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Number Systems  (Class XI)‏ Nita Arora (KHMS)‏ PGT Comp. Sc.
Why Binary? ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],0 1 No Yes False True Off On
Counting and Arithmetic ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Keeping Track of the Bits ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Numbers: Physical Representation ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Number System ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
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
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
Positional Notation:  Octal ,[object Object],Sum for Base 10  Evaluate Value Place 4 x 1 2 x 8 6 x 64 4 16 384 8 0 8 1 8 2 1 8 64 64’s place 8’s place 1’s place
Positional Notation:  Hexadecimal ,[object Object],4 x 1 0 x 16 7 x 256 6 x  4,096 Evaluate 4 0 1,792 24,576 Sum for Base 10 16 0 16 1 16 2 16 3 Place 1 16 256 4,096 Value 4,096’s place 256’s place 1’s place 16’s place
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
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
Range of Possible Numbers ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
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
Base or Radix ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Number of Symbols  vs. Number of Digits ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
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 )‏
Addition Largest Single Digit Problem Base   1 +0   6 +9   6 +1   6 +3 1 Binary F Hexadecimal 7 Octal 9 Decimal
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
Binary Arithmetic 1 1 0 0 0 0 0 1 0 1 1 0 1 + 1 0 1 1 0 1 1 1 1 1 1 1
Converting from Base 10 ,[object Object],256 64 4 2 1 16 4,096 65,536 16 1 8 512 4,096 32,768 8 1 2 8 16 32 64 128 256 2 0 1 3 4 5 6 7 8 Power Base
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
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
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
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
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
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
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
From Base 16 to Base 2 ,[object Object],[object Object],[object Object],[object Object],0111 0110 1111 0001 Base  2 7 6 F 1 Base 16
Fractions ,[object Object],[object Object],[object Object],[object Object],[object Object]
Decimal Fractions ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
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
Mixed Number Conversion ,[object Object],[object Object],[object Object],[object Object]

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Number System

  • 1. Number Systems (Class XI)‏ Nita Arora (KHMS)‏ PGT Comp. Sc.
  • 2.
  • 3.
  • 4.
  • 5.
  • 6.
  • 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
  • 9.
  • 10.
  • 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
  • 13.
  • 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
  • 15.
  • 16.
  • 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
  • 20. Binary Arithmetic 1 1 0 0 0 0 0 1 0 1 1 0 1 + 1 0 1 1 0 1 1 1 1 1 1 1
  • 21.
  • 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
  • 29.
  • 30.
  • 31.
  • 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
  • 33.