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Number Systems. Decimal, Binary, and Hexadecimal. 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). Decimal Number System.

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## Number Systems

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**Number Systems**Decimal, Binary, and Hexadecimal**Base-N Number System**• Base N • N Digits: 0, 1, 2, 3, 4, 5, …, N-1 • Example: 1045N • Positional Number System • Digit do is the least significant digit (LSD). • Digit dn-1 is the most significant digit (MSD).**Decimal Number System**• Base 10 • Ten Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 • Example: 104510 • Positional Number System • Digit d0 is the least significant digit (LSD). • Digit dn-1 is the most significant digit (MSD).**Binary Number System**• Base 2 • Two Digits: 0, 1 • Example: 10101102 • Positional Number System • Binary Digits are called Bits • Bit bo is the least significant bit (LSB). • Bit bn-1 is the most significant bit (MSB).**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**Hexadecimal Number System**• Base 16 • Sixteen Digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F • Example: EF5616 • Positional Number System**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.**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**Quiz 1-A (Practice)**• Assemble in groups of 4 • Question: Convert the following binary number into its decimal equivalent: 110102**Quiz 1-A (Practice)**THINK One Unit (e.g. 30 Seconds)**Quiz 1-A (Practice)**PAIR Two Units (e.g. 60 Seconds)**Quiz 1-A (Practice)**SHARE 1.5 units (e.g. 45 Seconds)**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!**Quiz 1-A Solution**• Convert the following number into base 10 decimal:**Quiz 1-B**• Convert the following number into base 10 decimal: 1010116**Collaborative Learning**• Think for 30 seconds • Pair for 1 minute • Share for 45 seconds • Report**Quiz 1-B Solution**• Convert the following number into base 10 decimal: 1010116 = 1·164 + 0·163 + 1·162 + 0·161 + 1·160 = 164 + 162 + 160 = 65,536 + 256 + 1 = 65,793**Binary Addition**• Single Bit Addition Table 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 10 Note “carry”**Hex Addition**• 4-bit Addition 4 + 4 = 8 4 + 8 = C 8 + 7 = F F + E = 1D Note “carry”**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**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**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**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 = −1·23 + 0·22 + 0·21 + 0·20 = −8**Negating in Two’s Complement**• Theorem: To negatea two’s complementnumber, just complement it and add 1. • Proof (for the case of 3-bit numbers XYZ):**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 - (2n-1 – 1) to 2n-1 - 1 where n=number of digits • Example: Let n=4: Range is –7 to 7 or • 1111 to 0111**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 - 2n-1 to 2n-1 -1 where n=number of digits • Example: Let n=4: Range is –8 to 7 Or 1000 to 0111**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**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**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.**Example**• Let N=4: two’s-complement • What is the decimal equivalent of 01012 Since msb is 0, number is positive 01012 = 4+1 = +510 • What is the decimal equivalent of 11012 = • Since MSB is one, number is negative • Must calculate its 2’s complement • 11012 = −(0010+1)= − 00112 or −310**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.**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 1 0 - 1 = 1 with a “borrow”**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)**Binary Subtraction - Example**• Let n=4, A=01002 (410), and B=00102 (210) • Let’s find A+B, A-B and B-A 0 1 0 0 + 0 0 1 0 (4)10 A+B (2)10 0 11 0 6**Binary Subtraction - Example**0 1 0 0 - 0 0 1 0 (4)10 A-B (2)10 0 1 0 0 + 1 1 1 0 (4)10 A+ (-B) (-2)10 10 0 1 0 2 “Throw this bit” away since n=4**Binary Subtraction - Example**0 0 1 0 - 0 1 0 0 (2)10 B-A (4)10 0 0 1 0 + 1 1 0 0 (2)10 B + (-A) (-4)10 1 1 1 0 -2 1 1 1 02 = - 0 0 1 02 = -210**“16’s Complement” method**• The 16’s complement of a 16 bit Hexadecimal number is just: • =1000016 – N16 • Q: What is the decimal equivalent of B2CE16 ?**16’s Complement**• Since sign bit is one, number is negative. Must calculate the 16’s complement to find magnitude. • =1000016 – B2CE16 = ????? • We have 10000 - B2CE**16’s Complement**FFF10 - B2CE 4 D 3 2**16’s Complement**• So, 1000016 – B2CE16 = 4D3216 = 4×4,096 + 13×256 + 3×16 + 2 = 19,76210 • Thus, B2CE16 (in signed-magnitude)represents -19,76210.**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?**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**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**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**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)·2n n

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