Vedic Mathematics/Why Does It Work?

Introduction

It should be understood that there is no magic in the many techniques described in the other chapters, indeed there is no magic in mathematics in general, it can be argued that mathematics is the purest of all sciences as there is no opinion and mathematics needs no experiments or interpretation of results; things are either true, (i.e. they are proven to be true), or they are not. That being the case, there must be sound reasons why all the previously described techniques work.
The reason some of the techniques work is simply that they perform a well understood algorithm (e.g. long multiplication) in a more efficient way, (often due to particular problem properties, e.g. the technique of multiplying any number by 11), even if it is difficult to see this at first. Other techniques work by making use of less widely understood mathematical laws, (e.g. algebra, quadratic equations, modular or 'clock' arithmetic, etc.). In either case, it is not necessary to know why a technique works to be able to use it, (much like you don't need to know how a car works to be able to drive one). It is for this reason, as well as to make the previous chapters more immediately usable, that the description of why each technique works has been omitted.
However, for those that are curious and want to investigate further, this chapter describes why many of the Vedic mathematics techniques work. Remember that some of the descriptions below will require knowledge of areas of mathematics that you may not be familiar with. Hopefully this will give you the impetus to investigate these areas and expand your mathematical knowledge, (this is a very rewarding way to discover new aspects of a subject). However even if this is not the case, you can (and should) still use the techniques and be happy in the knowledge that even if you don't know how the techniques work, they will still improve your numerical and arithmetic skills.
Think of this section as an appendix, useful for further study, but not essential to the understanding of the main theme of the book.

Addition and Subtraction Techniques

Multiplication Techniques

Multiplying by 11

When multiplying by 11 using long multiplication a pattern to the working out can be discovered e.g.

 46     876     4386      432672
 11x     11x      11x         11x
 --     ---     ----     -------
 46     876     4386      432672
460+   8760+   43860+    4326720+
---    ----    -----    --------
506    9636    48246     4759392
---    ----    -----    --------

You can see that in the addition section of each long multiplication above, each column apart from the first and last is the sum of the original digit in the column and the next one (to the right). Once you know this you can just write down the result of multiplying any number by 11.
Working from right to left:

1. Write the rightmost digit down.
2. Add each pair of digits and write the result down right to left (carrying digits where necessary).
3. Finally write down the left most digit.

e.g.

• Multiply 712x11

${\begin{matrix}&7&&1&&2\\&\swarrow \searrow &+&\swarrow \searrow &+&\swarrow \searrow \\7&&8&&3&&2\end{matrix}}$

712x11=7832

The reason for working from right to left instead of the more usual left to right is so any carries can be added in as you go along. e.g.

• Multiply 8738x11

${\begin{matrix}&8&&7&&3&&8\\&\swarrow \searrow &+&\swarrow \searrow &+&\swarrow \searrow &+&\swarrow \searrow \\9&\leftarrow _{1}&6&\leftarrow _{1}&1&\leftarrow _{1}&1&&8\end{matrix}}$

8738x11=96118

Multiplying numbers close to a power of 10

In the techniques section it is shown that the Vertically and Crosswise sutra can be used to easily multiply numbers that are close to 100. It is then shown that the same technique can be used to multiply any numbers near a power of 10, and that in fact the general technique will work for any numbers near any base, the key factor being that the technique is useful if the initial subtractions result in numbers that are easier to multiply. To understand why this technique works, you need a basic understanding of algebra, and quadratic equations.

Consider two numbers A and B that are to be multiplied together and a third number X that is close to both numbers (we will call X the 'base' ). We assume that the numbers A and B are difficult to multiply and so we are looking for an easier alternative that only involves addition, subtraction and the multiplication of easier (e.g. smaller or simpler) numbers. The key is to realise that since X is close to both numbers we can generate smaller numbers (that are hopefully easier to work with) related to A and B by subtracting each from X (we will call these smaller numbers a and b). i.e.

 ${\begin{aligned}&a=X-A\\&b=X-B\\Thus:\\&A=X-a\quad (1)\\&B=X-b\quad (2)\end{aligned}}$

We can multiply A and B by substituting for them using equations (1) and (2) above, i.e.

 ${\begin{aligned}AB&=(X-a)(X-b)\\&=X^{2}-aX-bX+ab\\&=X(X-a-b)+ab\end{aligned}}$

Now we have something we can work with! You can see from the equation above that we can replace the multiplication of A and B with some subtractions of small numbers (X-a-b) a multiplication of the result of this subtraction by the 'base' number X and then the addition of a small multiplication ab. (Remember a and b are small because X is close to A and B and a=X-A, b=X-B). The only multiplication that might be difficult is the multiplication of X by the result of the initial subtraction (X-a-b), however if we choose X carefully (e.g. by making X a power of 10) we can make sure that this multiplication is simple too. With this knowledge we can now make sense of the Vertically and Crosswise multiplication technique. i.e.

${\begin{matrix}A&\longrightarrow &(X-A)\\\\B&\longrightarrow &(X-B)\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}A&\longrightarrow &a\\\\B&\longrightarrow &b\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}A&&a\\&&\downarrow \\B&&b\\\hline &&ab\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}\quad \quad A&&a\\&\nwarrow &\\\quad \quad B&&b\\\hline (A-b)&&ab\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}\quad \quad A&&a\\&&\\\quad \quad B&&b\\\hline (X-a-b)&&ab\end{matrix}}$

Perhaps the cleverest bit is that if the base number X is an appropriate power of 10, both the multiplication of (X-a-b) by X and the subsequent addition of ab is handled automatically by the positional shift of the digits caused by appending the ab digits to the end of the (X-a-b) digits. The only potential problem left is if the product ab is equal to or larger than the base X. In this case the positional shift of the (X-a-b) digits will be one too many, so instead the leading digit of the product ab must be 'carried' and then added to the (X-a-b) value.