Emmy Noether

Emmy Noether

Emmy Noether was an influential German mathematician known for her groundbreaking contributions to abstract algebra and theoretical physics. Albert Einstein described her as the most important woman in mathematical history. She lived and worked in Germany during the same period that Ramanujan worked in England and India, but whereas Ramanujan’s contributions were in analysis and number theory, hers were in abstract algebra and in the application of algebra to theoretical physics. She revolutionized the theories of rings, fields and algebras. In physics Noether’s theorem explains the fundamental connection between symmetry and conservation laws. The innovative approach to modern abstract algebra of Emmy Noether not only produced major new results, but also inspired highly productive work by students and colleagues who emulated her techniques.

Amalie Emmy Noether to give her the full name was born on March 23, 1882 in Erlangen in Germany. Her father Max Noether was professor of mathematics at the University of Erlangen while her mother Ida Amalia was the daughter of a wealthy Jewish family of Cologne. She was better known as Emmy. She loved to dance and enjoyed music. She attended the Municipal school for the Higher Education of Daughters until she was 18. Emmy Noether showed early proficiency in French and English. In the spring of 1900 she took the examination for teachers of these languages and received an overall score of very good. Her performance qualified her to teach languages at schools reserved for girls, but she chose instead to continue her studies at the University of Erlangen. This was an unconventional decision because as late as 1900, women were not allowed to enroll in Universities in Germany. Professors frequently refused permission for women event to attend their lectures, and only very rarely women allowed taking university examination. The obstacle in a way could not deter her to enroll in a university, she had to get permission of the professors to take an entrance exam -- she did and she passed, after sitting in on mathematics lectures at the University of Erlangen. She was then allowed to audit courses -- first at the University of Erlangen and then the University of Göttingen, neither of which would permit a woman to attend classes for credit. Finally, in 1904, the University of Erlangen decided to permit women to enroll as regular students, and Emmy Noether returned there. She began to focus solely on mathematics. Under the supervision of Paul Gordan she wrote her dissertation, Über die Bildung des Formensystems der ternären biquadratischen Form (On Complete Systems of Invariants for Ternary Biquadratic Forms). Her dissertation in algebraic math earned her a doctorate summa cum laude in 1908.

Having obtained her doctorate, Noether was well qualified for a position at a university, but the persistent sexiest atmosphere in Germany prevented the brilliant young woman from being able to even apply for a job. This depressed her greatly, but she began helping her father in his research. She also started publishing papers on her own, which were so well received that she was invited to join a number of European Mathematical Societies, including the German Mathematical Association, which had been founded by George Cantor, but still she could not obtain a paying position at a University in Germany. By 1915 Noether was a famous mathematician in her own right and her papers were read with interest throughout the world. They dealt primarily with algebra.

In 1915, Emmy Noether's mentors, Felix Klein and David Hilbert, invited her to join them at the Mathematical Institute in Gottingen, again without compensation. There, she pursued important mathematical work that confirmed key parts of the general theory of relativity. Noether arrived at Gottingen and began her work on invariance in mathematical physics. Meanwhile, Klein rallied to get her appointed a professor at Gottingen, but he had to struggle with the administration until 1919, when his request was finally granted. She became a privatdozent allowing her to teach students and students would pay her directly. In 1922, the University gave her a position as an adjunct professor with a small salary and no tenure or benefits. Soon after arriving at Göttingen, however, she demonstrated her capabilities by proving the theorem now known as Noether's theorem, which shows that a conservation law is associated with any differentiable symmetry of a physical system. 

American physicists Leon M. Lederman and Christopher T. Hill argue in their book Symmetry and the Beautiful Universe that Noether's theorem is "certainly one of the most important mathematical theorems ever proved in guiding the development of modern physics, possibly on a par with the Pythagorean Theorem". Her work went far beyond mathematical physics. She made important contributions to Galois Theory, to many other areas of abstract algebra and to topology. Noether was, in fact the greatest algebraist of her time.

Noether’s groundbreaking work in algebra began in 1920 with a paper on non commutative fields. Her work earned her enough recognition that she was invited as a visiting professor in 1928-1929 at the University of Moscow and in 1930 at the University of Frankfurt. In 1932 Emmy Noether and Emil Artin received the Ackermann Teubner Memorial Award for their contributions to mathematics.  Though she was never able to gain a regular faculty position at Göttingen, she was one of many Jewish faculty members who were purged by the Nazis in 1933.  

In America, the Emergency Committee to Aid Displaced German Scholars obtained for Emmy Noether an offer of a professorship at Bryn Mawr College in America, and they paid, with the Rockefeller Foundation, her first year's salary. The grant was renewed for two more years in 1934. This was the first time that Emmy Noether was paid a full professor's salary and accepted as a full faculty member. In 1934, Noether began lecturing at the Institute for Advanced Study in Princeton upon the invitation of Abraham Flexner and Oswald Veblen. She also worked with and supervised Abraham Albert and Harry Vandiver. Her time in the United States was pleasant, surrounded as she was by supportive colleagues and absorbed in her favorite subjects. 

In April 1935 doctors discovered a tumor in Noether's pelvis. She was admitted to hospital for surgery to remove a uterine tumor. Although the operation was successful but on April 14, she fell unconscious and she developed a high fever resulting to her death.

Weyl said to her funeral – The memory of her work in science and of her personality among her fellows will not soon pass away. She was a great mathematician, the greatest, I firmly believe, her sex has ever produced, and a great woman.

Diophantus

Diophantus

Diophantus of Alexandria was a Greek mathematician. He was the first Greek mathematician who recognized fractions as numbers. He allowed positive rational numbers for the coefficients and solutions. He played a major role in the development of algebra and was a considerable influence on later number theorists. Diophantine analysis, which is closely related to algebraic geometry, has experienced a resurgence of interest in the past half century is due to him.

Little is known about Diophantus except the fact that he was born around 200 AD in Alexandria and died at the age of 84 in 284 AD. The claims that Diophantus lived from about 200 to 284 and spent time at Alexandria are based on detective work in finding clues to the times he flourished in his and others’ writing. Theon of Alexandra quoted Diophantus in 365 and his work was the subject of a commentary written by Theon’s daughter Hypatia at the beginning of 5^th century, which unfortunately is lost. Michael Psellus, an 11^th century Byzantine scholar mentioned that Diophantus dealt with Egyptian arithmetic which later led to other people working and developing the same branch of mathematics. The most details we have of Diophantus’s life come from the Greek Anthology which estimates his life span to be around 84 years. This anthology says----

 His boyhood lasted ¹/₆th of his life; he married after ¹/₇th more; his beard grew after ¹/₁₂th more, and his son was born 5 years later; the son lived to half his father's age, and the father died 4 years after the son.

Diophantus is best known for his famous work Arithmetica, written on the solution of algebraic equations and on the theory of numbers. The Arithmetica is a collection of 130 problems giving numerical solutions of determinate equations (those with a unique solution), and indeterminate equations. The method of solving indeterminate equation is known as Diophantine analysis. An equation is indeterminate if it has more than one variable. The Pythagorean Triplets containing the integers x, y and z satisfying the equation x² + y² = z² is the example of Diophantine analysis. Historically, the problem of solving Diophantine equation has been to find expressions that show the relationship of the integral values of x and y that satisfy an indeterminate equation ax + by = c, where the coefficients a, b and c are integers. Such equations have infinite number of solutions.

The Arithmetica had 13 volumes but only six of the original 13 books were thought to have survived. However, an Arabic manuscript in the library Astan-i-Quds (The Holy Shrine Library) in Meshed Iran has a title claiming it is a translation of Arithmetica and was discovered in 1968. In Arithmetica he considered the solutions of linear and quadratic equation. Diophantus looked at three types of quadratic equations ax² + bx = c, ax² = bx + c and ax² + c = bx. Since there was no symbol of zero when the book was written so Diophantus took all coefficients to be positive. Diophantus considered negative or irrational square root solutions "useless", "meaningless", and even "absurd". There is no evidence that suggests Diophantus even realized that there could be two solutions to a quadratic equation. Besides that, he also solved problem from Simultaneous equations.

Looking at the work of Diophantus it was quite clear that he was fully aware of the fact that every number can be written as the sum of four squares which was later proved by Lagrange. Here are two examples of the problems solved in Arithmetica.

1.      Find four numbers, the sum of every arrangement three at a time being given: say 22, 24 , 27 and 20. (Answer:- 9, 7 , 4 and 11)

2.      Divide a number, such as 13 which is the sum of two squares 4 and 9, into two other squares. (Answer:- 324/25 and 1/25)

In the Arithmetica, Diophantus stated certain theorem without proofs. It is believed that his proofs were found in his lost book The Porism, a collection of lemmas. One lemma states that the difference of the cubes of two rational numbers is equal to the sum of the cubes of two other rational numbers. For any given numbers a, b there exist numbers c, d such that a³ - b³= c³ + d³.

Diophantus' work has had a large influence in history. Editions of Arithmetica exerted a profound influence on the development of algebra in Europe in the late sixteenth and through the seventeenth and eighteenth centuries. Diophantus and his works have also influenced Arab mathematics and were of great fame among Arab mathematicians. Diophantus' work created a foundation for work on algebra and in fact much of advanced mathematics is based on algebra. European mathematicians did not learn of the gems in Diophantus's Arithmetica until Regiomontanus wrote in 1463:-No one has yet translated from the Greek into Latin the thirteen Books of Diophantus, in which the very flower of the whole of arithmetic lies hid...

Bombelli translated much of the work in 1570 but it was never published. Bombelli did borrow many of Diophantus's problems for his own Algebra. The most famous Latin translation of the Diophantus's Arithmetica is due to Bachet in 1621. It is that edition which Fermat studied. Certainly Fermat was inspired by this work which has become famous in recent years due to its connection with Fermat's Last Theorem. Diophantus earliest work was probably his short essay on polygonal numbers, containing ten propositions in which he employed the classical method in which numbers are represented by line segments.  The topic on polygonal numbers was of great interest to Pythagoreans.

Diophantus worked before the introduction of modern algebraic notation, but he moved from rhetorical algebra to syncopated algebra, where abbreviations are used. Diophantus made important advancement in mathematical notation. 

He was the first person to use algebraic notation and symbolism. Before him everyone wrote out equations completely. Diophantus introduced an algebraic symbolism that used an abridged notation for frequently occurring operations, and an abbreviation for the unknown and for the powers of the unknown. 

Mathematical historian Kurt Vogel states: “The symbolism that Diophantus introduced for the first time, and undoubtedly devised himself, provided a short and readily comprehensible means of expressing an equation... Since an abbreviation is also employed for the word ‘equals’, Diophantus took a fundamental step from verbal algebra towards symbolic algebra.”

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Augustin Louis Cauchy

Augustin Louis Cauchy

The greatest French Mathematician who has been described as the first revolution of rigour in mathematics is undoubtedly Augustin Louis Cauchy. He contributed in a major way to almost every branch of mathematics. It is humorously said—Cauchy did not so much master mathematics, mathematics mastered Cauchy.

Augustin Louis Cauchy was born in Paris on August 21, 1789. He was christened Augustin because he was born in August and Louis after his father’s name Louis Francois. His father Louis Francois was a parliamentary lawyer and a lieutenant of police in Paris. His mother Marie- Madeleine came from a very reputed Parisian family. Cauchy’s childhood fell in the bloodiest period of the revolution. In 1794, the Cauchy family fled to Arcueil to escape the terror. Louis Francois could hardly manage to feed his family and as a consequence Cauchy grew up delicate and underdeveloped physically.

Two years later the Cauchy family returned to Paris where Louis Francois began to rebuild his carrier under the new regime and was appointed as secretary general to the newly constituted Senate of which Laplace was chancellor. Louis had a great friendship with Lagrange’s and Laplace and they used to visit Cauchy’s house every now and then. Lagrange was impressed by the ability of Augustin and on his advice he was enrolled at the Ecole Centrale de Pantheon to study humanities. Pointing to Louis, Lagrange once said--- he will supplement all of us so far as we are mathematicians. Don’t let him touch a mathematical book till he is 17.

At the age of 13, Augustin Cauchy entered the Central school of Pantheon. He won the grand prize for the best student instituted by Napoleon. On leaving the school in 1804, Cauchy won the sweepstakes and a special prize in mathematics. For the next ten months he studied mathematics intensively with a good tutor and at the age of 16, he entered the Ecole Polytechnic. By 1810, he was a qualified Junior Engineer.

In March 1810, he left Paris to work on the construction of a naval base at Cherbourg, the Port Napoleon.  Cauchy took 4 books with him to Cherbourg-- The Mecanique celeste of Laplace, The Traite des fonctions analytiques of Lagrange, The Imitation of Christ by Thomas a Kempls and the work of Virgil. He remained there for 3 years where he used to do some mathematical research in part time.

In December 1810, he had begun to go over again all branches of mathematics, beginning with Arithmetic and finishing with Astronomy. Some of his discoveries were sufficiently significant to attract the attention of learned society in Paris particularly the memoir on polyhedral and that on symmetric functions, which included the germs of the fundamental ideas that eventually blossomed into Group Theory. He also extended the formula of Euler connecting the numbers(E), faces(F) and vertices(V) of a polyhedron given as E+2 =F+V

He also developed the theory of determinants but he got his due recognition as a brilliant mathematician when he submitted a work on Calculation of Definite Integrals to the Academy in 1814. In 1815, Cauchy created a sensation by proving one of the great theorems which Fermat has bequeathed to a baffled posteiry.

Every positive integers is a sum of three triangles, four squares, five pentagons, six hexagons and so on---

In 1816, he received the Grand Prize offered by the Academy for a theory of the propagation of waves on the surface of a heavy fluid of indefinite depth. In 1816 itself he was elected a member of the Paris Academy. This success made him the enemies of those great mathematician likes Monge and Carnot who were displaced from Academy. In the same year, he was appointed Assosiate Professor of Analysis at the Ecole Polytechnic.

The discovery with which Cauchy’s name is most firmly associated is his fundamental theorems in Complex Analysis. The first pivotal theorem proved by Cauchy, now known as Cauchy's integral theorem, was the following:

Where  f(z) is a complex-valued function holomorphic on and within the non-self-intersecting closed curve C (contour) lying in the complex plane. The contour integral is taken along the contour C.

 He is therefore known as the father of Complex Analysis.  In 1818, Cauchy married with Aloise de Bure. She was a close relative of the publisher who published most of Cauchy's works. They had two daughters Marie Francoise Alicia and Marie Mathilde born in 1819 and 1823 respectively. Encouraged by Laplace, Cauchy in 1821 wrote up for publication of his lecture on Analysis given in Ecole Polytechnic. This book contains the definition of limits and continuity and convergence of infinite series which we still read in the book of Calculus.

In 1825, Cauchy started his own monthly journal Exercises de mathemaliques. Its pages were exclusively filled by Cauchy himself. Cauchy's writings covered notable topics including: the theory of series, where he developed the notion of convergence and discovered many of the basic formulas for q-series. The theory of numbers and complex quantities; he was the first to define complex numbers as pairs of real numbers. 

The turmoil in 1830 in Paris had a great impact on the mind of Cauchy. Cauchy lost most of his positions at the institutes because he refused to take the oath of allegiance to the new king, Louis-Philippe.  He left for Switzerland and remain there in self imposed exile for eight years where he was offered the Post of Professor in Mahematical Physics by Charles Albert, the King of Sardinia.  He taught in Turin during 1832-1833. In 1831, he had been elected a foreign member of the Royal Swedish Academy of Sciences. In 1833, Cauchy was appointed tutor of Charles’ grandson.

Cauchy went back to Paris in 1838 when he finished his work with Charles X in Prague, and resumed his involvement with the Academy. At the time, because Cauchy was a mathematician, he was exempted from the oath of allegiance. After the establishment of the Second Republique in 1848, Cauchy resumed his position at the Sorbonne. Cauchy continued with his writings and publications through the remainder of his life. The long hour of work made him ill. On medical advice he left for Paris on May 12, 1857 but his health suddenly detorieted and he died on 23 rd May 1857 at the age of 67. His name is one of the 72 names inscribed on the Eiffel Tower.

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Casting Out Nines

                          Casting Out Nines

When you solve a sum whether it is addition, subtraction, multiplication or division, a doubt constantly haunts you. To avoid any mistake you begin to check every step and thus lose your precious time. Don’t you think that there should be a single method which is equally beneficial in all the basic mathematical operations? The answer to your question is YES, there is a method to check all the four fundamental mathematical operation i.e. addition, subtraction, multiplication or division and that too in few seconds. This wonder method is called Casting out Nines or Nines- Remainder Method.

How does this method work?

This is a big question. But you will be surprised to know that this method is so simple that even a primary school going kid can understand and check his/her calculation.

 Casting out Nines literally means to throw nines. Now let us focus on its working.

v  Add the digits of a number across, dropping out 9, to get a single figure. If it is not a single figure, add the digits obtained so as to get a single figure between  0 to 8.

v  9 is not taken into account in this process, as a digit sum of 9 is the same as a digit sum of zero.

Before coming to the actual problem, shouldn’t we understand the basic meaning of above two points? Let us take some examples to understand what meaning the above line has.

Example: Find the digit sum of 54653

Verification: Digit sum of 54653= 5 + 4 + 6 + 5 +3 = 23

                Since 23 is a double figure number so to get a single figure, we have to sum it again.

                 Digit sum of 23 = 2 + 3 = 5

The digit sum of 54653 can be done in other way very easily. As discussed above, we need not take 9 into account

Digit sum of 54653 = 5   +   4    +    6 +      5   +      3

The two group of number 5+ 4 and 6+ 3 can easily be left out while finding the digit sum of 54653, as their sum is equal to 9.

Example: Find the digit sum of 438219

Verification: Add all the digits

                        4 + 3 + 8 + 2 + 1 + 9 = 27

                 Digit sum of 27 = 2 + 7 = 9 

Check for Addition

Whatever we do to the numbers, we also do to their digit sum; then the result that we have from the digit sum of the numbers, must be equal to the digit sum of answer.

Example: Verify 47385 + 69384 + 58769+ 38173 + 29464 = 243144

Verification:                                                   Digit sum of number

                        47385                                                    0             

                        69384                                                    3

                        58769                                                    8

                        38173                                                    4

               +       29464                                                    3

                       243144                                               ?

LH.S. = Digit sum of 243144 = 2 + 4 + 3 + 1 + 4 + 4 = 0

R.H.S. = Sum of the Digit sum of number = 0 + 3 + 8 + 4 + 3 = 0

Since LHS = RHS

Result Verified

Example: Verify   87643 + 84397+ 38549 + 29765 = 240354

Verification                                                                    Digit sum of number

                        87643                                                                    1

                        84397                                                                    4

                        38549                                                                    2

                        29765                                                                    2

                        240354                                                                  ?

Check for subtraction

The process of subtraction is same as applied in addition. Remember that the value of the digit sum of minuend should be greater than that of subtrahend.

Example: Verify the result, 8934 – 6758 = 2176

Verification:                                                                Digit sum of number

                        8934                                                       6

                 -    6758                                                        8

                      2176                                                         ?

Since the digit sum of Minuend is less than the digit sum of Subtrahend, hence we need to replace the digit sum of 8934, in such a way that the final digit sum remains the same.

Here the value of the digit sum of 8934 = 6

This digit sum 6 can also be written in so many ways as the digit sum of 15, 24, 33, 42, 51 and 60 also gives the same value 6. Therefore, the need of the hour is to replace the digit sum of 8934 with any of the given value 15, 24, 33,42,51,60.

Digit sum of number

                        8934                                       15 or 24 or 33 or 42 or 51 or 60

                 -    6758                                                        8

                      2176                                                         ?

LHS = Digit sum of 2176 = 2 + 1 + 7+ 6 = 7

RHS = Digit sum of (15 – 8) = 7

          Or, Digit sum of (24 – 8 = 16) = 1 + 6 = 7

          Or, Digit sum of (33 – 8 = 25) = 2 + 5 = 7

          Or, Digit sum of (42 – 8 = 34) = 3 + 4 = 7

          Or, Digit sum of (51 – 8 = 43) = 4 + 3 = 7

          Or, Digit sum of (60 – 8 = 52) = 5+ 2 = 7

In all the above cases

LHS = RHS

LH.S. = Digit sum of 240354 = 2 + 4 + 0 + 3 + 5 + 4 = 0

R.H.S. = Sum of the Digit sum of number = 1 + 4 + 2 + 2 = 0

Check for Multiplication

Multiplication is the most error prone fundamental operation in mathematics. Students always have doubts about the accuracy of their result and waste time in re-checking every operation again and again. This method will prove a panacea for all those who are not very much sure about their result. Let us take few examples to understand the modus operandi of this method.

We know,

Multiplicand X Multiplier = Product

 Example: Verify 12876 x 43853 = 564651228

Verification:-          

 Digit sum of Multiplicand = 1 + 2 + 8 + 7 + 6 = 6

 Digit sum of Multipliers =4 + 3 + 8 + 5 + 3 = 5                              

LHS = Digit sum of Multiplicand and Multiplier taken together (6 x 5= 30) = 3 RHS= Digit sum of Product =5 + 6 + 4 + 6 + 5+ 1 + 2 + 2 + 8 = 3

Since LHS = RHS

Result Verified

Example: Verify 5972 X 4853 = 29882116

Verification:-          

 Digit sum of Multiplicand = 5 + 9 + 7 + 2 = 5

 Digit sum of Multipliers = 4 + 8+ 5 + 3 = 2                     

LHS = Digit sum of Multiplicand and Multiplier taken together (5 X 2= 10)= 1 RHS=  Digit sum of Product = 2 + 9 + 8 + 8 + 2 + 1 + 1 + 6 = 1

Since LHS = RHS

Result Verified

Example: Verify 12589476 x 43256853 = 54058115267908

Verification:-          

 Digit sum of Multiplicand =1 + 2 + 5 + 8 + 9 + 4+ 7 + 6 = 6

 Digit sum of Multipliers =4 + 3 +2 + 5+ 6+ 8 + 5 + 3 = 5                             

LHS = Digit sum of Multiplicand and Multiplier taken together (6 x 5= 30) = 3 RHS= Digit sum of Product = 5+ 4+ 0+ 5+ 8+1+1+ 5+ 2+ 6+ 7+ 9+ 0+ 8 = 7

Since LHS ≠ RHS

Result Incorrect

Check for Division

We Know

Dividend = Divisor x Quotient + Remainder

The casting out nines method described in the very beginning will suffice to check the division operation effectively. You have only one thing to do-

Find the digit sum of Dividend, Divisor, Quotient and Remainder and put the value of digits sum in LHS and RHS. If the same digit sum is obtained in both the side, it ultimately tells you that you have performed the right operation. Let us take some examples to understand how effectively this method work for Division.

Example: Verify 876543 ÷ 123, Q = 7126 and R = 45

Verification:-

Here Dividend = 876543

Digit sum of Dividend = 8 + 7 + 6 + 5 + 4 + 3 = 6

Divisor = 123

Digit sum of Divisor = 1 + 2 + 3 = 6

Quotient = 7126

Digit sum of quotient = 7 + 1 + 2 + 6= 7

Remainder = 45

Digit sum of Remainder = 4 + 5 = 0

Putting the digit sum value in the given formulae, we get,

L.H.S = Digit sum of Dividend = 6

R.H.S = Divisor x Quotient + Remainder

          = 6 X 7 + 0 = 42

Digit sum of 42 = 6

Hence LHS = RHS

Result Verified

Example: Verify 8765 ÷ 243, Q = 36 and R = 23

Finding percentage in your head

Finding Percentage in your head

Percentage is merely a two place decimal without the decimal point shown. Percent has its origin from the Latin word per centum meaning per hundred. It can best be defined as: - A fraction whose denominator is 100 is called a percentage and the numerator of the fraction is called the rate percent.

The importance of percentage can be estimated with the fact that whether you go to Bank for taking loan or go to shopping mall for buying something you always feel the importance of calculating percentage. Let’s take some example to understand the importance of percent in daily life.

a) If the bank offers you the interest rate of home loan @10% per annum it simply means that you have to pay ` 10 per ` 100.

b) Suppose while reading newspaper early in the morning you get an ad claiming that you will get 50% + 50% off on buying a pair of jeans and in the evening you rush to the shop in a hope you will get 100% discount on jeans then you are really thinking it other way because you have miscalculated the percentage discount the company or shop is offering you.

c) We all know that in order to get 1^st Division in School/ College examination you need to acquire 60% marks of the total.  

It is not all, in every competitive examination you appear; you encounter some problem on percentage directly or indirectly.

This altogether shows that it is an important tool for our life. Here I shall not deal with the typical problem asked in examination involved percentage which comes in the form of Profit and Loss, Simple or Compound Interest etc. but I shall simply show you the way how you can calculate the simple percentage.

This clearly shows the importance of percentage. 1 % of a number is 1/100 of the number; 15% of a number is 15/100 and so on. This reminds me a question which is crucial in primary level and is used to some extent in senior or competitive level – what percent one number is of other. Let me explain it before I move to the main discussion.

What percent one number is of other?

Follow these simple steps and get the answer in your mind.

·         Put the number that follows the word what percent is in the numerator of the fraction

·         Place the other number to the denominator of the fraction.

·         Reduce the fraction in simplest part if possible and finally multiply the result by 100.

Example; - What percent is 16 of 80?

Solution: - Place 16 in numerator and 80 at denominator and multiply the fraction by 100.

Example: What percent is 36 of 4?

Solution: - Place 36 at the numerator and 4 at the denominator and multiply the result by 100

Let’s learn some special technique to find a certain percentage of a number. These method will give you an edge to do the calculation fast.

Finding 2 ½ % of a number

Suppose you are asked to find 2 ½ % of a number you will simply convert it into a fraction and multiply the number by 5/2 and divide it by 100 but this will take undoubtedly a minute from you. Let’s learn some simple trick and do it in your mind.

1.      Divide the number whose 2 ½ % you are going to calculate by 4

2.      Move decimal point one place to left.

Hey, did you get the answer? Of course, YES. Isn’t it super simple?

Let me take some examples to put focus on its working.

Example: Find 2 ½ % of 86

Solution: -  

Divide 86 by 4; 86 ÷ 4 = 21.5                                                                                                             Move decimal point one place to left = 2.15

Hence, 86 × 2 ½ % = 2.15

Example: Find 2 ½ % of 648

Solution: -  

Divide 648 by 4; 648 ÷ 4 = 162                                                                                                          Move decimal point one place to left = 16.2

Hence, 648 × 2 ½ % = 16.2

Finding 5% of a number

Let me ask you a simple question: - Are you comfortable dividing a number by 2? Your answer is a big YES. Finding 5% of a number is as simple as dividing a number by 2. Let’s see how it works.

1.      Divide the number by 2

2.      Move the decimal point one place to left

Example: - Find 5% of 850

Solution: - Divide 850 by 2; 850 ÷ 2 = 425

                  Move decimal point one place to left = 42.5

Hence, 850 × 5% = 42.5

Example: - Find 5% of 326

Solution: - Divide 326 by 2; 326 ÷ 2 = 163

                  Move decimal point one place to left = 16.3

Hence, 326 × 5% = 16.3

Finding 10% of a number

Finding 10% of a number is a child’s play. You follow the simple steps and get the answer in second.

1.       Simply Move the decimal point one place to left

Example: - Find 10% of 729

Solution: - Move decimal point one place to left = 72.9

Hence, 729 ×10% = 72.9

Example: - Find 10% of 2549

Solution: - Move decimal point one place to left = 254.9

Hence, 2549 ×10% = 254.9

Finding 15% of a number

If I ask you to multiply a number by 2 then you will comfortably do it. If I again ask you to divide a number by 2 then also you will get your answer correct. That all you have to do in order to find 15% of a number. Follow the steps.

1.      Divide the number by 2

2.      Multiply the result obtained by 3

3.      Move the decimal point one place to left

Example: - Find 15% of 43

Solution: - Divide 43 by 2; 43 ÷ 2 = 21.5

                  Multiply it by 3 = 21.5 × 3 = 64.5

                  Move decimal point one place to left = 64.5

Hence, 43 × 15% = 64.5

Example: - Find 15% of 438

Solution: - Divide 483 by 2; 438 ÷ 2 = 219

                  Multiply it by 3 = 219 × 3 = 657

                  Move decimal point one place to left = 65.7

Hence, 438 × 15% = 65.7

Finding 20% of a number

 Follow the steps.

1.      Divide the number by 5

Example: - Find 20% of 43

Solution: - Divide 43 by 5; 43 ÷ 5 = 8.6

Hence, 43 × 20% = 8.6

Example: - Find 20% of 348

Solution: - Divide 348 by 5; 348 ÷ 5 = 69.6

Hence, 348 × 20% = 69.6

Finding 25% of a number

Follow the steps.

1.      Divide the number by 4

Example: - Find 25% of 86

Solution: - Divide 86 by 4; 86 ÷ 4 = 21.5

Hence, 86 × 25% = 21.5

Example: - Find 25% of 484

Solution: - Divide 484 by 4; 484 ÷ 5 = 121

Hence, 484 × 25% =121

Finding 33 ¹/₃ % of a number

Follow the steps.

1.      Divide the number by 3

Example: - Find 33 ¹/₃% of 69

Solution: - Divide 69 by 3; 69 ÷ 3 = 23

Hence, 69 × 33 ¹/₃ % = 23

Example: - Find 33 ¹/₃ % of 921

Solution: - Divide 921 by 3; 921 ÷ 3 = 307

Hence, 921 × 33 ¹/₃ % = 307

Finding 40 % of a number

Follow the steps.

1.      Multiply the number by 4

2.      Move decimal one point left

Example: - Find 40% of 24

Solution: - Multiply the number by 4:   24×4 = 96

                  Move decimal one point left = 9.6

Hence, 24 × 40 % = 9.6

Example: - Find 40% of 49

Solution: - Multiply the number by 4:   49×4 = 196

                  Move decimal one point left = 19.6

Hence, 49 × 40 % = 19.6

Finding 45% of a number

Multiplying a number by 9 is very easy. Leave the unit digit apart and subtract 1 more than the remaining digits from the original digit and place it in the LHS, in the RHS place the complement of unit digit.

Example: - 112 x 9 =?

Solution: - 112 – (11+ 1) / Complement of unit digit 2 = 1008

In order to find 45% of a number follow the steps.

1.      Divide the number by 2

2.      Multiply the result obtained by 9

3.      Move the decimal point one place to left

Example: - Find 45% of 36

Solution: - Divide 36 by 2; 36 ÷ 2 = 18

                  Multiply it by 9 = 18 ×9 = 162

                  Move decimal point one place to left = 16.2

Hence, 36 × 45% = 16.2

Example: - Find 45% of 640

Solution: - Divide 640 by 2; 640 ÷ 2 = 320

                  Multiply it by 9 = 320 ×9 = 2880

                  Move decimal point one place to left = 288

Hence, 460 × 45% = 288

Finding 50% of a number

It is as simple as asking a Grade 7 student to read the table of 2. Simply divide the number whose 50% you are intend to find by 2 and you get the answer.

Example: - Find 50% of 630

Solution: - Divide 630 by 2: 630÷2 = 315

Example: - Find 50% of 6850

Solution: - Divide 6850 by 2: 6850÷2 = 3425

Finding 55% of a number

Multiplying a number by 11 is very easy. In Multiplication chapter I have described a simple rule to multiply any number by 11. Simply put two zeros (along both side one each) with the number and keep adding from right to left to get the answer.

Example: - 112 ×11 =?

Solution: - 0(112)0 = 0+1 / 1+1 / 1+2 / 2+0 = 1232

In order to find 50% of a number follow the steps.

1.      Divide the number by 2

2.      Multiply the result obtained by 11

3.      Move the decimal point one place to left

Example: - Find 55% of 36

Solution: - Divide 36 by 2; 36 ÷ 2 = 18

                  Multiply it by 11 = 18 ×11 = 198

                  Move decimal point one place to left = 19.8

Hence, 36 × 55% = 19.8

Example: - Find 55% of 580

Solution: - Divide 640 by 2; 580 ÷ 2 = 290

                  Multiply it by 11 = 290 × 11 = 3190

                  Move decimal point one place to left = 319

Hence, 460 × 55% = 319

Read more on percentage in my book MATHS MADE EASY published by Rupa Publication

http://www.rupapublications.co.in/books/maths-made-easy

Send your comments at

Rajesh Kumar Thakur

rkthakur1974@gmail.com