1 Review on Number Systems Decimal, Binary, and Hexadecimal

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).

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

4 Binary Number System  Base 2  Two Digits: 0, 1  Example:  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).

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

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  A 1011 B 1100 C 1101 D 1110 E 1111 F

7 Binary Addition Single Bit Addition Table = = = = 10 N ote “carry”

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

9 Hex Digit Addition Table ABCDEF ABCDEF ABCDEF ABCDEF ABCDEF ABCDEF ABCDEF ABCDEF ABCDEF ABCDEF ABCDEF AABCDEF BBCDEF A CCDEF A1B DDEF A1B1C EEF A1B1C1D FF A1B1C1D1E

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  

11 Why choose 2’s complement?

12 2’s Complements  2’s complement  To calculate the 2’s complement just calculate the 1’s complement, then add  =  Handy Trick: Leave all of the least significant 0’s and first 1 unchanged, and then “flip” the bits for all other digits.  Eg: >

13 Complements  Note the 2’s complement of the 2’s complement is just the original number N  EX: let N =  (2’s comp of N) = M =  (2’s comp of M) = = N

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 0 = −8

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):

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 where n=number of digits  Example: Let n=4: Range is –7 to 7 or  1111 to 0111

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

18 Signed Numbers – 4-bit example Decimal 2’s comp Sign-Mag Pos 0

19 Signed Numbers-4 bit example Decimal 2’s comp Sign-Mag N/A (= +0) 1000

20 Signed Numbers-8 bit example

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.

22 Examples  Let N=4: two’s-complement  What is the decimal equivalent of Since MSB is 0, number is positive = 4+1 =  What is the decimal equivalent of =  Since MSB is one, number is negative  Must calculate its 2’s complement  = −(0010+1)= − or −3 10

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.

24 Arithmetic Subtraction  Borrow Method  This is the technique you learned in grade school  For binary numbers, we have  = = = = 1 w ith a “borrow” 1

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)

26 Binary Subtraction - Example  Let n=4, A= (4 10 ), and B= (2 10 )  Let’s find A+B, A-B and B-A  (4) 10  (2) A+B

27 Binary Subtraction - Example  (4) 10  (2) A-B  (4) 10  (-2) 10 A+ (-B) “Throw this bit” away since n=4

28 Binary Subtraction - Example  (2) 10  (4) B-A  (2) 10  (-4) 10 B + (-A) = = -2 10

29 “16’s Complement” method  The 16’s complement of a 16 bit Hexadecimal number is just: = – N 16  Q: What is the decimal equivalent of B2CE 16 ?

30 16’s Complement  Since sign bit is one, number is negative. Must calculate the 16’s complement to find magnitude – B2CE 16 = ?  We have B2CE

31 16’s Complement FFF B2CE 23D4

32 16’s Complement  So, – B2CE 16 = 4D32 16 = 4×4, × × = 19,  Thus, B2CE 16 (in signed-magnitude) represents -19,

33 Why does 2’s complement work?

34 Sign Extension

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?

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

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

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

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

40 Number Conversions

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 n2n

42 Decimal to Binary Conversion Suppose x = Subtract 1024: (2 10 ) = 540  n=10 or 1 in the (2 10 )’s position Thus: = ( ) 2 Subtract 512: (2 9 ) = 28  n=9 or 1 in the (2 9 )’s position Subtract 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

43 Decimal to Binary Conversion Method II: Use repeated division by radix. 2 | R = 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

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

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

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