Principle of Mathematical Induction: Proofs and Examples, Summaries of Linear Algebra

Download Principle of Mathematical Induction: Proofs and Examples and more Summaries Linear Algebra in PDF only on Docsity! MATHEMATICS 235 Notes MODULE-III Algebra-I Principle of Mathematical Induction 10 PRINCIPLE OF MATHEMATICAL INDUCTION In your daily life, you must be using various kinds of reasoning depending on the situation you are faced with. For instance, if you are told that your friend just has a child, you would know that it is either a girl or a boy. In this case, you would be applying general principles to a particular case. This form of reasoning is an example of deductive logic. Now let us consider another situation. When you look around, you find students who study regularly, do well in examinations, you may formulate the general rule (rightly or wrongly) that “any one who studies regularly will do well in examinations”. In this case, you would be formulating a general principle (or rule) based on several particular instances. Such reasoning is inductive, a process of reasoning by which general rules are discovered by the observation and consideration of several individual cases. Such reasoning is used in all the sciences, as well as in Mathematics. Mathematical induction is a more precise form of this process. This precision is required because a statement is accepted to be true mathematically only if it can be shown to be true for each and every case that it refers to. In the present chapter, first of all we shall introduce you with a statement and then we shall introduce the concept of principale of Mathematical induction, which we shall be using in proving some statements. OBJECTIVES After studying this lesson, you will be able to:  To check whether the given sentence is a statement or not.  state the Principle of Mathematical Induction;  verify the truth or otherwise of the statement P(n) for n = 1;  verify P(k+1) is true, assuming that P(k) is true;  use principle of mathematical induction to establish the truth or otherwise of mathematical statements; EXPECTED BACKGROUND KNOWLEDGE  Number System  Four fundamental operations on numbers and expressions. MATHEMATICS Notes MODULE-I Algebra-I 236 Principle of Mathematical Induction 10.1 WHAT IS A STATEMENT ? In your daily interactions, you must have made several assertions in the form of sentences. Of these assertions, the ones that are either true or false are called statement or propositions. For instance, “I am 20 years old” and “If x = 3, then x2 = 9” are statements, but ‘When will you leave ?’ And ‘How wonderful!’ are not statements. Notice that a statement has to be a definite assertion which can be true or false, but not both. For example, ‘x - 5 = 7’ is not a statement, because we don't know what x, is. If x = 12, it is true, but if x = 5, ‘it is not true. Therefore, ‘x – 5 = 7’ is not accepted by mathematicians as a statement. But both ‘x – 5 = 7  x = 12’ and x – 5 = 7 for any real number x’ are statements, the first one true and the second one false. Example 10.1 Which of the following sentences is a statement ? (i) India has never had a woman President. , (ii) 5 is an even number. (iii) xn > 1 , (iv) (a + b)2 = a2 + 2ab + b2 Solution : (i) and (ii) are statements, (i) being true and (ii) being false. (iii) is not a statement, since we can not determine whether it is true or false, unless we know the range of values that x and n can take. Now look at (iv). At first glance , you may say that it is not a statement, for the very same reasons that (iii) is not. But look at (iv) carefully. It is true for any value of a and b. It is an identity. Therefore, in this case, even though we have not specified the range of values for a and b, (iv) is a statement. Some statements, like the one given below are about natural numbers in general. Let us look at the statement : 1 2 1 2     ... ( )n n n This involves a general natural number n. Let us call this statement P (n) [P stands for proposition]. Then P (1) would be 1(1 1)1 2   Similarly, P (2) would be the statement , 2(2 1)1 2 2    and so on. Let us look at some examples to help you get used to this notation. Example 10.2 If P (n) denotes 2n > n–1, write P (1), P (k) and P (k+1), where k N . Solution : Replacing n by 1, k and k + 1, respectively in P (n), we get MATHEMATICS 239 Notes MODULE-III Algebra-I Principle of Mathematical Induction Let us, therefore, assume that P(k) is true, i.e., 1 + 2 + 3 ... + k = k 2 (k+1) ....(i) Now, P(k + 1) is 1 + 2 + 3 + ... + k + (k + 1) = ( )( )k k 1 2 2 It will be true, if we can show that LHS = RHS The LHS of P(k + 1) = (1 + 2 + 3 ... + k) + (k + 1) = k 2 (k+1) + (k + 1) ....[From (i)] = (k + 1) 1 2 k     = ( )( )k k 1 2 2 = RHS of P (k +1) So, P(k + 1) is true, if we assume that P(k) is true. Since P(1) is also true, both the conditions of the principle of mathematical induction are fulfilled, we conclude that the given statement is true for every natural number n. As you can see, we have proved the result in three steps – the basic step [i.e., checking (i)], the Induction step [i.e., checking (ii)], and hence arriving at the end result. Example 10.5 For every natural number n, prove that  2 1 2 1n nx y  is divisible by (x + y), where x, y .N Solution: Let us see if we can apply the principle of induction here. Let us call ( )P n the statement ‘  2 1 2 1n nx y  is divisible by (x + y)’, Then P(1) is ‘ 2 1 2 1x y  is divisible by (x + y)’, i.e., ‘(x + y) is divisible by (x + y)’, which is true. Therefore, P(1) is true. Let us now assume that P(k) is true for some natural number k, i.e., 2 1 2 1( )k kx y  is divisible by (x + y). This means that for some natural number t, 2 1 2 1 ( )k kx y x y t    Then, 2 1 2 1( )k kx x y t y    We wish to prove that P (k +1) is true, i.e., ‘ 2( 1) 1 2( 1) 1[ ]k kx y    is divisible by (x+y)’ is true. Now, 2( 1) 1 2( 1) 1 2 1 2 1k k k kx y x y        2 1 2 2 1k kx y    2. 2 1 2 1k kx x y   MATHEMATICS Notes MODULE-I Algebra-I 240 Principle of Mathematical Induction 2. 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 2 2 1 2 2 2 2 1 [( ) ] ( ) ( ) ( ) ( ) ( )[ ( ) ] k k k k k k k x x y t y y x x y t x y y x x y t x y y y x x y t y x y x y x t x y y                            which is divisible by (x + y). Thus, P (k+1) is true. Hence, by the principle of mathematical induction, the given statement is true for every natural number n. Example 10.6 Prove that 2n > n for every natural number n. Solution: We have P(n) : 2n > n. Therefore, P(1) : 21 > 1, i.e., 2 > 1, which is true. We assume P(k) to be true, that is, 2k > k ... (i) We wish to prove that P(k + 1) is true, i.e. 2k + 1 > k + 1. Now, multiplying both sides of (i) by 2, we get, 2k + 1 > 2k 2k + 1 > k + 1, since k > 1. Therefore, P(k + 1) is true. Hence, by the principle of mathematical induction, the given statement is true for every natural number n. Sometimes, we need to prove a statement for all natural numbers greater than a particular natural number, say a (as in Example 10.7 below). In such a situation, we replace P(1) by P(a + 1) in the statement of the principle. Example 10.7 Prove that n2 > 2(n + 1) for all n  3, where n is a natural number. Solution: For n  3, let us call the given statement, P(n) : n2 > 2 (n + 1) Since we have to prove the given statement for n  3, the first relevant statement is P(3). We, therefore, see whether P(3) is true. P(3) : 32 > 2  4, i.e. 9 > 8. So, P (3) is true. Let us assume that P(k) is true, where k  3, that is , k2 > 2(k + 1) ........ (i) We wish to prove that P (k + 1) is true. P (k + 1) : (k + 1)2 > 2(k + 2) LHS of P (k + 1) = (k + 1)2 = k2 + 2k + 1 MATHEMATICS 241 Notes MODULE-III Algebra-I Principle of Mathematical Induction > 2 (k + 1) + 2k + 1 ... [ By (i) ] > 3 + 2k +1, since 2 (k + 1) > 3. = 2 (k + 2), Thus, (k + 1)2 > 2(k + 2). Therefore, P(k + 1) is true. Hence, by the principle of mathematical induction, the given statement is true for every natural number n 3. Example 10.8 Using principle of mathematical induction, prove that        15 7 35 35 nnn is a natural number for all natural numbers n. Solution : Let P(n) :        15 7 35 35 nnn be a natural number.        15 7 3 1 5 1:)1(P is a natural number. or, 1 15 15 15 753 15 7 3 1 5 1    , which is a natural number . P(1) is true. Let P(k) :        15 7 35 35 kkk is a natural number be true ... (i) Now 15 )1(7 3 )1( 5 )1( 35      kkk =            15 7 15 7133 3 11510105 5 1 232345 kkkkkkkkk =                15 7 3 1 5 1232 15 7 35 234 35 kkkkkkk =   1232 15 7 35 234 35        kkkkkkk ...(ii) By (i), 15 7 35 35 kkk  is a natural number. also kkkk 232 234  is a natural number and 1 is also a natural number.  (ii) being sum of natural numbers is a natural number. MATHEMATICS Notes MODULE-I Algebra-I 244 Principle of Mathematical Induction (3) a + ar + ar2 +....+ arn-1 = 1 )1(   r ra n ,where r > 1 and a is a real number. (4) (x2n –1) is divisible by (x +1) Nx . (5) (102n–1 + 1) is a multiple of 11, where n N (6) (4.102n + 9.102 n–1 +5) is a multiple of 99. where n N (7) (1 +x)n > 1 + nx, where x > 0 and n N (8)  2 3 4 11.2 2.2 3.2 4.2 .2 1 .2n nn n        , where n N (9)   2 3 3 3 3 1 1 2 3 2 n n n             , n N (10)      1 1 1 1 3.5 5.7 7.9 2 1 2 3 3 2 3 n n n n          where n N MATHEMATICS 245 Notes MODULE-III Algebra-I Principle of Mathematical Induction ANSWERS CHECK YOUR PROGRESS 10.1 1. (b), (e) and (f) are statements; (a) is not, since we have not given the range of values of n, and therefore we are not in a position to decide, if it is true or not. (c) is subjective and hence not a mathematical statement. (d) is a question, not a statement. Note that (f)is universally false. 2. P(1) : 6 is a factor of 13 + 5.1, P(2) : 6 is a factor of 23 + 5.2 P(k) : 6 is a factor of k3 + 5k, P(k+1) : 6 is a factor of (k + 1)3 + 5(k + 1) 3. (a) P(1) : 2  2, P(k) : 2k  k + 1, P(k + 1) : 2k+1  k + 2 (b) P(1) : 1 + x  1 + x, P(k) : (1 + x)k  1 + kx P(k + 1): (1 + x)k + 1 1 + (k + 1)x (c) P(1): 6 is divisible by 6. P(k) : k(k + 1)(k + 2) is divisible by 6. P(k + 1): (k + 1) (k + 2) (k + 3) is divisible by 6 (d) P(1): (x – y) is divisible by (x – y). P(k): (xk –yk) is divisible by (x – y) P(k + 1): (xk + 1 –yk + 1) is divisible by (x – y) (e) P(1): ab = ab, P(k): (ab)k = a k b k P(k+1): (ab) k+1 = a k+1. b k+1 (f) P(1) : 15 7 3 1 5 1  is a natural number. P(k): 15 7 35 35 kkk  is a natural number. P(k + 1): 15 )1(7 3 )1( 5 )1( 35      kkk is a natural number. 4. (a) P(1): 2 1 21 1   , 3 2 32 1 21 1:)2(     P MATHEMATICS Notes MODULE-I Algebra-I 246 Principle of Mathematical Induction 1 1 1( ) : ... 1 2 2 3 ( 1) 1 kP k k k k         2 1 )2)(1( 1 )1( 1... 21 1:)1(          k k kkkk kP (b) 2(1) : 1 1P  , 2(2) : 1 3 2P   2( ) : 1 3 5 ... (2 1)P k k k        2( 1) : 1 3 5 ... (2 1) 2( 1) 1 ( 1)P k k k k           (c) 2(1) : 1 2 1(2)P   , 2(2) : (1 2) (2 3) 2(3)P     2( ) : (1 2) (2 3) ... ( 1) ( 1) .P k k k k k        2( 1) : (1 2) (2 3) ... ( 1)( 2) ( 1)( 2)P x k k k k           (d) 1 1(1) : 1 3 3 P   , 1 1 2(2) : 1 3 3 5 5 P     1 1 1( ) : ... 1 3 3 5 (2 1)(2 1) 2 1 kP k k k k          1 1 1 1( 1) : ... 1 3 3 5 (2 1)(2 3) 2 3 kP k k k k           