Addition & Subtraction - Other Number System

Subtraction of Octal number, Hexadecimal and Decimal can be done using compliments too. Therefore, a few examples are given below for Octal - 7 and 8 compliment, Hexadecimal - 15 and 16 compliments and for Decimal - 9 and 10 compliments.

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Octal Number

Example : Using 7 Compliments

(1546)₈ - (467)₈


Minuend : 1546 and Subtrahend : 467. Minuend contain 4 digits where Subtrahend having 3, so add a leading 0 to Subtrahend 0467. Now 7'th compliment mean subtract the Subtrahend from 7's.


  7 7 7 7
  0 4 6 7
-------------
  7 3 1 0
-------------

Now add the Subtrahend to the minuend

   1
   1 5 4 6
+ 7 3 1 0
-------------
1 1 0 5 6         (Discard the Carry [1])
-------------
    1 0 5 6
            1  ( Adding Carry digits [Refer to Binary subtraction])
-------------
    1 0 5 7

Result of : (1546)₈ - (467)₈ = (1057)₈

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Example : Using 8 Compliments

(1546)₈ - (467)₈

Minuend : 1546 and Subtrahend : 467. Minuend contain 4 digits where Subtrahend having 3, so add a leading 0 to Subtrahend 0467. Now 7'th compliment mean <> subtract the Subtrahend from 7's.

  7 7 7 7
  0 4 6 7
-------------
  7 3 1 0
+        1     (Add one to the 7 compliment)
-------------
  7 3 1 1    Now add the Subtrahend to the minuend
-------------  

           1
   1 5 4 6
+ 7 3 1 1
-------------
1  1 0 5 7  (discard the Carry digits)
-------------

Result of : (1546)₈ - (467)₈ = (1057)₈

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Hexadecimal Number

Subtraction of Hexadecimal number can be done with with 15 or 16 compliments

Example : Using 15 Compliments

(15CB)₁₆ - (2BC)₁₆ = 1 5 12 11 - 2 11 12

Minuend : 15CB and Subtrahend : 2BC Minuend contain 4 digits where Subtrahend having 3, so add a leading 0 to Subtrahend 0 2 11 12. Now 15'th compliment mean  subtract the Subtrahend from 15's.

15 15 15 15
  0   2 11 12
-------------
 15 13  4  3    Now add the Subtrahend to the minuend
-------------

1 1
   1  5 12  11
 15 13   4   3
---------------
1 1 3 0 14
              1 ( Remove the extra carry digits and add it to the remainder)
-------------
1 3 0 15 = 130F

Result of : (15CB)₁₆ - (2BC)₁₆ = (130F)₁₆

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Example : Using 16 Compliments

(15CB)₁₆ - (2BC)₁₆ = 1 5 12 11 - 2 11 12

Minuend : 15CB and Subtrahend : 2BC Minuend contain 4 digits where Subtrahend having 3, so add a leading 0 to Subtrahend 0 2 11 12. Now 15'th compliment mean subtract the Subtrahend from 15's.

15 15 15 15
  0   2 11  12
---------------
 15 13  4   3
+               1 (Add 1 [16 Compliment])
---------------
 15 13  4  4  Now add the Subtrahend to the minuend
---------------
  1   1
  1   5 12 11
15 13  4    4
--------------
1 1  3   0  15 ( Discard the Carry digit)
--------------
1 3 0 15 = 130F

Result of : (15CB)₁₆ - (2BC)₁₆ = (130F)₁₆

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Decimal Number

We can do it for our (yes our) Decimal number system too. Subtraction of decimal number can be done with with 9 or 10 compliments

Example : Using 9 Compliments

(1497)₁₀ - (879)₁₀

Minuend : 1497 and Subtrahend : 879 Minuend contain 4 digits where Subtrahend having 3, so add a leading 0 to Subtrahend 0879 . Now 9'th compliment mean subtract the Subtrahend from 9's.

 9 9 9 9
 0 8 7 9
----------
 9 1 2 0                 Now add the Subtrahend to the minuend
----------
     1
 1  4  9  7
 9  1  2  0
--------------
1 0 6 1 7
            1      ( Remove the extra carry digits and add it to the remainder) +1
---------------
6 1 8

Result of : (1497)₁₀ - (879)₁₀=(618)₁₀

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Example : Using 10 Compliments

(1497)₁₀ - (879)₁₀

Minuend : 1497 and Subtrahend : 879 Minuend contain 4 digits where Subtrahend having 3, so add a leading 0 to Subtrahend 0879 . Now 9'th compliment mean subtract the Subtrahend from 9's.

  9 9 9 9
  0 8 7 9
----------
  9 1 2 0       (add 1 to make it 10 compliment)
        + 1
----------
  9 1 2 1        Now add the Subtrahend to the minuend
----------

      1
   1 4 9 7
   9 1 2 1
--------------
1 0 6 1 8      (Discard the carry digits)
-------------
      6 1 8 
-------------

Result of : (1497)₁₀ - (879)₁₀= (618)₁₀

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We yet to finish with number system, but promise you, next one will be the last
in these series!.

Arithmetic Operation -Octal & Hexadecimal

Octal Addition and Subtraction : Radix or Base for Octal numbers are 8 and consists 0 to 7. While adding we follow following rules:

Addition

We already know that 8 is the base or radix of the Octal Number System, after adding of two digit if you get below 8 then write down it, but if it is more than 7 then divide the result by 8 and write down the remainder in the place of result and the quotient as carry numbers.

Example-1

(447)₈ + (622)₈
                      
                                                                    Rules

       1                     1) 7 + 2 = 9 (above 7) 9/8 Quotient=1 Remainder=1 (Carry = 1)
 ------------            2) 4 + 2 + 1(carry)=7 is below 8 (Carry = 0)
    4 4 7                   3) 4 + 6 = 10 (above 7) 10/8 Quotient=1 Remainder=2(Carry = 1)
    6  2 2
 ------------
  1 2 7 1
 ------------
  Answer :  (447)₈ + (622)₈= (1271)₈

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Example-2

(237.54)₈ + (753.74)₈
    
                                                            Rules

    1 1 1 1                1) 4 + 4 = 8 (8/8 Quotient=1 Remainder=0)(Carry = 1)
---------------          2) 5 + 7 + 1 (carry) = 13(13/8 Quotient=1 Remainder=5)Carry = 1)
    3 7 . 5 4               3) 7 + 2 + 1 (carry) = 10(10/8 Quotient=1 Remainder=2)(Carry = 1)
    7 2 . 7 4              4) 3 + 7 + 1 (carry) = 11(11/8 Quotient=1 Remainder=3)(Carry = 1)
---------------
 1 3 2 . 5 0
---------------

  Answer :  (37.54)₈ + (72.74)₈= (132.50)₈

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Subtraction

Octal Subtraction : While subtracting if minuend digit smaller than subtend then add 8 to the minuend digit and then subtract and in the process 1 should be borrow from the minuend digit at the end.

Example-1

(345)₈ - (126)₈

                                                    Rules
     1                     1) 5 is less than 6 so 5+8 = 13 - 6= 7 (Borrow = 1)
------------          2) 4 -1(borrow) = 3 is bigger than subtend so 3 - 2=1(Borrow = 0)
   3 4 5                 3) 3 -1 = 2 (Borrow = 0)
   1 2 6
 ------------
   2 1 7
 ------------
  Answer :  (345)₈ - (126)₈ = (217)₈

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Example-2

(246.24)₈ - (157.56)₈

                                                           Rules


   1 1 1    1           1) 4 + 8 - 6 = 6 (Borrow = 1)
 --------------      2) 2-1(borrow) =1 + 8 - 5 = 4 (Borrow = 1)
    2 4 5 .2 4         3) 5-1(borrow) =4 + 8 - 7 = 5 (Borrow = 1)
   1 5 7 . 5 6         4) 4-1(borrow) =3 + 8 - 5 = 6 (Borrow = 1)
 ---------------       5) 2-1(borrow) =1 - 1 = 0 (Borrow = 0)
   0 6 5 . 4 6
 ---------------

  Answer :  (246.24)₈ - (157.56)₈ = (65.56)₈

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Hexadecimal Addition and Subtraction : Radix or Base for Octal numbers are 16 and consists 0 to 15 (0 - 9, A, B, C, D, E, F). While adding we follow following rules:

Addition

We already know that 16 is the base or radix of the Hexadecimal Number System, after adding of two digit if you get below 16 then write down it, but if it is more than 15 then divide the result by 15 and write down the remainder in the place of result and the quotient as carry numbers. Replace Alphabet with its equivalent decimal numbers. e.g.. A = 10 or E = 14

Example-1

B83)₁₆ + (A2D)₁₆ = (11 8 3)₁₆ + (10 2 13)₁₆

                                                             Rules

         1                 1) 3 + 13 = 16(above 15) 16/16 Quotient=1 Remainder=0(Carry=1)
 ------------          2) 8 + 2 + 1(carry)= 11 is below 15 (Carry=0)
    11   8   3         3) 10 + 11 = 21(above 15) 21/16 Quotient=1 Remainder=5(Carry=1)
    10   2 13
 ------------
  1 5   11 0  = 1 5 B 0
 ------------

  Answer :(B83)₁₆ + (A2D)₁₆ = (15B0)₁₆

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Example-2

(AB.CD)₁₆ + (D5.2F)₁₆ = (10 11.12 13)₁₆ + (13 5.2 15)₁₆

                                                      Rules

   1          1                1) 13 + 15 = 28 (28/16 Quotient=1 Remainder=12(Carry = 1))
 -----------------        2) 12 + 2 + 1(carry) = 15 is below 15 (Carry = 0))
   10 11 .12 13          3) 11 + 5 = 16 (16/16 Quotient=1 Remainder=0 (Carry = 1))
   13  5  .  2 15         4) 10 + 13 + 1(carry)= 24 (24/16 Quotient=1 Remainder=8 (Carry = 1))
-----------------
  1 8 0  . 15 12       = 180.FC₁₆
------------
  Answer : (AB.CD)₁₆ + (D5.2F)₁₆  = (180.FC)₁₆

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Subtraction

Hexadecimal Subtraction : While subtracting if minuend digit smaller than subtend then add 16 to the minuend digit and then subtract and in the process 1 should be borrow from the minuend

digit at the end.

Example-1

(BA3)₁₆ - (AB4)_{16 =} (11 10 3)₁₆ - (10 11 5)₁₆)

                                                             Rules

         1              1) 3 is less than 5 so 3 + 16 = 19 - 4= 15 (F) (Borrow = 1)
-------------       2) 10 - 1(borrow) = 9 + 16 = 25 - 11 = 14 (E) (Borrow = 1)
  11 10  3           3) 11 -1 = 10 = 10 - 10 = 0 (Borrow = 0)
  10 11  4
------------
   0   E   F
------------

  Answer : (BA3)₁₆ - (AB4)₁₆ = (EF)₁₆

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Example-2

(C5.A2)₁₆ - (A7.B3)₁₆ =  ((12 5.10 2)₁₆ - (10 7.11 3)₁₆)

                                               Rules

   1    1    1                 1)  2 + 16 - 3 = 15 (F) (Borrow = 1)
---------------              2)  10-1(borrow) =9 + 16 - 11 = 14 (E) (Borrow = 1)
  12  5 . 10  2             3)  5-1(borrow) =4 + 16 - 7 = 13 (D) (Borrow = 1)
  10  7 . 11  3             4)  12-1(borrow) =11 - 10 = 1
---------------
  1  D .  E   F
---------------
  Answer : (C5.A2)₁₆ - (A7.B3)₁₆ = (17.EF)₈

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After fortnight coming back with some more with system only. !

Binary Addition & Subtraction - Back to the basic

Binary Addition  : Radix or Base for Binary numbers are 2 and having two digits 0 and 1. so while adding we follow following rules:

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1
1 + 1 = 0 and carry 1 (Decimal 2 and Binary 10)

1 + 1 + 1 = 1 and carry 1 (Decimal 3 and Binary 11)

1 + 1 + 1 + 1 = 0 carry 10 (Decimal 4 and Binary 100)

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Example : 1				Example : 2
Carry >>	1 1 1 1		Carry >>	1  1 1 1      1
	1 1 1 0 1			1 0 1 0 . 1 0 1
	0 1 1 1 1			1 1 0 1 . 1 0 1
Answer :	1 0 1 1 0 0		Answer :	1 1 1 0 0 . 0 1 0

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Binary Subtraction
While subtracting we follow the following rules:

0 - 0     = 0

0 + 1     = 1  borrow 1

1 + 0     = 1

1 + 1     = 0

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Example : 1				Example : 2
Borrow >>	1 1 1		Borrow >>	1
	1 1 1 0 1			1 0 1 0 . 1 0 1
	0 1 1 1 1			1 0 0 1 . 1 0 0
Answer :	0 1 1 1 0		Answer :	0 0 0 1 . 0 0 1

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Subtraction often is very tricky, you can solve it by using 1's Compliment or 2's Compliment

Like you have 50 - 12 = 38 same 50 + (-12) = 38

Compliment means Invert or all 0's change to 1 and 1's become 0. So Rules for 1's Compliment is :

Make the both side equal by placing leading 0's. Replace all the 1's with 0's and the 0's with 1's, then add both the numbers.

If there was 4 bits and after adding you getting five, the leading bits known as carry bits. If there is a carry bits then remove that from leading position and add to the result of the addition and write the answer down.

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 Example :(10010)₂ - (1001)₂ Minuend : 10010 and Subtrahend : 1001. Minuend contain 5 bits where Subtrahend having 4, so it will be 01001 (both side with same numbers of digits and done by adding leading zeros) 01001 (decimal 9) 0110 (one's complement) Add

Example : 1			Addition
Carry >>			1 1 1
	18		1 0 0 1 0
	+ -9		1 0 1 1 0
			1 0 1 0 0 0
			0 1 0 0 0
			+             1
Answer :	+9		0 1 0 0 1

1 is a carry bit because after adding two four bits number we are getting 5 bits, remove the carry from there and add to the remaining part of the result.

Result of : (10010)₂ - (1001) = (1001)₂ or 18 + (-9) = 9

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The two's complement of a binary number is obtained by:

1. Replacing all the 1's with 0's and the 0's with 1's (1's Compliment).

2. Adding 1 to the result of 1's compliment using the rules of binary addition.

Now you have the two's complement. The decimal subtraction 18 - 9 = 9 is the same as adding (18) + (-9) = 9

1. Convert the number to be subtracted to its two's complement:(10010)₂ - (1001)₂ = (10010)₂ - (01001)₂, 01001 (decimal 9) 10110 (one's complement) + 00001 (add 1 (two's complement)) 10111 (We get)

10111 now represents -9.

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Example : 1			Addition
Carry >>			1 1 1
	18		1 0 0 1 0
	+ -9		1 0 1 1 1
Answer :	+9		1 0 1 0 0 1

Note that the final carry 1 is ignored and dropped permanently. So you get :Result of : (10010)₂ - (1001) = (1001)₂ or 18 + (-9) = 9

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If no carry that indicate a negative result. When a subtraction produce negative result then you have to re-compliment the result of addition to get the correct answer.
(1000)₂ - (1001)₂ = (1000)₂ - (0110)₂, (one's complement).

Example : 1			Addition
Carry >>
	8		1 0 0 0
	+ -9		0 1 1 0
			1 1 1 0

If no carry bit then re-compliment the result 1110 = 0001 (after 1's compliment)   Result of : (1000)₂ - (1001) = (-1)₂ or 8 + (-9) = -1  

Answer :  (1000)₂ - (1001)₂ = (-0001)₂

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With above example with 2's compliment :(1000)₂ - (1001)₂ 1001 (decimal 9) after 1'compliment we get 0110 + 1 (2's Compliment ) ------- 0111

Example : 1			Addition
Carry >>
	8		1 0 0 0
	+ -9		0 1 1 1
			0 1 1 1

If no carry bit then re-compliment the result

0111 = 1111 (after 1's compliment) 1111+1 = 0001(2's compliment)


The result will be -0001 Result of : (1000)₂ - (1001) = (-1)₂ or 8 + (-9) = -1

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 Binary Multiplication

Binary multiplication is same as normal multiplication only we have to know the rules of addition properly.
(1010.11)₂ x (101.10)₂

Example : 1
	1 0 1 0 . 1 1
	x                   1 0 1 . 1 0
	0 0 0 0 0 0
	1 0 1 0 1 1
	1 0 1 0 1 1
	0 0 0 0 0 0
	1 0 1 0 1 1
	1 1 1 0 1 1 . 0 0 1 0

Answer : (1010.11)₂ x (101.11)₂= (1javascript:void(0)11011.0010)₂

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Binary Division

Binary division too is same as decimal division but binary subtraction rules going used here.
(110011)₂ / (111)₂

Example : 1
	1 1 1	1 1 0 0 1 1	1
		1 1 1
		1 0 0 0	1
		1 1 1
		1 1 1	1
		1 1 1
		0 0 0
			1 1 1

Answer : (110011)₂ / (111)₂= (111)₂

!!!Hope it will be helpful!!!

Number Systems-II - This one not that systematic, do not get lost.

Last time we have seen the methods and way to convert the Decimal numbers to Binary, Octal and Hexadecimal. Today I will tell you how you can convert the Binary, Octal and Hexadecimal numbers to Decimal

Binary to Decimal

(101)₂=(?)₁₀  (Convert binary 101 to decimal). Now we have 101 means three digit, so let us consider n=3.

101 = 1 x 2^n-1+ 0 x 2^n-2 + 1 x 2^n-3

= 1 x 2^3-1 + 0 x 2^3-2 + 1 x 2^n-3

= 1 x 2² + 0 x 2¹ + 1 x 2⁰

= 1 x 4 + 0 x 2 + 1 x 1

= 4 + 0 + 1 = 5

= (101)₂ = (5)₁₀

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For fractional Numbers :

(.11)₂=(?)₁₀

Here 'n' is 2,
.11 = 1 x 2^n-(n+1) + 1 x 2^n-(n+2)
= 1 x 2^-1 + 1 x 2^n-2
= 1/2¹ + 1/2²
= 1/2 +1/4
= 3/4 = .75
= (.11)₂ - (.75)₁₀

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Another example : (11.011)₂ - (?)₁₀
= 1 x 2¹+ 1 x 2⁰             + 0 x 2^-1 + 1 x 2^-2 + 1 x 2^-3
 = 1 x 2¹ + 1 x 2⁰           + 0 x 2^-1 + 1 x 2^-2 + 1 x 2^-3
 = 2 + 1                         + 1/4 + 1/8
= 3.3/8  = 3.375
= (11.011)₂ - (3.375)₁₀

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Easy way out:

8 4 2 1   2 4
1 0 0 1 . 1 1
Check the numbers on the top, its 2⁰ onwards 2^n-1, values written on the top for whole number right most digit to start with and for fractional numbers left most digit. For whole numbers add the values for non zero elements like 1 and 8 because 4 and 2 having a '0' below it. So by adding 8 + 1 =9 and divide the values for fractional numbers: 1/2 + 1/4 = 3/4 = .75.

(1001.11)₂=(9.75)₁₀

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Octal to Decimal

if you understand the conversion of Binary to decimal then other number system will be same process with different power.

Octal to Decimal
(56.34)₈=(?)₁₀

Two digit means n=2,
 = 5 x 8^n-1+ 6 x 8^n-2          + 3 x 8^n-(n+1)+ 4 x 8^n-(n+2)
= 5 x 8¹+ 6 x 8⁰                 + 3/8¹+ 4 x 8²
= 5 x 8 + 6 X 1                  + 3/8 + 4/ 64
= 40 + 6                             + 3/8 + 4/ 64
= 46.4375
= (56.34)₈ = (46.4375)₁₀

---

Hexadecimal to Decimal

(75.55)₁₆=(?)₁₀
Two digit means n=2, so
 = 7 x 16^n-1+ 5 x 16^n-2         + 5 x 16^n-(n+1)+ 5 x 16^n-(n+2)
= 7 x 16¹+ 5 x 16⁰               + 5/16¹+ 5 x 16²
= 7 x 16 + 5 x 1                   + 5/16 + 5/256
= 112 + 5 + 5/16 + 5/256
= 117.332031

(75.55)₁₆ = (117.332031)₁₀

---

Another Example :

(3A.B)₁₆=(?)₁₀

Here 'A to F' to be considered as 10 to 15 and as a single unit
3 x 16^n-1+ 10 x 16^n-2        +11 x 16^n-(n+1)
= 3 x 16 + 10 x 1              + 11/16
=  48 + 10                        + 11/16
=   58.6875
= (3A.B)₁₆ = (58.6875)₁₀

---

Octal to Binary and Binary to Octal

Last number of Octal number system is 7. 7 equivalent to binary is 111, that means three digit. so while converting Octal to binary just write down the binary value of the octal number.

(56.34)₈=(?)₂

   5     6   .  3    4
101 110   011 100(56.34)₈=(101110.011100)₂

---

Same for Binary to Octal, only first you have to three digits of binary nose, where for whole number start from last digit and fractional numbers start from first digit. At the end if numbers of digits less than three than put a leading zero for whole number and a trailing zero for fractional part.

(1011011101.1101101)₂=(?)₈

001 011 011 101 . 110 110 100

1      3      3     5   .   6    6     4

(1011011101.1101101)₁=(13335.664)₈

---

Hexadecimal to Binary and Binary to Hexadecimal

Last number of Octal number system is F means 15. 15 equivalent to binary is 1111, that means four digit. so while converting Hexadecimal to binary just write down the binary value of the Hexadecimal number.

(7A.3B)₁₆=(?)₂

0111 1010   . 0011 1011

7         A      .    3       B

(7A.3B)₁₆=(01111010.00111011)₂

---

Same for Binary to Hexadecimal, only first you have to group four bits for whole numbers right to left and for fractions left to right  At the end if numbers of digits less than three than put a zero at leading position for whole numbers and a trailing zero for fractional part.
(1010011100.1101101)₂=(?)₈

 0010 1001 1100  .  1101 1010

2          9       12    .    13      10

(1011011101.1101101)₂=(29C.DA)₁₆

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Number System-I, Very Systematic

As promised, coming back after a week. this time we will talk about Number System and its conversation.

Positional number systems

Our decimal number system is known as a positional number system, because the value of the number depends on the position of the digits. For example, the number 156 has a very different value than the number 165, although the same digits are used in both numbers.

In a positional number system, the value of each digit is determined by which place it appears in the full number. The lowest place value is the rightmost position, and each successive position to the left has a higher place value.

In our decimal number system, the rightmost position represents the "ones" column, the next position represents the "tens" column, the next position represents "hundreds", etc. Therefore, the number 156 represents 1 hundred and 5 tens and 6 ones, whereas the number 165 represents 1 hundreds and 6 tens and 5 one.

Other number systems use different bases. The binary number system uses base 2, so the place values of the digits of a binary number correspond to powers of 2. The octal number system uses base 8, so the place values of the digits of a octal number correspond to powers of 8. The Hexadecimal number system uses base 16, so the place values of the digits of a binary number correspond to powers of 16. However, Hexadecimal number system also represent by alphabets. From 0 to 9 as numbers and 10 to 15 numbers are represent by alphabet A-F.

Decimal to Binary

(75.625)₁₀ -> (?)₂ First take the whole number and divide by 2, the radix or base of binary number or to which number system you want to convert.

Integer Part :

Notes	Process	Quotient	Remainder
Start by dividing the number by 2 In this case, 75 divided by 2 Quotient is 37 and the remainder is 1	75 / 2	37	1
Then, divide the quotient again by 2, and repeat it .	37 /2	18	1
	18 / 2	9	0
	9 / 2	4	1
	4 / 2	2	0
	2 / 2	1	0
Stop because the result is already 0 (0 divided by 2 will always be 0). Write down the numbers in remainders column in reverse order, last digit to the first.	1 / 2	0	1

Decimal Part

Notes	Process	Result	Whole number
Multiply the decimal part by 2 and The whole number part of the result is the first binary digit to the right of the point, put away it and continue multiple the fraction part	.625 * 2	1.25	1
No whole number indicate a '0'	.25 * 2	.50	0
When there are no fraction left, the conversion is complete. Here you have to write the values top to bottom order	.50 * 2	1.00	1
Sometime you will find the process going infinite nos. of times, just do till five or six places and write result with [approx]

The result : (75.625)₁₀ -> (1001011.101)₂

The process for converting decimal to octal or decimal to Hexadecimal will be same only the radix or base will change from 2 to 8 or 16.

Decimal to Octal

(75.625)₁₀ -> (?)₈ First take the whole number and divide by 2, the radix or base of binary number or to which number system you want to convert.

Integer Part :

Notes	Process	Quotient	Remainder
Start by dividing the number by 8 In this case, 75 divided by 8 Quotient is and the remainder is	75 / 8	9	3
Then, divide the quotient again by 8, and repeat it .	9 /8	1	1
Stop because the result is already 0 (0 divided by 8 will always be 0). Write down the numbers in remainders column in reverse order, last digit to the first.	1 / 8	0	1

Decimal Part

Notes	Process	Result	Whole number
Multiply the decimal part by 8 and The whole number part of the result is the first binary digit to the right of the point, put away it and continue multiple the fraction part	.625 * 8	5.000	5
When there are no fraction left, the conversion is complete. Here you have to write the values top to bottom order
Sometime you will find the process going infinite nos. of times, just do till five or six places and write result with [approx]

Result : (75.625)₁₀ = (113.5)₈

Decimal to Hexadecimal

(75.625)₁₀ -> (?)₁₆ First take the whole number and divide by 2, the radix or base of binary number or to which number system you want to convert.

Integer Part :

Notes	Process	Quotient	Remainder
Start by dividing the number by 16 In this case, 75 divided by 16 Quotient is and the remainder is	75 / 16	4	11
Then, divide the quotient again by 16, and repeat it .	4 / 16	0	4
Stop because the result is already 0 (0 divided by 8 will always be 0). Write down the numbers in remainders column in reverse order, last digit to the first.

Decimal Part

Notes	Process	Result	Whole number
Multiply the decimal part by 16and The whole number part of the result is the first binary digit to the right of the point, put away it and continue multiple the fraction part	.625 * 16	12.000	12
When there are no fraction left, the conversion is complete. Here you have to write the values top to bottom order
Sometime you will find the process going infinite nos. of times, just do till five or six places and write result with [approx]

Result : (75.625)₁₀ -> (4B.C)₁₆

Following chart show the relation between the four number system.

Decimal	Binary	Octal	Hexadecimal
0	0000	0	0
1	0001	1	1
2	0010	2	2
3	0011	3	3
4	0100	4	4
5	0101	5	5
6	0110	6	6
7	0111	7	7
8	1000	10	8
9	1001	11	9
10	1010	12	A
11	1011	13	B
12	1100	14	C
13	1101	15	D
14	1110	16	E
15	1111	17	F
16	10000	20	10

After fortnight I will back with converting other number system to decimal.