Matematik Penulis: SITI HADIJAH BINTI NORSANI FAIRUZAH BINTI ABU SAMAH SITI HUZAIFAH BINTI MOHAMMAD NOR SHARIDAH BINTI MOHD ROSLAN MOHAMAD HANIFAH BIN ABD GHANI MATHEMATICAL COMPUTING FOR POLYTECHNIC
MATHEMATICAL COMPUTING FOR POLYTECHNIC SITI HADIJAH BINTI NORSANI FAIRUZAH BINTI ABU SAMAH SITI HUZAIFAH BINTI MOHAMMAD NOR SHARIDAH BINTI MOHD ROSLAN MOHAMAD HANIFAH BIN ABD GHANI
MATHEMATICAL COMPUTING Editor NOR ‘AINAA SYUHADA BINTI MOHD NOOR Writer SITI HADIJAH BINTI NORSANI FAIRUZAH BINTI ABU SAMAH SITI HUZAIFAH BINTI MOHAMMAD NOR SHARIDAH BINTI MOHD ROSLAN MOHAMAD HANIFAH BIN ABD GHANI Issue in 2021 All rights reserved. No part of this book may be reproduced in any form on electronic or mechanical means, including information storage and retrieval systems without permission in writing from the publisher. ([email protected]/[email protected]/[email protected]/[email protected]/ [email protected]/[email protected]) Published by: Politeknik Muadzam Shah, Lebuhraya Tun Abdul Razak, 26700 Muadzam Shah, Pahang Darul Makmur
NO TOPIC PAGE ABSTRACT CONTENTS 1 NUMBERING SYSTEM (Prepared by: Siti Hadijah binti Norsani) 1.1 Describe the concepts of numbering system 2 1.1.1 Define numbering system: decimal, binary, octal and hexadecimal 1.1.2 Define data organization 1.1.3 Define unit, number, base/radix, positional notation, and most and least significant digits 1.1.4 Identify binary system 1.1.5 State binary to decimal and decimal to binary 1.2 Discuss octal numbering system 10 1.2.1 Convert octal to decimal and decimal to octal 1.2.2 Convert octal to binary and binary to octal 1.3 Simplify hexadecimal numbering system 20 1.3.1 Convert hexadecimal to decimal and decimal to hexadecimal 1.3.2 Convert hexadecimal to binary and binary to hexadecimal 1.3.3 Convert hexadecimal to octal and octal to hexadecimal 1.4 Apply binary arithmetic 32 1.4.1 Use binary arithmetic operations: addition, subtraction andmultiplication 1.4.2 Add and subtract in binary, octal, and hexadecimal number systems 2 BASIC ALGEBRA (Prepared by: Fairuzah binti Abu Samah) 2.1 Express basic algebra 41 2.1.1 Simplify algebraic expression 2.1.2 Simplify algebraic expression by using: a. addition b. subtraction c. multiplication d. division 2.2 Show quadratic equations 49 2.2.1 Solve quadratic equations by using: a. factorization b. quadratic formula CONTENT
3 REAL AND COMPLEX NUMBER SYSTEM (Prepared by: Siti Huzaifah binti Mohammad) 3.1 Describe the concepts of real number 57 3.1.1 Identify type of real numbers 3.1.2 Define real number lines 3.1.3 Define graph of inequalities 3.2 Describe the concepts of complex number 60 3.2.1 State that complex numbers that derived by combining the real partswith the imaginary parts 3.2.2 Identify that product of two imaginary number is real number 3.2.3 State the conjugate of complex numbers 3.2.4 Describe the operations such as addition, subtraction, multiplication,division, conjugate and equivalent complex numbers 3.3 Solve the complex numbers using Argand's Diagram 70 3.3.1 Explain graphical representation of complex number through Argand’s Diagrams 3.3.2 Draw a straight line in an Argand’s Diagram to represent a complexnumber 3.3.3 Use Argand’s Diagrams to find the modulus and argument 3.4 Apply the concepts of complex numbers in other forms 74 3.4.1 Describe complex numbers in the form of polar, trigonometric andexponential 4 DIFFERENTIATION & INTEGRATION (Prepared by: Nor Sharidah binti Mohd Roslan) 4.1 Describe Differentiation 87 4.1.1 Limits and differentiation 4.1.2 Techniques of Differentiation 4.1.3 Rule of Differentiation: a. Chain Rule b. Product Rule c. Quotient Rule 4.1.4 Calculate gradients of tangent to a curve by applying thedifferentiation techniques 4.1.5 Compute the tangent and normal equations 4.2 Clarify Integration 111 4.2.1 Describe the integration as the inverse of differentiation. 4.2.2 Indefinite integral of algebraic function
4.2.3 Indefinite integral of a algebraic function with initial conditions given 4.2.4 Definite integrals 4.2.5 Solve the problems related to definite integrals 5 MATRICES AND LINEAR ALGEBRA (Prepared by: Mohamad Hanifah bin Abd Ghani) 5.1 Construct matrix 126 5.1.1 Define matrix 5.1.2 Identify characteristics of a matrix: element, order 5.1.3 Identify different types of matrices and their basic properties 5.1.4 Write identity matrix 5.1.5 Describe basic operations on matrices 5.1.6 Calculate the operations of addition, subtraction and multiplicationinvolving matrices 5.1.7 Compute the determinants of square matrices of orders 2 and 3 anduse the properties of determinants 5.1.8 Write the transposition of a matrix 5.1.9 Calculate the determinant of a matrix 5.1.10 Identify singular and non-singular matrices 5.1.11 Simplify the inverse of a matrix 5.2 Explain system of linear equations 146 5.2.1 Explain the procedure to solve linear equations 5.2.2 Solve systems of linear equations 5.2.3 Solve simultaneous equations with three variables using Cramer’sRule REFERENCES
ABSTRACT This module was written by the lecturers from the Department of Mathematics, Science and Computer, Polytechnic Muadzam Shah and developed based on the latest syllabus of Mathematical computing course for polytechnic. This module contains five (5) chapters; Numbering System, Basic Algebra, Real & Complex Number System, Differentiation & Integration and Matrices & Linear Algebra. Hopefully, it can be used as a main reference for polytechnic student especially in Diploma of Digital Technology program and also for newly trained lecturers in this field. Keywords: Mathematical Computing, Polytechnic
1 1.1 Describe the concepts of numbering system 1.1.1 Define numbering system: decimal, binary, octal and hexadecimal 1.1.2 Define data organization: 1.1.3 Define unit, number, base/radix, positional notation, and most and least significant digits. 1.1.4 Identify binary system. 1.1.5 State binary to decimal and decimal to binary 1.2 Discuss octal numbering system 1.2.1 Convert octal to decimal and decimal to octal 1.2.2 Convert octal to binary and binary to octal 1.3 Simplify hexadecimal numbering system 1.3.1 Convert hexadecimal to decimal and decimal to hexadecimal 1.3.2 Convert hexadecimal to binary and binary to hexadecimal 1.3.3 Convert hexadecimal to octal and octal to hexadecimal 1.4 Apply binary arithmetic 1.4.1 Use binary arithmetic operations: addition, subtraction and multiplication 1.4.2 Add and subtract in binary, octal, and hexadecimal number systems. CHAPTER 1: NUMBERING SYSTEM Today's numbers, also called Hindu-Arabic numbers, are a combination of just 10 symbols or digits: 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0. These digits were introduced in Europe within the XII century by Leonardo Pisano (also known as Fibonacci), an Italian mathematician. L. Pisano was educated in North Africa, where he learned and later carried to Italy the now popular Hindu-Arabic numerals. (https://www.archimedes-lab.org/numeral.html)
CHAPTER 1: NUMBERING SYSTEM 2 1.1.1 Define numbering system: decimal, binary, octal and hexadecimal Number systems are the techniques to represent numbers in the computer system architecture. Every value that you are saving or getting into or from computer memory has a defined number system. Computer architecture supports the following number systems. i. Decimal number system (base 10) ii. Binary number system (base 2) iii. Octal number system (base 8) iv. Hexadecimal number system (base 16) Number system Definition Decimal system Decimal number system has only ten digits from 0 to 9. Every number represents with 0,1,2,3,4,5,6, 7, 8 and 9 in this number system. The base of decimal number system is 10, because it has only 10 digits. Binary system Binary number system has only two digits that are 0 and 1. Every number represents with 0 and 1 in this number system. The base of binary system is 2, because it has only two digits. Octal system Octal number system has only eight digits from 0 to 7. Every number represents with 0,1,2,3,4,5,6 and 7 in this number system. The base of octal number system is 8, because it has only 8 digits. Hexadecimal system Hexadecimal number system has sixteen alphanumeric values from 0 to 9 and A to F. Every number represents with 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F in this number system. The base of hexadecimal number system is 16, because it has 16 alphanumeric values. Here A is 10, B is 11, C is 12, D is 13, E is 14 and F is 15. 1.1.2 Define data organization Inside today’s computers, data is represented as 1’s and 0’s. These 1’s and 0’s might be stored magnetically on a disk, or as a state in a transistor, core, or vacuum tube. To perform useful operations on these 1’s and 0’s one have to organize them togetherinto patternsthatmake up codes. The patterns are represent in data organization that consists of bits, nibbles, bytes, word and doubleword. Data organization, in broad terms, refers to the method of classifying and organizing data sets to make them more useful. A BITS (short for binary digit) is the smallest unit of data in a computer. A bit has a single binary value, either 0 or 1. Although computers usually provide instructions that can test and manipulate bits, they generally are designed to store data and execute instructions in 1.1 Describe the concepts of numbering system
CHAPTER 1: NUMBERING SYSTEM 3 Write 8 bytes of data in terms of nibbles. Answer: 1 = 2 So, 8 = 8 × 2 = 16 Write 3 words of data in terms of bits. Answer: 1 = 16 So, 3 = 3 × 16 = 48 bit multiples called BYTE. In most computer systems, there are eight bits in a byte. The value of a bit is usually stored as either above or below a designated level of electrical charge in a single capacitor within a memory device. Half a byte (four bits) is called a NIBBLE. Two bytes (16-bits) form a WORD and four bytes (32-bits) form a DOUBLE WORD. In such systems, instruction lengths are sometimes expressed as double-word (32 bits in length) or word (16 bits in length). Summary: The relationship between bit, nibble, byte, word and double word. Bit Nibble Byte Word Double Word 1 bit 4 bits 8 bits 2 Nibbles 16 bits 4 Nibbles 2 bytes 32 bits 8 Nibbles 4 bytes 2 words Example 1 Example 2
CHAPTER 1: NUMBERING SYSTEM 4 1.1.3 Define unit, number, base/radix, positional notation, and most and least significant digits. For example, Base 10 (Decimal) — Represent any number using 10 digits [0 to 9] Base 2 (Binary) — Represent any number using 2 digits [0 and 1] Base 8 (Octal) — Represent any number using 8 digits [0 to 7] Base 16(Hexadecimal) — Represent any number using 10 digits and 6 characters [0 to 9, A, B, C, D, E, F] A positional system is a numeral system in which the contribution of a digit to the value of a number is the product of the value of the digit by a factor determined by the position of the digit. In early numeral systems, a digit has only one value. In modern positional systems, such as the decimal system, the position of the digit means that its value must be multiplied by some value. For example: in 645, the three identical symbols represent six hundreds, four tens, and five units, respectively, due to their different positions in the digit string. (https://en.wikipedia.org/wiki/Positional_notation) In mathematics, a “base” or a “radix” is the number of different digits or combination of digits and letters that a system of counting uses to represent numbers. ~Wikipedia~
CHAPTER 1: NUMBERING SYSTEM 5 Most Significant Digit (MSD) The MSD in a number is the digit that has the greatest effect on that number. MSD is the highest power of base weighting. The digits on the left hand side are called the high-order digits (higher powers of 10). Least Significant Digit (LSD) The LSD in a number is the digit that has the least effect on that number. LSD is the lowest power of base weighting. The digits on the right hand side are called low-order digits (lower power of 10). 1.1.4 Identify binary system The binary number is a method of representing numbers that counts by using combinations of only two numerals: zero (0) and one (1). The base of binary system is 2. The 0s and 1s in binary represent OFF or ON respectively. In a transistor, an “0” represents no flow of electricity, and “1” represents electricity being allowed to flow. Computers use the binary number system to manipulate and store all of their data including numbers, words, videos, graphics, and music. Each binary digit is also called a bit. Binary number system is also positional value system, where each digit had a value expressed in power of 2, as displayed here. 2 2 2 1 2 0 . 2 -1 2 -2
CHAPTER 1: NUMBERING SYSTEM 6 Example of binary number: 1001112 1011.012 Convert 110012 to decimal number. Answer: = (1 × 2 4 ) + (1 × 2 3 ) + (0 × 2 2 ) + (0 × 2 1 ) + (1 × 2 0 ) = 16 + 8 + 0 + 0 + 1 = 2510 Convert 101.0012 to decimal number. Answer: = (1 × 2 2 ) + (0 × 2 1 ) + (1 × 2 0 ) + (0 × 2 −1 ) + (0 × 2 −2 ) + (1 × 2 −3 ) = 4 + 0 + 1 + 0 + 0 + 0.125 = 5.12510 1.1.5 State binary to decimal and decimal to binary The weighting for the binary number up to 4 decimal places and 2 decimal places after the binary point (.). weight 2 3 2 2 2 1 2 0 . 2 -1 2 -2 8 4 2 1 . 0.5 0.25 Convert binary to decimal Example 3 Binary point Any binary number can be converted to its decimal equivalent simply by summing together the weights of the various positions in the binary number which contain a 1. Example 4 Example 5 Binary point
CHAPTER 1: NUMBERING SYSTEM 7 1. Convert 110102 to decimal number. Answer: 2610 2. Convert 10102 to decimal number. Answer: 1010 3. Convert 10.1012 to decimal number. Answer: 2.62510 4. Convert 0.0102 to decimal number. Answer: 0.2510 5. Convert 11111.0112 to decimal number. Answer: 31.37510 Convert decimal to binary Exercise 1 You can use the following steps to convert decimal number to binary number. 1. Divide the number by 2. 2. Get the integer quotient for the next iteration. 3. Get remainder for the binary digit. 4. Repeat the steps until the quotient is equal to 0.
CHAPTER 1: NUMBERING SYSTEM 8 Convert 3510 to binary number. Answer: remainder 35 2 17 1 2 8 1 2 4 0 2 2 0 2 1 0 2 0 1 Convert 12.12510 to binary number. Answer: remainder 12 2 6 0 2 3 0 2 1 1 2 0 1 remainder 0.125 × 2 = 0.25 0 0.25 × 2 = 0.5 0 0.5 × 2 = 1.0 1 (*stop when the fractional part is zero) So, the answer is: 12.12510 = 1100.0012 Example 6 So, the answer is: 3510 = 1000112 Example 7 1210 = 11002 So, 0.12510 = 0.0012 Read from bottom to top Read from top to bottom Read from bottom to top
CHAPTER 1: NUMBERING SYSTEM 9 1. Convert 4110 to binary number. Answer: 1010012 2. Convert 3610 to binary number. Answer: 1001002 3. Convert 52.12510 to binary number. Answer: 110100.0012 4. Convert 0.62510 to binary number. Answer: 0.1012 5. Convert 3018.710 to binary number. Answer: 101111001010.101102 Exercise 2
CHAPTER 1: NUMBERING SYSTEM 10 Convert 2058 to decimal number. Answer: = (2 × 8 2 ) + (0 × 8 1 ) + (5 × 8 0 ) = 128 + 0 + 5 = 13310 Convert 327.258 to decimal number. Answer: = (3 × 8 2 ) + (2 × 8 1 ) + (7 × 8 0 ) + (2 × 8 −1 ) + (5 × 8 −2 ) = 192 + 16 + 7 + 0.25 + 0.0781 = 215.328110 Convert 53.0728 to decimal number. Answer: = (5 × 8 1 ) + (3 × 8 0 ) + (0 × 8 −1 ) + (7 × 8 −2 ) + (2 × 8 −3 ) = 40 + 3 + 0 + 0.109 + 0.004 = 43. 11310 1.2.1 Convert octal to decimal and decimal to octal. The octal number system is the base-8 number system. It has eight possible digits: 0, 1, 2, 3, 4, 5, 6, and 7. The digit positions in an octal number have weight as follows: Weight 8 3 8 2 8 1 8 0 . 8 -1 8 -2 512 64 8 1 . 0.125 0.0156 1.2 Discuss octal numbering system Octal point Example 8 Example 9 Example 10
CHAPTER 1: NUMBERING SYSTEM 11 1. Convert 1468 to decimal number. Answer:10210 2. Convert 75308 to decimal number. Answer: 392810 3. Convert 2058 to decimal number. Answer: 13310 4. Convert 42.078 to decimal number. Answer: 34.109410 5. Convert 0.258 to decimal number. Answer: 0.328110 6. Convert 6432.0508 to decimal number. Answer: 3354.078110 7. Convert 73.6358 to decimal number. Answer: 59.806610 Exercise 3
CHAPTER 1: NUMBERING SYSTEM 12 Convert 22610 to octal number. Answer: remainder 226 8 28 2 8 3 4 8 0 3 Convert decimal to octal There are two direct methods are available for converting a decimal number into octal number: Converting with Remainders and Converting with Division. Converting with Remainders (For integer part) This is a straightforward method which involve dividing the number to be converted. Let decimal number is N then divide this number from 8 because base of octal number system is 8. Note down the value of remainder, which will be: 0, 1, 2, 3, 4, 5, 6, or 7. Again divide remaining decimal number till it became 0 and note every remainder of every step. Then write remainders from bottom to up (or in reverse order), which will be equivalent octal number of given decimal number. This is procedure for converting an integer decimal number, algorithm is given below. i. Take decimal number as dividend. ii. Divide this number by 8 (8 is base of octal so divisor here). iii. Store the remainder in an array (it will be: 0, 1, 2, 3, 4, 5, 6 or 7 because of divisor 8). iv. Repeat the above two steps until the number is greater than zero. v. Print the array in reverse order (which will be equivalent octal number of given decimal number). Note that dividend (here given decimal number) is the number being divided, the divisor (here base of octal, i.e., 8) in the number by which the dividend is divided, and quotient (remaining divided decimal number) is the result of the division. Example 11 So, the answer is: 22610 = 3428 Read from bottom to top
CHAPTER 1: NUMBERING SYSTEM 13 Convert 0.312510 to octal number. Answer: remainders 0.3125 × 8 = 2.5 2 0.5 × 8 = 4.0 4 (*stop when the fractional part is zero) Converting with Remainders (For integer part) Let decimal fractional part, then multiply the decimal part with 8 because base of octal number system is 8. Then note the value of integer on the right. Next multiply the remaining fraction number with 8 again. Repeat the same step until you reach fraction number 0 and note every integer part of result of every step. Then write down the results of integer part, which will be equivalent fraction octal number of given decimal number. To procedure for converting a fractional decimal number, the algorithm is given below. i. Take decimal number as multiplicand. ii. Multiple this number by 8 (8 is base of octal so multiplier here). iii. Store the value of integer part of result in an array (it will be: 0, 1, 2, 3, 4, 5, 6, and 7 because of multiplier 8). iv. Repeat the above two steps until the number became zero. v. Print the array (which will be equivalent fractional octal number of given decimal fractional number). Note that a multiplicand (here decimal fractional number) is that to be multiplied by multiplier (here base of octal, i.e., 8) So, the answer is: 0.312510 = 0.248 Example 12 Read from top to bottom
CHAPTER 1: NUMBERING SYSTEM 14 1. Convert 16810 to octal number. Answer: 2508 2. Convert 972110 to octal number. Answer: 227718 3. Convert 0.12510 to octal number. Answer: 0.18 4. Convert 0.85510 to octal number. Answer: 0.665608 5. Convert 85.312510 to octal number. Answer: 125.248 Exercise 4
CHAPTER 1: NUMBERING SYSTEM 15 Convert 4628 to binary number. Answer: 1. Separate the digits of the given octal number, if it contains more than 1 digit. 2. Find the equivalent binary number for each digit of octal number. Add 0's to the left if any of the binary equivalent is shorter than 3 bits. 4 6 2 100 110 010 3. Write the all group's binary numbers together, maintaining the same group order provides the equivalent binary for the given octal number. 100110010 Result: 4628 = 1001100102 1.2.2 Convert octal to binary and binary to octal. Octal to Binary conversion This conversion can be done by finding the binary equivalent for an each digit of the octal number, combining them together in the same order. This steps may useful to know how to perform octal to binary number conversion. Each digit of octal can be represent in group of 3 bits in binary. Octal Number Binary equivalent 0 000 1 001 2 010 3 011 4 100 5 101 6 110 7 111 Example 13 Step 1 •Separate the digits of the given octal number, if it contains more than 1 digit. Step 2 •Find the equivalent binary number for each digit of octal number. Add 0's to the left if any of the binary equivalent is shorter than 3 bits. Step 3 •Write the all group's binary numbers together, maintaining the same group order provides the equivalent binary for the given octal number.
CHAPTER 1: NUMBERING SYSTEM 16 Convert 51308 to binary number. Answer: Ans: 51308 = 1010010110002 Convert 16.058 to binary number. Answer: Ans: 16.058 = 1110.0001012 5 1 3 0 101 001 011 000 1 6 . 0 5 001 110 . 000 101 1. Convert 1168 to binary number. Answer:10011102 2. Convert 45308 to binary number. Answer: 1001010110002 3. Convert 0.0358 to binary number. Answer: 0.0000111012 4. Convert 40.1178 to binary number. Answer: 100000.0010011112 Example 14 Example 15 Exercise 5
CHAPTER 1: NUMBERING SYSTEM 17 Binary to Octal conversion The binary to octal conversion can be done by grouping of bits method. Follow the following steps. Step 1 •Separate every three digits in the binary number starting from the right hand side. Step 2 •Add 0's to the left, if the last group doesn't contain 3 bits. Step 3 •Find the equivalent octal number for each group. Step 4 •Write the all groups octal numbers together, maintaining the group order provides the equivalent octal number for the given binary. 5. Convert 103.508 to binary number. Answer: 1000011.1010002 6. Convert 75.028 to binary number. Answer: 111101.0000102 7. Convert 254618 to binary number. Answer: 101011001100012 8. Convert 5.2438 to binary number. Answer: 101.0101000112
CHAPTER 1: NUMBERING SYSTEM 18 Convert 11010011102 to octal number. Answer: 1. Separate the digits of the given binary number into groups from right to left side, each containing 3 bits. 1 101 001 110 2. Add 0’s to the left, if the last group doesn’t contain 3 bits. 001 101 001 110 3. Find equivalent octal number for each group. 001 101 001 110 1 5 1 6 4. Write the all group’s octal numbers together, maintaining the group order provides the equivalent octal number for the given binary. Result: 11010011102 = 15168 Convert 10010100011012 to octal number. Answer: Ans: 10010100011012 = 112158 Convert 10110.01100112 to octal number. Answer: Ans: 10110.01100112 = 26.3148 001 001 010 001 101 1 1 2 1 5 010 110 . 011 001 100 2 6 . 3 1 4 Example 16 Add 0’s Example 17 Example 18
CHAPTER 1: NUMBERING SYSTEM 19 1. Convert 1101100012 to octal number. Answer: 6618 2. Convert 111000112 to octal number. Answer: 3438 3. Convert 10001010102 to octal number. Answer: 10528 4. Convert 11.0101112 to octal number. Answer: 3.278 5. Convert 111011.01012 to octal number. Answer: 73.248 6. Convert 11011.12 to octal number. Answer: 33.48 7. Convert 1110011.1110102 to octal number. Answer: 163.728 8. Convert 1111.12 to octal number. Answer: 17.48 Exercise 6
CHAPTER 1: NUMBERING SYSTEM 20 The base is 16. Hexadecimal number uses 16 digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F. Hexadecimal numbers are compact and easy to read. It is very easy to convert numbers from binary system to hexadecimal system and vice-versa, every nibble (4 bits) can be converted to a hexadecimal digit using this table: Table 1: Decimal, Hexadecimal and Binary table. Decimal (Base 10) Hexadecimal (Base 16) Binary (Base 2) 0 0 0000 1 1 0001 2 2 0010 3 3 0011 4 4 0100 5 5 0101 6 6 0110 7 7 0111 8 8 1000 9 9 1001 10 A 1010 11 B 1011 12 C 1100 13 D 1101 14 E 1110 15 F 1111 1.3.1 Convert hexadecimal to decimal and decimal to hexadecimal. Hexadecimal to decimal conversion The weighting for the hexadecimal number up to 4 decimal places and 2 decimal places after the point (.). weight 163 162 161 160 . 16-1 16-2 4096 256 16 1 . 0.0625 0.0039 1.3 Simplify hexadecimal numbering system Point
CHAPTER 1: NUMBERING SYSTEM 21 Convert 3A216 to decimal number. Answer: = (3 × 162 ) + (10 × 161 ) + (2 × 160 ) = 768 + 160 + 2 = 93010 Convert 80.C516 to decimal number. Answer: = (8 × 161 ) + (0 × 160 ) + (12 × 16−1 ) + (5 × 16−2 ) = 128 + 0 + 0.75 + 0.0195 = 128.769510 1. Convert 68716 to decimal number. Answer:167110 2. Convert 1AD16 to decimal number. Answer: 42910 3. Convert E2.0516 to decimal number. Answer: 226.019510 Example 19 Example 20 Exercise 7
CHAPTER 1: NUMBERING SYSTEM 22 Convert 52910 to hexadecimal number. Answer: remainder 529 16 33 1 16 2 1 16 0 2 So, the answer is: 52910 = 21116 Decimal to hexadecimal conversion 4. Convert 60C.1A16 to decimal number. Answer: 1548.101610 5. Convert B11.D216 to decimal number. Answer: 2833.820310 6. Convert 8071.216 to decimal number. Answer: 32881.12510 Example 21 Read from bottom to top
CHAPTER 1: NUMBERING SYSTEM 23 Convert 668.4510 to hexadecimal number. Answer: remainder 668 16 41 12 (C) 16 2 9 16 0 2 remainders 0.45 × 16 = 7. 7 0.2 × 16 = 3. 3 **note: stop when the fractional part is all zero or repeated So, the answer is: 668.4510 = 29C.7316 1. Convert 70510 to hexadecimal number. Answer: 2C116 2. Convert 248610 to hexadecimal number. Answer: 9B616 3. Convert 304.3510 to hexadecimal number. Answer: 130.5916 Example 22 66810 = 29C16 0.4510 = 0.7316 Exercise 8
CHAPTER 1: NUMBERING SYSTEM 24 Convert 29616 to its binary equivalent. Answer: 1. Separate the digits of the given hexadecimal number, if it contains more than 1 digit. 2 9 6 2. Find the equivalent binary number for each digit of hexadecimal number. Add 0's to the left if any of the binary equivalent is shorter than 4 bits. 2 9 6 0010 1001 0110 3. Write the all group's binary numbers together, maintaining the same group order provides the equivalent binary for the given hexadecimal number. 001010010110 Result: 29616 = 10100101102 1.3.2 Convert hexadecimal to binary and binary to hexadecimal. Hexadecimal to Binary conversion This conversion can be done by finding the binary equivalent for an each digit of the hexadecimal number, combining them together in the same order. This steps may useful to know how to perform hexadecimal to binary number conversion. Step 1 •Separate the digits of the given hexadecimal number, if it contains more than 1 digit. Step 2 •Find the equivalent binary number for each digit of hexadecimal number. Add 0's to the left if any of the binary equivalent is shorter than 4 bits. Step 3 •Write the all group's binary numbers together, maintaining the same group order provides the equivalent binary for the given hexadecimal number. 4. Convert 9531.812510 to hexadecimal number. Answer: 253B.D16 Example 23
CHAPTER 1: NUMBERING SYSTEM 25 Convert 4A716 to its binary equivalent. Answer: 4 A 7 0100 1010 0111 Ans: 4A716 = 100101001112 Convert 59C.0416 to its binary equivalent. Answer: 5 9 C . 0 4 0101 1001 1100 . 0000 0100 Ans: 59C.0416 = 10110011100.000001002 1. Convert D01216 to its binary equivalent. Answer: 11010000000100102 2. Convert 863.916 to its binary equivalent. Answer: 100001100011.10012 3. Convert 0.7A416 to its binary equivalent. Answer: 0.0111101001002 Example 24 Example 25 Exercise 9
CHAPTER 1: NUMBERING SYSTEM 26 Convert 1100.12 to its hexadecimal equivalent. Answer: 1100 . 1000 C . 8 Ans: 1100.12 = C.816 Convert 1000100.01102 to its hexadecimal equivalent. Answer: 0100 0100 . 0110 4 4 . 6 Ans: 1000100.01102 = 44.616 Hexadecimal to Binary conversion Grouping the binary position in 4 bit groups, starting from the least significant position. 4. Convert 6C.0116 to its binary equivalent. Answer: 1101100.000000012 5. Convert F2916 to its binary equivalent. Answer: 1111001010012 6. Convert 543.E16 to its binary equivalent. Answer: 010101000011.11102 Example 26 Example 27
CHAPTER 1: NUMBERING SYSTEM 27 1. Convert 100101.102 to its hexadecimal equivalent. Answer: 25.816 2. Convert 11100100.10112 to its hexadecimal equivalent. Answer: E4.B16 3. Convert 10010111.0112 to its hexadecimal equivalent. Answer: 97.616 4. Convert 111100110.1112 to its hexadecimal equivalent. Answer: 1E6.E16 5. Convert 10101.012 to its hexadecimal equivalent. Answer: 15.416 6. Convert 1001011.12 to its hexadecimal equivalent. Answer: 4B.816 Exercise 10
CHAPTER 1: NUMBERING SYSTEM 28 Convert B5A16 to its octal equivalent. Answer: 1. Separate the digits of given hexadecimal number, if it contains more than 1 digit. B 5 A 2. Find the equivalent binary number for each digit of hexadecimal number. Add 0’s to the left if any of the binary equivalent is shorter than 4 bits. B 5 A 1011 0101 1010 3. Write the all groups binary numbers together, maintaining the same group. 1011010110102 4. Separate the binary digits into groups, each containing 3 bits or digits from right to left. Add 0’s to the left, if the last group contains less than 3 bits. 101 101 011 010 5. Find the octal equivalent for each group. 101 101 011 010 5 5 3 2 6. Write all octal equivalent for each digit together where keeping the same order provides the octal equivalent for the given hexadecimal. 55318 Answer: B5A16 = 55328 1.3.3 Convert hexadecimal to octal and octal to hexadecimal. Hexadecimal to Octal conversion To perform hexadecimal to octal number conversion, we can follow these steps: 1. Separate the digits of given hexadecimal number, if it contains more than 1 digit. 2. Find the equivalent binary number for each digit of hex number. Add 0’s to the left if any of the binary equivalent is shorter than 4 bits. 3. Write the all groups binary numbers together, maintaining the same group. 4. Separate the binary digits into groups, each containing 3 bits or digits from right to left. Add 0’s to the left, if the last group contains less than 3 bits. 5. Find the octal equivalent for each group. 6. Write all octal equivalent for each digit together where keeping the same order provides the octal equivalent for the given hexadecimal. Example 28
CHAPTER 1: NUMBERING SYSTEM 29 Convert 4D816 to its octal equivalent. Answer: 4 D 8 0100 1101 1000 4D816 = 0100110110002 010 011 011 000 2 3 3 0 Ans: 0100110110002 = 23308 1. Convert C53E16 to octal number. Answer: 1424768 2. Convert 31F.516 to octal number. Answer: 1437.248 3. Convert 2E.3216 to octal number. Answer: 56.1448 Example 29 Exercise 11
CHAPTER 1: NUMBERING SYSTEM 30 Convert 7258 to its hexadecimal equivalent. Answer: 1. Separate the digits of given octal number, if it contains more than 1 digit. 7 2 5 2. Find the equivalent binary number for each digit of octal number. Add 0’s to the left if any of the binary equivalent is shorter than 3 bits. 7 2 5 111 010 101 3. Write the all groups binary numbers together, maintaining the same group. 1110101012 4. Separate the binary digits into groups, each containing 4 bits or digits from right to left. Add 0’s to the left, if the last group contains less than 4 bits. 1 1101 0101 5. Find the octal equivalent for each group. 0001 1101 0101 1 D 5 6. Write all octal equivalent for each digit together where keeping the same order provides the octal equivalent for the given hexadecimal. 1D516 Answer: 7258 = 1D516 Octal to Hexadecimal conversion To perform octal to hexadecimal number conversion, we can follow these steps: 1. Separate the digits of given octal number, if it contains more than 1 digit. 2. Find the equivalent binary number for each digit of octal number. Add 0’s to the left if any of the binary equivalent is shorter than 3 bits. 3. Write the all groups binary numbers together, maintaining the same group. 4. Separate the binary digits into groups, each containing 4 bits or digits from right to left. Add 0’s to the left, if the last group contains less than 4 bits. 5. Find the hex equivalent for each group. 6. Write all hex equivalent for each digit together where keeping the same order provides the hex equivalent for the given octal number. Example 30
CHAPTER 1: NUMBERING SYSTEM 31 Convert 1048 to its hexadecimal number. Answer: 1 0 4 001 000 100 1048 = 0010001002 0000 0100 0100 0 4 4 Ans: 1048 = 4416 1. Convert 63328 to hexadecimal number. Answer: CDA16 2. Convert 267.38 to hexadecimal number. Answer: B7.616 3. Convert 110.78 to hexadecimal number. Answer: 48.E16 Example 31 Exercise 12
CHAPTER 1: NUMBERING SYSTEM 32 Add the 10112 + 1002 in binary number. Answer: 1 0 1 12 + 0 1 0 02 1 1 1 12 10112 + 1002 = 11112 Add the 10102 + 1102 in binary number. Answer: 1 0 1 02 + 0 1 1 02 1 0 0 0 02 10102 + 1102 = 100002 1.4.1 Use binary arithmetic operations: Addition, subtraction and multiplication. Binary Addition The rules for adding binary numbers are as follows: 1.4 Apply binary arithmetic 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 carry 1 Example 32 Example 33 Carry
CHAPTER 1: NUMBERING SYSTEM 33 1. Add the 10012 + 11012 in binary number. Answer: 101102 2. Add the 01012 + 10002 in binary number. Answer: 11012 3. Add the 11012 + 00112 in binary number. Answer: 100002 4. Add the 01102 + 10012 in binary number. Answer: 11112 Exercise 13
CHAPTER 1: NUMBERING SYSTEM 34 Subtract the 10112 - 1002 in binary number. Answer: 1 0 1 12 - 0 1 0 02 0 1 1 12 10112 - 1002 = 01112 Subtract the 11012 - 00102 in binary number. Answer: 1 1 0 12 - 0 0 1 02 1 0 1 12 11012 - 00102 = 10112 Binary Subtraction The rules for subtracting binary numbers are as follows: 1. Subtract the 10012 - 01012 in binary number. Answer: 01002 0 - 0 = 0 0 - 1 = 1 borrow 1 1 - 0 = 1 1 - 1 = 0 Example 34 Borrow Example 35 Borrow Exercise 14
CHAPTER 1: NUMBERING SYSTEM 35 Binary Multiplication The rules for multiplying binary numbers are as follows: 2. Subtract the 11002 - 00102 in binary number. Answer: 10102 3. Subtract the 11102 - 10012 in binary number. Answer: 1012 4. Subtract the 11012 - 00112 in binary number. Answer: 10102 0 x 0 = 0 0 x 1 = 0 1 x 0 = 0 1 x 1 = 1
CHAPTER 1: NUMBERING SYSTEM 36 Multiplying 10102 and 1012 in binary number. Answer: 1 0 1 0 x 1 0 1 1 0 1 0 0 0 0 0 + 1 0 1 0 1 1 0 0 1 0 10102 × 1012 = 1100102 1. Multiplying 10012 and 1012 in binary number. Answer: 1011012 2. Multiplying 01012 and 0112 in binary number. Answer: 11112 3. Multiplying 11102 and 1002 in binary number. Answer: 1110002 Example 36 Exercise 15
CHAPTER 1: NUMBERING SYSTEM 37 Solve 11012 + DAD16 (Give your answer in octal number). Answer: Convert 11012 to octal number 001 101 1 5 Convert DAD16 to octal number D A D 1101 1010 1101 Convert to octal 110 110 101 101 6 6 5 5 DAD16 = 66558 Then, solve the problem when the base are same: 1 +1 58 + 6 6 5 58 6 6 7 28 So, the answer is 66728 1.4.2 Add and subtract in binary, octal and hexadecimal number systems. To solve the arithmetic process in a numbering system, the base of each number system that involved in the operation must be the same. 11012 = 158 Example 37
CHAPTER 1: NUMBERING SYSTEM 38 Solve 2468 - 5910 (Give your answer in hexadecimal number). Answer: Convert 2468 to decimal number = (2 × 8 2 ) + (4 × 8 1 ) + (6 × 8 0 ) = 128 + 32 + 6 = 16610 Solve the problem: 16610 - 5910 = 10710 Convert 10710 to hexadecimal number So, the answer is 6B16 **** Or, you can solve with other method **** Convert 2468 to hexadecimal by converting to binary first. 2 4 6 010 100 110 0000 1010 0110 0 A 6 2468 = A616 Convert 5910 to hexadecimal remainder 59 16 3 11 (B) 16 0 3 5910 = 3B16 A -1 Carry 16616 - 3 B16 6 B16 So, the answer is 6B16 remainder 107 16 6 11 (B) 16 0 6 Example 38
CHAPTER 1: NUMBERING SYSTEM 39 1. Solve 17210 + 2638 (Give your answer in hexadecimal number). Answer: 15F16 2. Solve 112 + 778 -5010 (Give your answer in octal number). Answer: 208 Exercise 16
40 2.1 Express basic algebra 2.1.1 Simplify algebraic expression 2.1.2 Simplify algebraic expression by using: a. Addition and Subtraction b. Multiplication and Division 2.2 Show quadratic equations 2.2.1 Solve quadratic equations by using: a. Factorization b. Quadratic formula CHAPTER 2: BASIC ALGEBRA The word algebra comes from the Arabic (al-jabr lit. "The reunion of broken parts") from the title of the book Ilm al-jabr wa'l-muḳābala by the Persian mathematician and astronomer alKhwarizmi. The word entered the English language during the fifteenth century, from either Spanish, Italian, or Medieval Latin. (https://en.wikipedia.org/wiki/Algebra)
CHAPTER 2: BASIC ALGEBRA 41 2.1.1 Simplify algebraic expression A constant is a quantity or parameter that does not change its value whatever the value of the variables, under a given set of conditions. It is written as a number on its own or sometimes as a letter for example a, b or c to stand for a fixed number. A variable is an element, feature, or factor that is liable to vary or change. It is written in a symbol usually a letter like x or y. Coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression. An operator is a symbol or function denoting an operation (e.g. ×, +,÷, −). An expression is a collection of symbols that jointly express a quantity. A term is the quantities in a ratio, series or mathematical expression that is wrote as either a single number or a variable, or numbers and variables multiplied together. Expression Number of terms Variable Coefficient Ex: 2 + 3 − 3 , , 2, 3,-1 4 − 3 ________________ ________________ ________________ −12 − 7 + 6 ________________ ________________ ________________ 3 ________________ ________________ ________________ 2 + 3 − ________________ ________________ ________________ 2 + 3 − 11 ________________ ________________ ________________ 2.1 Express basic algebra Try this! 7 Coefficient An expression in single term with factors 7 and x. Variable Coefficient Variable x Variable y Coefficient An expression in two terms. 7 + 4
CHAPTER 2: BASIC ALGEBRA 42 Simplify each of the following: a. 2 3 + 2 + 5 + 3 − 4 2 + 1 c. (−7 − 2) − (10 + 3) b. ( − 5) + (2 + 3) Answer: a. 2 3 + 2 + 5 + 3 − 4 2 + 1 Group together the like terms. = 2 3 + 3 + 2 − 4 2 + 5 + 1 Simplify. = 3 3 − 3 2 + 6 Like terms and not like terms. Like terms are terms that contain the same variables raised to the same power. Not like terms are terms that contain more than one variable. Like Terms Not Like Terms + 3 7 − 12 2 + 10 7 + 4 2 7 2 − 3 2 7 + 4 + −2 + 5 + 11 3 − 32 + 1 2 3 − 3 3 + 3 4 3 1 2 3 − 1 2 3 2.1.2 Simplify algebraic expression by using addition, subtraction, multiplication and division. a) Addition and Subtraction Important! We can only add and subtract like terms. Example 1 Exercise 1 b. ( − 5) + (2 + 3) = − 5 + 2 + 3 = + 2 − 5 + 3 = 3 − 2 c. (−7 − 2) − (10 + 3) = −7 − 2 − 10 − 3 = −7 − 10 − 2 − 3 = −17 − 5
CHAPTER 2: BASIC ALGEBRA 43 1. 5w² − 1 + 7w + 2w² − w − 6w²= 2. 2 2 + 4 + 1 − 2 − 2 − 7 = 3. 3 + + 2 5 + 5 = 4. (4 − 3) + (7 + 6) = 5. (2 + ) + (4 − 7) − (5 + 7) = 6. ( 4 + 4 − 3) − (2 4 − 4 + 2) = 7. 2 − ( + 10) − = 8. ( 2 + ) − (10) + ( 2 ) = 9. 10. b) Multiplication and Division 2 3 + + 1 2 = 1 2 2 2 + 2 2 2 − 5 2 2 =