MATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:

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1 MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3, } has a smallest element This axiom can be thought of a instance of the extremal principle which will be discussed later Suppose that k is an integer and we want to prove that P (n) is true for every positive integer n, where P (n) is a proposition (statement) which depends on a positive integer n Proving P (), P (), P (3) individually, would take an infinite amount of time Instead, we can use the Induction Principle: Theorem (Induction Principle) Assume that k is an integer and P (n) is a proposition for all n k () Suppose that P (k) is true, and () for any integer m k for which P (m) is true, P (m + ) is true Then P (n) is true for all integers n k For k =, the induction principle can be compared to an infinite sequence of dominos tiles, numbered,,3, etc If the m-th domino tile falls, it will hit the (m + )-th domino tile and the (m + )-th domino tile will fall as well If the first domino tile falls, then all domino tiles will fall down (Here P (n) is the statement: the n-th domino tile falls down ) Proof of the Induction Principle Let S to be the set of all nonnegative integers i for which P (k + i) is false If S is nonempty, then S has a smallest element, say l by Axiom By property ()), P (k) is true and l > 0 The statement P (k + l ) is true because l is the minimal element of S Therefore, statement P (k + l) is true because of property () Hence l S Contradiction! We conclude that S is empty, and P (n) is true for all n k A typical example of the induction principle is the following:

2 Example 3 Prove that () n = for every positive integer n n(n + ) Proof We prove () by induction on n For n = we check that = Suppose that () is true for n = m Then ( + ) m + (m + ) = ( m) + (m + ) = m(m + ) (m + )(m + ) = + (m + ) = So () is true for n = m + Now () is true for all positive integers n by the induction principle Remark 4 When the German mathematician Carl Friedrich Gauss ( ) was 0 years old, his school teacher gave the class an assignment to add all the numbers from to 00 Gauss gave the answer almost immediately: 5050 This is how (we think) he did it: Write the numbers from to 00 from left to right Write under that the numbers from to 00 in reverse order } 0 0 {{ 0 } 00 Each of the 00 column sums is 0 This shows that and ( ) = = This easily generalizes to a proof of () 00 0 = 50 0 = 5050 A formula similar to () exists for the sums of squares, namely () n = Example 5 Give and prove a formula for n 3 n(n + )(n + ) 6 We have seen similar examples, namely () and () We can also add the formula n 0 = n Let p k (n) = k + k + 3 k + + n k where k Z 0 The examples so far suggest that p k (n) is a polynomial of degree k + (and that the leading coefficient is ) Let us assume that p k+ 3(n) is a polynomial of degree 4

3 Since p 3 (0) is an empty sum, we have that p 3 (0) = 0 We can write p 3 (n) = an 4 + bn 3 + cn + dn for certain real numbers a, b, c, d We have (3) n 3 = p 3 (n) p 3 (n ) = a(n 4 (n ) 4 )+b(n 3 (n ) 3 )+c(n (n ) )+d(n (n )) = = a(4n 3 6n + 4n ) + b(3n 3n + ) + c(n ) + d = = n 3 (4a) + n ( 6a + 3b) + n(4a 3b + c) + ( a + b c + d) Comparing coefficients in (3) gives us the linear equations: (4) (5) (6) (7) = 4a 0 = 6a + 3b 0 = 4a 3b + c 0 = a + b c + d We solve the system of equations and find a =, b =, c = and d = 0 We now should 4 4 conjecture the following formula: n 3 = 4 n4 + n3 + 4 n Finding this formula was the hard part It is now not so hard to prove this formula by induction: Proof We will prove that (8) n 3 = 4 n4 + n3 + 4 n by induction on n The case n = 0 is clear, because both sides of the equation are equal to 0 If (8) is true for n = m, then From this follows that (m ) 3 = 4 (m )4 + (m )3 + 4 (m ) (m ) 3 + m 3 = 4 (m )4 + (m )3 + 4 (m ) + m 3 = = 4 (m4 4m 3 + 6m 4m + ) + (m3 3m + 3m ) + 4 (m m + ) + m 3 = = 4 m4 + m3 + 4 m, so (8) is true for n = m By induction follows that (8) is true for all n Z 0 Notice that 4 m4 + m3 + 4 m = ( n(n + )) which leads to the following aesthetic formula: Example 6 What is the value of n 3 = ( n) ? 3

4 Let us compute the partial sums Perhaps we will find a pattern + 3 = = 3, = 3 + = 8 + = 9 = 3 4, = = = 6 0 = 4 5 A pattern emerges Namely, it seems that (9) Proof By induction on n we prove: (0) For n = we check If (0) is true for n = m, then n(n + ) = n n(n + ) = n + = m(m + ) + (m + )(m + ) = = ( ) + ( m + m + m + ) = m + Hence (0) is true for n = m + By induction, (0) is true for all integers n We have = lim ( ) = n n + Example 7 (UMUMC, 988) Let S n be the set of all pairs (x, y) with integral coordinates such that x 0, y 0 and x + y n Show that S n cannot be covered by the union of n straight lines First we should try a few small cases, say n = 0,,, 3, 4: n=0 n= n= n=3 n=4 Notice that S n is a subset of S n+ This will be helpful for our induction proof: 4

5 Proof We prove the statement by induction on n, the case n = 0 being trivial Suppose that one needs at least n + lines to cover S n Define C n+ = S n+ \ S n The set C n+ consists of n + points on the line x + y = n + Suppose that k lines l, l,, l k cover S n+ case : One of the lines is equal to the line x + y = n + Without loss of generality we may assume that l k is equal to the line x + y = n + Then l, l,, l k cover S n because l k S n = From the induction hypothesis follows that k n +, so k n + case : None of the lines are equal to the line x+y = n+ Then each of the lines intersects the line x + y = n + in at most one point, and therefore it intersects the set C n+ in at most one point Since C n+ has n + elements, there must be at least n + lines So in both cases we conclude that one needs at least n + lines to cover S n+ Strong Induction The following example illustrates that sometimes one has to make a statement stronger in order to be able to prove it by induction Example 8 Prove that () , 999, 000, 000 < 000 Since 000 =, 000, 000 one might suggest that n n < n for all n Let us try to prove () We can check () for small n (which gives some validity to our conjecture that this inequality holds) Suppose that () holds for n = m: () We have to prove () for n = m + : (3) If we divide (3) by () we obtain (4) m m < m m + m + < m + m + m + m m + If () and (4) are true, then (3) is true By squaring (4) we see that (4) is equivalent to ( ) m + m m + m + and to (5) (m + ) (m + )(m) So if (5) is true then our induction proof is complete Unfortunately (5) is not true and we are stuck Sometimes it is easier to prove a stronger statement by induction: 5

6 Proof We prove (6) 3 4 n n by induction on n The case n = is clear because < 3 < n + Suppose that (6) is true for n = m: (7) 3 4 m m < m + Since (m + )(m + 3) = (m + ) < (m + ) we have that ( ) m + < m + m + m + 3 and m + m + (8) m + < m + 3 Multiplying (7) by (8) yields 3 (9) 4 m + m + <, m + 3 so (6) is true for n = m + This shows that (6) is true for all positive integers n In particular, for n = 500, 000 we get , 999, 000, 000 < <, 000, Below is a trickier proof of Example 8 Proof Let and Clearly A < B because It follows that and A < 000 A = B = , , 000, , 000, , 000, 00 < 3, 3 4 < 4 5,, 999, 999, 000, 000 A < AB = <, 000, 000, 000, 00, 000, 00 <, 000, 000 Example 9 Prove that every integer n is a product of prime numbers 6

7 Proof Let Q(n) be the statement: every integer r with r n is a product of prime numbers We use induction on n to prove that Q(n) holds for all integers n For n = the statement is true because is a prime number Suppose that Q(m) is true We will prove Q(m + ) Suppose that r m + If r m then r is a product of prime numbers because Q(m) is true Suppose that r = m + If m + is a prime number, then m + is a product of prime numbers and we are done Otherwise, m + can be written as a product ab with a, b m Because Q(m) is true, both a and b are products of prime numbers Hence m + = ab is a product of prime numbers We have shown that Q(n) holds for all n In particular, every integer r is a product of prime numbers because Q(r) is true 3 Induction in definitions We can also use induction in a definition For example, the Fibonacci numbers is a sequence of numbers F 0, F, F, defined by F 0 = 0, F = and F n+ = F n + F n, n By (strong) induction on n we can prove that F n is well-defined for all integers n 0 The first few Fibonacci numbers are: 0,,,, 3, 5, 8, 3,, 34, 55, 89, The sum notation is an example of a recursive definition function If a, b are integers and a b + then we define b f(n) as follows and (0) a f(n) = 0 b b f(n) = f(b) + f(n) if b a One can then formally prove by induction that c b f(n) = f(n) + c n=b+ f(n) if a, b, c Z and a b c (Induction on c Start with c = b) Similarly we have the product notation a f(n) = 7 Suppose that f(n) is some

8 and b b f(n) = f(b) f(n) if b a For nonnegative integers m and n with m n we define a binomial coefficient by ( ) { n = ) ( if m = 0 or m = n m + n ) if 0 < m < n ( n m If we arrange the binomial coefficients in a triangular shape, we get Pascal s triangle: ( 0 ( ) 0) ( ( 0 ) ( ) ) ( ( 0 3 ) ( 3 ) ( 3 ) ) ( 3 ( 0 4 ) ( 4 ) ( 4 ) ( 4 ) 3) ( ) m After evaluating the binomial coefficients we get: Exercise * Prove that for all positive integers n Exercise * Show that for all n N Exercise 3 * Prove that 4 Exercises n = n(n + )(n + ) n n = n + + n n n ( ) ( ) m + i m + n + = m m + i=0 for all nonnegative integers m and n Exercise 4 * Prove that Exercise 5 * (a + b) n = a n + 0 a n b + 8 a n b + + b n n

9 (a) Prove that + x + x + + x n = xn+ x for every real number x and every positive integer n (b) If x is a real number with x < then + x + x + = x Exercise 6 ** Show that the sum of the squares of two consecutive Fibonacci numbers is again a Fibonacci number Exercise 7 ** Cut out a corner of a n n chess board (n ) Show that the remainder of the chess board can be covered with L-shaped tiles (see picture) The case n = is shown below Exercise 8 ** Find and prove a formula for Exercise 9 ** Show that as follows: Define f(x) = ( + x) n = n 4 = m n! m! (n m)! + 0 x + + and consider f (k) (0) (f (k) (x) is the k-th derivative of f(x)) x n n Exercise 0 ** Define a sequence a, a, by a = 5 and a n+ = a n for n Give an explicit formula for a n and prove it Exercise (Division with remainder) ** Suppose that n is a nonnegative integer, and m is a positive integer Prove that there exist integers q and r with n = qm + r and 0 r < m Exercise ** Give and prove a formula for n (n + ) (n + ) 9

10 Exercise 3 (Expansion in base b) *** Suppose that n is a positive integer, and b is an integer Show that there exist an nonnegative integer m, and integers a 0, a,, a m {0,,,, b } such that () n = a m b m + a m b m + + a 0 and a m 0 Moreover, show that m and a 0, a,, a m are uniquely determined by n (We will write (a m a m a 0 ) b for the right-hand side in ()) Exercise 4 (Zeckendorf s Theorem) *** Suppose that n is a positive integer Show that we can write n = F i + F i + + F ik where k is a positive integer, i and i j i j + for j =, 3,, k Also show that k and i,, i k are uniquely determined by n Exercise 5 (Putnam 985, B) *** Define polynomials f n (x) for n 0 by f 0 (x) =, f n (0) = 0 for n, and d dx (f n+(x)) = (n + )f n (x + ) for n 0 Find, with proof, the explicit factorization of f 00 () into powers of distinct primes Exercise 6 (Putnam 987, B) *** Let r, s and t be integers with 0 r, 0 s and r + s t Prove that ( s ( s ( s ( 0) t + ( ) t ) + + s) t + ( t ) = r) (t + s) ( t s) r+ r+s r Exercise 7 **** Let n = k Prove that we can select n integers from any (n ) integers such that their sum is divisible by n Exercise 8 *** Find the sum of all fractions /xy such that gcd(x, y) =, x n, y n, x + y > n Exercise 9 **** Find and prove a formula for π sin n (x) dx 0 Exercise 0 **** Suppose that a, a,, a n are positive integers such that a a a n Prove that a + a + + a n = implies that a n < n! Exercise *** Generalize example 7: Let S n,d R d be the set of all integer vectors (x,, x d ) such that x i 0 for all i and x + x + + x d n Show that S n,d cannot be covered with n hyperplanes in R d 0

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