MATH 55 - HOMEWORK 6 SOLUTIONS. 1. Section = 1 = (n + 1) 3 = 2. + (n + 1) 3. + (n + 1) 3 = n2 (n + 1) 2.

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1 MATH 55 - HOMEWORK 6 SOLUTIONS Exercise Section 5 Proof (a) P () is the statement ( ) 3 (b) P () is true since ( ) 3 (c) The inductive hypothesis is P (n): ( ) n(n + ) n 3 (d) Assuming the inductive hypothesis, we aim to show P (n + ): ( ) (n + )(n + ) n 3 + (n + ) 3 (e) Suppose n is such that ( ) n(n + ) n 3 Then by our inductive hypothesis, we have ( ) n(n + ) n 3 + (n + ) 3 + (n + ) 3 Now we compute ( ) n(n + ) + (n + ) 3 n (n + ) + (n + ) 3 n (n + ) + (n + )3 (n + (n + ))(n + ) (n + n + )(n + ) (n + ) (n + ) ( ) (n + )(n + ) This shows ( ) (n + )(n + ) (n + ) 3 (f) We showed P () and ( n)(p (n) P (n + )) by universal generalization It follows by mathematical induction that ( n)p (n) Exercise 0 Proof Notice we have

2 MATH 55 - HOMEWORK 6 SOLUTIONS From this, it seems reasonable to conjecture that for all n, P (n): n(n + ) n n + (b) Let s prove it P () was proven in (a) Now assume P (n) as the inductive hypothesis and let s show P (n + ): Now by the inductive hypothesis, Now we calculate (n + )(n + ) + (n + )(n + ) n + n n(n + ) + (n + )(n + ) n n + + (n + )(n + ) n n + + (n + )(n + ) n(n + ) (n + )(n + ) + (n + )(n + ) n + n + (n + )(n + ) (n + ) (n + )(n + ) (n + ) (n + ) It follows that n(n + ) + (n + )(n + ) n + n +, which establishes P (n + ) Therefore, by mathematical induction, P (n) is true for all positive integers n Exercise 3 9 Proof (a) P () is the statement: (b) P () is true as (c) The inductive hypothesis is P (n): + < + 5 < n < n (d) We need to prove, assuming the inductive hypothesis, that P (n + ): (e) By the inductive hypothesis, we have n + (n + ) < n n + (n + ) < n + (n + ) An inequality we will find useful is that (n + ) + n < n(n + ) To see this, notice that (n + ) + n n n + n n n < n n n(n + ) Using this inequality, we can establish P (n + ), since n + (n + ) (n + ) + n n(n + ) n(n + ) < n(n + ) n + (f) We showed P () and ( n )[P (n) P (n + )] by universal generalization By mathematical induction, this establishes ( n)p (n) Exercise 36 Proof First we will prove this in the case n We need to show And this is true as which is clearly divisible by Next we assume that n is such that n+ + 5 n and we will aim to show n+ + 5 (n+) Notice that n+ + 5 (n+) n n n+ + ( + )5 n ( n+ + 5 n ) + 5 n

3 MATH 55 - HOMEWORK 6 SOLUTIONS 3 Now ( n+ + 5 n ) as ( n+ + 5 n ) by the inductive hypothesis Furthermore, clearly divides 5 n It follows by the equation above that n+ +5 (n+) Therefore, by mathematical induction, n+ +5 n for all positive integers n Exercise 5 9 Proof The problem with this proof is in the inductive step In particular, specializing to the case when k, it is easy to see that the proof is false: suppose you have a set of k + horses Then the set of the first k and the set of the last last k horses - these are each just one horse! Exercise 6 6 Proof First, we make the following observation: if p is a prime, a and b are integers and p ab then p a or p b To see this, consider the prime factorization of a and the prime factorization of b The set of primes occuring in the prime factorization of ab is the union of the primes occuring in the prime factorization of a and the prime factorization of b, thus p must appear in the prime factorization of a or that of b Now we prove the statement by induction The base case is trivial: if p a then p a Now suppose that for any positive integers a,, a n, if p a a n then p a i for some i Then suppose a,, a n+ are positive integers and p a a n+ We can think of the product a a n+ as the product of a a n and a n+ Then by our initial observation, either p a a n or p a n+ In the latter case, we re done In the former case, it follows from the induction hypothesis that p a i for some i,, n This means we ve shown that if p a a n+ then p a i for some i,, n + Exercise Proof (a) To see P (8), P (9), P (0), P (), we just check: Section (b) The inductive hypothesis is that P (k) is true for all k satisfying 8 k n - ie k cent postage can be paid for with only and 7 cent stamps for all k between 8 and n (inclusive) (c) In the inductive step, we need to show that if n is some number with 8 n and k cent postage can be paid for with only and 7 cent stamps for all k between 8 and n (inclusive), then n + cent postage can be paid for using only and 7 cent postage formally, this says P (8) P (9) P (n) P (n + ) (d) Suppose n and P (k) holds for all k satisfying 8 k n We want to show P (n + ) is true As n we know n + It follows that (n + ) 8 and therefore by the inductive hypothesis, (n + ) cent postage can be paid for using only and 7 cent stamps Then adding one more cent stamp gives postage for an n + cent delivery (e) We established that P (8) is true and for all n, if P (k) is true for all k satisfying 8 k n then P (n + ) is true It follows by strong induction that P (n) is true for all integers n 8 Exercise 9 Proof Let P (n) be the statement ( b Z )( n b ) First, we show P (): if b is an arbitrary positive integer then b and hence is not equal to so P () is true Next suppose n is a number so that P (k) is true for all k n Then suppose towards contradiction that P (n + ) is false, ie there is a b Z + so that (n+) b Then (n + ) b so (n + ) is even in other words, there is some positive integer j so that j n + Notice that necessarily j < n + Now we have b (n + ) (j) j, so b j, so b is also even Write b k Now we have ( ) n + b ( ) j k ( ) j k This shows that there is a positive integer b - namely k - so that j b - in other words, P (j) is false This contradicts our assumption that P (k) is true for all k satisfying k n Therefore P (n + ) must be true This completes the induction Now we explain why we may conclude that is irrational If not, p q for some integers p and q Because > 0, we know p and q may be both taken to be positive integers Then p q so P (p) must be false But we proved P (p) so is irrational Exercise 3

4 MATH 55 - HOMEWORK 6 SOLUTIONS Proof Note that P () is true as 0 Next, suppose n is such that P (k) is true for all k n We want to show n + can be written as a sum of powers of If n + is even then n+ is a positive integer n By the inductive hypothesis, there are nonnegative integers i < < i l so that (n + ) i + i + + i l Then n + ( i + i + + i l ) i+ + i+ + + i l+, which shows n + can be written as a sum of distinct powers of If n + is odd, then n is even so if we write (using the inductive hypothesis) n i + i + + i l, then we know that i j for all j (otherwise n would be odd) Hence n + + i + + i l 0 + i + i + + i l, so n + can be written as a sum of distinct powers of Exercise Proof Let P (n) be the predicate which asserts of n that the sum of the products computed after all the splittings is n(n )/ Note that P () is true as if there s pile of, you never split so you tally 0 products, which is expected by the formula ( )/ 0 Now suppose n is such that P (k) is true for all k n Now suppose we re given a pile of n + stones Consider any splitting of n + into two piles of s and t stones respectively So s + t n + This splitting yields the product st Any further splittings will be of the size s pile or the size t pile So the outcome of the tally will be st plus the sum of the splittings of the size s pile and the sum of the splittings of the size t pile By the inductive hypothesis, this yields a tally of st + s(s ) + t(t ) Now using that n + s + t and therefore that s + t n, we compute st + s(s ) + t(t ) st + s(s ) + t(t ) which shows that P (n + ) is true Therefore, by strong induction, P (n) holds for all n st + st + s(s ) + t(t ) s(s + t ) + t(s + t ) (s + t)(s + t ) n(n + ), Exercise 5 36 Proof (a) The set S of positive integers of the form as + bt for integers s, t is non-empty: it contains, for example, a + b (b) S is a non-empty subset of Z + therefore has a least element c (c) Write c as + bt for some integers s, t Then if d is a positive integer so that d a and d b, we can find α and β in Z + so that αd a and βd b Then (sα + tβ)d (sαd + tβd) (sa + tb) c This shows d c (d) We will show c a and c b Suppose c a Then by Euclid s algorithm, we can find an integer q and an integer r with 0 < r < c so that a qc + r Write c sa + tb Then r a qc a qsa qtb ( qs)a + ( qt)b, so r S But c is the least element of S, a contradiction So c a The argument that c b is entirely similar (e) The solutions to (c) and (d) show that c has the property that it is a common divisor of a and b and any other common divisor of a and b must also divide c this shows that c is a greatest common divisor of a and b To see that it is unique, suppose c was another greatest common divisor of a and b Since c divides a and b we know that c c by (c), so 0 < c < c As c is a gcd of a and b, we may, by Bezout s theorem, write c sa + tb for some integers s and t So c S This contradicts our assumption that c is least in S Therefore c is unique with the property that c a and c b and for any common divisor d of a and b, d c

5 MATH 55 - HOMEWORK 6 SOLUTIONS 5 3 Section 53 Exercise 3 Proof (a) f(n + ) f(n) f(n ) So we compute f() f() f(0) 0 f(3) f() f() 0 f() f(3) f() 0 f(5) f() f(3) ( ) 0 (b) f(n + ) f(n)f(n ) So we compute f() f()f(0) f(3) f()f() f() f(3)f() f(5) f()f(3) (c) f(n + ) f(n) + f(n ) 3 So we compute f() f() + f(0) 3 f(3) f() + f() 3 5 f() f(3) + f() 3 33 f(5) f() + f(3) 3 (d) f(n + ) f(n)/f(n ) f(n + ) f(n)f(n ) So we compute f() f()/f(0) f(3) f()/f() f() f(3)/f() f(5) f()/f(3) Exercise 3 6 Proof (a) Valid We ll show f(n) ( ) n This works for n as ( ) 0 Suppose for some fixed n, f(n) ( ) n Then we have f(n + ) f(n) ( )( ) n ( ) n+, which proves, by induction, that f(n) ( ) n for all n (b) Valid We have k if n 3k f(n) 0 if n 3k + k+ if 3k + Notice that, by Euclid s algorithm, every nonnegative integer may be written n 3k + i for i {0,, }, so this defines a function with domain the nonnegative integers Suppose it has been established that f(n) agrees with the above formula for some n Then consider the case of n + Then n + 3 so (n + ) 3 0 If (n + ) 3 3k then we have n + 3k + 3 3(k + ) so we have f(n + ) f(n + 3) k+ which agrees with the formula above If n+ 3 3k + for some k then n+ 3(k +)+ and f(n+) f(n+ 3) 0 0 so agrees with the formula above Finally, if n + 3 3k + for some k, then n + 3(k + ) + and we have f(n + ) f(n + 3) k+ k+ which agrees with the formula above This finishes the argument (c) Not valid The problem states the the function should have domain the set of nonnegative integers and range the integers The recursive definition f(n) f(n + ) is equivalent to the equation f(n) f(n + ) and since f() this implies f() which is not an integer (d) Not valid if f(0) 0 and f(n) f(n ) for n then f() 0 but we re told f() (e) Not valid Consider the case when n 3 Then n is odd and so f(3) f() Then f() f(0) So f(3) But f() f(0) Then f() f() 8 Hence f(5) f() 8 and f(5) f(3) We claim { k+ if x k f(x) k+ if x k +

6 6 MATH 55 - HOMEWORK 6 SOLUTIONS In other words, f(x) x + As f(0) 0, this formula works for n 0 Suppose n is such that f(k) agrees with the above formula for all k n, and we ll prove it for n + If n + is odd, then n+ n so f(n + ) f(n) and this agrees with the above formula Furthermore, if n + then n+ n n+ so + n + f(n ) so if n + then f(n + ) f(n ) and this agrees with the above formula Exercise 33 8 Proof (a) a and a n+ + a n (b) a 0 and a and a n a n for n (c) a and a n+ (n+)an n (d) a and a n+ a n + n + Exercise 3 Proof We want to show that f + + f n f n f n+ where f n denotes the nth Fibonacci number For n, this is true as f f so f f f Now suppose it s been shown for n and we ll prove it for n + By the inductive hypothesis, we have f + f + + f n + f n+ f n f n+ + f n+ Then we note that f n + f n+ f n+ by definition of the Fibonacci sequence This entails which completes the induction step Exercise 35 f n f n+ + f n+ f n+ (f n + f n+ ) f n+ f n+, Proof We want to show f n+ f n f n ( ) n For n this is true as f 3 3 and f f so f 3 f f ( ) Now suppose it has been established for n and we will show it s true for n + Using that f n+ f n+ + f n and f n+ f n + f n, we have f n+ f n f n+ (f n+ + f n )f n f n+ f n + f n f n+ f n+ f n f n+ (f n+ f n ) So by the induction hypothesis, we have This completes the induction step Exercise 36 0 Proof Now we define max(a ) a and max(a, a ) by f n f n+ f n (f n+ f n f n) f n+ f n f n+ (f n+ f n f n) ( )( ) n ( ) n+ { a if a max(a, a ) a otherwise Then we set max(a,, a n+ ) max(max(a,, a n ), a n+ ) Similarly we define min a a and min(a, a ) by a { a if a min(a, a ) a otherwise Then we define min(a,, a n+ ) min(min(a,, a n ), a n+ ) a