# binary numbers 1. outcome familiar with the binary system binary to decimal and decimal to binary...

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Binary Numbers 1 Slide 2 Outcome Familiar with the binary system Binary to Decimal and decimal to binary Arithmetic and logic operation in binary system Logic gates Half Adder and Full Adder Hexadecimal system Slide 3 Reading Goldsmiths Study guide: mathematics for computing http://www.math.grin.edu/~rebelsky/Courses/152/97F/Readings/student-binary http://en.wikipedia.org/wiki/Binary_numeral_system http://www.cut-the-knot.org/do_you_know/BinaryHistory.shtml http://www.binarymath.info/ Slide 4 The Decimal Number System (cont) 4 The decimal number system is also known as base 10. The values of the positions are calculated by taking 10 to some power. Why is the base 10 for decimal numbers? o Because we use 10 digits, the digits 0 through 9. Slide 5 The Binary Number System base 2 5 The decimal number system is a positional number system with a base 10. Example: 5623 5623 = 5000 + 600 + 20 + 3 = 5 x 10 3 + 6 x10 2 + 2 x 10 1 + 3 x 10 0 5000 600203 5 x 10 3 6 x10 2 2 x 10 1 3 x 10 0 Slide 6 The Binary Number System 6 The binary number system is also known as base 2. The values of the positions are calculated by taking 2 to some power. Why is the base 2 for binary numbers? o Because we use 2 digits, the digits 0 and 1. Slide 7 The Decimal Number System - base 10 7 The decimal number system is a positional number system with a base 10. Example: 1011 1011 2 = 1000 + 000 + 10 + 1 = 1 x 2 3 + 0 x2 2 + 1 x 2 1 + 1 x 2 0 = 11 10 1000 000101 1 x 2 3 0x2 2 1 x 2 1 1x 2 0 Slide 8 Why Bits (Binary Digits)? Computers are built using digital circuits Inputs and outputs can have only two values True (high voltage) or false (low voltage) Represented as 1 and 0 Can represent many kinds of information Boolean (true or false) Numbers (23, 79, ) Characters (a, z, ) ASCII, UNICODE Pixels Sound Can manipulate in many ways Read and write Logical operations Arithmetic 8 Slide 9 Base 10 and Base 2 Base 10 Each digit represents a power of 10 54173 10 = 5 x 10 4 + 4 x 10 3 + 1 x 10 2 + 7 x 10 1 + 3 x 10 0 Base 2 Each bit represents a power of 2 10101 2 = 1 x 2 4 + 0 x 2 3 + 1 x 2 2 + 0 x 2 1 + 1 x 2 0 = 21 10 9 Slide 10 Converting from Binary to Decimal 10 1 0 0 1 1 0 1 1 X 2 0 = 1 2 6 2 5 2 4 2 3 2 2 2 1 2 0 0 X 2 1 = 0 1 X 2 2 = 4 2 0 = 1 1 X 2 3 = 8 2 1 = 2 0 X 2 4 = 0 2 2 = 4 0 X 2 5 = 0 2 3 = 8 1 X 2 6 = 64 2 4 = 16 77 10 2 5 = 32 2 6 = 64 Slide 11 Converting from Binary to Decimal (cont) Practice conversions: Binary Decimal 110010 100111 1010101 11 Slide 12 Converting From Decimal to Binary (cont) 12 Make a list of the binary place values up to the number being converted. Perform successive divisions by 2, placing the remainder of 0 or 1 in each of the positions from right to left. Continue until the quotient is zero. Example:42 10 2 5 2 4 2 3 2 2 2 1 2 0 32 16 8 4 2 1 1 0 1 0 1 0 42/2 = 21 and R = 0 21/2 = 10 and R = 1 10/2 = 5 and R = 0 5/2 = 2 and R = 1 2/2 = 1 and R = 0 1/2 = 0 and R = 1 42 10 = 101010 2 Slide 13 Example 17 10 We repeatedly divide the decimal number by 2 and keep remainders 17/2 = 8 and R = 1 8/2 = 4and R = 0 4/2 = 2and R = 0 2/2 = 1and R = 0 1/2 = 0and R = 1 The binary number representing 17 is 10001 Slide 14 Converting From Decimal to Binary (cont) Practice conversions: Decimal Binary 59 82 175 14 Slide 15 15 Fractional Numbers Decimal 456.78 10 = 4 x 10 2 + 5 x 10 1 + 6 x 10 0 + 7 x 10 -1 +8 x 10 -2 Binary 1011.11 2 = 1 x 2 3 + 0 x 2 2 + 1 x 2 1 + 1 x 2 0 + 1 x 2 -1 + 1 x 2 -2 = 8 + 0 + 2 + 1 + 1/2 + = 11 + 0.5 + 0.25 = 11.75 10 Slide 16 16 Binary Fractional to decimal nNumbers (cont) Example1 1011.11 2 = 1 x 2 3 + 0 x 2 2 + 1 x 2 1 + 1 x 2 0 + 1 x 2 -1 + 1 x 2 -2 = 8 + 0 + 2 + 1 + 1/2 + = 11 + 0.5 + 0.25 = 11.75 10 Example 2: 111.11 2 = 1 x 2 2 + 1 x 2 1 + 1 x 2 0 + 1 x 2 -1 + 1 x 2 -2 = 4 + 2 + 1 + 1/2 + = 7.75 10 Example3: 11.011 2 = 1 x 2 1 + 1 x 2 0 + 0x 2 -1 + 1 x 2 -2 + 1 x 2 -3 = 2 +1 +0 + +1/8 = 3.375 10 Slide 17 17 Fractional numbers Examples: 7.75 10 = (?) 2 1.Conversion of the integer part: same as before repeated division by 2 7 / 2 = 3 (Q), 1 (R) 3 / 2 = 1 (Q), 1 (R) 1 / 2 = 0 (Q), 1 (R) 7 10 = 111 2 2.Conversion of the fractional part: perform a repeated multiplication by 2 and extract the integer part of the result 0.75 x 2 =1.50 extract 1 0.5 x 2 = 1.0 extract 1 0.75 10 = 0.11 2 0.0 stop Combine the results from integer and fractional part, 7.75 10 = 111.11 2 How about choose some of Examples: try 5.625 write in the same order 421 1/21/41/8 =0.5 =0.25=0.125 Slide 18 18 Fractional Numbers (cont.) Exercise 3: Convert (0.8125) 10 to its binary form Solution: 0.8125 x 2 = 1.625 extract 1 0.625 x 2 = 1.25 extract 1 0.25 x 2 = 0.5 extract 0 0.5 x 2 = 1.0 extract 1 0.0 stop (0.8125) 10 = (0.1101) 2 Slide 19 19 Representing fraction with error 0.6 x 2 = 1.2 extract 1 0.2 x 2 = 0.4 extract 0 0.4 x 2 = 0.8 extract 0 0.8 x 2 = 1.6 extract 1 0.6 x 2 = (0.6) 10 = (0.1001 1001 1001 ) 2 Example: Convert (0.6) 10 to its binary form Slide 20 20 Fractional Numbers (cont.) Errors One source of error in the computations is due to back and forth conversions between decimal and binary formats Example: (0.6) 10 + (0.6) 10 = 1.2 10 Since (0.6) 10 = (0.1001 1001 1001 ) 2 Lets assume a 8-bit representation: (0.6) 10 = (0.1001 1001) 2, therefore 0.60.10011001 + 0.6 +0.10011001 1.00110010 Lets reconvert to decimal system: (1.00110010) b = 1 x 2 0 + 0 x 2 -1 + 0 x 2 -2 + 1 x 2 -3 + 1 x 2 -4 + 0 x 2 -5 + 0 x 2 -6 + 1 x 2 -7 + 0 x 2 -8 = 1 + 1/8 + 1/16 + 1/128 = 1.1953125 Error = 1.2 1.1953125 = 0.0046875 Slide 21 Bits, Bytes, and Words A bit is a single binary digit (a 1 or 0). A byte is 8 bits A word is 32 bits or 4 bytes Long word = 8 bytes = 64 bits Quad word = 16 bytes = 128 bits Programming languages use these standard number of bits when organizing data storage and access. 21 Slide 22 Adding Two Integers: Base 10 From right to left, we add each pair of digits We write the sum, and add the carry to the next column 22 198 +264 Sum Carry 011 +001 Sum Carry 2121 6161 4040 0101 0101 1010 Slide 23 Example 10011110 1101111 + 111 1101 -------------------- ------------------- = 101 0 0 101 = 1111100 Slide 24 Binary subtraction Slide 25 Binary subtraction (Cont) Slide 26 Slide 27 Exercise 10010101 - 11011 =? 10000001 - 111 = ? Slide 28 Spring 2007, Jan. 17ELEC 2200 (Agrawal)28 Binary Multiplication 1 0 0 0 two = 8 ten multiplicand 1 0 0 1 two = 9 ten multiplier ____________ 1 0 0 0 0 0 0 0partial products 0 0 0 0 1 0 0 0 ____________ 1 0 0 1 0 0 0 two = 72 ten Slide 29 Spring 2007, Jan. 17ELEC 2200 (Agrawal)29 Binary Division 1 3 Quotient 1 1 / 1 4 7Divisor / Dividend 1 1 3 7 Partial remainder 3 3 4 Remainder 0 0 0 0 1 1 0 1 1 0 1 1 / 1 0 0 1 0 0 1 1 1 0 1 1 0 0 1 1 1 0 1 0 1 1 0 0 1 1 1 1 1 0 1 1 1 0 0 Slide 30 30 Bitwise Operators: Shift Left/Right Shift left (): Divide by powers of 2 Shift some # of bits to the right For unsigned integer, fill in blanks with 0 What about signed integers? Varies across machines Can vary from one machine to another! 0011010053 11010000532 Slide 31 Boolean Algebra to Logic Gates Logic circuits are built from components called logic gates. The logic gates correspond to Boolean operations +, *,. Binary operations have two inputs, unary has one OR + AND * NOT Slide 32 AND A B A*B Logic Gate: Series Circuit: ABABA*B000 010 100 111 Truth Table: A*B Slide 33 A B A+B Logic Gate: Parallel Circuit: A B ABA+B 000 011 101 111 Truth Table: A+B OR Slide 34 NOT ALogic Gate: (also called an inverter) aA 01 10 Truth Table: A or A Slide 35 n -input Gates Because + and * are binary operations, they can be cascaded together to OR or AND multiple inputs. A B C A B C A+B+C A B A B C ABC Slide 36 NAND and NOR Gates NAND and NOR gates can greatly simplify circuit diagrams. As we will see, can you use these gates wherever you could use AND, OR, and NOT. NAND NORAB A B 001 011 101 110 AB 001 010 100 110 Slide 37 XOR and XNOR Gates XOR is used to choose between two mutually exclusive inputs. Unlike OR, XOR is true only when one input or the other is true, not both. XOR XNOR AB A B 000 011 101 110 AB 001 010 100 111 Slide 38 Binary Sums and Carries abSumabCarry 000000 011010 101100 110111 38 XORAND 0100 0101 + 0110 0111 1010 1100 69 103 172 Slide 39 Spring 2007, Jan. 17ELEC 2200 (Agrawal)39 Design Hardware Bit by Bit Adding two bits: abhalf_sumcarry_out 0000 0110 1010 1101 Half-adder circuit a b half_sum carry_out XOR AND Slide 40 Half Adder (1-bit) ABS(um)C(arry) 0000 0110 1010 1101 Half Adder AB S C Slide 41 Half Adder (1-bit) ABS(um)C(arry) 0000 0110 1010 1101 A B Sum Carry Slide 42 Full Adder CinABS(um)Cout 00000 00110 01010 01101 10010 10101 11001 11111 Full Adder AB S Cout Carry In (Cin) Slide 43 Full Adder A B Cin Cout S H.A. Slide 44 Full Adder Cout S Half Adder S C A B Half Adder S C A B B A Cin Slide 45 4-bit Ripple Adder using Full Adder Full Adder AB Cin Cout S S0 A0B0 Full Adder AB Cin Cout S S1 A1B1 Full Adder AB Cin Cout S S2 A2B2 Full Adder AB Cin Cout S S3 A3B3 Carry A B S C Half Adder A B Cin Cout S H.A. Full Adder Slide 46 Working with Large Numbers 46 0 1 0 1 0 0 0 0 1 0 1 0 0 1 1 1 = ? Humans cant work well with binary numbers; there are too many digits to deal with. Memory addresses and other data can be quite large. Therefore, we sometimes use the hexadecimal number system. Slide 47 The Hexadecimal Number System The hexadecimal number system is also known as base 16. The values of the positions are calculated by taking 16 to some power. Why is the base 16 for hexadecimal numbers ? Because we use 16 symbols, the digits 0 and 1 and the letters A through F. 47 Slide 48 The Hexadecimal Number System (cont) Binary Decimal Hexadecimal 0 0 0 1010 10 A 1 1 1 1011 11 B 10 2 2 1100 12 C 11 3 3 1101 13 D 100 4 4 1110 14 E 101 5 5 1111 15 F 110 6 6 111 7 7 1000 8 8 1001 9 9 48 Slide 49 The Hexadecimal Number System (cont) Example of a hexadecimal number and the values of the positions: 3 C 8 B 0 5