Vedic Mathematics
Vedic Mathematics is the name given to the ancient system of Mathematics which was rediscovered from the Vedas between 1911 and 1918 by Sri Bharati Krsna Tirthaji (1884-1960). According to his research all of mathematics is based on sixteen Sutras or word-formulae. For example, 'Vertically and Crosswise` is one of these Sutras. These formulae describe the way the mind naturally works and are therefore a great help in directing the student to the appropriate method of solution.
Perhaps the most striking feature of the Vedic system is its coherence. Instead of a hotch-potch of unrelated techniques the whole system is beautifully interrelated and unified: the general multiplication method, for example, is easily reversed to allow one-line divisions and the simple squaring method can be reversed to give one-line square roots. And these are all easily understood. This unifying quality is very satisfying, it makes mathematics easy and enjoyable and encourages innovation.
In the Vedic system 'difficult' problems or huge sums can often be solved immediately by the Vedic method. These striking and beautiful methods are just a part of a complete system of mathematics which is far more systematic than the modern 'system'. Vedic Mathematics manifests the coherent and unified structure of mathematics and the methods are complementary, direct and easy.
The simplicity of Vedic Mathematics means that calculations can be carried out mentally (though the methods can also be written down). There are many advantages in using a flexible, mental system. Pupils can invent their own methods, they are not limited to the one 'correct' method. This leads to more creative, interested and intelligent pupils.
Interest in the Vedic system is growing in education where mathematics teachers are looking for something better and finding the Vedic system is the answer. Research is being carried out in many areas including the effects of learning Vedic Maths on children; developing new, powerful but easy applications of the Vedic Sutras in geometry, calculus, computing etc.
But the real beauty and effectiveness of Vedic Mathematics cannot be fully appreciated without actually practising the system. One can then see that it is perhaps the most refined and efficient mathematical system possible.
The Vedic Mathematics Sutras
This list of sutras is taken from the book Vedic Mathematics, which includes a full list of the sixteen Sutras in Sanskrit, but in some cases a translation of the Sanskrit is not given in the text and comes from elsewhere.
This formula 'On the Flag' is not in the list given in Vedic Mathematics, but is referred to in the text.
The Main Sutras
| By one more than the one before. |
| All from 9 and the last from 10. |
| Vertically and Cross-wise |
| Transpose and Apply |
| If the Samuccaya is the Same it is Zero |
| If One is in Ratio the Other is Zero |
| By Addition and by Subtraction |
| By the Completion or Non-Completion |
| Differential Calculus |
| By the Deficiency |
| Specific and General |
| The Remainders by the Last Digit |
| The Ultimate and Twice the Penultimate |
| By One Less than the One Before |
| The Product of the Sum |
| All the Multipliers |
The Sub Sutras
| Proportionately |
| The Remainder Remains Constant |
| The First by the First and the Last by the Last |
| For 7 the Multiplicand is 143 |
| By Osculation |
| Lessen by the Deficiency |
| Whatever the Deficiency lessen by that amount and set up the Square of the Deficiency |
| Last Totalling 10 |
| Only the Last Terms |
| The Sum of the Products |
| By Alternative Elimination and Retention |
| By Mere Observation |
| The Product of the Sum is the Sum of the Products |
| On the Flag |
Vedic maths comes from the Vedic tradition of India. The Vedas are the most ancient record of human experience and knowledge, passed down orally for generations and written down about 5,000 years ago. Medicine, architecture, astronomy and many other branches of knowledge, including maths, are dealt with in the texts. Perhaps it is not surprising that the country credited with introducing our current number system and the invention of perhaps the most important mathematical symbol, 0, may have more to offer in the field of maths.
The remarkable system of Vedic maths was rediscovered from ancient Sanskrit texts early last century. The system is based on 16 sutras or aphorisms, such as: "by one more than the one before" and "all from nine and the last from 10". These describe natural processes in the mind and ways of solving a whole range of mathematical problems. For example, if we wished to subtract 564 from 1,000 we simply apply the sutra "all from nine and the last from 10". Each figure in 564 is subtracted from nine and the last figure is subtracted from 10, yielding 436.
The whole approach of Vedic maths is suitable for slow learners, as it is so simple and easy to use.
The sutra "vertically and crosswise" is often used in long multiplication. Suppose we wish to multiply
32 by 44. We multiply vertically 2x4=8.
Then we multiply crosswise and add the two results: 3x4+4x2=20, so put down 0 and carry 2.
Finally we multiply vertically 3x4=12 and add the carried 2 =14. Result: 1,408.
32 by 44. We multiply vertically 2x4=8.
Then we multiply crosswise and add the two results: 3x4+4x2=20, so put down 0 and carry 2.
Finally we multiply vertically 3x4=12 and add the carried 2 =14. Result: 1,408.
We can extend this method to deal with long multiplication of numbers of any size. The great advantage of this system is that the answer can be obtained in one line and mentally. By the end of Year 8, I would expect all students to be able to do a "3 by 2" long multiplication in their heads. This gives enormous confidence to the pupils who lose their fear of numbers and go on to tackle harder maths in a more open manner.
All the techniques produce one-line answers and most can be dealt with mentally, so calculators are not used until Year 10. The methods are either "special", in that they only apply under certain conditions, or general. This encourages flexibility and innovation on the part of the students.
Multiplication can also be carried out starting from the left, which can be better because we write and pronounce numbers from left to right. Here is an example of doing this in a special method for long multiplication of numbers near a base (10, 100, 1,000 etc), for example, 96 by 92. 96 is 4 below the base and 92 is 8 below.
We can cross-subtract either way: 96-8=88 or 92-4=88. This is the first part of the answer and multiplying the "differences" vertically 4x8=32 gives the second part of the answer.
How to find
Squares of Numbers ending in 5
in less than 3 seconds?
in less than 3 seconds?
35² 45² 85² etc.
Lets take for example
35
X 35
______________
X 35
______________
That the units place (first column on the right) adds up to 5 + 5 = 10 and
that the Ten's place (second column from the right) is the same ( 3 and 3)
that the Ten's place (second column from the right) is the same ( 3 and 3)
When this is true what we do is
3 5
x 3 5
________
12
________
12
Apply the word formula ‘By one more than the One before’
to the ten’s place.
to the ten’s place.
So what’s 1 more than 3? It is 4. So we multiply
3 with 4 to get 12. That’s the first part of our answer.
3 with 4 to get 12. That’s the first part of our answer.
To get the second part of our answer
3 5
x 3 5
________
1225
________
1225
We multiply 5 with 5 to get 25.
That’s the second part of our answer.
That’s the second part of our answer.
So our answer becomes 1225.
Now try
85
x 85
__________
__________
Solution
85
x 85
________
7225
________
7225
We multiply 5 with 5 to get 25. And
multiply 8 with the next number 9 to
get 72. So our answer becomes 7225.
Simple.
multiply 8 with the next number 9 to
get 72. So our answer becomes 7225.
Simple.
Now try 45² 55² 65² 95² 105² 125² etc.
in the same way.
in the same way.
Now what do we do with these sums?
26 48 67
x24 x 42 x 63
_____ _____ _____
_____ _____ _____
26 48 67
x 24 x 42 x 63
_____ _____ _____
_____ _____ _____
Note that the units add up to 10. (Numbers in red colour
7 + 3 = 10. 8 + 2 = 10. 6 + 4 = 10.) and the ten's place (second
column from the right) is same (6 and 6. 4 and 4. 2 and 2.)
7 + 3 = 10. 8 + 2 = 10. 6 + 4 = 10.) and the ten's place (second
column from the right) is same (6 and 6. 4 and 4. 2 and 2.)
So we can apply the same rule 'By one more than the One before'.
Lets see what we get.
Lets see what we get.
26 48 67
x 24 x 42 x 63
______ ______ ______
6|24 20|16 42|21
______ ______ ______
6|24 20|16 42|21
In 48x42we have 4 multiplied with the next
number 5 which gives us 20 and 8x2 which
gives us 16. Therefore, our answer is 2016.
number 5 which gives us 20 and 8x2 which
gives us 16. Therefore, our answer is 2016.
How to multiply numbers with
11-19 in less than 5 seconds?
11-19 in less than 5 seconds?
Multiply numbers with 11
Lets take for example 32 x 11
Step 1 : Spli 3 and 2 into left and right parts.
3 2
3 _ 2
3 _ 2
Step 2 : Simply add 3 + 2 and place 5 in between
3 2
\/
3 5 2
\/ 3 5 2
Step 3 : Our final answer is 352.
3 2
3 5 2
3 2
3 5 2
Try with other numbers now.
Now lets take for example 1234 x 11
The method is called Adding to the Neighbour.
In 1234 the neighbours of 2 are 1 & 3 and the
neighbours of 3 are 2 & 4 respectively.
In 1234 the neighbours of 2 are 1 & 3 and the
neighbours of 3 are 2 & 4 respectively.
Step 1 : We place two strokes beside
1234 and place two zeros.
1234 and place two zeros.
0 | 1 2 3 4 | 0
Step 2 : Now we start adding to the neighbour.
First we add 4 to 0
First we add 4 to 0
0 | 1 2 3 4 | 0
(4 + 0 = 4)
(4 + 0 = 4)
Gives us : 4
Step 3 : Add 3 to 4 gives us 7.
( 4 + 0 = 4 )
( 3 + 4 = 7 )
( 3 + 4 = 7 )
0 | 1 2 3 4 | 0
Gives us : 74
Step 4 : Add 2 to 3 gives us 5.
( 4 + 0 = 4 )
( 3 + 4 = 7 )
( 2 + 3 = 5 )
( 4 + 0 = 4 )
( 3 + 4 = 7 )
( 2 + 3 = 5 )
0 | 1 2 3 4 | 0
Gives us : 5 74
Gives us : 5 74
Step 5 : Add 1 to 2 gives us 3.
( 4 + 0 = 4 )
( 3 + 4 = 7 )
( 2 + 3 = 5 )
( 1 + 2 = 3 )
( 4 + 0 = 4 )
( 3 + 4 = 7 )
( 2 + 3 = 5 )
( 1 + 2 = 3 )
0 | 1 2 3 4 | 0
Gives us : 3 5 74
Step 6 : Finally Add 0 to 1 gives us 1.
( 4 + 0 = 4 )
( 3 + 4 = 7 )
( 2 + 3 = 5 )
( 1 + 2 = 3 )
( 0 + 1 = 1 )
( 4 + 0 = 4 )
( 3 + 4 = 7 )
( 2 + 3 = 5 )
( 1 + 2 = 3 )
( 0 + 1 = 1 )
0 | 1 2 3 4 | 0
Gives us our answer: 1 3 5 74
Simple isn't it? Now try 12345 x 11 on your own.
Now lets take for example 777 x 11
This sum will involve a carryover
Step 1 : We place two strokes beside
777 and place two zeros.
777 and place two zeros.
0 | 777 | 0
Gives us : 7
Step 2 : Now we start adding to the neighbour.
First we add 7 to 0.
( 7 + 0 = 7 )
First we add 7 to 0.
( 7 + 0 = 7 )
0 | 777 | 0
Step 3 : Add 7 to 7 gives us 14. We write
4 and take 1 as carry to the next step.
( 7 + 0 = 7 )
( 7 + 7 = ¹4 )
4 and take 1 as carry to the next step.
( 7 + 0 = 7 )
( 7 + 7 = ¹4 )
0 | 7 7 7 | 0
Gives us : ¹ 4 7
Step 4 : Add 7 to 7 gives us 14. We add 1 from carryover
gives us 15. We write 5 and again take 1 as carry over to
the next step.
( 7 + 0 = 7 )
( 7 + 7 = ¹4 )
( 7 + 7 = 14 + 1 = ¹5 )
gives us 15. We write 5 and again take 1 as carry over to
the next step.
( 7 + 0 = 7 )
( 7 + 7 = ¹4 )
( 7 + 7 = 14 + 1 = ¹5 )
Gives us : ¹ 5 4 7
Step 5 : Finally we add 0 to 7 and then add 1 of the carry
over to give us 8.
over to give us 8.
( 7 + 0 = 7 )
( 7 + 7 = ¹4 )
( 7 + 7 = 14 + 1 = ¹5 )
( 0 + 7 = 7 + 1 = 8 )
( 7 + 7 = ¹4 )
( 7 + 7 = 14 + 1 = ¹5 )
( 0 + 7 = 7 + 1 = 8 )
Gives us our final answer : 8 5 4 7
Simple isn't it? Now any number multiplied with 11.
No comments:
Post a Comment