Discrete Math, Spring 202 - Solutions to Problems V Suppose we have statements P, P 2, P 3,, one for each natural number In other words, we have the collection or set of statements {P n n N} a Suppose further that if P n is false for some n then also P m is false for some m less than n Does it follow that P n is false, for all n? b Suppose instead that, for any n, if P m is true for all natural numbers m less than n, then P n is true Does it follow that P n is true, for all n? a Consider S = {n N P n is false} By hypothesis, S has no least element By the well-ordering principle, it follows that S is empty Hence P n is true, for all n So it certainly does not follow that P n is false, for all n b We use strong induction to show that P n is true, for all n Thus we have to show: i P is true; ii for n >, if P m is true for all m < n, then P n is true Condition ii holds by hypothesis, so we only have to prove i The case n = of our hypothesis says that if P m is true for all m < then P is true Well there are no natural numbers less than one, and so P m is true for all m < (To put it another way, there is no m < for which P m is false as there are no natural numbers m < at all the condition is vacuously satisfied) Thus P is true Hence, by strong induction, P n is true, for all n 2 Show that the sum of the first n odd natural numbers is n 2, for any natural number n We need to show that n i= (2i ) = n2, for any natural number n Writing P (n) for this identity, we ll prove by induction that P (n) holds for all n It s clear that P () holds: 2 = 2 Suppose P (k) holds for a given natural number k, ie, k i= (2i ) = k2 Then k+ (2i ) = (2i ) + (2k + ) i= i= = k 2 + 2k + (using P (k)) = (k + ) 2 Thus P (k) implies P (k + ) for any natural number k By induction, n i= (2i ) = n2, for all natural numbers n 3 Show that n i= ( )i+ i 2 = ( ) n+ n i= i, for all n N We prove this by induction using n i = n(n + )/2 ( ) i= which we proved in class The base case ( ) + 2 = ( ) + is clear Assume then that the identity holds for a natural number k We need to show it also holds for k + We have k+ ( ) i+ i 2 = ( ) i+ i 2 + ( ) k+2 (k + ) 2 i= i=
2 By our inductive hypothesis, this is ( ) k+ i + ( ) k+2 (k + ) 2 = ( ) k+ k(k + )/2 + ( ) k+2 (k 2 + 2k + ), i= using ( ) We rewrite the right side as ( ) k+2 (k 2 + 2k + k 2 /2 k/2) = ( ) k+2 (k 2 /2 + 3k/2 + ) = ( ) k+2 (k + )(k + 2)/2 k+ = ( ) k+2 i, using ( ) once more Thus the identity holds for k + By induction, it therefore holds for all natural numbers n 4 Consider the Fibonacci sequence,, 2, 3, 5, 8,, given by f =, f 2 =, f n = f n + f n 2, for n > 2 a Show that f 3n is even, for all natural numbers n b Show that f 3n and f 3n 2 are both odd, for all natural numbers n a We show by induction that f 3n is even, for all n Clearly, f 3 = 2 is even Assume that f 3k is even for a natural number k Then f 3(k+) = f 3k+3 = f 3k+2 + f 3k+ = (f 3k+ + f 3k ) + f 3k+ = 2f 3k+ + f 3k, so that f 3(k+) is also even By induction, we conclude that f 3n is even, for all n b We show by induction that f 3n is odd, for all n (The proof that f 3n 2 is odd, for all n, is effectively identical) First, it s obvious that f 3 = f 2 is odd Next, suppose f 3k is odd, for a natural number k Then f 3(k+) = f 3k+2 = f 3k+ + f 3k = (f 3k + f 3k ) + f 3k = 2f 3k + f 3k, so that f 3(k+) is also odd Hence, by induction, f 3n is odd, for all n 5 Consider 2! + 2 3! + + n (n + )! For n =, 2, 3, this is /2, 5/6, 23/24 respectively Find the general statement suggested by these special cases and prove it i=
3 The general statement is 2! + 2 3! + + n (n + )! (n + )! = (n + )! (Note (n + )! = 2, 6, 24 for n =, 2, 3 respectively) We ll prove the general statement using induction The base case n = is clear: 2! = 2! 2! Now suppose the statement holds for a natural number k Then 2! + 2 3! + + k (k + )! + k + (k + )! = + k + (k + 2)! (k + )! (k + 2)! (k + 2)((k + )! ) + (k + ) = (k + 2)! (k + 2)! =, (k + 2)! and so the statement also holds for k + Thus, by induction, the statement holds for all natural numbers n 6 Consider! + 2! 2 + + n! n For n =, 2, 3, 4, this equals, 5, 23, 9 respectively Find the pattern suggested by these special cases and show it holds in all cases The general statement is! + 2! 2 + + n! n = (n + )! which we now prove by induction The base case! = 2! is clearly true For the inductive step, we assume for some k Then! + 2! 2 + + k! k = (k + )!,! + 2! 2 + + k! k + (k + )! (k + ) = (k + )! + (k + )! (k + ) and so the statement also holds for k + natural numbers n = (k + 2) (k + )! = = (k + 2)!, By induction, the statement holds for all 7 Prove the inequality + 2 + + n > n, for any n > Using n = n/ n, we rewrite the inequality as + + + > n, for n >, 2 n n or equivalently n n n + + + > n, for n > 2 n
4 Now each term on the left hand side is and the first is surely > (since n > ) and thus the sum of the n terms must be > n We can also prove the inequality by induction The base case n = 2 is the statement + 2 > 2 Multiplying both sides by 2, we see that this holds if and only if 2 + > 2 which is true and so we ve established the base case Suppose now that for some natural number k 2 Then + + + > k, 2 k + + + + 2 k k + > k + It suffices therefore to show that k + k + > k +, for k 2 k + Multiplying both sides by k +, we see that this is equivalent to k + k + > k +, for k 2 But k + > k, for k > 0, so certainly k + k + > k k + = k +, for k 2 This establishes the inductive step and hence the inequality holds for all n 2 8 Suppose we have a collection of real numbers {a m,n m, n N} (that is, we re given a real number for each ordered pair of natural numbers) that satisfies a a, = 0, b if a m,n = 0, for some m and n, then a m,n+ = 0, c if a m,n = 0, for some m and all n, then a m+, = 0 Show that a m,n = 0, for all m and n Here s a way of thinking about the given conditions Imagine the numbers a m,n listed in an array with infinitely many rows and columns: a, a,2 a,n a m, a m,2 a m,n a m+, a m+,2 a m+,n Condition a says the (, ) entry is zero; b says that if the n-th entry in row m is zero then the n + -st entry in row m is also zero; c says that if all the entries in a row are zero then the first entry in the next row is also zero
5 Let P (m) be the statement a m,n = 0, for all n N In other words, P (m) says that each entry in the m-th row is zero We ll show by induction that P (m) holds for all m, so that a m,n = 0 for all m and n We first prove P () using induction (a proof by induction inside a proof by induction) We have a, = 0 If a,l = 0, then also a,l+ = 0 (by b with m =, n = l) Hence (by induction) a,n = 0, for all n Now suppose P (k) holds, ie, a k,n = 0, for all n Then c gives a k+, = 0 Further, by b, if a k+,l = 0, then a k+,l+ = 0 Hence (by induction) a k+,n = 0, for all n That is, P (k + ) holds By induction, P (n) holds, for all n