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 The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and th...

 The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and the National Anthem; (b) to cherish and follow the noble ideals which inspired our national struggle for freedom; (c) to uphold and protect the sovereignty, unity and integrity of India; (d) to defend the country and render national service when called upon to do so; (e) to promote harmony and the spirit of common brotherhood amongst all the people of India transcending religious, linguistic and regional or sectional diversities, to renounce practices derogatory to the dignity of women; (f) to value and preserve the rich heritage of our composite culture; (g) to protect and improve the natural environment including forests, lakes, rivers and wild life and to have compassion for living creatures; (h) to develop the scientific temper, humanism and the spirit of inquiry and reform; (i) to safeguard public property and to abjure violence; (j) to strive towards excellence in all spheres of individual and collective activity so that the nation constantly rises to higher levels of endeavour and achievement; (k) who is a parent or guardian to provide opportunities for education to his child or, as the case may be, ward between the age of six and fourteen years.  The Coordination Committee formed by GR No.Abhyas - 2116/(Pra.Kra 43/16) SD - 4 Dated 25.4.2016 has given approved to prescribe this textbook in its meeting held on 30.01.2020 and it has been decided to implement it from the educational year 2020 - 21. Mathematics and Statistics (Arts and Science) Part - I STANDARD - XII Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. - 411 004 Download DIKSHA App on your smartphone. If you scan the Q.R.Code on this page of your textbook, you will be able to access full text and the audio-visual study material relevant to each lesson provided as teachaing and learning aids. First Edition : 2020 © Maharashtra State Bureau of Textbook Production and Curriculum Second Reprint : 2022 Research, Pune- 411 004. Maharashtra State Bureau of Textbook Production and Curriculum Research reserves all rights relating to the book. No part of this book should be reproduced without the written permission of the Director, Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. Mathematics and Statistics Co-ordinator (Arts and Science) Ujjwala Shrikant Godbole Commitee Members I/C Special Officer for Mathematics Dr. Mangla Narlikar (Chairman) Dr. Sharad Gore Member Cover Shri. Prahallad Chippalagatti Member & Shri. Prasad Kunte Member Computer Drawings Shri. Sujit Shinde Member Shri. Sandip Koli, Artist, Mumbai Smt. Prajakti Gokhale Member Shri. Ramakant Sarode Member Smt. Ujjwala Godbole (Member -Secretary) Production Sachin Mehta Mathematics and Statistics Chief Production Officer (Arts and Science) Sanjay Kamble Study Group Members Production Officer Dr. lshwor Patil Dr. Pradeep Mugale Prashant Harne Dr. Pradnyankumar Bhojnkar Shri. Uday Mahindrakar Asst. Production Officer Shri. Prafullachandra Pawar Shri. Balkrishna Mapari Shri. Sachin Batwal Shri. Devanand Bagul Shri. Swapnil Shinde Smt. Swati Pawar Paper Shri. Pradipkumor Bhavsar Smt. Mahadevi Mane 70 GSM Cream wove Shri. Amit Totade Shri. Dilip Panchamia Smt. Gouri Prachand Shri. Pramod Deshpande Print Order No. Shri. Sharadchandra Walagade Shri. Milind Patil Shri. Dhananjay Panhalkar Smt. Supriya Abhyankar Printer Smt. Deepti Kharpas Smt. Nileema Khaladkor Typesetter Publisher Vivek Uttam Gosavi, Controller M/s. Shree Graphic Centre, Maharashtra State Textbook Bureau, Mumbai Prabhadevi Mumbai- 400 025  The Constitution of India Preamble WE, THE PEOPLE OF INDIA, having solemnly resolved to constitute India into a SOVEREIGN SOCIALIST SECULAR DEMOCRATIC REPUBLIC and to secure to all its citizens: JUSTICE, social, economic and political; LIBERTY of thought, expression, belief, faith and worship; EQUALITY of status and    of opportunity; and to promote among them all FRATERNITY assuring the dignity of the individual and the unity and integrity of the Nation; IN OUR CONSTITUENT ASSEMBLY this twenty-sixth day of November, 1949, do HEREBY ADOPT, ENACT AND GIVE TO OURSELVES THIS CONSTITUTION. NATIONAL ANTHEM  PREFACE Dear Students, Welcome to Standard XII, an important milestone in your life. Standard XII or Higher Secondary School Certificate opens the doors of higher education. Alternatively, you can pursue other career paths like joining the workforce. Either way, you will find that mathematics education helps you considerably. Learning mathematics enables you to think logically, consistently, and rationally. The curriculum for Standard XII Mathematics and Statistics for Science and Arts students has been designed and developed keeping both of these possibilities in mind. The curriculum of Mathematics and Statistics for Standard XII for Science and Arts students is divided in two parts. Part I deals with topics like Mathematical Logic, Matrices, Vectors and Introduction to three dimensional geometry. Part II deals with Differentiation, Integration and their applications, Introduction to random variables and statistical methods. The new text books have three types of exercises for focussed and comprehensive practice. First, there are exercises on every important topic. Second, there are comprehensive exercises at the end of all chapters. Third, every chapter includes activities that students must attempt after discussion with classmates and teachers. Additional information has been provided on the E-balbharati website (www.ebalbharati.in). We are living in the age of Internet. You can make use of modern technology with the help of the Q.R. code given on the title page. The Q.R. code will take you to links that provide additional useful information. Your learning will be fruitful if you balance between reading the text books and solving exercises. Solving more problems will make you more confident and efficient. The text books are prepared by a subject committee and a study group. The books (Paper I and Paper II) are reviewed by experienced teachers and eminent scholars. The Bureau would like to thank all of them for their valuable contribution in the form of creative writing, constructive and useful suggestions for making the text books valuable. The Bureau hopes and wishes that the text books are very useful and well received by students, teachers and parents. Students, you are now ready to study. All the best wishes for a happy learning experience and a well deserved success. Enjoy learning and be successful. (Vivek Gosavi) Pune Director Date: 21 February 2020 Maharashtra State Bureau of Textbook Bharatiya Saur: 2 Phalguna 1941 Production and Curriculum Research, Pune.  Mathematics and Statistics Std. XII Part - I Competency Statement Sr.No. Area Topic Competency Statement 1. Mathematical Logic Mathematical Logic The student will be able to Identify statement in logic and truth value of it. Combine two or more statements Construct the truth table and examine logical equivalence of statement patterns Find dual and negation of statement pattern Study the applications of logic to switching circuits. 2. Matrices Matrices Identify orders and types of matrices Perform basic algebraic operations on matrices. Find the inverse of a matrix using elementary transformation and adjoint method Solve the system of linear equations using matrices. 3. Trigonometric Trigonometric Understand and write trigonometric Functions Equations equation Find the principal and general solution of a trigonometric equation. Solution of a triangle Solve triangle by using sine rule, co- sine rule and projection rule and find area of a triangle. Inverse trigonometric Understand inverse trigonometric function functions with domain and range. 4. Pair of straight lines Pair of straight lines Write and interpret the combined equation of two straight lines in plane. Find the point of intersection of two lines and calculate the acute angle between them Study the general second degree equation in x and y with reference to homogeneous part of it 5. Vectors Vectors Understand scalars and vectors and algebra of vectors. Write vectors of 2 or 3 dimensions understand the scalar and vector products Study applications of vectors to area of triangle. Interprete scalar triple product and its applications. Direction ratios and direction cosines of a vector 6. Line and Plane Line and Plane Find different forms of equation of line Find angle between two intersecting planes Find the angle between a line and a plane Find condition for perpendicularity and parallelness of planes Calculate distance of a point from a plane Find equation of a plane in different forms Find angle between two intersecting planes Find the angle between a line and a plane 7. Linear programming Linear programming Understand linear equations in one Problem Problem and two variables. Find graphical solution of linear inequation. Understand meaning and formulation of L.P.P. Find solution of L.P.P. by graphical methods.  INDEX Sr.No. Chapter Page No 1. Mathematical Logic 1 2. Matrices 36 3. Trigonometric Functions 65 4. Pair of Straight Lines 112 5. Vectors 133 6. Line and Plane 194 7. Linear Programming 227 Answers 246  1 Mathematical Logic Let’s study. 1 Statement and its truth value. 2. Logical connective, compound statements. 3. Truth tables negation of statements and compound statements. 4. Statement pattern, logical equivalence. 5. Tautology, contradiction and contingency. 6. Quantifiers and quantified statements, Duality. 7. Application of logic to switching circuits, switching table. Let’s recall. 1.1.1 Introduction : Mathematics is a logical subject and tries to be exact. For exactness, it requires proofs which depend upon proper reasoning. Reasoning requires logic. The word Logic is derived from the Greek word "LOGOS" which means reason. Therefore logic deals with the method of reasoning. In ancient Greece the great philosopher and thinker Aristotle started study of Logic systematically. In mathematics Logic has been developed by English Philosopher and mathematician George Boole (2 November 1815 - 8 December 1864) Language is the medium of communication of our thoughts. For communication we use sentences. In logic, we use the statements which are special sentences. 1.1.2 Statement : A statement is a declarative (assertive) sentence which is either true or false, but not both simultaneously. Statements are denoted by p, q, r,..... 1.1.3 Truth value of a statement : Each statement is either true or false. If a statement is true then its truth value is 'T' and if the statement is false then its truth value is F. Illustrations : 1) Following sentences are statements. i) Sun rises in the East. ii) 5 × 2 = 11 iii) Every triangle has three sides. 1  iv) Mumbai is the capital of Maharashtra. v) Every equilateral triangle is an equiangular triangle. vi) A natural number is an integer. 2) Following sentences are not statements. i) Please, give your Pen. ii) What is your name ? iii) What a beautiful place it is ! iv) How are you ? v) Do you like to play tennis ? vi) Open the window. vii) Let us go for tea viii) Sit down. Note : Interrogative, exclamatory, command, order, request, suggestion are not statements. 3) Consider the following. 3x i) −9 = 0 2 ii) He is tall. iii) Mathematics is an interesting subject. iv) It is black in colour. Let us analyse these statements. i) For x = 6 it is true but for other than 6 it is not true. ii) Here, we cannot determine the truth value. For iii) & iv) the truth value varies from person to person. In all the above sentences, the truth value depends upon the situation. Such sentences are called as open sentences. Open sentence is not a statement. Solved examples Q.1. Which of the following sentences are statements in logic ? Write down the truth values of the statements. i) 6 × 4 = 25 ii) x + 6 = 9 iii) What are you doing ? iv) The quadratic equation x2 – 5x + 6 = 0 has 2 real roots. v) Please, sit down vi) The Moon revolves around the earth. vii) Every real number is a complex number. viii) He is honest. ix) The square of a prime number is a prime number. 2 Solution : i) It is a statement which is false, hence its truth value is F. ii) It is an open sentence hence it is not a statement. iii) It is an interrogative hence it is not a statement. iv) It is a statement which is true hence its truth value is T. v) It is a request hence it is not a statement. vi) It is a statement which is true, hence its truth value is T. vii) It is a statement which is true, hence its truth value is T. viii) It is open sentence, hence it is not a statement. ix) It is a statement which is false, hence its truth value is F. 1.1.4 Logical connectives, simple and compound statements : The words or phrases which are used to connect two statements are called logical connectives. We will study the connectives 'and', 'or', 'if..... then', 'if and only if ', 'not''. Simple and Compound Statements : A statement which cannot be split further into two or more statements is called a simple statement. If a statement is the combination of two or more simple statements, then it is called a compound statement. "3 is a prime and 4 is an even number", is a compound statement. "3 and 5 are twin primes", is a simple statement. We describe some connectives. 1) Conjunction : If two statements are combined using the connective 'and' then it is called as a conjunction. In other words if p, q are two statements then 'p and q' is called as conjunction. It is denoted by 'p ∧ q' and it is read as 'p conjunction q' or 'p and q'. The conjunction p ∧ q is said to be true if and only if both p and q are true. Truth table for conjunction p q p∧q T T T T F F F T F F F F Table 1.1 2) Disjunction : If two statements are combined by using the logical connective 'or' then it is called as a disjunction. In other words if p, q are two staements then 'p or q' is called as disjunction. It is denoted by 'p ∨ q and it is read as 'p or q' or 'p disjunction q'. Truth table for disjunction p q p∨q T T T T F T The disjunction p ∨ q is false if and only if both p and q are false. F T T F F F Table 1.2 3 3) Conditional (Implication) : If two statements are combined by using the connective. 'if.... then', then it is called as conditional or implication. In other words if p, q are two state- ments then 'if p then q' is called as conditional. It is denoted by p → q or p ⇒ q and it is read as 'p implies q' or 'if p then q'. Truth table for conditional. p q p→q T T T The conditional statement p → q is False only if p is true T F F and q is false. Otherse it is true. Here p is called hypothesis or antecedent and q is called conclusion or consequence. F T T F F T Table 1.3 Note : The following are also conditional statement p → q i) p is sufficient for q ii) q is necessary for p iii) p implies q iv) q follows from p v) p only if q. 4) Biconditional (Double implication) : If two statements are combined using the logical connective 'if and only if ' then it is called as biconditional. In other words if p, q are two statements then 'p if and only if q' is called as biconditional. It is denoted by 'p ↔ q' or p ⇔ q. It is read as 'p biconditional q' or 'p if and only if q'. Truth table for biconditional. p q p↔q T T T T F F Biconditional statement p ↔ q is true if p and q have same F T F truth values. Otherwise it is False. F F T Table 1.4 5) Negation of a statement : For any given statement p, there is another statement which is defined to be true when p is false, and false when p is true, is called the negation of p and is denoted by ~p. Truth table for negation. p ~ p T F F T Table 1.5 Note : Negation of negation of a statement is the statement itself. That is, ~ (~ p) = p. 4  Solved examples Ex.1: Express the following compound statements symbolically without examining the truth values. i) 2 is an even number and 25 is a perfect square. ii) A school is open or there is a holiday. iii) Delhi is in India but Dhaka is not in Srilanks. iv) 3 + 8 ≥ 12 if and only if 5 × 4 ≤ 25. Solution : i) Let p : 2 is an even numder q : 25 is a perfect square. The symbolic form is p ∧ q. ii) Let p : The school is open q : There is a holiday The symbolic form is p ∨ q iii) Let p : Delhi is in India q : Dhaka is in Srilanka The symbolic form is p ∧ ~ q. iv) Let p : 3 + 8 ≥ 12; q : 5 × 4 ≤ 25 The symbolic form is p ↔ q Ex.2. Write the truth values of the following statements. i) 3 is a prime number and 4 is a rational number. ii) All flowers are red or all cows are black. iii) If Mumbai is in Maharashtra then Delhi is the capital of India. iv) Milk is white if and only if the Sun rises in the West. Solution : i) Let p : 3 is a prime number q : 4 is a rational number. Truth values of p and q are T and T respectively. The given statement in symbolic form is p ∧ q. The truth value of given statement is T. ii) Let p : All flowers are red ; q : All cows are black. Truth values of p and q are F and F respectively. The given statement in the symbolic form is p ∨ q \ p ∨ q ≡ F ∨ F is F \ Truth value of given statement is F. iii) Let p : Mumbai is in Maharashtra q : Delhi is capital of India Truth values of p and q are T and T respectively. The given statement in symbolic form is p → q 5  \ p → q ≡ T → T is T \Truth value of given statement is T iv) Let p : Milk is white; q : Sun rises in the West. Truth values of p and q are T and F respectively. The given statement in symbolic form is p ↔ q \ p ↔ q ≡ T ↔ F is F \Truth value of given statement is F Ex.3 : If statements p, q are true and r, s are false, determine the truth values of the following. i) ~ p ∧ (q ∨ ~ r) ii) (p ∧ ~ r) ∧ (~ q ∨ s) iii) ~ (p → q) ↔ (r ∧ s) iv) (~p → q) ∧ (r ↔ s) Solution : i) ~ p ∧ (q ∨ ~ r) ≡ ~ T ∧ (T ∨ ~ F) ≡ F ∧ (T ∨ T) ≡ F ∧ T ≡ F Hence truth value is F. ii) (p ∧ ~ r) ∧ (~ q ∨ s) ≡ (T ∧ ~ F) ∧ (~T ∨ F) ≡ (T ∧ T) ∧ (F ∨ F) ≡ T ∧ F ≡ F. Hence truth value is F. iii) [~(p → q)] ↔ (r ∧ s) ≡ [~ ( T → T )] ↔ (F ∧ F) ≡ (~ T) ↔ (F) ≡ F ↔ F ≡ T. Hence truth value is T iv) (~p → q) ∧ (r ↔ s) ≡ (~T → T) ∧ (F ↔ F) ≡ (F →T) ∧ T ≡ T ∧ T ≡ T. Hence truth value is T. Ex.4. Write the negations of the following. i) Price increases ii) 0! ≠ 1 iii) 5 + 4 = 9 Solution : i) Price does not increase ii) 0! = 1 iii) 5 + 4 ≠ 9 Exercise 1.1 Q.1. State which of the following are statements. Justify. In case of statement, state its truth value. i) 5 + 4 = 13. ii) x – 3 = 14. iii) Close the door. iv) Zero is a complex number. v) Please get me breafast. vi) Congruent triangles are similar. vii) x2 = x. 6  viii) A quadratic equation cannot have more than two roots. ix) Do you like Mathematics ? x) The sun sets in the west xi) All real numbers are whole numbers xii) Can you speak in Marathi ? xiii) x2 – 6x – 7 = 0, when x = 7 xiv) The sum of cuberoots of unity is zero. xv) It rains heavily. Q.2. Write the following compound statements symbollically. i) Nagpur is in Maharashtra and Chennai is in Tamilnadu ii) Triangle is equilateral or isosceles. iii) The angle is right angle if and only if it is of measure 90º. iv) Angle is neither acute nor obtuse. v) If D ABC is right angled at B, then m ∠A + m ∠C = 90º vi) Hima Das wins gold medal if and only if she runs fast. vii) x is not irrational number but is a square of an integer. Q.3. Write the truth values of the following. i) 4 is odd or 1 is prime. ii) 64 is a perfect square and 46 is a prime number. iii) 5 is a prime number and 7 divides 94. iv) It is not true that 5-3i is a real number. v) If 3 × 5 = 8 then 3 + 5 = 15. vi) Milk is white if and only if sky is blue. vii) 24 is a composite number or 17 is a prime number. Q.4. If the statements p, q are true statements and r, s are false statements then determine the truth values of the following. i) p ∨ (q ∧ r) ii) (p → q) ∨ (r → s) iii) (q ∧ r) ∨ (~ p ∧ s) iv) (p → q) ∧ ~r v) (~r ↔ p) → ~ q vi) [~p ∧ (~q ∧ r)] ∨ [(q ∧ r) ∨ (p ∧ r)] vii) [(~p ∧ q) ∧ ~r ] ∨ [(q → p) → (~s ∨ r)] viii) ~[(~p ∧ r) ∨ (s → ~ q)] ↔ (p ∧ r) Q.5. Write the negations of the following. i) Tirupati is in Andhra Pradesh ii) 3 is not a root of the equation x2 + 3x – 18 = 0 iii) 2 is a rational number. iv) Polygon ABCDE is a pentagon. v) 7+3>5 7  1.2 STATEMENT PATTERN, LOGICAL EQUIVALENCE, TAUTOLOGY, CONTRADICTION, CONTINGENCY. 1.2.1 Statement Pattern : Letters used to denote statements are called statement letters. Proper combination of statement letters and connectives is called a statement pattern. Statement pattern is also called as a proposition. p → q, p ∧ q, ~ p ∨ q are statement patterns. p and q are their prime components. A table which shows the possible truth values of a statement pattern obtained by considering all possible combinations of truth values of its prime components is called the truth table of the statement pattern. 1.2.2. Logical Equivalence : Two statement patterns are said to be equivalent if their truth tables are identical. If statement patterns A and B are equivalent, we write it as A ≡ B. 1.2.3 Tautology, Contradiction and Contingency : Tautology : A statement pattern whose truth value is true for all possible combinations of truth values of its prime components is called a tautology. We denote tautology by t. Statement pattern p ∨ ~ p is a tautology. Contradiction : A statement pattern whose truth value is false for all possible combinations of truth values of its prime components is called a contradiction. We denote contradiction by c. Statement pattern p ∧ ~ p is a contradiction. Contingency : A statement pattern which is neither a tautology nor a contradiction is called a contingency. p ∧ q is a contingency. Important table for all connectives : p q ~p p∧q p∨q p→q p↔q T T F T T T T T F F F T F F F T T F T T F F F T F F T T  In a statement pattern, different symbols are considered in the following priority ~, ∨, ∧, →, ↔ Solved Examples Ex.1.: Construct the truth table for each of the following statement patterns. i) p → (q → p) ii) (~ p ∨ q) ↔ ~ (p ∧ q) iii) ~ (~ p ∧ ~ q) ∨ q iv) [(p ∧ q) ∨r] ∧ [~ r ∨ (p ∧ q)] v) [(~ p ∨ q) ∧ (q → r)] → (p → r) 8 Solution : i) p → (q → p) p q q→p p → (q → p) T T T T T F T T F T F T F F T T Table 1.7 ii) (~p ∨ q) ↔ ~(p ∧ q) p q ~p ~p∨q p∧q ~ ( p ∧ q) (~ p ∨ q) ↔ ∧ (p ∧q) T T F T T F F T F F F F T F F T T T F T T F F T T F T T Table 1.8 iii) ~ (~ p ∧ ~ q) ∨ q p q ~p ~q (~p ∧~q) ~(~p ∧~ q) ~(~p ∧~ q) ∨ q T T F F F T T T F F T F T T F T T F F T T F F T T T F F Table 1.9 iv) [(p ∧ q) ∨ r] ∧ [~ r ∨ (p ∧ q) ] p q r ~r p ∧ q (p ∧ q) ∨ r ~r∨(p ∧ q) [p ∧ q) ∨ r] ∧ ( ~ r ∨ (p ∧ q) T T T F T T T T T T F T T T T T T F T F F T F F T F F T F F T F F T T F F T F F F T F T F F T F F F T F F T F F F F F T F F T F Table 1.10 9 v) [(~p ∨ q) ∧ (q → r)] → (p → r) p q r ~p ~p∨q q → r p → r (~p ∨ q) ∧ [(~p ∨ q) ∧ (q → r)] → (q → r) (q → r) T T T F T T T T T T T F F T F F F T T F T F F T T F T T F F F F T F F T F T T T T T T T T F T F T T F T F T F F T T T T T T T F F F T T T T T T Table 1.11 Ex.2: Using truth tables, prove the following logical equivalences i) (p ∧ q) ≡ ~ (p → ~ q) ii) (p ↔ q) ≡ (p ∧ q) ∨ (~ p ∧ ~ q) iii) (p ∧ q) → r ≡ p → (q → r) iv) p→(q ∨ r) ≡ (p → q) ∨ (p → r) i) Solution : (1) (p ∧ q) ≡ ~(p → ~q) I II III IV V VI p q p ∧ q p → ~q ~(p →~ q) ~q T T F T F T T F T F T F F T F F T F F F T F T F Table 1.12 Columns (IV) and (VI) are identical \ (p ∧ q) ≡ ~(p → ~q) ii) (p ↔ q) ≡ (p ∧ q) ∨ ( ~ p ∧ ~ q) I II III IV V VI VII VIII p q ~p ~q p↔q (p ∧ q) ~p ∧ ~ q (p ∧ q) ∨ (~p ∧ ~ q) T T F F T T F T T F F T F F F F F T T F F F F F F F T T T F T T Table 1.13 Columns V and VIII are identical \ (p ↔ q) ≡ (p ∧ q) ∨ (~p ∨ ~ q) (iii) (p ∧ q) → r ≡ p → (q → r) 10  I II III IV V VI VII p q r p∧q (p ∧ q) → r q→r p →( q → r) T T T T T T T T T F T F F F T F T F T T T T F F F T T T F T T F T T T F T F F T F T F F T F T T T F F F F T T T Table 1.14 Column (V) and (VII) are identical \ (p ∧ q) → r ≡ p → (q → r) (iv) p → (q ∨ r) ≡ (p → q) ∨ (p → r) I II III IV V VI VII VIII p q r q∨r p → (q ∨ r) p→q p→r (p → q) ∨ (p → r) T T T T T T T T T T F T T T F T T F T T T F T T T F F F F F F F F T T T T T T T F T F T T T T T F F T T T T T T F F F F T T T T Table 1.15 Columns V and VIII are identical \ p → (q ∨ r) ≡ (p → q) ∨ (p → r) Ex.3. Using truth tables, examine whether each of the following statement pattern is a tautology or a contradiction or contingency. i) (p ∧ q) ∧ (~ p ∨ ~ q) ii) [p ∧ (p → ~ q)] → q iii) (p → q) ∧ [(q → r) → (p → r)] iv) [(p ∨ q) ∨ r] ↔ [p ∨ (q ∨ r)] 11 Solution: i) (p ∧ q) ∧ (~ p ∨ ~ q) p q ~p ~q p∧q ~p ∨ ~ q (p ∧ q) ∧ (~p ∨ ~q) T T F F T F F T F F T F T F F T T F F T F F F T T F T F Table 1.16 All the truth values in the last column are F. Hence it is contradiction. ii) [p ∧ (p → ~q)] → q p q ~q p → ~q p ∧( p → ~ q) p ∧ (p → ~q) → q T T F F F T T F T T T F F T F T F T F F T T F T Table 1.17 Truth values in the last column are not identical. Hence it is contingency. iii) (p → q) ∧ [(q → r) → (p → r)] p q r p→q q → r p → r (q → r)→ (p→ r) (p → q) ∧(q → r) → (p → r) T T T T T T T T T T F T F F T T T F T F T T T F T F F F T F F F F T T T T T T T F T F T F T T T F F T T T T T T F F F T T T T T Table 1.18 Truth values in the last column are not same, hence it is contingency. iv) [(p ∨ q) ∨ r] ↔ (p ∨ (q ∨ r)] p q r p ∨ q (p ∨ q)∨ r q ∨ r p ∨ (q ∨ r) [(p ∨ q) ∨ r] ↔ [p ∨ (q ∨ r)] T T T T T T T T T T F T T T T T T F T T T T T T T F F T T F T T F T T T T T T T F T F T T T T T F F T F T T T T F F F F F F F T Table 1.19 All the truth values in the last column are T, hence it is tautology. 12  Exercie.1.2 Q.1. Construct the truth table for each of the following statement patterns. i) [(p → q) ∧ q] → p ii) (p ∧ ~ q) ↔ (p → q) iii) (p ∧ q) ↔ (q ∨ r) iv) p → [~ (q ∧ r)] v) ~ p ∧ [(p ∨ ~ q ) ∧ q] vi) (~ p → ~ q) ∧ (~ q → ~ p) vii) (q → p) ∨ (~ p ↔ q) viii) [p → (q → r)] ↔ [(p ∧ q) → r] ix) p → [~ (q ∧ r)] x) (p ∨ ~ q) → (r ∧ p) Q.2. Using truth tables prove the following logical equivalences. i) ~ p ∧ q ≡ (p ∨ q) ∧ ~ p ii) ~ (p ∨ q) ∨ (~ p ∧ q) ≡ ~ p iii) p ↔ q ≡ ~ [(p ∨ q) ∧ ~ (p ∧ q)] iv) p → (q → p) ≡ ~ p → (p → q) v) (p ∨ q ) → r ≡ (p → r) ∧ (q → r) vi) p → (q ∧ r) ≡ (p → q) ∧ (p → r) vii) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) viii) [~ (p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r ix) ~ (p ↔ q) ≡ (p ∧ ~ q) ∨ (q ∧ ~ p) Q.3. Examine whether each of the following statement patterns is a tautology or a contradiction or a contingency. i) (p ∧ q) → (q ∨ p) ii) (p → q) ↔ (~ p ∨ q) iii) [~ (~ p ∧ ~ q)] ∨ q iv) [(p → q) ∧ q)] → p v) [(p → q) ∧ ~ q] → ~ p vi) (p ↔ q) ∧ (p → ~ q) vii) ~ (~ q ∧ p) ∧ q viii) (p ∧ ~ q) ↔ (p → q) ix) (~ p → q) ∧ (p ∧ r) x) [p → (~ q ∨ r)] ↔ ~ [p → (q → r)] 13 1.3 QUANTIFIERS, QUANTIFIED STATEMENTS, DUALS, NEGATION OF COMPOUND STATEMENTS, CONVERSE, INVERSE AND CONTRAPOSITIVE OF IMPLICATION. 1.3.1 Quantifiers and quantified statements. Look at the following statements : p : "There exists an even prime number in the set of natural numbers". q : "All natural numbers are positive". Each of them asserts a condition for some or all objects in a collection. Words "there exists" and "for all" are called quantifiers. "There exists is called existential quantifier and is denoted by symbol $. "For all" is called universal quantifier and is denoted by ". Statements involving quantifiers are called quantified statements. Every quantified statement corresponds to a collection and a condition. In statement p the collection is 'the set of natural numbers' and the condition is 'being even prime'. What is the condition in the statement q ? A statement quantified by universal quantifier " is true if all objects in the collection satisfy the condition. And it is false if at least one object in the collection does not satisfy the conditon. A statement quantified by existential quantifier $ is true if at least one object in the collection satisfies the condition. And it is false if no object in the collection satisfies the condition. Ex.1. If A = {1, 2, 3, 4, 5, 6, 7}, determine the truth value of the following. i) ∃ x ∈ A such that x – 4 = 3 ii) ∀x∈A, x+1>3 iii) ∀ x ∈ A, 8 – x < 7 iv) ∃ x ∈ A, such that x + 8 = 16 Solution : i) For x = 7, x – 4 = 7 – 4 = 3 \ x = 7 satisfies the equation x – 4 = 3 \ The given statement is true and its truth value is T. ii) For x = 1, x + 1 = 1 + 1 = 2 which is not greater than or equal to 3 \ For x = 1, x + 1 > 3 is not true. \ The truth value of given statement is F. iii) For each x ∈ A 8 – x < 7 \ The given statement is true. \ Its truth value is T. iv) There is no x in A which satisfies x + 8 = 16. \ The given statement is false. \ Its truth value is F. 1.3.2 Dual : We use letters t and c to denote tautology and contradiction respectively. If two statements contain logical connectives like ∨, ∧ and letters t and c then they are said to be duals of each other if one of them is obtained from the other by interchanging ∨ with ∧ and ∧ with ∨ , t with c and c with t. The dual of i) p ∨ q is p ∧ q ii) t ∨ p is c ∧ p iii) t ∧ p is c ∨ p 14 Ex.1. Write the duals of each of the following : i) (p ∧ q) ∨ r ii) t ∨ (p ∨ q) iii) p ∧ [~ q ∨ (p ∧ q) ∨ ~ r] iv) (p ∨ q) ∧ t v) (p ∨ q) ∨ r ≡ p ∨ (q ∨ r ) vi) p ∧ q ∧ r vii) (p ∧ t) ∨ (c ∧ ~ q) Solution : i) (p ∨ q) ∧ r ii) c ∧ (p ∧ q) iii) p ∨ [(~ q ∧ (p ∨ q) ∧ ~ r] iv) (p ∧ q) ∨ c v) (p ∧ q) ∧ r ≡ p ∧ (q ∧ r ) vi) p∨q∨r vii) (p ∨ c) ∧ (t ∨ ~q) 1.3.3 Negations of compound statements : Negation of conjunction : When is the statment "6 is even and perfect number" is false? It is so, if 6 is not even or 6 is not perfect number. The negation of p ∧ q is ~ p ∨ ~ q. The negation of "6 is even and perfect number" is "6 not even or not perfect number". Activity : Using truth table verify that ~ (p ∧ q) ≡ ~ p ∨ ~ q Negation of disjunction : When is the statement "x is prime or y is even" is false? It is so, if x is not prime and y is not even. The negation of p ∨ q is ~ p ∧ ~ q. The negation of "x is prime or y is even" is "x is not prime and y is not even". Activity : Using truth table verify that ~ (p ∨ q) ≡ ~ p ∧ ~ q Note : '~ (p ∧ q) ≡ ~ p ∨ ~ q' and '~ (p ∨ q) ≡ ~ p ∧ ~ q' are called De'Morgan's Laws Negation of implication : Implication p → q asserts that "if p is true statement then q is true statement". When is an implication a true statement and when is it false? Consider the statement "If bakery is open then I will buy a cake for you." Clearly statement is false only when the bakery was open and I did not buy a cake for you. The conditional statement "If p then q" is false only in the case "p is true and q is false". In all other cases it is true. The negation of the statement "If p then q" is the statement "p and not q". i.e. p does not imply q Activity : Using truth table verify that ~ (p → q) ≡ p ∧ ~ q Negation of biconditional : The biconditional p ↔ q is the conjuction of statement p → q and q → p. \ p ↔ q ≡ (p → q) ∧ (q → p) \ The conditional statement p ↔ q is false if p → q is false or q → p false. The negation of the statement "p if and only if q " is the statement "p and not q, or q and not p". \ ~ (p ↔ q) ≡ (p ∧ ~ q) ∨ (q ∧ ~ p) Activity : Using truth table verify that ~ (p ↔ q) ≡ (p ∧ ~ q) ∨ (q ∧ ~ p) 1.3.4 Converse, inverse and contrapositive From implication p → q we can obtain three implications, called converse, inverse and contrapositive. q → p is called the converse of p → q ~ p → ~ q is called the inverse of p → q. ~ q → ~ p is called the contrapositive of p → q. 15 Activity : Prepare the truth table for p → q, q → p, ~ p → ~q and ~ q → ~p. What is your conclusion from the truth table ? i)........................ ≡......................... ii)........................ ≡......................... Ex.1) Write the negations of the following. i) 3+3 3 and 4 > 11 iii) The number is neither odd nor perfect square. iv) The number is an even number if and only if it is divisible by 2. Solution : i) Let p : 3 + 3 < 5 : q : 5 + 5 = 9 Given statement is p ∨ q and its negation is ~(p ∨ q) and ~(p ∨ q) ≡ ~ p ∧ ~ q \ The negation of given statement is 3 + 3 > 5 and 5 + 5 ≠ 9 ii) Let p : 7 > 3; q: 4 > 11 The given statement is p ∧ q Its negation is ~ (p ∧ q) and ~ (p ∧ q) ≡ ~ p ∨ ~ q \ The negation of given statement is 7 < 3 or 4 < 11 iii) Let p : The number is odd q : The number is perfect square Given statement can be written as 'the number is not odd and not perfect square' Given statement is ~ p ∧ ~ q Its negation is ~ (~ p ∧ ~ q) ≡ p ∨ q The negation of given statement is 'The number is odd or perfect square'. iv) Let p : The number is an even number. q : The number is divisible by 2 Given statement is p ↔ q Its negation is ~ (p ↔ q) But ~ (p ↔ q) ≡ (p ∧ ~ q) ∨ (q ∧ ~ p) \ The negation of given statement is 'A number is even but not divisible by 2 or a number is divisible by 2 but not even'. Negation of quantified statement : while doing the negations of quantified statement we replace the word 'all' by 'some', "for every" by "there exists" and vice versa. Ex.2. Write the negations of the following statements. i) All natural numbers are rational. ii) Some students of class X are sixteen year old. iii) ∃ n ∈ N such that n + 8 > 11 iv) ∀ x ∈ N, 2x + 1 is odd 16 Solution : i) Some natural numbers are not rationals. ii) No student of class X is sixteen year old. iii) ∀ n ∈ N, n + 8 < 11 iv) ∃ x ∈ N such that 2x + 1 is not odd Ex.3. Write the converse, inverse and contrapositive of the following statements. i) If a function is differentiable then it is continuous. ii) If it rains then the match will be cancelled. Solution : (1) Let p : A function is differentiable q : A function is continuous. \ Given statement is p → q i) Its converse is q → p If a function is continuous then it is differentiable. ii) Its inverse is ~ p → ~ q. If a function not differentiable then it is not continuous. iii) Its contrapositive is ~ q → ~ p If a function is not continuous then it is not differentiable. (2) Let p : It rains, q : The match gets cancelled. \ Given statement is p → q i) Its converse is q → p If the match gets cancelled then it rains. ii) Inverse is ~ p → ~ q If it does not rain then the match will not be cancelled. iii) Its contrapositive is ~ q → ~ p. If the match is not cancelled then it does not rain. Exercise 1.3 Q.1. If A = {3, 5, 7, 9, 11, 12}, determine the truth value of each of the following. i) ∃ x ∈ A such that x – 8 = 1 ii) ∀ x ∈ A, x2 + x is an even number iii) ∃ x∈ A such that x2 < 0 iv) ∀ x ∈ A, x is an even number v) ∃ x ∈ A such that 3x + 8 > 40 vi) ∀ x ∈ A, 2x + 9 > 14 Q.2. Write the duals of each of the following. i) p ∨ (q ∧ r) ii) p ∧ (q ∧ r) iii) (p ∨ q) ∧ (r ∨ s) iv) p∧~q v) (~ p ∨ q) ∧ (~ r ∧ s) vi) ~ p ∧ (~ q ∧ (p ∨ q) ∧ ~ r) vii) [~ (p ∨ q)] ∧ [p ∨~ (q ∧ ~ s)] viii) c ∨ {p ∧ (q ∨ r)} ix) ~ p ∨ (q ∧ r) ∧ t x) (p ∨ q) ∨ c 17 Q.3. Write the negations of the following. i) x + 8 > 11 or y – 3 = 6 ii) 11 < 15 and 25 > 20 iii) Qudrilateral is a square if and only if it is a rhombus. iv) It is cold and raining. v) If it is raining then we will go and play football. vi) 2 is a rational number. vii) All natural numbers are whole numbers. viii) ∀ n ∈ N, n2 + n + 2 is divisible by 4. ix) ∃ x ∈ N such that x – 17 < 20 Q.4. Write converse, inverse and contrapositive of the following statements. i) If x < y then x2 < y2 (x, y ∈ R) ii) A family becomes literate if the woman in it is literate. iii) If surface area decreases then pressure increases. iv) If voltage increases then current decreases. 1.4 SOME IMPORTANT RESULTS : 1.4.1. i) p → q ≡ ~ p ∨ q ii) p ↔ q ≡ (p → q) ∧ (q → p) iii) p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r) iv) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) I II III IV V VI VII VIII p q ~p p→q q→p p↔q ~p ∨ q (p → q) ∧ (q → p) T T F T T T T T T F F F T F F F F T T T F F T F F F T T T T T T Table 1.20 Columns (IV, VII) and (VI, VIII) are identical. \ p → q ≡ ~ p ∨ q and p ↔ q ≡ (p → q) ∧ (q → p) are proved. Activity : Prove the results (iii) and (iv) by using truth table. 18 1.4.2. Algebra of statements. Idempotent Law p ∧ p ≡ p, p∨p≡p Commutative Law p∨q≡q∨p p∧q≡q∧p Associative Law p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r ≡ p ∧ q ∧ r p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r ≡ p ∨ q ∨ r Distributive Law p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r) De Morgan's Law ~(p ∧ q) ≡ ~ p ∨ ~q, ~(p ∨ q) ≡ ~p ∧ ~q Identity Law p ∧ t ≡ p, p ∧ c ≡ c,p ∨ c ≡ p, p ∨ t ≡ t Complement Law p ∧ ~ p ≡ c, p ∨ ~ p ≡ t Absorption Law p ∨ (p ∧ q) ≡ p, p ∧ (p ∨ q) ≡ p Conditional Law p→q≡~p∨q Biconditional Law p ↔ q ≡ (p → q) ∧ (q → p) ≡ (~p ∨ q)∧ (~q ∨ p) Table 1.21 Solved Examples Ex.1. Write the negations of the following stating the rules used. i) (p ∨ q) ∧ (q ∨ ~ r) ii) (p → q) ∨ r iii) p ∧ (q ∨ r) iv) (~ p ∧ q) ∨ (p ∧ ~ q) v) (p ∧ q) → (~ p ∨ r) Solution : i) ~ [(p ∨ q) ∧ (q ∨ ~ r)] ≡ ~ (p ∨ q) ∨ ~ (q ∨ ~ r) [DeMorgan’s law] ≡ (~ p ∧ ~ q) ∨ (~ q ∧ r) [DeMorgan’s law] ≡ (~ q ∧ ~ p) ∨ (~ q ∧ r) [Commutative law] ≡ ~ q ∧ (~ p ∨ r) [Distributive law] ii) ~ [(p → q) ∨ r] ≡ ~ (p → q) ∧ ~ r [DeMorgan’s law] ≡ (p ∧ ~ q) ∧ ~ r [~ (p → q) ≡ p ∧ ~ q] iii) ~ [p ∧ (q ∨ r)] ≡ ~ p ∨ ~ (q ∨ r) [DeMorgan’s law] ≡ ~ p ∨ (~ q ∧ ~ r) [DeMorgan’s law] iv) ~ [(~ p ∧ q) ∨ (p ∧ ~ q)] ≡ ~ (~ p ∧ q) ∧ ~ (p ∧ ~ q) [DeMorgan’s law] ≡ (p ∨ ~ q) ∧ (~ p ∨ q) [DeMorgan’s law] ~ [(p ∧ q) → (~ p ∨ r)] ≡ (p ∧ q) ∧ ~ (~ p ∨ r) v) [~ (p → q) ≡ p ∧ ~ q] ≡ (p ∧ q) ∧ [p ∧ ~ r] [DeMorgan’s law] ≡ q ∧ p ∧ p ∧ ~ r [Associative law ≡ q ∧ p ∧ ~ r [Idempotent law] 19 Ex.2. Rewrite the following statements without using if...... then. i) If prices increase then the wages rise. ii) If it is cold, then we wear woolen clothes. Solution : i) Let p : Prices increase q : The wages rise. The given statement is p → q but p → q ≡ ~ p ∨ q The given statement can be written as 'Prices do not increase or the wages rise'. ii) Let p : It is cold, q : We wear woollen clothes. The given statement is p → q but p → q ≡ ~ p ∨ q The given statement can be written as It is not cold or we wear woollen clothes. Ex.3. Without using truth table prove that : i) p ↔ q ≡ ~ (p ∧ ~ q) ∧ ~ (q ∧ ~ p) ii) ~ (p ∨ q) ∨ (~ p ∧ q) ≡ ~ p iii) ~ p ∧ q ≡ (p ∨ q) ∧ ~ p Solution : i) We know that p ↔ q ≡ (p → q) ∧ (q → p) ≡ (~ p ∨ q) ∧ (~ q ∨ p) [Conditional law] ≡ ~ ( p ∧ ~ q) ∧ ~ (q ∧ ~p) [Demorgan’s law] ii) ~ (p ∨ q) ∨ (~ p ∧ q) ≡ (~ p ∧ ~ q) ∨ (~ p ∧ q) [Demorgan’s law] ≡ ~ p ∧ (~ q ∨ q) [Distributive law] ≡ ~ p ∧ t [Complement law] ≡ ~ p [Identity law] iii) (p ∧ q) ∧ ~ p ≡ ~ p ∧ (p ∨ q) [Commutative law] ≡ (~ p ∧ p) ∨ (~ p ∧ q) [Distributive law] ≡ c ∨ (~ p ∧ q) [Complement law] ≡ ~ p ∧ q [Identity law] 20  Exercise 1.4 Q.1. Using rules of negation write the negations of the following with justification. i) ~q→p ii) p∧~q iii) p∨~q iv) (p ∨ ~ q) ∧ r v) p → (p ∨ ~ q) vi) ~ (p ∧ q) ∨ (p ∨ ~ q) vii) (p ∨ ~ q) → (p ∧ ~ q) viii) (~ p ∨ ~ q)∨(p ∧ ~ q) Q.2. Rewrite the following statements without using if.. then. i) If a man is a judge then he is honest. ii) It 2 is a rational number then 2 is irrational number. iii) It f(2) = 0 then f(x) is divisible by (x – 2). Q.3. Without using truth table prove that : i) p ↔ q ≡ (p∧ q) ∨ (~ p ∧ ~ q) ii) (p ∨ q) ∧ (p ∨ ~ q) ≡ p iii) (p ∧ q) ∨ (~ p ∧ q) ∨ (p ∧ ~ q) ≡ p ∨ q iv) ~ [(p ∨ ~ q) → (p ∧ ~ q)] ≡ (p ∨ ~ q) ∧ (~ p ∨ q) Application of Logic to switching circuits : We shall study how the theory of Logic can be applied in switching network. We have seen that a logical statement can be either true or false i.e. it can have truth value either T or F. A similar situation exists in various electrical devices. For example, an electric switch can be on or off. In 1930 Claude Shannan noticed an analogy between operation of switching circuits and operation of logical connectives. In an electric circuit, switches are connected by wires. If the switch is 'on', it allows the electric current to pass through, it. If the switch is 'off', it does not allow the electric current to pass through it. We now define the term 'switch' as follows. Switch : A switch is a two state device used to control the flow of current in a circuit. We shall denote the switches by letters S, S1, S2, S3.... etc. Fig. 1.1 Fig. 1.2 21  In figure 1.2, we consider a circuit containing an electric lamp L, controlled by a switch S. When the switch S is closed (i.e. on), then current flows in the circuit and hence the lamp glows. When the switch S is open (i.e. off), then current does not flow in the circuit and subsequently the lamp does not glow. The theory of symbolic logic can be used to represent a circuit by a statement pattern. Conversely for given statement pattern a circuit can be constructed. Corresponding to each switch in the cirucit we take a statement letter in statement pattern. Switches having the same state will be denoted by the same letter and called equivalent switches. Switches having opposite states are denoted by S and S'. They are called complementary switches. In circuit we don't show whether switch is open or closed. In figure 1.3 switch S1 corresponds to statement letter p in the corresponding statement pattern. We write it as p : switch S1 and ~ p : switch S'1 The correspondence between switch S2 and statement letter q is shown as q : switch S2 and ~ q : switch S'2. We don't know the actual states of switches in the circuit. We consider all possible combinations of states of all switches in the circuit and prepare a table, called "Input Output table", which is similar to truth table of the corresponding statement pattern.  In an Input-output table we represent '1' when the state of the switch is 'on' and '0' when the state of the switch is 'off '. 1.5.1. Two switches in series. Two switches S1 and S2 connected in series and electric lamp 'L' as shown in fig 1.3. Fig. 1.3 Let p : The switch S1 q : The switch S2 L : The lamp L Input output table (switching table) for p ∧ q. p q p∧q 1 1 1 1 0 0 0 1 0 0 0 0 Table 1.22 22 1.5.2 Two switches in parallel : Two switches S1 and S2 are connected in parallel and electric lamp L is as shown in fig. 1.4 Let p : The switch S1 q : The switch S2 L : The lamp L Input - output table. for p ∨ q. p q p∨q 1 1 1 1 0 1 0 1 1 0 0 0 Table 1.23 Fig. 1.4 Q.1. Express the following circuits in the symbolic form of logic and write the input-output table. i) Fig. 1.5 ii) iii) Fig. 1.6 Fig. 1.7 23 Solution : i) Let p : The switch S1 q : The switch S2 L : The lamp L Given circuit is expressed as (p ∨ q) ∨ (~ p ∧ ~ q) Solution : p q ~p ~q p∨q ~ p ∧ ~ q (p ∨ q) ∨ (~p ~ q) 1 1 0 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 1 Table 1.24 ii) Let p : The switch S1 is closed q : The switch S2 r : The switch S3 L : The lamp L The symbolic form is [(p ∧ q) ∨ (~ p ∧ ~ q)] ∧ r p q r ~p ~q p ∧ q (p ∧ q) ∨ (~p ∧ ~q) (p ∧ q) ∨ (~p ∧ ~q) [(p ∧ q) ∨ (~p ∧ ~q)] ∧ r 1 1 1 0 0 1 0 1 1 1 1 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 1 0 1 1 1 0 0 0 1 1 0 1 1 0 Table 1.25 iii) Let p : The switch S1 q : The switch S2 r : The switch S3 L : The lamp L The symbolic form of given circuit is (p ∨ q) ∧ q ∧ (r ∨ ~ p) p q r ~p p ∨ q r ∨ ~p (p ∨ q) ∧ q (p ∨ q) ∧ q ∧ (r ∨ ~p) 1 1 1 0 1 1 1 1 1 1 0 0 1 0 1 0 1 0 1 0 1 1 0 0 1 0 0 0 1 0 0 0 0 1 1 1 1 1 1 1 0 1 0 1 1 1 1 1 0 0 1 1 0 1 0 0 0 0 0 1 0 1 0 0 Table 1.26 24 Ex.2. Construct switching circuits of the following. i) [(p ∨ (~ p ∧ q)] ∨ [(~ q ∧ r) ∨ ~ p] ii) (p ∧ q ∧ r) ∨ [~ p ∨ (q ∧ ~ r)] iii) [(p ∧ r) ∨ (~ q ∧ ~r)] ∨ (~ p ∧ ~ r) Solution : Let p : The switch S1 q : The switch S2 r : The switch S3 The circuits are as follows. Fig. 1.8 Fig. 1.9 Fig. 1.10 25 Ex.3. Give an alternative arrangement for the following circuit, so that the new circuit has minimum switches. Let p : The switch S1 Fig. 1.11 q: The switch S2 The symbolic form is (p∧~ q) ∨ (~p ∧ q) ∨ (~ p ∧ ~ q) Consider (p ∧~ q) ∨ (~ p ∧ q) ∨ (~ p ∧ ~ q) ≡ (p ∧ ~ q) ∨ [~ p ∧ (q ∨ ~ q)] [Distributive Law] ≡ (p ∧ ~ q) ∨ [~ p ∧ t] [Complement Law] ≡ (p ∧ ~ q) ∨ ~ p [Identity Law] ≡ ~ p ∨ (p ∧ ~ q) [Commutative Law] ≡ (~p ∨ p) ∧ (~p ∨ ~ q) [Distributive Law] ≡ t ∧ (~ p ∨ ~ q) [Law of Complement] ≡ ~ p ∨ ~ q [Identity Law] The alternative arrangement for the given circuit is as follows : Fig. 1.12 Ex.4. Express the following switching circuit in the symbolic form of Logic. Construct the switching table and interpret it. Fig. 1.13 26 Solution : Let p : The switch S1 q : The switch S2 The symbolic form of the given switching circuit is (p ∨ q) ∧ (~p) ∧ (~q) The switching table. p q ~p ~q p∨q (p ∨ q) ∧ (~ p) (p ∨ q) ∧ (~ p) ∧ (~ q) 1 1 0 0 1 0 0 1 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 1 1 0 0 0 Table 1.27 Last column contains all 0, lamp will not glow irrespective of the status of the switches. Ex.5. Simplify the given circuit by writing its logical expression. Also,write your conclusion. Fig. 1.14 Let p : The switch S1 q: The The switch S2 The logical expression for the given circuit is p ∧ (~ p ∨ ~ q) ∧ q Consider p ∧ (~ p ∨ ~ q) ∧ q ≡ [ p ∧ ( ~ p ∨ ~ q] ∧ q [Associative Law] ≡ [(p ∧ ~ p) ∨ (p ∨ ~ q)] ∧ q [Distributive Law] ≡ [c ∨ (p ∧ ~ q)] ∧ q [Complement Law] ≡ (p ∧ ~ q) ∧ q [Identity Law] ≡ p ∧ (~ q ∧ q) [Associative Law] ≡ p∧c [Complement Law] ≡ c [Identity Law] Conclusion : The lamp will not glow irrespective of the status of the switches. 27 Ex. 6 : In the following switching circuit, i) Write symbolic form ii) Construct switching table iii) Simplify the circuit Fig. 1.15 Solution : Let p : The switch S1. q : The switch S2. r : The switch S3 1) The symbolic form of given circuit is (p ∧ q) ∨ (~ p ∧ q) ∨ (r ∧ ~ q). ii) Switching Table : P q r ~ p ~ q ~ p ∧ q ~ p ∧ q r ∧ ~ q (p ∧ q) ∨ (~p ∧ q) ∨ (r ∧ ~ q) 1 1 1 0 0 1 0 0 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 1 0 0 1 0 1 0 1 0 1 0 0 1 0 1 0 0 1 1 1 0 0 1 1 0 0 0 1 1 0 0 0 0 Table 1.28 iii) Consider = (p ∧ q) ∨ (~p ∧ q) ∨ (r ∧ ~q) = [(p ∨ ~p) ∧ q] ∨ [(r ∧ ~q)] [Distributive Law] = (t ∧ q) ∨ (r ∨ ~q) [Complement Law] = q ∨ (r ∧ ~q) [Identity Law] = (q ∨ r) ∧ (q ∨ ~q) [Distributive Law] = (q ∨ r) ∧ t [Complement Law] = (q ∨ r) [Identify Law] Simplified circuit is : Fig. 1.16 28  Exercise 1.5 Q.1. Express the following circuits in the symbolic form of logic and writ the input-output table. (i) (ii) Fig. 1.17 Fig. 1.18 (iii) (iv) Fig. 1.19 Fig. 1.20 (v) (vi) Fig. 1.21 Fig. 1.22 29 Q.2. Construct the switching circuit of the following : i) (~p ∧ q) ∨ (p ∧ ~ r) ii) (p ∧ q) ∨ [~p ∧ (~ q ∨ p ∨ r)] iii) [(p ∧ r) ∨ (~q ∧ ~ r)] ∧ (~ p ∧ ~ r) iv) (p ∧ ~ q ∧ r) ∨ [p ∧ (~ q ∨ ~r)] v) p ∨ (~ p ) ∨ (~ q) ∨ (p ∧ q) vi) (p ∧ q) ∨ (~ p) ∨ (p ∧ ~q) Q.3. Give an alternative equivalent simple circuits for the following circuits : (i) (ii) Fig. 1.23 Fig. 1.24 Q.4. Write the symbolic form of the following switching circuits construct its switching table and interpret it. i) ii ) Fig. 1.25 Fig. 1.26 (iii) Fig. 1.27 Q.5. Obtain the simple logical expression of the following. Draw the corresponding switching circuit. i) p ∨ (q ∧ ~ q) ii) (~ p ∧ q) ∨ (~ p ∧ ~ q) ∨ (p ∧ ~ q)] iii) [p (∨ (~ q) ∨ ~r)] ∧ (p ∨ (q ∧ r) iv) (p ∧ q ∧ ~ p) ∨ (~ p ∧ q ∧ r) ∨ (p ∧ ~ q ∧ r) ∨ (p ∧ q ∧ r) 30  Let’s remember! 1) A declarative sentence which is either true or false, but not both simultaneously is called a statement. Sr. Connective Symbolic Name of Com- Hint in Negation No. Form pound Statement truth table i) and p∧q Conjunction T∧T≡T ~p ∨ ~q ii) or p∨q Disjunction F∨F≡F ~p ∧ ~ q iii) if... then p → q Conditional T→F≡F p∧~q iv) if and only p ↔ q Biconditional T ↔ T ≡ T (p ∧ ~q) ∨ (q ∧ ~p) if F↔F≡T Table 1.29 3) In the truth table of the statement pattern if all truth values in the last column a) are 'T' then it is tautology. b) are 'F' then it is contradiction. 4) In the truth table of the statement pattern if some entries are 'T' and some are 'F' then it is called as contingency. 5) The symbol ∀ stands for 'for all' or 'for every'. It is universal quantifier. The symbol ∃ stands for 'for some' or 'for one' or 'there exists at least one'. It is called as existential quantifier. 6) Algebra of statements. Idempotent Law p ∧ p ≡ p, p∨p≡p Commutative Law p∨q≡q∨p p∧q≡q∧p Associative Law p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r ≡ p ∧ q ∧ r p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r ≡ p ∨ q ∨ r Distributive Law p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r) De Morgan's Law ~(p ∧ q) ≡ ~ p ∨ ~q, ~(p ∨ q) ≡ ~p ∧ ~q Identity Law p ∧ t ≡ p, p ∧ c ≡ c,p ∨ c ≡ p, p ∨ t ≡ t Complement Law p ∧ ~ p ≡ c, p ∨ ~ p ≡ t Absorption Law p ∨ (p ∧ q) ≡ p, p ∧ (p ∨ q) ≡ p Conditional Law p→q≡~p∨q Biconditional Law p ↔ q ≡ (p → q) ∧ (q → p) ≡ (~p ∨ q)∧ (~q ∨ p) 7) If p → q is conditional then its converse is q → p, inverse is ~p → ~q and contrapositive is ~ q → ~p. 31 8) Switching circuits : i) Switches in series ii) Switches in parallel. Fig. 1.28 Fig. 1.29 Input-output table p q p∧q p∨q 1 1 1 1 1 0 0 1 0 1 0 1 0 0 0 0 Table 1.30 Miscellaneous Exercise 1 I]Select and write the correct answer from the given alternatives in each of the following questions : i) If p ∧ q is false and p ∨ q is true, the ________ is not true. A) p ∨ q B) p↔q C) ~p ∨ ~q D) q ∨ ~p ii) (p ∧ q) → r is logically equivalent to ________. A) p → (q → r) B) (p ∧ q) → ~r C) (~p ∨ ~q) → ~r D) (p ∨ q) → r iii) Inverse of statement pattern (p ∨ q) → (p ∧ q) is ________. A) (p ∧ q) → (p ∨ q) B) ~(p ∨ q) → (p ∧ q) C) (~p ∧ ~q) → (~p ∨ ~q) D) (~p ∨ ~q) → (~p ∧ ~q) iv) If p ∧ q is F, p → q is F then the truth values of p and q are ________. A) T, T B) T, F C) F, T D) F, F v) The negation of inverse of ~p → q is ________. A) q ∧ p B) ~p ∧ ~q C) p ∨ q D) ~q → ~p vi) The negation of p ∧ (q → r) is ________. A) ~p ∧ (~q → ~r) B) p ∨ (~q ∨ r) C) ~p ∧ (~q → ~r) D) ~p ∨ (q ∧ ~r) vii) If A = {1, 2, 3, 4, 5} then which of the following is not true? A) ∃ x ∈ A such that x + 3 = 8 B) ∃ x ∈ A such that x + 2 < 9 C) ∀ x ∈ A, x + 6 ≥ 9 D) ∃ x ∈ A such that x + 6 < 10 32 Q.2. Which of the following sentences are statements in logic? Justify. Write down the truth value of the statements : i) 4! = 24. ii) p is an irrational number. iii) India is a country and Himalayas is a river. iv) Please get me a glass of water. v) cos2q - sin2q = cos2q for all q∈R. vi) If x is a whole number the x + 6 = 0. Q.3. Write the truth values of the following statements : i) 5 is an irrational but 3 5 is a complex number. ii) " n ∈ N, n2 + n is even number while n2 – n is an odd number. iii) ∃ n ∈ N such that n + 5 > 10. iv) The square of any even number is odd or the cube of any odd number is odd. v) In ∆ABC if all sides are equal then its all angles are equal. vi) ∀ n ∈ N, n + 6 > 8. Q.4. If A = {1, 2, 3, 4, 5, 6, 7, 8, 9}, determine the truth value of each of the following statement : i) ∃ x ∈ A such that x + 8 = 15. ii) ∀ x ∈ A, x + 5 < 12. iii) ∃ x ∈ A, such that x + 7 ≥ 11. iv) ∀ x ∈ A, 3x ≤ 25. Q.5. Write the negations of the following : i) ∀ n ∈ A, n + 7 > 6. ii) ∃ x ∈ A, such that x + 9 ≤ 15. iii) Some triangles are equilateral triangle. Q.6. Construct the truth table for each of the following : i) p → (q → p) ii) (~p ∨ ~q) ↔ [~(p ∧ q)] iii) ~(~p ∧ ~q) ∨ q iv) [(p ∧ q) ∨ r] ∧ [~r ∨ (p ∧ q)] v) [(~p ∨ q) ∧ (q → r)] → (p → r) Q.7. Determine whether the following statement patterns are tautologies contradictions or contingencies : i) [(p → q) ∧ ~q)] → ~p ii) [(p ∨ q) ∧ ~p] ∧ ~q iii) (p → q) ∧ (p ∧ ~q) iv) [p → (q → r)] ↔ [(p ∧ q) → r] v) [(p ∧ (p → q)] → q vi) (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q) vii) [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r viii) (p → q) ∨ (q → p) 33 Q.8. Determine the truth values of p and q in the following cases : i) (p ∨ q) is T and (p ∧ q) is T ii) (p ∨ q) is T and (p ∨ q) → q is F iii) (p ∧ q) is F and (p ∧ q) → q is T Q.9. Using truth tables prove the following logical equivalences : i) p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q) ii) (p ∧ q) → r ≡ p → (q → r) Q.10. Using rules in logic, prove the following : i) p ↔ q ≡ ~ (p ∧ ~q) ∧ ~(q ∧ ~p) ii) ~p ∧ q ≡ (p ∨ q) ∧ ~p iii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p Q.11. Using the rules in logic, write the negations of the following : i) (p ∨ q) ∧ (q ∨ ~r) ii) p ∧ (q ∨ r) iii) (p → q) ∧ r iv) (~p ∧ q) ∨ (p ∧ ~q) Q.12. Express the following circuits in the symbolic form. Prepare the switching table : (i) (ii) Fig. 1.30 Fig. 1.31 Q.13. Simplify the following so that the new circuit has minimum number of switches. Also, draw the simplified circuit. (i) (ii) Fig. 1.32 Fig. 1.33 34 Q.14. Check whether the following switching circuits are logically equivalent - Justify. (A) i) ii) Fig. 1.34 Fig. 1.35 (B) i) ii) Fig. 1.36 Fig. 1.37 Q.15. Give alternative arrangement of the Q.16 Simplify the following switching following circuit, so that the new circuit circuit. has minimum switches. has minimum switches. Fig. 1.38 Fig. 1.39 Q.17. Represent the following switching circuit in symbolic form and construct its switching table. Write your conclusion from the switching table. Fig. 1.40  35  2 Matrices Let's Study. 2.1 Elementry transformations. 2.2 Inverse of a matrix 2.2.1 Elementary transformation Method 2.2.2 Adjoint method 2.3 Application of matrices. Solution of a system of linear equations 2.3.1 Method of Inversion 2.3.2 Method of Reduction Let’s learn. A matrix of order mxm is a square arrangement of m2 elements. The corresponding determinant of the same elements, after expansion is seen to be a value which is an element itself. In standard XI, we have studies the types of matrices and algebra of matrices namely addition, subtraction, multiplication of two matrices. The matrices are useful in almost every branch of science. Many problems in Statistics are expressed in terms of matrices. Matrices are also useful in Economics, Operation Research. It would not be an exaggeration to say that the matrices are the language of atomic Physics. Hence, it is necessary to learn the uses of matrices with the help of elementary transformations and the inverse of a matrix. 2.1 Elementary Transformation : Let us first understand the meaning and applications of elementary transformations. The elementary transformation of a matrix are the six operations, three of which are due to row and three are due to column. They are as follows : (a) Interchange of any two rows or any two columns. If we interchange the ith row and the jth row of a matrix then after this interchange the original matrix is transformed to a new matrix. This transformation is symbolically denoted as Ri ↔ Rj or Rij. The similar transformation can be due to two columns say Ck ↔ Ci or Cki. (Recall that R and C symbolically represent the rows and columns of a matrix.) 36  1 2  3 4  For example, if A = 3 4  then R1 ↔ R2 gives the new matrix   and C1↔C2 gives the   1 2   2 1 new matrix  .  4 3 3 4   2 1 3 4   2 1 Note that A ≠   and ≠   but we write A ~ 1 2  and A ~  4 3 1 2   4 3     Note : The symbol ~ is read as equivalent to. (b) Multiplication of the elements of any row or column by a non-zero scalar : If k is a non-zero scalar and the row Ri is to be multiplied by constant k then we multiply every element of Ri by the constant k and symbolically the transformation is denoted by kRi or Ri →kRi. 0 2 0 2 For example, if A =   then R2 → 4R2 gives A ~   3 4 12 16  Similarly, if any column of a matrix is to be multiplied by a constant then we multiply every element of the column by the constant. It is denoted as kCl or Cl → kCl. 0 2  0 2  For example, if A = 3 4  then Cl → –3 Cl gives A ~  −9 4      0 2  0 2 Can you say that A = 12 16  or A =  −9 4      (c) Adding the scalar multiples of all the elements of any row (column) to corresponding ele- ments of any other row (column). If k is a non-zero scalar and the k-multiples of the elements of Ri (Ci) are to be added to the elements of Rj (Cj) then the transformation is symbolically denoted as Rj → Rj + kRi , Cj → Cj + kCi  −1 4  For example, if A =   and k = 2 then Rl → Rl + 2R2 gives  2 5  −1 + 2 ( 2 ) 4 + 2 ( 5 )  A ~  2 5     3 14  i.e. A ~   2 5  (Can you find the transformation of A using C2 → C2 + (–3) C1 ?) Note (1) : After the transformation , Rj → Rj + kRi , Ri remains the same as in the original matrix. Similarly, with the transformation , Cj → Cj + kCi , Ci remains the same as in the original matrix. 37 Note (2) : After the elementary transformation, the matrix obtained is said to be equivalent to the original matrix.  1 0 Ex. 1 : If A =   , apply the transformation R1 ↔ R2 on A.  −1 3 Solution :  1 0 As A =  −1 3   R1 ↔ R2 gives  −1 3 A ~    1 0 Ex. 2 : If A =  1 0 2 , apply the transformation C1 → C1 + 2C3. 2 3 4   Solution : A =  1 0 2 2 3 4   C1 → C1 + 2C3 gives 1 + 2 ( 2 ) 0 2  A ~ 2 + 2 ( 4 ) 3 4    5 0 2 A ~ 10 3 4    1 2 −1 Ex. 3 : If A =   , apply R1 ↔ R2 and then C1 → C1 + 2C3 on A. 3 −2 5  Solution : 1 2 −1 A =   3 −2 5  38 R1 ↔ R2 gives 3 −2 5  1 2 −1 A ~   Now C1 → C1 + 2C3 gives  3 + 2 ( 5 ) −2 5  13 −2 5  A ~   \ A ~  −1 2 −1 1 + 2 ( −1) 2 −1   Exercise 2.1 Apply the given elementary transformation on each of the following matrices.  1 0 1 −1 3  1. A =   , R1 ↔ R2. 2. B =   R1→ R1 – R2.  −1 3 2 5 4 5 4   3 1 3. A =   , C1 ↔ C2; B=   R1 ↔ R2. 1 3   4 5 What do you observe? 1 2 −1 1 0 2  4. A =   , 2C2 B =   , –3R1. 0 1 3  2 4 5 Find the addition of the two new matrices. 1 −1 3   5. A =  2 1 0  , 3R3 and then C3 + 2C2.  3 3 1  1 −1 3   6. A =  2 1 0  , C3 + 2C2 and then 3R3.  3 3 1  What do you conclude from ex. 5 and ex. 6? 1 2  7. Use suitable transformation on   to convert it into an upper triangular matrix.  3 4  1 −1 8. Convert   into an identity matrix by suitable row transformations. 2 3  39  1 −1 2  9. Transform  2 1 3  into an upper triangular matrix by suitable column transformations.    3 2 4  2.2 Inverse of a matrix : Definition : If A is a square matrix of order m and if there exists another square matrix B of the same order such that AB = BA = I, where I is the identity matrix of order m, then B is called as the inverse of A and is denoted by A–1. Using the notation A-1 for B we get the above equation as AA–1 = A–1A = I. Hence, using the same definition we can say that A is also the inverse of B. \ B-1 = A 2 3  2 −3 2 3  2 −3 For example, if A =   and B =  −1 2  then AB =    −1 2  1 2    1 2    1 0  \ AB =   ==12I2 0 1   2 −3  2 3  1 0  and BA =  −1 2  1 2  = 0 1  = I 2      \ B = A-1 and A = B-1 1 2  If A =  2 4  , can you find a matrix X such that AX = I ? Justify the answer.   This example illustrates that for the existence of such a matrix X, the necessary condition is |A| ≠ 0, i.e. A is a non-singular matrix. Note that - (1) Every square matrix A of order m × m has its corresponding determinant; det A = |A| (2) A matrix is said to be invertible if its inverse exists. (3) A square matrix A has inverse if and only if |A| ≠ 0 Uniqueness of inverse of a matrix It can be proved that if A is a square matrix where |A| ≠ 0 then its inverse, say A-1, is unique. Theorem : Prove that if A is a square matrix and its inverse exists then it is unique. Proof : Let, ‘A’ be a square matrix of order ‘m’ and let its inverse exist. Let, if possible, B and C be the two inverses of A. Therefore, by definition of inverse AB = BA = I and AC = CA = I. 40  Now consider B = BI = B (AC) \ B = (BA) C = IC \ B = C Hence B = C i.e. the inverse is unique. The inverse of a matrix (if it exists) can be obtained by using two methods. (i) Elementary row or column transformation (ii) Adjoint method We now study these methods. 2.2.1 Inverse of a nonsingular matrix by elementary transformation : By definition of inverse of A, if A–1 exists then AA-1 = A-1 A = I. Let us consider the equation AA-1 = I. Here A is the given matrix of order m and I is the identity matrix of order ‘m’. Hence the only unknown matrix is A-1. Therefore, to find A-1, we have to first convert A into I. This can be done by using elementary transformations. Here we note that whenever any elementary row transformation is to be applied on the product AB = C of two matrices A and B, it is enough to apply it only on the prefactor, A. B remains un- changed. And apply the same row transformation to C.  −1 0  1 10  For example, if A =  1 2 and B =   then AB =   = C (say) 3 4   1 5  1 20     1 20  Now if we require C to be transformed to a new matrix by R1 ↔ R2 then C~   1 10  3 4  If the same transformation is used for A then A ~   and B remains unchanged, 1 2  3 4   −1 0  1 20  then the product AB =    1 5  = 1 10  = as required. (Verify the product.) 1 2      Hence, the equation AA–1 = I can be transformed into an equation of the type IA–1 = B, by applying same series of row transformations on both the sides of the equation. However, if we start with the equation A-1A = I (which is also true by the definition of inverse) then the transformation of A should be due to the column transformation. Apply column transformation to post factor and other side, where as prefactor remains unchanged. Thus, starting with the equation AA–1 = I , we perform a series of row transformations on both sides of the equation, so that ‘A’ gets transformed to I. Thus, A A-1 = I Row Row Transformations Transformations I A-1 = B \ A-1 = B 41 and for the equation A-1A = I, we use a series of column transformations. Thus A-1 A = I Column Column Transformation Transformation A -1 I = B \ A -1 = B Now if A is a given matrix of order ‘3’ and it is nonsingular then we consider  a11 a12 a13   a23  A =  a21 a22  a31 a32 a33  For reducing the above matrix to 1 0 0  0 1 0  I3 =   , the suitable row transformations are as follows : 0 0 1  (1) Reduce a11 to ‘1’. (2) Then, reduce a21 and a31 to ‘0’. (3) Reduce a