2. 1 Review on 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 Binary Addition Single Bit Addition Table 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 10 N ote “carry”

9. 8 Hex Addition 4-bit Addition 4 + 4 = 8 4 + 8 = C 8 + 7 = F F + E = 1D N ote “carry”

10. 9 Hex Digit Addition Table +0123456789ABCDEF 00123456789ABCDEF 1123456789ABCDEF10 223456789ABCDEF 11 33456789ABCDEF101112 4456789ABCDEF10111213 556789ABCDEF1011121314 66789ABCDEF101112131415 7789ABCDEF10111213141516 889ABCDEF1011121314151617 99ABCDEF101112131415161718 AABCDEF10111213141516171819 BBCDEF101112131415161718191A CCDEF101112131415161718191A1B DDEF101112131415161718191A1B1C EEF101112131415161718191A1B1C1D FF101112131415161718191A1B1C1D1E

11. 10 1’s Complements  1’s complement (or Ones’ 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

12. 11 Why choose 2’s complement?

13. 12 2’s 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

14. 13 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

15. 14 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

16. 15 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):

17. 16 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

18. 17 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

19. 18 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

20. 19 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

21. 20 Signed Numbers-8 bit example

22. 21 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.

23. 22 Examples  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

24. 23 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.

25. 24 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

26. 25 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)

27. 26 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

28. 27 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

29. 28 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

30. 29 “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 ?

31. 30 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

32. 31 16’s Complement FFF 1 0 - B2CE 23D4

33. 32 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.

34. 33 Why does 2’s complement work?

35. 34 Sign Extension

36. 35 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?

37. 36 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

38. 37 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

39. 38 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

40. 39 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

41. 40 Number Conversions

42. 41 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

43. 42 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

44. 43 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

45. 44 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

46. 45 Hexadecimal to Binary Conversion Example 1.Convert each hex digit to equivalent binary (1 E 9 C) 16 ( 0001 1110 1001 1100 ) 2

47. 46 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