumference of the disk or any other circular lamina. Radius for a disk is always same, irrespective of the location of point at circumference to which you are joining the center of disk. The area of disk is defined as the ‘measure of surface‘ surrounded by the round edge (circumference) of the disk. The area of a disk can be derived by breaking it into a number of identical parts of disk as units — calculating their areas and summing them up till disk is reformed. There are many ways to imagine a unit of disk. We can imagine the disk to be made up of several concentric very thin rings increasing in radius from zero to the radius of disc. In this method we can take an arbitrary ring, calculate its area and then in similar manner, induce areas of other rings — sum them till whole disk is obtained. Rings and Sections Mathematically, we can imagine a ring of with radius$ x$and thickness$ dx$, anywhere in the disk having the same center as disk, calculate its area and then sum up (integrate) it from$ x=0$to$ x=R$. Area of a thin ring is since$ \pi x dx$. And after integrating we get, area of disk$ A=2 \int_0^R \pi x dx$or$ A=\pi R^2$. There is another approach to achieve the area of a disk, A. An inscribed Triangle Imagine a disk is made up of a number equal sections or arcs. If there are$ n$number of arcs then interior angle of an arc is exactly$ \frac{2\pi}{n}$, since$ 2 \pi$is the total angle at the center of disk and we are dividing this angle into$ n$equal parts. If we join two ends of each sections –we can get$ n$identical triangles in which an angle with vertex O is$ \frac{2 \pi}{n}$. Now, if we can calculate the area of one such section, we can approach to the area of the disk intuitively. This approach is called the method of exhaustion. Let, we draw two lines joining center O of the disk and points A & B at circumference. It is clear that OB and OA are the radius of the disk. We joined points A and B in order to form a triangle OAB. Now consider that the disk is made up of n-number of such triangles. We see that there is some area remaining outside the line AB and inside the circumference. If we had this triangle thinner, the remaining area must be lesser. Area remaining after the Triangle So, if we increase the number of triangles in disk —-we decrease the remaining areas. We can achieve to a point where we can accurately calculate the area of disk when there are infinitely many such triangles or in other words area of one such triangle is very small. So our plan is to find the area of one triangle —sum it up to n — make$ n$tending to infinity to get the area of disk. It is clear that the sum of areas of all identical triangles like OAB must be either less than or equal to area of the disk. We can call triangles like OAB as inscribed triangles. Now, if we draw a radius-line OT’, perpendicular to AB at point T and intersecting the circumference at point T’, we can easily draw another triangle OA’B’ as shown in figure. AOB and A’OB’ are inscribed and superscribed triangles of disk with same angle at vertex O. So, it is clear that the angle A’OB’ is equal to the angle AOB. Triangle A’OB’ is larger than the circular arc OAB and circular arc OAB is larger than the inscribed triangle AOB. Also, the sum of areas of triangles identical to OA’B’ is either greater than or equal to area of the disk. ## Triangle Inequality Triangle inequality has its name on a geometrical fact that the length of one side of a triangle can never be greater than the sum of the lengths of other two sides of the triangle. If$ a$,$ b$and$ c$be the three sides of a triangle, then neither$ a$can be greater than$ b+c$, nor$ b$can be greater than$ c+a$and so$ c$can not be greater than$ a+b$. Triangle Consider the triangle in the image, side$ a$shall be equal to the sum of other two sides$ b$and$ c$, only if the triangle behaves like a straight line. Thinking practically, one can say that one side is formed by joining the end points of two other sides. In modulus form,$ |x+y|$represents the side$ a$if$ |x|$represents side$ b$and$ |y|$represents side$ c$. A modulus is nothing, but the distance of a point on the number line from point zero. ## On Ramanujan’s Nested Radicals Ramanujan (1887-1920) discovered some formulas on algebraic nested radicals. This article is based on one of those formulas. The main aim of this article is to discuss and derive them intuitively. Nested radicals have many applications in Number Theory as well as in Numerical Methods . The simple binomial theorem of degree 2 can be written as:$ {(x+a)}^2=x^2+2xa+a^2 \ \ldots (1)$Replacing$ a$by$ (n+a)$where$ x, n, a \in \mathbb{R}$, we can have$ {(x+(n+a))}^2= x^2+2x(n+a)+{(n+a)}^2$or,$ {(x+n+a)}^2 =x^2+2xn+2ax+{(n+a)}^2$Arranging terms in a way that$ {(x+n+a)}^2 =ax+{(n+a)}^2+x^2+2xn+ax=ax+{(n+a)}^2+x(x+2n+a)$Taking Square-root of both sides or, $ x+n+a=\sqrt{ax+{(n+a)}^2+x(x+2n+a)} \ \ldots (2)$Take a break. And now think about$ (x+2n+a)$in the same way, as:$ x+2n+a =(x+n)+n+a$. Therefore, in equation (2), if we replace$ x$by$ x+n$, we get$ x+2n+a=(x+n)+n+a=\sqrt{a(x+n)+{(n+a)}^2+(x+n)((x+n)+2n+a)}$or,$ x+2n+a=\sqrt{a(x+n)+{(n+a)}^2+(x+n)(x+3n+a)} \ \ldots (3)$Similarly,$ x+3n+a=\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)(x+4n+a)} \ \ldots (4)$and also,$ x+4n+a=\sqrt{a(x+3n)+{(n+a)}^2+(x+3n)(x+5n+a)} \ \ldots (5)$Similarly,$ x+kn+a=\sqrt{a(x+(k-1)n)+{(n+a)}^2+(x+(k-1)n)(x+(k+1)n+a)} \ \ldots (6)$where,$ k \in \mathbb{N}$. Putting the value of$ x+2n+a$from equation (3) in equation (2), we get:$ x+n+a=\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)(x+3n+a)}} \ \ldots (7)$Again, putting the value of$ x+3n+a$from equation (4) in equation (7), we get$ x+n+a =\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)(x+4n+a)}}} \ \ldots (8)$Generalizing$ the result for $k$ -nested radicals:
$x+n+a =\\ \sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)\sqrt{\ldots+(x+(k-2)n)\sqrt{a(x+(k-1)n)+{(n+a)}^2+x(x+(k+1)n+a)}}}}} \ \ldots (9)$
This is the general formula of Ramanujan Nested Radicals up-to $k$ roots.

Some interesting points
As $x,n$ and $a$ all are real numbers, thus they can be interchanged with each other.
i.e.,

etc.

Putting $n=0$ in equation (9)
we have
$x+a =\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{\ldots+x\sqrt{ax+{a}^2+x(x+a)}}}}} \ \ldots (11)$
or just, $x+a =\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{\ldots}}}} \ \ldots (12)$

Again putting $x=1 \ a=0$ in (9)

$1+n =\sqrt{{n}^2+\sqrt{n^2+(1+n)\sqrt{{n}^2+(1+2n)\sqrt{\ldots+(1+(k-2)n)\sqrt{{n}^2+1+(k+1)n}}}}} \ldots (13)$

Putting $x=1 \ a=0$ in equation (8)
$1+n =\sqrt{{n}^2+\sqrt{{n}^2+(1+n)\sqrt{{n}^2+(1+2n)(1+4n)}}} \ \ldots (14)$

Again putting $x=a=n$ =n(say) then
$3n=\sqrt{n^2+4{n}^2+n\sqrt{2n^2+4{n}^2+2n\sqrt{3n^2+4{n}^2+3n\sqrt{\ldots+(k-1)n\sqrt{kn^2+4{n}^2+(k+3)n^2}}}}}$
or, $3n=\sqrt{5{n}^2+n\sqrt{6{n}^2+2n\sqrt{7{n}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+4)n^2+(k+3)n^2}}}}} \ \ldots (15)$

Putting $n=1$ in (15)
$3=\sqrt{5+\sqrt{6+2\sqrt{7+3\sqrt{\ldots+(k-1)\sqrt{(2k+7)}}}}} \ \ldots (16)$

Putting $x=n \in \mathbb{N}$ and $a=0$ in (9) we get even numbers
$2n =\sqrt{{n}^2+n\sqrt{{n}^2+2n\sqrt{{n}^2+3n)\sqrt{\ldots+(k-1)n\sqrt{(k-1)n)+{n}^2+(k+2)n^2}}}}} \ \ldots (17)$

Similary putting $x=n \in \mathbb{N}$ and $a=1$ in (9) we get a formula for odd numbers:
$\tiny \dpi{150} 2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{kn+{(n+1)}^2+(k+2)n^2+n}}}}} \ \ldots (18)$
or,
$\tiny \dpi{150} 2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+3)n^2+(k+3)n+1}}}}} \ \ldots (19)$

## Calendar Formula: Finding the Week-days

Pope Gregory

This is the last month of the glorious prime year 2011. We are all set to welcome upcoming 2012, which is not a prime but a leap year. Calendars have very decent stories and since this blog is based on mathematical approach, let we talk about the mathematical aspects of calendars.

The international calendar we use is called Gregorian Calendar, said to be created by Pope Gregory XIII. Gregorian calendar was introduced in 80s of 16th century, to be accurate in $1582$ ,—as a correction to earlier Julian Calendar. Julian Calendar was introduced by Julius Caesar and was based on the fact that there were $365 \frac{1}{4}$ days in a year, with leap year every fourth year. Astronomical calculations told us that one year on earth (the time required for the earth to complete an orbit around the sun) was equal to $365.2422$ days —thus we can say that Julian Calendar hadn’t enough precise measure of dates. A difference of $365 \frac{1}{4}-365.2422 =0.0078$ days per year meant that the Julian Calendar receded a day from its astronomical data every $\frac{1}{0.0078}$ years (viz. Approx. $128.2$ years). More information on Julian Calendar can be found at Julian_Calendar’s Wikipedia Page.

The centuries old calendar came to an end as the accumulating inaccuracy caused the vernal equinox (the first day of Spring) to fall on March 11 instead of its proper day, March 21. The inaccuracy naturally persisted throughout the year, but at this season it meant that the Easter festival was celebrated at the wrong astronomical time. Pope Gregory XIII rectified the discrepancy in a new calendar, imposed on the predominantly Catholic Countries of Europe. He decreed that 10 years (11 March to 21 March) were to be omitted from the year $1582$ , by having October 15 of that year immediately follow October 4. At the same time, C. Clavius proposed the scheme for leap years —which must be divisible by 4, except for those marking centuries. Century years would be leap years only if they were disible by 400. This implies that the century years $1600, 2000, 2400$ are leap years, but $1700, 1800, 1900, 2100, 2200, 2300$ are not.

There are many tricks to determine the day of a week for a given day after the year 1600 in the Gregorian Calendar. But we shall use a number-theoretic method to determine it, as described in the book ‘ELEMENTARY NUMBER THEORY’ by David M. Burton.
We all know that the extra day of a leap year is added to February month of the year, so let us adopt the convenient fiction that each year ends at the end of February. The months for any year Y are:

[LIST A]$1. March$
2. April$3. May$
4. June$5. July$
6. August$7. September$
8. October$9. November$
10. December$11. January$
12. February.$It is clear that if we count for any year$ Y$, January and February must be in next year,$ Y+1$of Gregorian Calendar. We need another convenient notation as we denote days by numbers$ 0,1,2,3…6$as: [LIST B]$
0. Sunday$1. Monday$
2. Tuesday$3. Wednesday$
4. Thursday$5. Friday$
6. Saturday.$The number of days in a common year is$ 365$, and the number weeks thus are$ \frac{365}{7}$=52 weeks and 1 day while that in a leap year is$ 366$claiming the number of weeks being 52 with two extra days. We could write last sentence as this way too: Read More ## Equations- A Basic Introduction Applied mathematics is one which is used in day-to-day life, in solving tensions (problems) or in business purposes. Let me write an example: George had some money. He gave 14 Dollars to Matthew. Now he has 27 dollars. How much money had he? If you are familiar with day-to-day calculations –you must say that George had 41 dollars, and since he had 41, gave 14 to Matthew saving 27 dollars. That’s right? Off course! This is a general(layman) approach. ‘How will we achieve it mathematically?’ –we shall restate the above problem as another statement (meaning the same): George had some money$ x$dollars. He gave 14 dollars to Matthew. Now he has 27 dollars. How much money he had? Find the value of$ x$.$

This is equivalent to the problem asked above. I have just replaced ‘some money’ by ‘x dollars’. As ‘some’ senses as unknown quantity— $x$ does the same. Now all we need to get the value of x.
When solving for $x$ , we should have a plan like this:

 George had $x$ dollars. He gave to Matthew 14 dollars Now he must have $x-14$ dollars

But problem says that he has 27 dollars left. This implies that $x-14$ dollars are equal to 27 dollars.
i.e., $x-14=27$

$x-14=27$ contains an alphabet x which we assumed to be unknown–can have any certain value. Statements (like $x-14=27$ ) containing unknown quantities and an equality are called Equations. The unknown quantities used in equations are called variables, usually represented by bottom letters in English alphabet (e.g.,$x,y,z$ ). Top letters of alphabet ($a,b,c,d$ ..) are usually used to represent constants (one whose value is known, but not shown).

Now let we concentrate on the problem again. We have the equation x-14=27.
Now adding 14 to both sides of the equal sign:
$x-14 +14 =27 +14$
or, $x-0 = 41$ (-14+14=0)
or, $x= 41$ .
So, $x$ is 41. This means George had 41 dollars. And this answer is equal to the answer we found practically. Solving problems practically are not always possible, specially when complicated problems encountered —we use theory of equations. To solve equations, you need to know only four basic operations viz., Addition, Subtraction, Multiplication and Division; and also about the properties of equality sign.
We could also deal above problem as this way:
$x-14= 27$
or,$x= 27+14 =41$
-14 transfers to another side, which makes the change in sign of the value, i.e., +14.

When we transport a number from left side to right of the equal sign, the sign of the number changes and vice-versa. As here -14 converts into +14; +18 converts into -18 in example below:
$x+18 =32$
or, $x=32 -18 =14$ .
Please note, any number not having a sign before its value is deemed to be positive—e.g., 179 and +179 are the same, in theory of equations.
Before we proceed, why not take another example?

Marry had seven sheep. Marry’s uncle gifted her some more sheep. She has eighteen sheep now. How many sheep did her uncle gift?

First of all, how would you state it as an equation?
$7 + x = 18$
or, $+7 +x =18$ (just to illustrate that 7=+7)
or, $x= 18-7 =9$ .
So, Marry’s uncle gifted her 9 sheep. ///
Now tackle this problem,

Monty had some cricket balls. Graham had double number of balls as compared to Monty. Adam had also 6 cricket balls. They all collected their balls and found that total number of cricket balls was 27. How many balls had Monty and Graham?

As usual our first step to solve this problem must be to restate it as an equation. We do it like this:
Then Graham must had $x \times 2=2x$ balls.
The total sum=$x+2x+6=3x+6$
But that is 27 according to our question.
Hence, $3x+6=27$
or, $3x=27-6 =21$
or,$x=21 /3 =7$ .
Here multiplication sign converts into division sign, when transferred.

## Numbers – The Basic Introduction

If mathematics was a language, logic was the grammar, numbers should have been the alphabet.

There are many types of numbers we use in mathematics, but at a broader aspect we may categorize them in two categories:

1. Countable Numbers
2. Uncountable Numbers

The numbers which can be counted in nature are called Countable Numbers and the numbers which can not be counted are called Uncountable Numbers.

Well, this is not the correct way to classify the bunch of types of numbers. We have some formal names for special types of numbers, like Real numbers, Complex Numbers, Rational Numbers, Irrational Numbers etc.. We shall discuss these non-interesting numbers (let me say them non-interesting) at first and then some interesting numbers(those numbers are really interesting to learn). Although in this post I have concisely described the classification, I will rigorously discuss them later.
Let me start this discussion with the memorable quote by Leopold Kronecker:

“God created the natural numbers, and all the rest is the work of man.”

Actually, he meant to say that all numbers, like Real Numbers, Complex Numbers, Fractions, Integers, Non-integers etc. are made up of the numbers given by God to the human. These God Gifted numbers are actually called Natural Numbers. Natural Numbers are the numbers which are used to count things in nature.

Eight pens, Eighteen trees, Three Thousands people etc. are measure of natural things and thus ‘Eight’, ‘Eighteen’, ‘Three Thousands’ etc. are called natural numbers and we represent them numerically as ‘8’, ’18’, ‘3000’ respectively. So, if 8, 18, 3000 are used in counting natural things, are natural numbers. Similarly, 1, 2, 3, 4, and other numbers are also used in counting things —thus these are also Natural Numbers.

Let we try to form a set of Natural Numbers. What will we include in this set?

1? (yes!).
2? (yes).
3? (yes).
….
1785? (yes)
…and so on.

This way, after including all elements we get a set of natural numbers {1, 2, 3, 4, 5, …1785, …, 2011,….}. This set includes infinite number of elements. We represent this set by Borbouki’s capital letter N, which looks like $\mathbb{N}$ or bold capital letter N , where N stands for NATURAL. We will define the set of all natural numbers as:

$\mathbb{N} := \{ 1, 2, 3, 4, \ldots, n \ldots \}$ .

It is clear from above set-theoretic notation that $n$ -th element of the set of natural numbers is $n$ .
In general, if a number $n$ is a natural number, we right that $n \in \mathbb{N}$ .
Please note that some mathematicians (and Wolfram Research) treat ‘0’ as a natural number and state the set as $\mathbb{N} :=\{0, 1, 2, \ldots, n-1, \ldots \}$ , where $n-1$ is the nth element of the set of natural numbers; but we will use first notion since it is broadly accepted.

Now we shall try to define Integers in form of natural numbers, as Kronecker’s quote demands. Integers (or Whole numbers) are the numbers which may be either positives or negatives of natural numbers including 0.
Few examples are 1, -1, 8, 0, -37, 5943 etc.
The set of integers is denoted by $\mathbb{Z}$ or $\mathbf{Z}$ (here Z stands for ‘Zahlen‘, the German alternative of integers). It is defined by
$\mathbb{Z} := \{ \pm n: n \in \mathbb{N} \} \cup \{0\}$
i.e., $\mathbb{Z} := \{\ldots -3, -2, -1, 0, 1, 2, 3 \ldots \}$ . Kronecker’s quote was therefore, later modified as

“God created the integers, and all the rest is the work of man.”

Now please go a step back and again consider the statement of Kronecker. One may ask that how could we prepare the integer set $\mathbb{Z}$ by the set $\mathbb{N}$ of natural numbers?

The construction of $\mathbb{Z}$ from $\mathbb{N}$ is motivated from the requirement that every integer can be expressed as difference of two positive integers (i.e., Natural Numbers). Let $a,b,c,d \in \mathbb{N}$ and a relation ρ is defined on $\mathbb{N} \times \mathbb{N}$ by $(a,b) \rho (c,d)$ if and only if $a+d = b+c$ . The relation ρ is an equivalence relation and the equivalence classes under ρ are called integers and defined as $\mathbb{Z} := \mathbb{N} \times \mathbb{N} /\rho$ . Now we can define set of integers by an easier way, as $\mathbb{Z}:= \{a-b; \ a,b \in \mathbb{N}\}$ . Thus an integer is a number which can be produced by difference of two or more natural numbers. And similarly as converse definition, positive integers are called Natural Numbers.

After Integers, we head to rational numbers. Say it again– ‘ratio-nal numbers‘ –numbers of ratio.

A rational number $\frac{p}{q}$ is defined as a ratio of an integer p and a non-zero integer q. (Well that is not a perfect definition, but as an introduction it is great for understanding.) The set of rational numbers is defined by $\mathbb{Q}$ .
Once integers are formed, we can form Rational (and Irrational numbers: numbers which are not rational ) using integers.
We consider an ordered pair $(p,q):=\mathbb{Z} \times (\mathbb{Z} \setminus \{0 \})$ and another ordered pair $(r,s):=\mathbb{Z} \times (\mathbb{Z} \setminus \{0\})$ and define a relation ρ by $(p,q) \rho (r,s) \iff ps=qr$ for $p,q,r,s \in \mathbb{Z}, \ q, r \ne 0$ . Then ρ is an equivalence relation of rationality, class (p,q). The set $\mathbb{Z} \times (\mathbb{Z} \setminus \{0\})/\rho$ is denoted by $\mathbb{Q}$ (and the elements of this set are called rational numbers).
In practical understandings, the ratio of integers is a phrase which will always help you to define the rational numbers. Examples are $\frac{6}{19}, \ \frac{-1}{2}=\frac{-7}{14}, \ 3\frac{2}{3}, \ 5=\frac{5}{1} \ldots$ . Set of rational numbers includes Natural Numbers and Integers as subsets.
Consequently, irrational numbers are those numbers which can not be represented as the ratio of two integers. For example $\pi, \sqrt{3}, e, \sqrt{11}$ are irrationals.

# What is Logic?

If mathematics is regarded as a language, then logic is its grammar.

In other words, logical precision has the same importance in mathematics as grammatical accuracy in a language. As linguistic grammar has sentences, statements— logic has them too. After we discuss about Sentence & Statements, we will proceed to further logical theories .

# Sentences & Statements

A sentence is a collection of some words, those together having some sense.

For example:

1. Math is a tough subject.
2. English is not a tough subject.
3. Math and English both are tough subjects.
4. Either Math or English is tough subject.
5. If Math is a tough subject, then English is also a tough subject.
6. Math is a tough subject, if and only if English is a tough subject.

Just have a quick look on above collections of words. Those are sentences, as they yield some meanings too. First sentence is called Prime Sentence, i.e., sentence which either contains no connectives or, by choice, is regarded as “indivisible” . The five words

• not
• and
• or
• if …. then
• if and only if

and their combinations are called ‘connectives‘. The sentences (all but first) are called composite sentences, i.e., a declarative sentence (statement ) in which one or more connectives appear. Remember that there is no difference between a sentence and statement in general logic. In this series, sentences and statements would have the same meaning.

# Connectives

not: A sentence which is modified by the word “not ” is called the negation of the original sentence.
For example: “English is not a tough subject ” is the negation of “English is a tough subject“. Also, “3 is not a prime” is the negation of “3 is a prime“. Always note that negation doesn’t really mean the converse of a sentence . For example, you can not write “English is a simple subject ” as the negation of “English is a tough subject “.
In mathematical writings, symbols are often used for conciseness. The negation of sentences/statements is expressed by putting a slash (/) over that symbol which incorporates the principal verb in the statement.
For example: The statement $x=y$ (read ‘x is equal to y’) is negated as $x \ne y$ (read ‘x is not equal to y ‘). Similarly, $x \notin A$ (read ‘x does not belong to set A ‘) is the negation of $x \in A$ (read ‘x belongs to set A ‘).
Statements are sometimes represented by symbols like p, q, r, s etc. With this notation there is a symbol, $\not$ or ¬ (read as ‘not’) for negation. For example if ‘p’ stands for the statement “Terence Tao is a professor” then $\not p$ [or ¬p] is read as ‘not p’ and states for “Terence Tao is not a professor.” Sometimes ~p is also used for the negation of p.
and:
The word “and “ is used to join two sentences to form a composite sentence which is called the conjunction of the two sentences. For example, the sentence “I am writing, and my sister is reading ” is the conjunction of the two sentences: “I am writing ” and “My sister is reading “. In ordinary language (English), words like “but, while ” are used as approximate synonyms for “and “, however in math, we shall ignore possible differences in shades of meaning which might accompany the use of one in the place of the other. This allows us to write “I am writing but my sister is reading ” having the same mathematical meaning as above.
The standard notation for conjunction is \wedge , read as ‘and ‘. If p and q are statements then their conjunction is denoted by p \wedge q and is read as ‘p and q’.
or:
A sentence formed by connecting two sentences with the word “or ” is called the disjunction of the two sentences. For example, “Justin Bieber is a celebrity, or Sachin Tendulkar is a footballer .” is a disjunction of “Justin Bieber is a celebrity ” and “Sachin Tendulkar is a footballer “.
Sometimes we put the word ‘either ‘ before the first statement to make the disjunction sound nice, but it is not necessary to do so, so far as a logician is concerned. The symbolic notation for disjunction is $\vee$, read ‘or’. If p and q are two statements, their disjunction is represented by $p \vee q$ and read as p or q.

if….then: From two sentences we may construct one of the from “If . . . . . then . . .“; which is called a conditional sentence. The sentence immediately following IF is the antecedent, and the sentence immediately following THEN is the consequent. For example, “If 5 <6, 6<7, then 5<7 ” is a conditional sentence with “5<6, 6<7 ” as antecedent and “5<7 ” as consequent. If p and q are antecedent and consequent sentences respectively, then the conditional sentence can be written as:

“If p then q”.

This can be mathematically represented as $p \Rightarrow q$ and is read as “p implies q” and the statement sometimes is also called implication statement. Several other ways are available to paraphrase implication statements including:

1. If p then q
2. p implies q
3. q follows from p
4. q is a logical consequence of p
5. p (is true) only if q (is true)
6. p is a sufficient condition for q
7. q is a necessary condition for p

If and Only If : The phrase “if and only if ” (abbreviated as ‘iff‘) is used to obtain a bi-conditional sentence. For example, “A triangle is called a right-angled triangle, if and only if one of its angles is 90° .” This sentence can be understood in either ways: “A triangle is called a right-angled triangle if one of its angles is 90° ” and “One of angles of a triangle is 90° if the triangle is right-angled triangle. ” This means that first prime sentence implies second prime sentence and second prime sentence implies first one. (This is why ‘iff’ is sometimes called double-implication.)