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Number Systems & Operations

Number Systems & Operations. Part II. Arithmetic Operations with Signed Numbers. Because the 2’s complement form for representing signed numbers is the most widely used in computer systems. We’ll limit to 2’s complement arithmetic on: Addition Subtraction Multiplication Division.

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Number Systems & Operations

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  1. Number Systems & Operations Part II

  2. Arithmetic Operations with Signed Numbers • Because the 2’s complement form for representing signed numbers is the most widely used in computer systems. We’ll limit to 2’s complement arithmetic on: • Addition • Subtraction • Multiplication • Division

  3. Addition • 4 cases that can occur when 2 signed numbers are added: • Both numbers positive • Positive number with magnitude larger than negative number • Negative number with magnitude larger than positive number • Both numbers negative

  4. Addition • Both numbers positive: ex: 00000111 7 +00000100+4 00001011 11 • The sum is positive and is therefore in true (uncomplemented) binary.

  5. Addition • Positive number with magnitude larger than negative number: ex: 00001111 15 +11111010+ -6 1 00001001 9 • The final carry bit is discarded. The sum is positive and is therefore in true (uncomplemented) binary. Discard carry

  6. Addition • Negative number with magnitude larger than positive number: ex: 00010000 16 +11101000+ -24 11111000 -8 • The sum is negative and therefore in 2’s complement form.

  7. Addition • Both numbers negative: ex: 11111011 -5 +11110111+ -9 1 11110010 -14 • The final carry bit is discarded. The sum is negative and therefore in 2’s complement form. Discard carry

  8. Addition • Remark: • The negative numbers are stored in 2’s complement form so, as you can see, the addition process is very simple: Add the two numbers and discard any final carry bit.

  9. Addition • Overflow condition: • When two numbers are added and the number of bits required to represent the sum exceeds the number of bits in the two numbers, an overflow results as indicated by an incorrect sign bit. • An overflow can occur only when both numbers are + or -. ex: 01111101 125 +00111010+ 58 10110111 183 Magnitude incorrect Sign incorrect

  10. Addition • Numbers are added two at a time: • Computer add strings of numbers two numbers at a time. ex:add the signed numbers: 01000100, 00011011, 00001110, and 00010010 68 01000100 + 27+ 00011011Add 1st two numbers 95 01011111 1st sum + 14+ 00001110Add 3rd number 109 01101101 2nd sum + 18+ 00010010Add 4th number 127 01111111 Final sum

  11. Subtraction • Subtraction is a special case of addition. • Subtraction is addition with the sign of the subtrahend changed. • The result of a subtraction is called the difference. • The sign of a positive or negative binary is changed by taking it’s 2’s complement. _______________________________ Subtrahend = ตัวลบ, Minuend = ตัวตั้ง

  12. Subtraction • Since subtraction is simply an addition with the sign of the subtrahend changed, the process is stated as follows: • To subtract two signed numbers, take the 2’s complement of the subtrahend and add. Discard any final carry.

  13. Subtraction ex: Perform each of the following subtraction of the signed numbers: (a) 00001000 – 00000011 (b) 00001100 – 11110111 (c) 11100111 – 00010011 (d) 10001000 - 11100010 (a) 00001000 8 (b) 00001100 12 +11111101+ -3+00001001+ 9 100000101 5 00010101 21 (c) 11100111 -25 (d) 10001000 -120 +11101101+ -19+00011110+ 30 111010100 -44 10100110 -90

  14. Multiplication • The multiplication operation in most computer is accomplished using addition. • There are 2 basic methods: • Direct addition • Partial products

  15. Multiplication • Direct addition • Add the multiplicand a number of times equal to the multiplier. • For example, 8x3 = 8+8+8 = 24 • Disadvantage  lengthy operations • When two binary numbers are multiplied, both numbers MUST BE in true (uncomplemented) form. • Let’s try multiply the signed binary numbers: 010011012 and 000001002 using direct addition. _______________________________ Multiplier = ตัวคูณ, multiplicand = ตัวตั้ง

  16. Multiplication • Partial products • The multiplicand is multiplied by each multiplier digit beginning with the LSD. 239 multiplicand x 123multiplier 717 478 +239 .

  17. Multiplication • Partial products • The result of the multiplication of the multiplicand by a multiplier digit is called a partial product. 239 multiplicand x 123multiplier 717 1st part prod 478 2nd part prod +239 3rd part prod

  18. Multiplication • Partial products • Each successive partial product is shifted one place to the left and when all the partial products have been produced, they are added to get the final product. 239 multiplicand x 123multiplier 717 1st part prod 478 2nd part prod +239 3rd part prod 29397 Final Product

  19. Multiplication • The sign of the product of a multiplication depends on the signs of the multiplicand and the multiplier according to the following two rules: • If the signs are the same, the product is positive. • If the signs are different, the product is negative.

  20. Multiplication • Step 1: Determine if the signs of the multiplicand and multiplier are the same or different. This determines what the sign of the product will be. 01010011 (multiplicand) 11000101 (multiplier) Multiplicand  + Multiplier  - Product  -

  21. Multiplication • Step 2: Change any negative number to true form. Because most computers store negative numbers in 2’s comp, a 2’s comp operation is required to get the negative number into true form. 01010011 (multiplicand) 11000101 (multiplier) 00111011 (2’s comp)

  22. Multiplication 01010011 (multiplicand) 11000101 (multiplier) 1010011 x 0111011 1010011 + 1010011 . 11111001 + 0000000 . 011111001 + 1010011 . 1110010001 + 1010011 . 100011000001 + 1010011 . 1001100100001 + 0000000 . 1001100100001 • Step 3: Starting with the LSB (multiplier), generate the partial products. Shift each successive partial product one bit to the left. • Step 4: add each successive partial product to the sum of the previous partial products to get the final product.

  23. Multiplication 01010011 (multiplicand) 11000101 (multiplier) 1001100100001 (final) 0110011011111 (2’s comp) According to step 1  the sign bit of the product must be 1 (negative). Hence: 10110011011111 • Step 4: if the sign bit was determined in step 1 is negative, take the 2’s comp of the product. If positive, leave the product in true form. Attach the sign bit to the product.

  24. Division • The number in a division are the dividend, the devisor, and the quotient. dividend = quotient devisor • The division op in computer is accomplished using subtraction. Since subtraction is done with an adder, division can also be accomplished with an adder.

  25. Division • The quotient is the number of times that the divisor will go into the dividend. This means that the divisor can be subtracted from the dividend a number of times equal to the quotient. (let’s do 21/7) 21 - 7 14 - 7 7 - 7 0 The divisor was subtracted from the dividend 3 times before a remainder of zero was obtained. Therefore, the quotient is 3.

  26. Division • The sign of the quotient depends on the signs of the dividend and the divisor according to the following two rules: • If the signs are the same, the quotient is positive. • If the signs are different, the quotient is negative. • When 2 binary numbers are divided, both numbers MUST BE in true form.

  27. Division • Step 1: Determine if the signs of the dividend and divisor are the same or different. This determines what the sign bit of the quotient will be. Initialize the quotient to zero. 01100100 (dividend) 00011001 (divisor) Dividend  + Divisor  + Quotient  +

  28. Division • Step 2: Subtract the divisor from the dividend using 2’s complement addition to get the first partial remainder and add 1 to the quotient. • If this partial remainder is positive, go to step 3. • If the partial remainder is zero or negative, the division is complete. 01100100 (dividend) 00011001 (divisor) 01100100 + 11100111 (2’s comp) 101001011 Quotient = 00000000+00000001 = 00000001 Note: The final carries are discarded.

  29. Division 01100100 (dividend) 00011001 (divisor) 01001011 + 11100111 (2’s comp) 100110010 00110010 + 11100111 (2’s comp) 100011001 00011001 + 11100111 (2’s comp) 100000000 • Step 3: Subtract the divisor from the partial remainder and add 1 to the quotient. • If this partial remainder is positive, repeat for the next partial remainder. • If the result is zero or negative, the division is complete. Quotient = 00000001+00000001 = 00000010 Quotient = 00000010+00000001 = 00000011 Stop when the result is zero (or negative) Quotient = 00000011+00000001 = 00000100 Quotient = 4

  30. Hexadecimal and Octal Numbers

  31. Hexadecimal Numbers • We will call it for short as “hex”. • It has 16 characters. Digits 0-9 and letters A-F. • It used primarily as a compact way of displaying or writing binary numbers since it is very easy to convert between bin and hex. • Hex is widely used in computer and microprocessor applications.

  32. Hexadecimal Numbers

  33. Hexadecimal Numbers • If you see ‘h’ mixing in numbers (in the context of computer systems), please note that it’s most likely that the numbers are hexadecimal numbers. (Be careful. ‘h’ is not one of A-F using in hex). • For example 16h = 000101102 0Dh = 000011012

  34. Hexadecimal Numbers • Bin-to-Hex Conversion • Simply break the binary number into 4-bit groups, starting at the right-most bit and replace each 4-bit group with the equivalent hex symbol. (a) 1100101001010111 (b) 111111000101101001 110010100101011100111111000101101001 C A 5 73 F 1 6 9 = CA5716 = 3F16916

  35. Hexadecimal Numbers • Hex-to-Bin Conversion • Reverse the process (of bin-to-hex) and replace each hex symbol with the appropriate four bits. ex: Determine the binary numbers for the following hex numbers: (a) 10A4h (b) CF8Eh (c) 9742h 1 0 A 4 C F 8 E 9 7 4 2 0001000010100100 1100111110001110 1001011101000010

  36. Hexadecimal Numbers • Hex-to-Dec Conversion • 2 methods: • Hex-to-Bin first and then Bin-to-Dec. • Multiply the decimal values of each hex digits by its weight and then take the sum of these products.

  37. Hexadecimal Numbers • Hex-to-Dec Conversion • Hex-to-Bin first and then Bin-to-Dec ex: Convert the following hex numbers to decimal: (a) 1Ch 1Ch = 00011100 = 16+8+4 = 2810 (b) A85h A85h = 101010000101 = 2048+512+128+4+1 = 269310

  38. Hexadecimal Numbers • Hex-to-Dec Conversion • Multiply the decimal values of each hex digits by its weight and then take the sum of these products. ex: Convert the following hex numbers to decimal: (a) E5h E5h = (Ex16)+(5x1) = (14x16)+5 = 224+5 = 22910 (b) B2F8h B2F8h = (Bx4096)+(2x256)+(Fx16)+(8x1) = (11x4096)+(2x256)+(15x16)+(8x1) = 45,056+512+240+8 = 45,81610

  39. Hexadecimal Numbers • Dec-to-Hex conversion • Repeated division of a dec number by 16 ex: Convert the dec number 650 to hex 650/16 = 40.6250.625x16 = 10 = A 40/16 = 2.50.5x16 = 8 = 8 2/16 = 0.1250.125x16 = 2 = 2 LSD Hence 65010= 28Ah MSD Stop when whole number quotient is ZERO.

  40. Octal Numbers • Like the hex, the “oct” provides a convenient way to express binary numbers and codes. (btw, it’s not as commonly used as hex). • 8 digits: 0-7 • 0,1,2,3,4,5,6,7,10,11,12,13,14,15,16,17,20,… • Operations we learn about hex so far work the same way on oct just mark this: • Hex = 4 binary bits • Oct = 3 binary bits • Now, let’s crack the following exercises…

  41. Octal Numbers • Bin-to-Oct Conversion (a) 110101 (b) 101111001 (c) 100110011010 (d) 11010000100 • Oct-to-Bin Conversion (a) 138 (b) 258 (c) 1408 (d) 75268 • Oct-to-Dec Conversion (a) 23748 • Dec-to-Oct Conversion (a) 35910

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