Addition and subtraction in vedic maths

Vedic maths Addition

Usually if we want to add to numbers say 52 and 66 we would add the unit digits. and if there is any remainder we will bring it to the tens digit and atlast we will add the tens digit.

Now we will do it in the different way.

Let us take the same number 52 and 66.

The first step is to add the first number and the unit digit of the second number

ie. 52 + 6 = 58

The second step is to jump tens digit time.

our tens digit is 6.

so the answer is 118

Vedic Maths Subtraction

This Vedic Maths Subtraction method found as sutra in ancient vedas, is given below is very useful for such subtractions.

For example 1000 – 357 = ?

We simply take each figure in 357 from 9 and the last figure from 10.

Step 1. 9-3 = 6

Step 2. 9-5 = 4

Step 3. 10-7 = 3

So the answer is 1000 – 357 = 643

This always works for subtractions from numbers consisting of a 1 followed by

noughts: 100; 1000; 10,000 etc.

Similarly 10,000 – 1049 = 8951 (subtraction from 10000)

9-1 = 8

9-0 = 9

9-4 = 5

10-9 = 1

So answer is 8951

For 1000 – 83, in which we have more zeros than figures in the numbers being subtracted, we simply suppose 83 is 083.

So 1000 – 83 becomes 1000 – 083 = 917

Vedic maths multiplication

The Vedic method (the general method, at least) is based on the Urdhva-Tiryagbhyam sutra. A very terse sutra, it simply translates in English to say “vertically and crosswise”. The sutra is rather vague, so the technique, as well as an algebraic analysis of the technique, is presented in the following steps:

An Algebraic Perspective

All numbers n the base 10 number system (and number systems of any other base, for that matter) consist of a number of digits. Each digit represents a multiple times a power of 10 (or whatever the number system’s base is). So, for example, given a number like 52, we could rewrite it as 5*10+2.

Algebraically speaking, we can express any 3-digit number as: ax+b (where a, and b are integers).

So, suppose we wanted to multiply 2 2-digit numbers. We can express them in polynomial form. Then, by foiling:

(ax+b)(cx+d) = acx^{2}+(ad+bc)x+bd

Algebraic Multiplication for Higher Numbers of Digits

So, for brief review, 2-digit by 2-digit algebraic multiplication goes as follows (x in all of the following examples is the base of the number system being used, which is usually 10):

(ax+b)(cx+d) = acx^{2}+(ad+bc)x+bd

Expanding to 3-digit by 3-digit algebraic multiplication:

(ax^{2}+bx+c)(dx^{2}+ex+f) = adx^{4}+(ae+bd)x^{3}+(af+be+cd)x^{2}+(bf+ec)x+cf

Now, 4-digitby 4-digit algebraic multiplication (ax^{3}+bx^{2}+cx+d)(ex^{3}+fx^{2}+gx+h)=aex^{6}+(af+be)x^{5}+(ag+bf+ce)x^{4}+(ah+bg+cf+de)x^{3}+(bh+cg+df)x^{2}+(ch+dg)x+dh