2. 1 Number Systems Decimal, Binary, and Hexadecimal

3. 2 Base-N Number System Base N N Digits: 0, 1, 2, 3, 4, 5, …, N-1 Example: 1045 N Positional Number System Digit d o is the least significant digit (LSD). Digit d n-1 is the most significant digit (MSD).

4. 3 Decimal Number System Base 10 Ten Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 Example: 1045 10 Positional Number System Digit d 0 is the least significant digit (LSD). Digit d n-1 is the most significant digit (MSD).

5. 4 Binary Number System Base 2 Two Digits: 0, 1 Example: 1010110 2 Positional Number System Binary Digits are called Bits Bit b o is the least significant bit (LSB). Bit b n-1 is the most significant bit (MSB).

6. 5 Definitions nybble = 4 bits byte = 8 bits (short) word = 2 bytes = 16 bits (double) word = 4 bytes = 32 bits (long) word = 8 bytes = 64 bits 1K (kilo or “kibi”) = 1,024 1M (mega or “mebi”) = (1K)*(1K) = 1,048,576 1G (giga or “gibi”) = (1K)*(1M) = 1,073,741,824

7. 6 Hexadecimal Number System Base 16 Sixteen Digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F Example: EF56 16 Positional Number System 0000 0 0001 1 0010 2 0011 3 0100 4 0101 5 0110 6 0111 7 1000 8 1001 9 1010 A 1011 B 1100 C 1101 D 1110 E 1111 F

8. 7 Collaborative Learning Learning methodology in which students are not only responsible for their own learning but for the learning of other members of the group.

9. 8 Think - Pair - Share (TPS) Quizzes Think – Pair – Share –Think individually for one time units –Pair with partner for two time units –Share with group for one and half time units –Report results

10. 9 Quiz 1-A (Practice) Assemble in groups of 4 Question: Convert the following binary number into its decimal equivalent: 11010 2

11. 10 Quiz 1-A (Practice) THINK One Unit (e.g. 30 Seconds)

12. 11 Quiz 1-A (Practice) PAIR Two Units (e.g. 60 Seconds)

13. 12 Quiz 1-A (Practice) SHARE 1.5 units (e.g. 45 Seconds)

14. 13 Quiz 1-A (Practice) Report Write names of all group members and the consensus answer on one sheet of paper. All sheets will be collected. One will be picked at random to read to the class. All papers will be graded!

15. 14 Quiz 1-A Solution Convert the following number into base 10 decimal:

16. 15 Quiz 1-B Convert the following number into base 10 decimal: 10101 16

17. 16 Collaborative Learning Think for 30 seconds Pair for 1 minute Share for 45 seconds Report

18. 17 Quiz 1-B Solution Convert the following number into base 10 decimal: 10101 16 = 1·16 4 + 0·16 3 + 1·16 2 + 0·16 1 + 1·16 0 = 16 4 + 16 2 + 16 0 = 65,536 + 256 + 1 = 65,793

19. 18 TPS Quiz 2

20. 19 Binary Addition Single Bit Addition Table 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 10 N ote “carry”

21. 20 Hex Addition 4-bit Addition 4 + 4 = 8 4 + 8 = C 8 + 7 = F F + E = 1D N ote “carry”

22. 21 Hex Digit Addition Table +0123456789ABCDEF 00123456789ABCDEF 1123456789ABCDEF10 223456789ABCDEF 11 33456789ABCDEF101112 4456789ABCDEF10111213 556789ABCDEF1011121314 66789ABCDEF101112131415 7789ABCDEF10111213141516 889ABCDEF1011121314151617 99ABCDEF101112131415161718 AABCDEF10111213141516171819 BBCDEF101112131415161718191A CCDEF101112131415161718191A1B DDEF101112131415161718191A1B1C EEF101112131415161718191A1B1C1D FF101112131415161718191A1B1C1D1E

23. 22 TPS Quiz 3

24. 23 Complements 1’s complement –To calculate the 1’s complement of a binary number just “flip” each bit of the original binary number. –E.g. 0  1, 1  0 – 01010100100  10101011011

25. 24 Complements 2’s complement –To calculate the 2’s complement just calculate the 1’s complement, then add 1. 01010100100  10101011011 + 1= 10101011100 –Handy Trick: Leave all of the least significant 0’s and first 1 unchanged, and then “flip” the bits for all other digits. Eg: 01010100100 -> 10101011100

26. 25 Complements Note the 2’s complement of the 2’s complement is just the original number N –EX: let N = 01010100100 –2’s comp of N = M = 10101011100 –2’s comp of M = 01010100100 = N

27. 26 Two’s Complement Representation for Signed Numbers Let’s introduce a notation for negative digits: –For any digit d, define d = −d. Notice that in binary, where d  {0,1}, we have: Two’s complement notation: –To encode a negative number, we implicitly negate the leftmost (most significant) bit: E.g., 1000 = (−1)000 = −1·2 3 + 0·2 2 + 0·2 1 + 0·2 0 = −8

28. 27 Negating in Two’s Complement Theorem: To negate a two’s complement number, just complement it and add 1. Proof (for the case of 3-bit numbers XYZ):

29. 28 Signed Binary Numbers Two methods: –First method: sign-magnitude Use one bit to represent the sign –0 = positive, 1 = negative Remaining bits are used to represent the magnitude Range - (2 n-1 – 1) to 2 n-1 - 1 where n=number of digits Example: Let n=4: Range is –7 to 7 or 1111 to 0111

30. 29 Signed Binary Numbers Second method: Two’s-complement Use the 2’s complement of N to represent -N Note: MSB is 0 if positive and 1 if negative Range - 2 n-1 to 2 n-1 -1 where n=number of digits Example: Let n=4: Range is –8 to 7 Or 1000 to 0111

31. 30 Signed Numbers – 4-bit example Decimal 2’s comp Sign-Mag 7 0111 0111 6 0110 0110 5 0101 0101 4 0100 0100 3 0011 0011 2 0010 0010 1 0001 0001 0 0000 0000 Pos 0

32. 31 Signed Numbers-4 bit example Decimal 2’s comp Sign-Mag -8 1000 N/A -7 1001 1111 -6 1010 1110 -5 1011 1101 -4 1100 1100 -3 1101 1011 -2 1110 1010 -1 1111 1001 -0 0000 (= +0) 1000

33. 32 Notes: “Humans” normally use sign-magnitude representation for signed numbers –Eg: Positive numbers: +N or N – Negative numbers: -N Computers generally use two’s-complement representation for signed numbers –First bit still indicates positive or negative. –If the number is negative, take 2’s complement to determine its magnitude Or, just add up the values of bits at their positions, remembering that the first bit is implicitly negative.

34. 33 Example Let N=4: two’s-complement What is the decimal equivalent of 0101 2 Since msb is 0, number is positive 0101 2 = 4+1 = +5 10 What is the decimal equivalent of 1101 2 = Since MSB is one, number is negative Must calculate its 2’s complement 1101 2 = −(0010+1)= − 0011 2 or −3 10

35. 34 Very Important!!! – Unless otherwise stated, assume two’s- complement numbers for all problems, quizzes, HW’s, etc. The first digit will not necessarily be explicitly underlined.

36. 35 TPS Quizzes 5-7

37. 36 Arithmetic Subtraction Borrow Method –This is the technique you learned in grade school –For binary numbers, we have – 0 - 0 = 0 1 - 0 = 1 1 - 1 = 0 0 - 1 = 1 w ith a “borrow” 1

38. 37 Binary Subtraction Note: –A – (+B) = A + (-B) –A – (-B) = A + (-(-B))= A + (+B) –In other words, we can “subtract” B from A by “adding” –B to A. –However, -B is just the 2’s complement of B, so to perform subtraction, we 1. Calculate the 2’s complement of B 2. Add A + (-B)

39. 38 Binary Subtraction - Example Let n=4, A=0100 2 (4 10 ), and B=0010 2 (2 10 ) Let’s find A+B, A-B and B-A 0 1 0 0 + 0 0 1 0  (4) 10  (2) 10 0 11 0 6 A+B

40. 39 Binary Subtraction - Example 0 1 0 0 - 0 0 1 0  (4) 10  (2) 10 10 0 1 0 2 A-B 0 1 0 0 + 1 1 1 0  (4) 10  (-2) 10 A+ (-B) “Throw this bit” away since n=4

41. 40 Binary Subtraction - Example 0 0 1 0 - 0 1 0 0  (2) 10  (4) 10 1 1 1 0 -2 B-A 0 0 1 0 + 1 1 0 0  (2) 10  (-4) 10 B + (-A) 1 1 1 0 2 = - 0 0 1 0 2 = -2 10

42. 41 “16’s Complement” method The 16’s complement of a 16 bit Hexadecimal number is just: =10000 16 – N 16 Q: What is the decimal equivalent of B2CE 16 ?

43. 42 16’s Complement Since sign bit is one, number is negative. Must calculate the 16’s complement to find magnitude. =10000 16 – B2CE 16 = ????? We have 10000 - B2CE

44. 43 16’s Complement FFF 1 0 - B2CE 23D4

45. 44 16’s Complement So, 10000 16 – B2CE 16 = 4D32 16 = 4×4,096 + 13×256 + 3×16 + 2 = 19,762 10 Thus, B2CE 16 (in signed-magnitude) represents -19,762 10.

46. 45 Sign Extension

47. 46 Sign Extension Assume a signed binary system Let A = 0101 (4 bits) and B = 010 (3 bits) What is A+B? –To add these two values we need A and B to be of the same bit width. –Do we truncate A to 3 bits or add an additional bit to B?

48. 47 Sign Extension A = 0101 and B=010 Can’t truncate A!! Why? –A: 0101 -> 101 –But 0101 <> 101 in a signed system –0101 = +5 –101 = -3

49. 48 Sign Extension Must “sign extend” B, so B becomes 010 -> 0010 Note: Value of B remains the same So 0101 (5) +0010 (2) -------- 0111 (7) Sign bit is extended

50. 49 Sign Extension What about negative numbers? Let A=0101 and B=100 Now B = 100  1100 Sign bit is extended 0101 (5) +1100 (-4) ------- 10001 (1) Throw away

51. 50 Why does sign extension work? Note that: (−1) = 1 = 11 = 111 = 1111 = 111…1 –Thus, any number of leading 1’s is equivalent, so long as the leftmost one of them is implicitly negative. Proof: 111…1 = −(111…1) = = −(100…0 − 11…1) = −(1) So, the combined value of any sequence of leading ones is always just −1 times the position value of the rightmost 1 in the sequence. 111…100…0 = (−1)·2 n n

52. 51 Number Conversions

53. 52 Decimal to Binary Conversion Method I: Use repeated subtraction. Subtract largest power of 2, then next largest, etc. Powers of 2: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2 n Exponent: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, n 2 10 2n2n 2929 2828 2020 2727 21212 2323 2626 2424 2525

54. 53 Decimal to Binary Conversion Suppose x = 1564 10 Subtract 1024: 1564-1024 (2 10 ) = 540  n=10 or 1 in the (2 10 )’s position Thus: 1564 10 = (1 1 0 0 0 0 1 1 1 0 0) 2 Subtract 512:540-512 (2 9 ) = 28  n=9 or 1 in the (2 9 )’s position Subtract 16: 28-16 (2 4 ) = 12  n=4 or 1 in (2 4 )’s position Subtract 8:12 – 8 (2 3 ) = 4  n=3 or 1 in (2 3 )’s position Subtract 4:4 – 4 (2 2 ) = 0  n=2 or 1 in (2 2 )’s position 2 8 =256, 2 7 =128, 2 6 =64, 2 5 =32 > 28, so we have 0 in all of these positions

55. 54 Decimal to Binary Conversion Method II: Use repeated division by radix. 2 | 1564 782R = 0 2|_____ 391 R = 0 2|_____ 195 R = 1 2|_____ 97 R = 1 2|_____ 48 R = 1 2|_____ 24 R = 0 2|__24_ 12 R = 0 2|_____ 6 R = 0 2|_____ 3 R = 0 2|_____ 1 R = 1 2|_____ 0 R = 1  Collect remainders in reverse order 1 1 0 0 0 0 1 1 1 0 0

56. 55 Binary to Hex Conversion 1.Divide binary number into 4-bit groups 2. Substitute hex digit for each group 1 1 0 0 0 0 1 1 1 0 0 0 Pad with 0’s If unsigned number 61C 16 Pad with sign bit if signed number

57. 56 Hexadecimal to Binary Conversion Example 1.Convert each hex digit to equivalent binary (1 E 9 C) 16 ( 0001 1110 1001 1100 ) 2

58. 57 Decimal to Hex Conversion Method II: Use repeated division by radix. 16 | 1564 97R = 12 = C 16|_____ 6 R = 1 16|_____ 0 R = 6  N = 61C 16