- The document discusses number systems and bases, including binary, decimal, octal, and hexadecimal.
- It explains positional notation and how numbers are represented in different bases using place values that are powers of the base.
- The range of numbers that can be represented depends on the base and number of digits used. More digits allow larger numbers to be represented.
1. Number Systems (Class XI) Nita Arora (KHMS) PGT Comp. Sc.
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7. Positional Notation: Base 10 43 = 4 x 10 1 + 3 x 10 0 Sum Evaluate Value Place 3 40 3 x1 4 x 10 1 10 10 0 10 1 1’s place 10’s place
8. Positional Notation: Base 10 527 = 5 x 10 2 + 2 x 10 1 + 7 x 10 0 500 5 x 100 100 10 2 Sum Evaluate Value Place 7 20 7 x1 2 x 10 1 10 10 0 10 1 1’s place 10’s place 100’s place
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11. Positional Notation: Binary 1101 0110 2 = 214 10 Sum for Base 10 Evaluate Value Place 0 x 1 1 x 2 1 x 4 0 x 8 1 x16 0 x 32 1 x 64 1 x 128 0 2 4 0 16 0 64 128 1 2 4 8 16 32 64 128 2 0 2 1 2 2 2 3 2 4 2 5 2 6 2 7
12. Estimating Magnitude: Binary 1101 011 02 = 214 10 1101 011 02 > 192 10 (128 + 64 + additional bits to the right) Sum for Base 10 Evaluate Value Place 0 x 1 1 x 2 1 x 4 0 x 8 1 x16 0 x 32 1 x 64 1 x 128 0 2 4 0 16 0 64 128 1 2 4 8 16 32 64 128 2 0 2 1 2 2 2 3 2 4 2 5 2 6 2 7
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14. Decimal Range for Bit Widths 38+ 19+ 9+ 6 4+ 3 2+ 1+ 0+ Digits Approx. 2.6 x 10 38 128 Approx. 1.6 x 10 19 64 4,294,967,296 (4G) 32 1,048,576 (1M) 20 65,536 (64K) 16 1,024 (1K) 10 256 8 16 (0 to 15) 4 2 (0 and 1) 1 Range Bits
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17. Counting in Base 2 Decimal Number Equivalent Binary Number 10 1 x 2 1 1 x 2 3 1010 9 1 x 2 0 1 x 2 3 1001 8 1 x 2 3 1000 7 1 x 2 0 1 x 2 1 1 x 2 2 111 6 1 x 2 1 1 x 2 2 110 5 1 x 2 0 1 x 2 2 101 4 1 x 2 2 100 3 1 x 2 0 1 x 2 1 11 2 0 x 2 0 1 x 2 1 10 1 1 x 2 0 1 0 0 x 2 0 0 1’s (2 0 ) 2’s (2 1 ) 4’s (2 2 ) 8’s (2 3 )
18. Addition Largest Single Digit Problem Base 1 +0 6 +9 6 +1 6 +3 1 Binary F Hexadecimal 7 Octal 9 Decimal
19. Addition Answer Carry Problem Base Carry the 2 Carry the 16 Carry the 8 Carry the 10 10 1 +1 Binary 10 6 +A Hexadecimal 10 6 +2 Octal 10 6 +4 Decimal
22. From Base 10 to Base 2 10 0 1 0 64 6 42/32= 1 Integer Remainder 1 0 1 0 1 2 4 8 16 32 2 1 2 3 4 5 Power Base 42 10 = 101010 2 10/16 = 0 10 10/8 = 1 2 2/4 = 0 2 2/2 = 1 0 0/1 = 0 0
23. From Base 10 to Base 2 Most significant bit 1 2 ) ( 0 2 2 ) 42 Base 10 101010 Base 2 ( 1 5 2 ) ( 0 10 2 ) ( 1 21 2 ) ( 0 Least significant bit 42 2 ) Remainder Quotient
24. From Base 10 to Base 16 5,735 10 = 1667 16 103 – 96 = 7 1,639 –1,536 = 103 5,735 - 4,096 = 1,639 Remainder 7 103 /16 = 6 1,639 / 256 = 6 5,735 /4,096 = 1 Integer 7 6 6 1 1 16 256 4,096 65,536 16 0 1 2 3 4 Power Base
25. From Base 10 to Base 16 5,735 Base 10 1667 Base 16 0 16 ) ( 1 Most significant bit 1 16 ) ( 6 22 16 ) ( 6 358 16 ) ( 7 Least significant bit 5,735 16 ) Quotient Remainder
26. From Base 10 to Base 16 8,039 Base 10 1F67 Base 16 0 16 ) ( 1 Most significant bit 1 16 ) ( 15 31 16 ) ( 6 502 16 ) ( 7 Least significant bit 8,039 16 ) Quotient Remainder
27. From Base 8 to Base 10 3,584 x 7 512 8 3 = 3,763 10 3 48 128 Sum for Base 10 x 3 x 6 x 2 1 8 64 8 0 8 1 8 2 Power 7263 8
28. From Base 8 to Base 10 = 3,763 10 7263 8 + 3 = + 6 = + 2 = 3,763 3760 464 58 x 8 470 x 8 56 7 x 8
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32. Fractions: Base 10 and Base 2 .101011 2 = 0.671875 10 .2589 10 .008 8 x 1/1000 1/1000 10 -3 .2 2 x 1/10 1/10 10 -1 Sum Evaluate Value Place .0009 .05 9 x1/1000 5 x 1/100 1/10000 1/100 10 -4 10 -2 0 x 1/16 1/16 2 -4 0.03125 1 x 1/32 1/32 2 -5 0.015625 0.125 .5 Sum 1 x 1/64 1x 1/8 0 x 1/4 1 x 1/2 Evaluate 1/64 1/8 1/4 1/2 Value 2 -6 2 -3 2 -2 2 -1 Place