What can you prove by induction?

MEI CONFERENCE 013 What can you prove by induction? Martyn Parker M.J.Parker@keele.ac.uk

Contents Contents iii 1 Splitting Coins.................................................. 1 Convex Polygons................................................ 3 3 Tower of Hanoi................................................. 4 4 Money...................................................... 6 5 Coins rearranging.............................................. 8 6 Goodstein s Theorem............................................. 9 A Induction Further details 1 1 Induction at A-level............................................... 1 What is mathematical induction?....................................... 14 3 Why do we need all the bits?......................................... 18 4 Examples.................................................... 5 Propositions that look like they can be proved by induction........................ 31 B Solutions and further details 34

1 SPLITTING COINS SPLITTING COINS Suppose your start with seven coins. Suppose you start with seven coins. Split the coins into two groups anyway you want. Multiply the two numbers in each pile together. Keep doing this until you all the groups contain only 1 coin. Finally add up all these numbers. For example, we could split the coins as follows 5 = 10. 1 1 = 1. 3 = 6. 1 1 = 1. 1 =. 1 1 = 1. We get 10+1+6+1++1 = 1. QUESTION 1 Complete the table below computing the final total a few times splitting the piles in different ways. How does the final total depend on the way you split the piles at each stage? Contents Page 1

Coins 1 0 3 4 5 6 7 1, 8 9 10 Final total Contents Page

CONVEX POLYGONS CONVEX POLYGONS The sum of the interior angles in any n-sided convex polygon is exactly 180(n ) degrees, for all n 3. Let P(n) be the proposition that sum of the interior angles in any n-sided convex polygon is exactly 180(n ) degrees. Base case n = 3. A 3-sided polygon is a triangle, whose interior angles were shown always to sum to 180 degrees by Euclid. Induction hypothesis Suppose that P(k) holds for some k 3. That is, the interior angles in any k-sided convex polygon is exactly 180(n ) degrees. Induction step We must show that P(k +1) is true. That is, the interior angles of any k +1-sided convex polygon is exactly 180(k + 1 ) = 180(k 1) degrees, Let X be any (k + 1)-vertex convex polygon, say with successive vertices x 1, x,..., x k+1. Let Y be the polygon with vertices x 1, x,..., x m. That is, Y is obtained by cutting out one vertex from X. Now Y is also a convex polygon (proof?), so by the induction hypothesis P(k), the sum of the interior angles of Y is 180(k ). Now let T be the triangle with vertices x k, x k+1, x 1. The sum of the interior angles in X is the sum of those in Y plus the sum of those in T. (Proof?) So the sum of the interior angles in X is 180(k )+180 = 180((k + 1) ) = 180(k 1). Since X was arbitrary, we conclude that the sum of the interior angles of any (k + 1)-sided convex polygon is 180((k )+1) = 180(k 1). That is, P(k + 1) holds. Conclusion Thefore, P(1) is true and if P(k) is true, then P(k + 1) is true. Therefore, by the princple of mathematical induction, P(n) is true for all natural numbers n. Contents Page 3

3 TOWER OF HANOI TOWER OF HANOI Suppose there are n different sized discs which can be placed in three heaps 1, and 3. A disc may be moved legally from the top of one heap to the top of another heap provided that it is not placed on top of a smaller disc. Initially the discs are all on heap 1; with the largest at the bottom and in decreasing order of size up the pile. The other two piles are empty. Prove there exists a sequence of legal moves which will transfer all the discs to a different heap. Let P(n) be the proposition that there exists a sequence of legal moves that transfer n discs from heap 1 to a different heap. Base case Given one disc on heap 1, we move it to either heap or heap 3 in one move. Therefore, P(1) is true. Induction hypothesis: Suppose P(k) is true for some k 1. That is, there exists a sequence of legal moves that transfer k discs from heap 1 to either heap or heap 3. (Remember there is nothing special about the numbering of a the heaps.) Induction step We need to prove that P(k + 1) is true; that is, there exists a sequence of legal moves that transfer k + 1 discs from heap 1 to a different heap. By the induction hypothesis there is a sequence of legal moves that transfer the top k discs on heap 1 to a different heap, in particular heap. Now move the largest disc from heap to heap 3. By the induction hypothesis again we can move the k discs on heap to heap 3 in a sequence of legal moves. Contents Page 4

We have now performed a sequence of legal moves that transfer the k + 1 discs from heap 1 to a different heap. Therefore, P(k + 1) is true. Conclusion: Hence, P(1) is true and if P(k) is true, then P(k + 1) is true. Therefore, by the priciple of mathematical induction P(n) is true for all natural numbers n. Contents Page 5

4 MONEY MONEY Suppose you have an infinite supply of p and 5p coins. Prove that you can obtain n pence using only p and 5p coins for all n 4. Base case When n = 4 we use two p coins. Therefore, P(4) is true. Induction hypothesis Assume that P(k) holds for some k 4. That is, k pence can be obtained using only p and 5p coins. Induction step We must show that P(k + 1) is true; that is, k + 1 pence can be obtained using only p and 5p coins. Firstly suppose there is a 5p coin used to create the k pence. Remove this 5p coin and replace it with three p coins. This will give k + 1 pence as required. Contents Page 6

Now suppose there are no 5p coins used to create the k pence (for example if k = 4). In this case we remove two p coins which must be present since k 4 and replace them with one 5p coin. This will give k + 1 pence as required. Hence, P(k + 1) is true. Conclusion Therefore P(4) is true and if P(k) is true, then P(k + 1) is true. Hence by the Principle of Mathematical Induction, the proposition P(n) is true for all value of the natural number n 4. Contents Page 7

5 COINS REARRANGING COINS You have a collection of coins. You want to arrange them in continuous rows so each coin touches coins below. How many different ways are there to arrange the coins when you have coins, 3 coins, 4 coins, and so on. Some initial arrangements are: The sequence goes: 1,,3,5,8,1,18,6,38,53,... You can look this sequence up on the Internet using The On-Line Encyclopedia of Integer Sequences (http://oeis.org) and all you get is: Number of stacks, or arrangements of n pennies in contiguous rows, each touching in row below We cannot prove anything by induction since we do not have a closed form for the number of ways given n coins. Contents Page 8

6 GOODSTEIN S THEOREM The hereditary representation of: 66 in base is 66 = 8 + 3 + = +1 + +1 +. 35 in base is 35 = +1 + +1. Take a number in hereditary base two notation and increase the base from to 3: +1 + +1 + goes to 3 33 +1 + 3 3+1 + 3. + +1 goes to 3 33 + 3+1. Subtract one. 3 33 +1 + 3 3+1 + 3 goes to 3 33 +1 + 3 3+1 +. 3 33 + 3+1 goes to 3 33 + 3. Now increase the base by 1 and then subtract 1. 3 33 +1 + 3 3+1 + goes to 4 44 +1 + 4 4+1 + 1. 3 33 + 3 goes to 4 44 + 3. Amazingly this sequence will always converge to zero! Starting with 4 we reach a maximum of 3 10 4065310 1. Stay there for 3 10 4065309 steps, then decrease to zero. Starting with 19, after 7 steps we have a number of the order 10 369693099. Amazingly to the power of 11 it is mathematical impossible to prove this result by induction. (It is independent of Peano arthimetic.) Contents Page 9

PART A INDUCTION FURTHER DETAILS

1 INDUCTION AT A-LEVEL TYPICAL INDUCTION QUESTION Prove n r =1 r = 1 n(n + 1). There are a number of motivational arguments for this result. ARGUMENT 1 This creates a rectangle which is 10 units by 11 units. The number of squares is equal to 10 11. However, we only want half of them so the final total is 10 11. This method clearly generalises to n. n Let s = r. Then r =1 ARGUMENT s = 1+ + 3+... +98 +99 +100 s = 100+ 99+ 98+... +3 + +1 + s = 101 +101 +101... +101 +101 +101 Simplifying gives s = 100 101 since there are 100 lots of 101 on the right-hand side. Thus This clearly generalises to n. s = 100 101. FORMAL ARGUMENT PROOF BY INDUCTION At A-level, after introducting these motivational arguments, we then tell them to prove the result by mathematical induction. When asked how convincing the proof by induction is, the response is typically as follows: How convinced are you? (1 high, 3 low) Argument Student Academic Early Student 1 Rank 3 Rank 1/ Rank 1/ Rank Rank 1/ Rank 1/ 3 Rank 1 Rank 3 Rank 3 A. Induction Further details Page 1

QUESTION Prove by induction that for all natural numbers n. 1 + 3 4 +...+( 1) (n 1) n (n 1) n(n + 1) = ( 1). A typical solution to this question would consist of: Check n = 1, properly! Assume true for n = k. Do some algebra to to get n = k + 1 out. Add some memorised lines at the end. Smile and move on. The solution is pretty straight forward, but we are we doing? A. Induction Further details Page 13

WHAT IS MATHEMATICAL INDUCTION? TYPICAL INDUCTION QUESTION Prove by induction that for all natural numbers n. 1 + 3 4 +...+( 1) (n 1) n (n 1) n(n + 1) = ( 1). What are we REALLY being asked to do? Statement 1: 1 (1 1) 1(1+1) = ( 1). Statement : 1 ( 1) (+1) = ( 1). Statement 3: 1 + 3 (3 1) 3(3+1) = ( 1). The proposition is asking us to verify an infinite number of statements. TYPICAL INDUCTION QUESTION Prove by induction that for all natural numbers n. n > n Statement 1: 1 > 1. Statement : >. Statement 3: 3 > 3. Statement 4: 4 > 4.... TYPICAL INDUCTION QUESTION Prove the following by induction: for all natural numbers n. d dx (xn ) = nx n 1 d Statement 1: (x) = 1. dx Statement : Statement 3: Statement 4:... d dx (x ) = x. d dx (x3 ) = 3x. d dx (x4 ) = 4x 3. A. Induction Further details Page 14

TYPICAL INDUCTION QUESTION Prove by induction that is divisible by 5, for all natural numbers n. 7 n n Statement 1: 7 1 1 is divisible by 5. Statement : 7 is divisible by 5. Statement 3: 7 3 3 is divisible by 5. Statement 4: 7 4 4 is divisible by 5.... TYPICAL INDUCTION QUESTION Prove by induction that you can obtain n pence using only p and 5p coins for all n 4. 4 pence can be obtained using only p and 5p coins. 5 pence can be obtained using only p and 5p coins. 6 pence can be obtained using only p and 5p coins. 7 pence can be obtained using only p and 5p coins.... TYPICAL INDUCTION QUESTION For n 3, the sum of the angles in a convex n-gon is 180(n ) degrees. A convex 3-gon has angle sum 180 degrees. A convex 4-gon has angle sum 360 degrees. A convex 5-gon has angle sum 540 degrees. A convex 6-gon has angle sum 70 degrees.... TYPICAL INDUCTION QUESTION Every natural number is the sum of distinct, non-consecutive Fibonacci numbers. The number 1 is the sum of distinct, non-consecutive Fibonacci numbers. The number is the sum of distinct, non-consecutive Fibonacci numbers. The number 3 is the sum of distinct, non-consecutive Fibonacci numbers. The number 4 is the sum of distinct, non-consecutive Fibonacci numbers.... A. Induction Further details Page 15

TYPICAL INDUCTION QUESTION If x 1, x,..., x n are real numbers then x 1 + x + + x n x 1 + x +...+ x n x 1 x 1. x 1 + x x 1 + x. x 1 + x + x 3 x 1 + x + x 3.... TYPICAL INDUCTION QUESTION (n r )!r! divides n! for all 0 r n. (So the binomial coefficients are always integers, which is NOT obvious from just the formula.) (1 r)r! divides 1! for all 0 r 1. ( r)r! divides! for all 0 r. (3 r)r! divides 3! for all 0 r 3.... TYPICAL INDUCTION QUESTION Suppose there are n different sized discs which can be placed in three heaps 1, and 3. A disc may be moved legally from the top of one heap to the top of another heap provided that it is not placed on top of a smaller disc. Initially the discs are all on heap 1; with the largest at the bottom and in decreasing order of size up the pile. The other two piles are empty. Prove there exists a sequence of legal moves which will transfer all the discs to heap. Given two integers m and n, with n 0, there exist unique integers q and r such that m = bn + r and 0 r < n. Prime factorisation. Remainder and Factor theorem. For x 1,..., x n R and positive, x 1 + x +...+ x n n and so on... n x 1 x... x n. PROPOSTIONS Let P(n) be the proposition that We want to show: 1 + 3 4 +...+( 1) n 1 n n 1 n(n + 1) = ( 1). A. Induction Further details Page 16

Or as symbols {n N P(n) is true} = N. How can we show that a set S is equal to the set of natural numbers, N? The following conditions allow us to show that a set S is equal to the set of natural numbers, N: If the set S contains 1. If when the set S contains k N then the set S also contains k + 1. Then S = N. (Whether we start from 1 or 0 is the source of many a mathematical punch-up.) It is important to realise that we have an if then statement and we never declare that P(k) IS true! So we never write just P(k) is true, without a suppose. These conditions are axioms, they cannot be derived. PROOF BY INDUCTION To prove that a set S is equal to the set of natural numbers, N, we check: If the set S contains 1. If when the set S contains k N then the set S also contains k + 1. For proof by induction the set S is the set {n N P(n) is true. }. A. Induction Further details Page 17

3 WHY DO WE NEED ALL THE BITS? What happens if we do not perform all the steps required in a proof by mathematical induction? EXAMPLE Let P(n) be the proposition that n + 1 < n for n 1. Suppose P(k) is true for some k 1. Then k + 1 < k. We need to prove that P(k + 1) is true; that is, k + < k + 1. k + = (k + 1)+1 < k + 1 By the induction hypothesis = k + 1. Therefore, P(k + 1) is true. Therefore, by the principle of mathematical induction P(n) is true for all natural numbers n. In this example we have only prove that if n + 1 < n is true for any value of n, then it is true for all values of n. However, we have not verified that n +1 < n is true for any value of n. This means we cannot deduce that n +1 < n for all values of n. Since there is no base case the proof fails. QUESTION 3 Suppose P(n) is the statement n + 5n + 1 is even. Prove that if P(k) is true then P(k + 1) is true. Suppose P(n) is the statement n + 5n + 1 is odd. Prove that if P(k) is true then P(k + 1) is true. QUESTION 4 In fact, n + 5n + 1 is odd for all natural numbers n! Prove this. ERMINTRUDE THE COW Let P(n) be the proposition that for any group of n cows they are all the same colour Clear a single cow is the same colour as itself, so P(1) is true. Suppose P(k) is true, so ANY collection of k cows are the same colour. Consider some collection of k + 1 cows. A. Induction Further details Page 18

Send one cow away. The remaining k cows are the same colour. Bring the cow back and send a different cow away. We now have a collection of k cows, which must all the same colour. A. Induction Further details Page 19

Bring the cow back and we now have k + 1 cows all of the same colour. We have proven that P(1) is true and that P(k) P(k + 1) for k. The mistake here is that P(k) does not imply P(k + 1) for any k 1. In particular, it fails when k =, indeed the implication fails only for this case. FIBONACCI NUMBERS The Fibonacci number are defined by F 1 = 0, F = 1 and F n+1 = F n + F n 1 for all n. Let r be the positive real solution of r = r + 1; that is, r = 1+ 5. Let P(n) be the proposition that F n = r n for each Fibonacci number smaller than n. Suppose P(k) is true. Then F k = r k. We need to show that P(k + 1) is true; that is F k+1 = r k 1. We have F k+1 = F k + F k 1 = r k + r k 3 = r k 3 (r + 1) = r k 3 r since r = r + 1 = r k 1. Therefore, P(k + 1) is true. But we didn t check P(1) which is false, F 1 = 1, which is not r 1 = r 1. QUESTION 5 Where is the mistake in the following proof by induction? Let F 1 = 0 and F = 1 and define F n+1 = F n + F n 1 for all n 1. Let P(n) be the proposition that F r is even 0 r n for all natural numbers n. Clearly P(1) is true, since F 1 = 0 is even. A. Induction Further details Page 0

Suppose P(k) is true. That is, all F r are even for 0 r k. Now F k+1 = F k + F k 1 is the sum of two even numbers so is even. Therefore, P(k + 1) is true. Since P(1) is true and if P(k) is true, then P(k + 1) is true. Therefore, by the principle of mathematical induction P(n) is true for all natural numbers n. QUESTION 6 What is wrong with the following proof? Let P(n) be the proposition that n n + 41 is prime for all natural numbers n. We have 1 1+41 = 41, +41 = 43, 3 3+41 = 47, 4 4+41 = 53, 5 5+41 = 61. Each of these numbers is prime and this pattern clearly continues for each n, so n n +41 is prime for all natural numbers n. QUESTION 7 What is wrong with the following proof by induction? Let P(n) be the proposition that if n straight lines are drawn across a circular disc, such that no three meet in the same point, then they divide the disc into n 1 parts. Counting the number of regions gives: Points Number of regions 1 1 3 4 4 8 5 16 Clearly the pattern for the number of regions continues and we have n 1 regions. SUMMARY We need to show that the set of values for which P(n) is true is the same as the set of natural numbers. We need to verify that P(1) is true. We need to check that if P(k) is true, then P(k + 1) is true. This all has to be written formally. A. Induction Further details Page 1

4 EXAMPLES TILING A GRID Consider the following tiles: For all n 0 there exists a tiling, using only the four L-shaped tiles above, of a n n square grid with one square removed. For any of these infinitely many boards we are going to show how if precisely one square is removed at random from such a board, you can always tile what remains using the below tiles! With a by board this is easy: whichever square is removed you are left with 3 squares which form an L-shape! We re now going to show you how it s done with an 8 by 8 board. Imagine a square has been removed at random. The trick is to break the board up into four 4 by 4 boards as indicated by the red lines (notice that 4 is one power of two down from 8). Three of the 4 by 4 boards do not have a square removed. Place a tile which covers the middle corner square of each of the 4 by 4 boards which have not yet had a square removed. We now just have the problem of dealing with 4 by 4 boards each with one square removed! So, using the same idea as before, break each of these 4 by 4 boards up into four by boards. In each of the 4 by 4 boards, three of the by boards have no square removed. So place a tile covering the middle corner square of each by board which does not yet have a square removed. We are left with sixteen by boards, all with a square missing, so slot in one L-shaped tile for each by board! The logic behind proving the statement for an 8 8 board is We can prove the statement for any board; Hence we can prove the statement for any 4 4 board; A. Induction Further details Page

Hence we can prove the statement for any 8 8 board. We need to formalise how we would continue this line of reasoning for any n n board. Written out properly Let P(n) be the proposition that for any n a n n with one square removed can be covered using only the four tiles Suppose n = 1 then we have a board. Removing one square will leave exactly one of the tiles given in the list. Hence, the board can be covered with on of the tiles in the list. Suppose that P(k) is true for some k 1; that is, a k k board with one square removed can be tiled using only the four given tiles. Consider a k+1 k+1 board. This board consists of 4 copies of a k k board. We remove one tile, this must be contained in one of the four k k boards. At the meeting point of the remaining three boards we can use one of the four permitted tiles to cover the three squares at the meeting point. Hence, we now have four k k boards each with one tile missing. By the induction hypothesis each of these boards can be tiled using only the four permitted tiles. Hence, Therefore, the entire k+1 k+1 board can be tiled using only the four permitted tiles. Therefore, P(k +1) is true. Hence, by the Principle of Mathematical Induction the proposition P(n) is true for all values of the natural number n. TOWER OF HANOI Suppose there are n different sized discs which can be placed in three heaps 1, and 3. A disc may be moved legally from the top of one heap to the top of another heap provided that it is not placed on top of a smaller disc. Initially the discs are all on heap 1; with the largest at the bottom and in decreasing order of size up the pile. The other two piles are empty. Prove there exists a sequence of legal moves which will transfer all the discs to heap 3. A solution to this problem when there are three discs is given by: A. Induction Further details Page 3

The proposition we wish to prove is, P(n): The minimum number of moves required to move n disc from the heap 1 to heap 3 is n 1. Base case In the case of the tower of hanoi, the starting point is n = 1. In this case there is one disc and hence one move is required to move this disc from heap 1 to heap 3. In this case P(1) is the statement that the minimum number of moves to move the roof from the left tower to the right tower is 1 1 = 1 moves. Therefore, P(1) is true. Induction hypothesis Suppose that P(k) is true for some k 1. In other words, the minimum number of moves to required to move k discs from the heap 1 to heap 3 is k 1. Induction step We now need to show that this supposition allows us to deduce that the minimum number of moves with k + 1 discs is k+1 1. Suppose there are k +1 discs on heap 1. The induction hypothesis allows us to move the top k disc to heap 3 in a minimum of k 1 moves. But it actually tells us more than this, it tells us we can move the top k disc from any heap to any different heap in a minimum of k 1 moves. Therefore, by the induction hypothesis we can move the k discs on the heap 1 to heap in k 1 moves. We can now move the k + 1th disc (the largest disc) to heap 3. Once we have done this the induction hypothesis tells us that we can move the k disc on heap to heap 3 in a minimum of k 1 moves. To complete the induction step we just need to add up the number of moves and check it is what we expect from P(k + 1). Indeed, the total number of moves is: k 1 + 1 + k 1 = k 1 = k+1 1. Move k discs to heap Move large disc to heap 3 Move k discs from heap to heap 3 Conclusion Therefore P(1) is true and if P(k) is true, then P(k + 1) is true. Hence by the Principle of Mathematical Induction, the proposition P(n) is true for all value of the natural number n. The number of subsets of a set with n elements is n. INTEGRATION Let P(n) be the proposition that a set with n elements has n subsets. Base case (n = 0 or n = 1 both work) Suppose n = 1. Since any 1-element set has subsets, namely the empty set and the set itself, and 1 =, the statement P(1) is true. Induction hypothesis Suppose the proposition P(k) is true for some k 1; that is, any k-element set has k subsets. Induction step We need to show that P(k + 1) is true; that is, any set with k + 1 elements has k+1 subsets. Let A be a set with k + 1 elements. Let a be an element of A. Consider A = A {a}. (That is, the set A is the set A without the element a.) The set A has k elements. Any subset of A either contains a or it does not. Those subsets of A are exactly the subsets of A. By the induction hypothesis there are k such subsets. Any subset of A that contains a must have the form B {a}. That is, it is a subset of A with a included. There are k subsets of A and thus k subsets of A that contain a. Therefore, there are k + k = k+1 subsets of A. This shows that P(k + 1) is true. Conclusion Hence, P(1) is true and if P(k) is true, then P(k + 1) is true. Therefore, by the Principle of Mathematical Induction the proposition P(n) is true for all natural numbers n. DIFFERENTIATION Prove the following by induction: for all natural numbers n. d dx (xn ) = nx n 1 A. Induction Further details Page 4

Let P(n) be the proposition that d dx (xn ) = nx n 1. Base case When n = 1, we have d dx (x1 ) = 1. Moreover, 1 x 1 1 = 1. Therefore, the proposition is true when n = 1 and P(1) is true. Induction hypothesis Suppose that P(k) is true for some k 1; that is, d dx (xk ) = kx k 1. Induction step We need to show that P(k + 1) is true; that is, Using the product rule we have d dx (xk+1 ) = (k + 1)x k. d d dx (xk+1 ) = dx (xk x) d = dx (xk ) x + x k d dx (x) = k x k 1 x + x k 1 = x k (k + 1) Here the induction hypothesis is used for the third equality. Hence, P(k + 1) is true. Conclusion Hence, P(1) is true and if P(k) is true, then P(k + 1) is true. Therefore, by the Principle of Mathematical Induction the proposition is true for all natural numbers n. INEQUALITIES Prove the following by induction: n + 1 < n for all natural numbers n 3. Let P(n) be the proposition that n + 1 < n. Base case When n = 3, we have 3+1 = 7. Moreover, 3 = 8. Therefore, the proposition is true when n = 3. Induction hypothesis Suppose the P(k) is true for some k 3; that is, Induction step When n = k + 1 we need to show that When n = k + 1. k + 1 < k k + 3 = (k + 1)+1 < k+1 (k + 1)+1 = k + +1 = k + 1+ < k + < k + k = k = k+1 Here the induction hypothesis is used for the first inequality. The second inequality follows since < k for k 3. Hence, P(k + 1) is true. Conclusion Hence, P(1) is true and if P(k) is true, then P(k + 1) is true. Therefore, by the Principle of Mathematical Induction the proposition is true for all natural numbers n 3. A. Induction Further details Page 5

DIVISIBILITY Prove by mathematical induction that f (n) = 7 n n is divisible by 5 for all natural numbers n. Let P(n) be the proposition that f (n) = 7 n n is divisible by 5. Base case Suppose n = 1, then f (1) = 7 1 1 = 5. This is clearly divisible by 5. Therefore, the proposition is true when n = 1. Induction hypothesis Suppose the proposition P(k) is true for some k 1; that is, f (k) = 7 k k is divisible by 5. Thus f (k) = 7 k k = 5M for some natural number M. Induction step Consider the case when n = k + 1, we need to show that is divisible by 5. When n = k + 1 it follows that f (k + 1) = 7 k+1 k+1 f (k + 1) f (k) = 7 k+1 k+1 (7 k k ) = 7 k+1 7 k k+1 + k = 7 k (7 1) k ( 1) = 6 7 k k = (5+1) 7 k k = 5 7 k + 7 k k = 5 7 k + f (k). Hence f (k + 1) f (k) = 5 7 k + f (k). By the induction hypothesis f (k) is divisible by 5, which implies f (k + 1) f (k) is divisible by 5. So f (k + 1) f (k) = 5N for some natural number N. Thus f (k + 1) = 5N + f (k). By the induction hypothesis f (k) is divisible by 5, hence f (k + 1) is divisible by 5. Hence P(k + 1) is true. Conclusion Hence P(1) is true and if P(k) is true, then P(k + 1) is true. Therefore, by the Principle of Mathematical Induction the proposition P(n) is true for all natural numbers n. SEQUENCES A sequence (u n ) is defined by u 1 = 7 and u n+1 = 7u n 3. Prove the following using mathematical induction: for all natural numbers n. u n = (13 7n 1 )+1 Let P(n) be the proposition that u n = (13 7n 1 )+1. Base case Suppose n = 1, then u 1 = (13 70 )+1 = 7. This is just the value of u 1 as defined. Therefore, the proposition is true when n = 1. Induction hypothesis Suppose the proposition P(k) is true for some k 1; that is, u k = (13 7k 1 )+1 A. Induction Further details Page 6

Induction step Consider the case when n = k + 1, we need to show that The definition of the sequence (u n ) implies that: u k+1 = (13 7k )+1. u k+1 = 7u k 3 = 7( (13 7k 1 )+1 ) 3 = = = 13 7 k + 7 3 13 7 k + 7 6 13 7 k + 1. The second equality follows from the induction hypothesis. Hence, P(k + 1) is true. Conclusion Hence, P(1) is true and if P(k) is true, then P(k + 1) is true. Therefore, by the Principle of Mathematical Induction the proposition P(n) is true for all natural numbers n. INTEGRATION Show that for all n 0 Let P(n) be the proposition that x n e x dx = n!. 0 x n e x dx = n! 0 Base case Suppose n = 0. We have 0 e x dx = [ e x ] 0 = 0+1 = 1 = 0!. Therefore, P(1) is true. Induction hypothesis Suppose the proposition P(k) is true for some k 0; that is, x k e x dx = k! 0 Induction step We need to show that P(k + 1) is true; that is, x k+1 e x dx = (k + 1)! 0 The induction step is performed using integration by parts (which is exactly how you would derive the result in the first place)! Therefore, P(k + 1) is true. 0 + e x (k + 1)x k dx 0 = 0+(k + 1) e x x k dx 0 = (k + 1)k! By the induction hypothesis. x k+1 e x dx = [x k+1 ( e x )] 0 = (k + 1)! Conclusion Hence, P(1) is true and if P(k) is true, then P(k + 1) is true. Therefore, by the Principle of Mathematical Induction the proposition P(n) is true for all natural numbers n. A. Induction Further details Page 7

CONVERGENCE OF SEQUENCES 1. Prove that the cube root of an irrational number is irrational.. You are given that 3 5 is irrational. Let a n = 5 1 3 n. Prove that the sequence (a n ) is irrational for all n N. 3. Deduce that given m Z there exists a sequence of irrational numbers converging to m. (You may assume 1 that as n tend to infinity that 3 n 0.) 1. Let y be irrational. Suppose for a contradiction that y 1 3 = p p3 where p, q Z and q 0. Then y = is a rational q q 3 number since p 3 and q 3 0 are integers. This contradiction establishes the result.. Let P(n) be the proposition that a n = 5 1 3 n is irrational. Base case When n = 1 we have a 1 = 3 5. We are given that this number is irrational, hence P(1) is true. Induction hypothesis Suppose P(k) is true for some k 1; that is, then a k = 5 1 3 k is irrational. Induction step We need to prove that a k+1 = 5 1 3 k+1 is irrational. Now 5 1 3 k+1 = (5 1 1 3 k 3 ) by the indices laws. By the induction hypothesis 5 1 3 k is irrational and we have proved that the cube root of an irrational number is irrational. Therefore, 5 1 3 k+1 is irrational. Hence P(k) is true. Conclusion Hence, P(1) is true and if P(k) is true, then P(k + 1) is true. Therefore, by the Principal of Mathematical Induction P(n) is true for all n N. 3. Let m Z. Define u n = ma n. Then as n it follows that a n 1 and so u n m. This proof has combined several things we ve seen before: direct proof, proof by contradiction, proof by induction and then combining results to prove a new result. The fact we have assumed 1 3 n 0 as n. This is used at A-level when we sum an infinite geometric series and deduce that when 1 < x < 1. a(1 x n+1 ) 1 x a 1 x FIBONACCI NUMBERS Every positive integer can be represented as the sum of one or more distinct Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers. (In fact, it is unique, but we don t prove this, although it can be done by induction.) An example and non-example of such expressions are: 100 = 89+8+3, yes100 = 89+8++1. no Let P(n) be the proposition that each integer r with 1 r n can be represented as the sum of one of more distinct Fibonacci numbers, with the sum not including any two consecutive numbers. Base case The true of P(1) is clear, since r = n = 1 is a Fibonacci number. Induction hypothesis Now suppose that P(k) is true for some k 1. That is, for every positive integer r with 1 r k can be represented as the sum of distinct Fibonacci numbers and the sum does not include any two consecutive Fibonacci numbers. A. Induction Further details Page 8

Induction step We need to show that P(k+1) is true. That is, for every positive interger r with 1 r k+1 can be represented as the sum of distinct Fibonacci numbers and the sum does not include any two consecutive Fibonacci numbers. Clearly, we just need to check the case r = k + 1 (the induction hypothesis deals with the other cases.) If k + 1 is a Fibonacci number then we are done and P(k + 1) is true. Now suppose k + 1 is not a Fibonacci number, then there exists i such that F i < k + 1 < F i+1. Now k+1 = F i +n for some positive integer n. If we can show that n is the sum of distinct Fibonacci numbers that do not include F i 1 then we are done. Now n = k +1 F i. The largest that n can be is k, so 0 < n k. (If n = 0, then k +1 = F i and if n = k, then F i = 1 and so k + 1 = and the proposition is clearly two.) So we may now assume n = k + 1 F i < k. But by the induction hypothesis we know that n is the sum of distinct Fibonacci numbers with the sum not including two consecutive Fibonacci numbers. We finally need to check that the sum n does not include F i 1 for otherwise k +1 = F i +n would contain the consecutive Fibonacci numbers F i and F i 1. Suppose F i 1 was contained in the representation of n, then since F i+1 = F i + F i 1 we have k + 1 = F i + n = F i + F i 1 + m = F i+1 + m > F i+1. This is a contraction, therefore, the representation of n does not contain F i 1. Therefore P(k + 1) is true. Conclusion Since P(1) is true and the truth of P(k) implies the truth of P(k + 1) the principle of mathematical induction implies that P(n) holds for all natural numbers n. INFINITELY MANY PRIMES The nth Fermat number, F n, is defined by for n N. F n = n + 1 1. It is claimed that F 0 F 1 F...F n 1 = F n for all n N. Prove by induction that Equation (A.1) holds for all natural numbers. (A.1). Hence, deduce that no two Fermat numbers have a common factor greater than 1. 3. Hence, prove that there are infinitely many prime numbers. [Hint: You will find it useful to recall that if n is an odd number then the only common factor of n and n is 1.] 1. Let P(n) be the proposition that F 0 F 1 F...F n 1 = F n. Base case When n = 1 we need to check that F 0 = F 1. We have F 0 = 3 and F 1 = 5. Hence, since 3 = 5, P(1) is true. Induction hypothesis Suppose P(k) is true for some k 1; that is, F 0 F 1 F...F k 1 = F k. Induction step We need to show that F 0 F 1 F...F k 1 F k = F k+1 A. Induction Further details Page 9

By the induction hypothesis F 0 F 1 F...F k 1 F k = (F k )F k Thus if P(k) is true it follows that P(k + 1) is true. = ( k + 1 )( k + 1) = ( k 1)( k + 1) = k k 1 = k+1 1 = k+1 + 1 = F k+1. Conclusion Hence, P(1) is true and if P(k) is true, then P(k +1) is true. Therefore it follows by the Principle of Mathematical Induction that P(n) is true for all natural numbers n.. Suppose F a and F b have a common factor m. We may assume that a < b (otherwise just flip the two numbers round). By our proved result F 0 F 1 F...F a...f b 1 = F b. So m divides F 0 F 1 F...F a...f b 1 and thus m divides F b. This implies that F b and F b have a common factor m. Since F b is odd, it follows that the only common factor of F b and F b is 1. Therefore, m = 1. 3. Any Fermat number must either be prime or can be factorised into prime factors. Since no two Fermat numbers have a common factor greater than 1, each Fermat number must consist of primes or prime factors different to all other Fermat numbers. Clearly there are an infinite number of Fermat numbers so there must be an infinite number of primes. To prove the fact given in the question, suppose n and n are divisible by p. Then n = pr for some integer r and n = ps for some integer s. Subtracting gives = pr ps = p(r s). Since p and r s are integers either p = 1 and r s = or p = and r s = 1. Suppose p =, then an even number divides into the odd number n. This contradiction shows that p = 1 giving the result. PIRATES GOLD Some pirates have got hold of some gold bars. The gold bars are 1 units in size. The pirates have boxes which are n in size. The pirates wonder how many different ways there are to arrange the gold in:, 3,, 5 and n boxes. One arrangement in a 8 box is: It shoud be clear that that this actually can be proved by induction: So, we get the relation This relation can be proved by induction. P n = P n 1 + P n A. Induction Further details Page 30

5 PROPOSITIONS THAT LOOK LIKE THEY CAN BE PROVED BY INDUCTION GOLDBACH CONJECTURE Every even integer greater than can be expressed as the sum of two primes. Comments: Examples are: 4 = +, 10 = 3+7 = 5+5, 100 = 47+53. Not asserting uniqueness or distinct primes. Proved by hand up to 10 5 by Nils Pipping in 1938. Extended by computer up to 10 18. Unproved generally as of when I wrote this sentence. A. Induction Further details Page 31

PART B SOLUTIONS AND FURTHER DETAILS

SOLUTION TO QUESTION 1: Coins 1 0 1 3 3 4 6 5 10 6 15 7 1 8 8 9 36 10 45 Final total We might conjecture that: the final total is always 1 n(n 1). How do we prove this? There are 1 n(n 1) handshakes (edges) between the coins. To get each coin on its own we need to remove each edge. We now prove this by mathematical induction. Let P(n) be the proposition that given any pile of n coins, then no matter how the piles of split the final total is always f (n) = 1 n(n 1). When n = 1, we have one coin, there is no second coin so the final total will be 0. Moreover, 1 1 0 = 0. Therefore, P(1) is true. Suppose P(k) is true for all i k with k 1. That is, We need to show that P(k + 1) is true; that is, f (i) = 1 i(i 1). f (k + 1) = 1 k(k + 1). Suppose we have k + 1 coins. Let 1 i k. Then split the (k + 1) coins into a pile of i and (k + 1) i coins. The product is now given by i[(k + 1) i]. Since i k and (k + 1) i k we have So the final product will be f (i) = 1 i(i 1) f (k + 1 i) = 1 (k + 1 i)(k i). i(k + 1 i) +f (i) + f (k + 1 i) = i(k + 1 i) + 1 i(i 1) + 1 (k + 1 i)(k i) Product for current pile = 1 [ik + i i + i i + k ki + k i ik + i ] = 1 (k + 1)k. Therefore, P(k + 1) is true. Since P(1) is true and if P(k) is true, then P(k + 1) is true, by mathematical induction P(n) is true for all natural numbers n. In a very real sense, the formal proof by induction is far less convincing that the argument using the graph! SOLUTION TO QUESTION : Let P(n) be the proposition that 1 + 3 4 +... + ( 1) n 1 n n 1 n(n + 1) = ( 1). B. Solutions and further details Page 34

When n = 1, then Also 1 = 1. Therefore, P(1) is true. Assume P(k) is true for some k 1; that is, We need to prove that P(k + 1) is true; that is, We have (1 1) 1(1 + 1) ( 1) = 1 = 1. 1 + 3 4 +... + ( 1) k 1 k k 1 k(k + 1) = ( 1). 1 + 3 4 +... + ( 1) k 1 k + ( 1) k (k + 1) k (k + 1)(k + ) = ( 1). 1 + 3 4 +... + ( 1) k 1 k +( 1) k (k + 1) First k terms k 1 k(k + 1) = ( 1) + ( 1) k (k + 1) (by the induction hypothesis) k (k + 1) = ( 1) [( 1) 1 k + (k + 1)] k (k + 1) = ( 1) [ k + k + ] k (k + 1)(k + ) = ( 1). Therefore, P(k + 1) is true. Since P(1) is true and if P(k) is true, then P(k +1) is true, the principle of mathematical induction implies P(n) is true for all natural numbers n. SOLUTION TO QUESTION 3: Suppose P(k) is true; that is k + 5k + 1 is even. We need to show that P(k + 1) is true; that is, is even. Since (k + 1) + 5(k + 1) + 1 = k + k + 1 + 5k + 5 + 1 = k + 7k + 7 k + 7k + 7 = k + 5k + 1 + (k + 6) = k + 5k + 1 +(k + 3) Assumed to be even even it follows that k + 7k + 7 is even. Indeed, since we ve assumed P(k) is true, k + 5k + 1 is even and (k + 3) is even and the sum of two even numbers is even. Therefore, P(k + 1) is true. Suppose P(k) is true; that is k + 5k + 1 is odd. We need to show that P(k + 1) is true; that is, is odd. Since (k + 1) + 5(k + 1) + 1 = k + k + 1 + 5k + 5 + 1 = k + 7k + 7 k + 7k + 7 = k + 5k + 1 + (k + 6) = k + 5k + 1 +(k + 3) Assumed to be odd even it follows that k + 7k + 7 is odd. Indeed, by the induction hypothesis k + 5k + 1 is odd and (k + 3) is even and the sum of an even number and an odd number is odd. Therefore, P(k + 1) is true. SOLUTION TO QUESTION 4: Suppose n is even. Then n = m for some m Z. Thus n + 5n + 1 = (m) + 5(m) + 1 = 4m + 10m + 1 = (m + 5m) + 1 Since m + 5m is an integer, (m + 5m) + 1 is odd and it follows that n + 5n + 1 is odd. Suppose that n is odd. Then n = m + 1 for some m Z. Then n + 5n + 1 = (m + 1) + 5(m + 1) + 1 = 4m + 14m + 7 = (m + 7m + 3) + 1. Since m + 7m + 3 is an integer, (m + 7m + 3) + 1 is odd. Therefore, n + 5n + 1 is odd. SOLUTION TO QUESTION 5: The statement F k+1 = F k + F k 1 is false when k = 1. Since F k+1 = F = 1. B. Solutions and further details Page 35

We have proved P(1); P(1) and P() implies P(3); P() and P(3) implies P(4). Since P() is not true and hence we have not verified that P(k) implies P(k + 1). SOLUTION TO QUESTION 6: Except, 41 41 + 41 = 41 clearly is not prime! Let P(n) be the proposition that n n + 41 is prime for all natural numbers n. When n = 1, 1 1 + 41 = 41 is prime. Therefore, P(1) is true. Suppose P(k) is true; that is, k k + 41 is prime. We need to prove that P(k + 1) is true; that is, (k + 1) (k + 1) + 41 is prime. (k + 1) (k + 1) + 41 = k + k + 1 k 1 + 41 = (k k + 41) + k. Now (k k+41) is prime, but this does not imply the resulting expression is prime. As we have seen, let k = 40 to generate a counter-example. SOLUTION TO QUESTION 7: In fact we actually have the following: In fact it gets worse, the actual formula is Points Number of regions 1 1 3 4 4 8 5 16 6 31 7 57 ( n 4 ) + (n ) + 1. You need to know that for a planar graph with V vertices, E edges and F faces, then V E + F =. (This is Euler s formula.) The number of faces F is equal to the number of regions, except there is an additional region outside the circle. Now we find V. Note there are n points on the circle. Now there are also some vertices where the edges cross. Now we create a new vertex when two edges cross: to create an edge we need to select of the n points. So to create two edges we need to select 4 of the n points. There are ( n ) ways to do 4 this. Therefore, there are n + ( n 4 ) vertices. We now need to determine E. Firstly there are n edges which are the circular arcs. Next, at each of the ( n 4 ) interior vertices there are 4 edges. This gives 4(n 4 ) edges. Next there are ( n ) chords. (Each chord is determined by selecting of the n points.) We have counted ( n ) + 4(n 4 ) edges. But each has been counted twice. So there are n + (n ) + (n 4 ) edges. Euler s formula now gives F = + E V = + n + ( n ) + (n 4 ) (n + (n 4 )) = + (n ) + (n 4 ). But remember we have counted the face exterior to the circle, so the number of regions is F 1. B. Solutions and further details Page 36