Vinculum number is nothing but the mixture of positive digit & negative digit in same number. Where as we represent negative number by having bar over them.
Here 5 is a negative number & others are positive, so this number is a Vinculum Number.
We can simplify this number as,
In Vedic Mathematics by using Vinculum number we can write the higher number which is greater than 5 (6,7,8,9) in terms of lower number (0,1,2,3,4).
This is because it is easier to work with lower number, especially in Multiplication.
For subtraction also vinculum number is very useful.
A.How to get Vinculum number from Normal number
Here the Vedic sutra “ Nikhilam Navatashcaramam Dashatsh” means “ All from 9 and last from 10” and “Ekadhikena Purvrna” means “one more than previous” are useful.
Ex. i) 36
we convert this number in vinculum number step by step:
We move from Right to Left (R>>L).
Step 1: Identify the digit from right hand side which is greater than 5.
Here the number is 6.
Step 2: Here we apply sutra “ All from 9 & last from 10”. As 6 is our last digit we subtract it from 10. We get 4, write it as
Step 3: Now move towards left hand side but there is no digit that is greater than 5. (So, all from 9 is not applicable here.)
Step 4: Now we use “one more than previous”
Here previous number is 3, So 3 + 1 =4.
Ex. ii) 274
Step 1: As 7 is the last higher number (Because 4 is less than 5), we subtract it from 10.
We get 3, write it as
Step 2: Now move towards left, there is no higher number. So, take 2 as a previous digit, & one more than previous is 2 + 1= 3.
Step 3: Number 4 is unaffected, so it stays as it is.
Ex. iii) 379
Step 1: The last higher number is 9. So, 10 -9 = 1. We write it as
Step 2: Here in left side we have second higher digit 7, for this we apply first part of the “Nikhilam sutra…. All from 9…”. So, 9 – 7 =2. We write it as
Step 3: Add 1 to the previous number 3,we get 4.
Ex. iv) 483962
Step 1: Here last higher number is 6, so subtract it from 10 we get 4. We write it as .
Step 2: we have second higher number 9, so subtract it from 9, 9-9 =0.
Step 3: Move towards left, number 3 is there which is not a higher number, so we add 1 to the previous number. We get 4.
Step 4: Now we have one higher number, 8. solve it separately. Consider it as last higher number, subtract it from 10, we get 2. Write down it as
Step 5: Previous digit is 4, one more than 4 is 5.
Step 6: Last digit 2 is unaffected so we write down as it is.
B)Vinculum to Normal Number:
Here also again we start from right to left.
Step 1: Find first bar digit and take it’s 10’s complement.
Step 2: If next digit is again a bar digit take it’s 9 complement, continue until a non-bar digit is obtained.
Step 3: Decrease immediate non bar digit by 1.
Step 1: First bar digit is 3, so take its 10’s compliment. 10-3 =7
Step 2: Here only one bar digit is given, so step 2 is not applicable here.
Step 3: Immediate digit is 5 here, so decrease it by 1, 5-1 =4
Step1: First bar digit is 6, It’s 10 complement no. is, 10-6=4.
Step 2: Next bar digit is 4, we take it’s 9 compliment , 9-4 =5.
Step 3: Decrease immediate no. by 1, 7-1 = 6
i) Subtraction using Vinculum number
We have a difficulty whenever it comes to subtract large number from small number. Here Vinculum method is useful for Subtraction.
Ex. i) 457657 – 228294
Step 1: First we write one number below other number.
Step 2: Subtract the below number from upper no. as it is. If bigger number is subtracting from smaller number, then just perform bigger number – Small number & give bar over it.
Step 3: Now convert this vinculum number in to normal number.
ii)Multiplication by using Vinculum:
Ex.i) 212 x 167