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BINARY NUMBER SYSTEM - PLACE VALUE CHART, CONVERSION TO AND FROM BASE-10 SYSTEM
There we have seen base-10 number system But, computers work internally with only two symbols Thus, the base-2 number system is the basis for digital computers. The two base symbols or digits used in binary system are Let us learn how to write numbers using the binary number system. This system is analogous to the decimal number system There, value of the place becomes ten times, Place value rules in Binary number SystemThe value of the right extreme place is one (1) or unity. Value of the place increases as it moves to the left. Value of the place becomes two times, as we move one place to the left. So the value of the place second from right is two times one and is equal to two. The value of the place third from right is two times two and is equal to four. The value of the place, fourth from right is two times four and is equal to eight. The value of the place, fifth from right is two times eight and is equalto sixteen. Thus, the next place values are thirty two, sixty four, Conversion of base-two numerals into base-ten numeralsThe following examples will make the process clear. Example 1 Binary number SystemThe following table shows the place value chart for 1001
In the above example, the value of 1001 Let us see how the single digit numbers in decimal number system Representation of single digit numbers
|
| Decimal System | Binary System |
| 0 | 0 |
| 1 | 1 |
| 2 | 10 |
| 3 | 11 |
| 4 | 100 |
| 5 | 101 |
| 6 | 110 |
| 7 | 111 |
| 8 | 1000 |
| 9 | 1001 |
Note :
(i) subscript (2) indicates that the number is in binary number system.
(ii) where no subscript (base) is shown, it should be taken as 10.
Example 2 Binary number System
Write 10010(2) in the decimal number system.
Solution :
| 16 | 8 | 4 | 2 | 1 |
| 1 | 0 | 0 | 1 | 0 |
The value of 10010(2)
= 1(16) + 0(8) + 0(4) + 1(2) + 0(1)
= 16 + 0 + 0 + 2 + 0 = 18. Ans.
Example 3 of Binary number System
Write 1110011(2) in the decimal system.
Solution :
| 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| 1 | 1 | 1 | 0 | 0 | 1 | 1 |
The value of 1110011(2)
= 1(64) + 1(32) + 1(16) + 0(8) + 0(4) + 1(2) + 1(1)
= 64 + 32 + 16 + 0 + 0 + 2 + 1 = 115. Ans.
Conversion of base-ten numerals into base-two numerals
We use division method. We successively divide by 2 and take
the remainder 0 or 1 in successive places starting from units' place.
We continue the process till the quotient is 0.
The following examples will make the process clear.
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Example 4 of Binary number System
Write 36 in the binary system.
Solution :
2 | 36 ------ 2 | 18 - 0 Units' place ------ 2 | 9 - 0 Twos' place ------ 2 | 4 - 1 Fours' place ------ 2 | 2 - 0 Eights' place ------ 2 | 1 - 0 Sixteens' place ------ | 0 - 1 Thirty two's place
Thus, 36 = 100100(2)
Example 5 of Number Systems : Binary System
Write 101 in the binary system.
Solution :
2 | 101 ------- 2 | 50 - 1 Units' place ------- 2 | 25 - 0 Twos' place ------- 2 | 12 - 1 Fours' place ------- 2 | 6 - 0 Eights' place ------- 2 | 3 - 0 Sixteens' place ------- 2 | 1 - 1 Thirty two's place ------- | 0 - 1 Sixty four's place
Thus, 101 = 1100101(2)
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Example 6 of Binary number System
Write 1227 in the binary system.
Solution :
2 | 1227 -------- 2 | 613 - 1 units' place -------- 2 | 306 - 1 Twos' place -------- 2 | 153 - 0 Fours' place -------- 2 | 76 - 1 Eights' place -------- 2 | 38 - 0 Sixteens' place -------- 2 | 19 - 0 Thirty twos' place -------- 2 | 9 - 1 Sixty Fours' place -------- 2 | 4 - 1 One hundred twenty eights' place -------- 2 | 2 - 0 Two hundred fifty six' place -------- 2 | 1 - 0 Five Hundred twelve's place -------- | 0 - 1 One thousand twenty four's place
Thus, 1227 = 10011001011(2)
Exercise of Binary number System
(1) Write the following numbers in decimal system.
(i) 1001(2) (ii) 11100(2)(iii) 1000101(2)
(iv) 110101(2)(v) 10001110(2) (vi) 1110001111(2)
(2) Write the following numbers in binary system.
(i) 47 (ii) 89 (iii) 721 (iv) 123 (v) 2049 (vi) 1123
For answers see at the bottom of the page.
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Answers to Exercise of Binary number System
(1) (i) 9 (ii) 28 (iii) 69 (iv) 53 (v) 142 (vi) 911
(2) (i) 101111 (ii) 1011001 (iii) 1011010001 (iv) 1111011
(v) 100000000001 (vi) 10001100011
