martyn parker - meimei.org.uk/files/conference13/k2_2013.pdfthe sum of the interior angles in any...

MEI CONFERENCE 2013

What can you prove by induction?

Martyn Parker

[email protected]

mailto:[email protected]

Contents

Contents iii1 Splitting Coins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Convex Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Tower of Hanoi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 Money . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Coins — rearranging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 Goodstein’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

A Induction — Further details 121 Induction at A-level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 What is mathematical induction? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Why do we need all the bits? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 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

5×2 = 10.

1×1 = 1.

3×2 = 6.

1×1 = 1.

2×1 = 2.

1×1 = 1.

We get 10+1+6+1+2+1 = 21.

QUESTION 1

Complete the table below computing the final total a few times splitting the piles in different ways. How does thefinal total depend on the way you split the piles at each stage?

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Coins Final total

1 0234567 21,8910

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2 CONVEX POLYGONS

CONVEX POLYGONS

The sum of the interior angles in any n-sided convex polygon is exactly 180(n−2) 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 −2)degrees.

Base case n = 3. A 3-sided polygon is a triangle, whose interior angles were shown always to sum to 180 de-grees by Euclid.

Induction hypothesis Suppose that P(k) holds for some k ≥ 3. That is, the interior angles in any k-sided convexpolygon is exactly 180(n−2) degrees.

Induction step We must show that P(k+1) is true. That is, the interior angles of any k+1-sided convex polygonis exactly 180(k +1−2) = 180(k −1) degrees,

Let X be any (k +1)-vertex convex polygon, say with successive vertices x1, x2, . . . , xk+1.

Let Y be the polygon with vertices x1, x2, . . . , xm . 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 interiorangles of Y is 180(k −2).

Now let T be the triangle with vertices xk , xk+1, x1. The sum of the interior angles in X is the sum of thosein Y plus the sum of those in T . (Proof?)

So the sum of the interior angles in X is

180(k −2)+180 = 180((k +1)−2) = 180(k −1).

Since X was arbitrary, we conclude that the sum of the interior angles of any (k +1)-sided convex polygonis 180((k −2)+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 mathe-matical induction, P(n) is true for all natural numbers n.

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3 TOWER OF HANOI

TOWER OF HANOI

Suppose there are n different sized discs which can be placed in three heaps 1, 2 and 3. A disc may be movedlegally 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 adifferent heap.

Let P(n) be the proposition that there exists a sequence of legal moves that transfer n discs from heap 1 to adifferent heap.

Base case Given one disc on heap 1, we move it to either heap 2 or heap 3 in one move. Therefore, P(1) istrue.

Induction hypothesis: Suppose P(k) is true for some k ≥ 1. That is, there exists a sequence of legal movesthat transfer k discs from heap 1 to either heap 2 or heap 3. (Remember there is nothing special about thenumbering 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 thattransfer 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 adifferent heap, in particular heap 2.

Now move the largest disc from heap 2 to heap 3.

By the induction hypothesis again we can move the k discs on heap 2 to heap 3 in a sequence of legalmoves.

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We have now performed a sequence of legal moves that transfer the k +1 discs from heap 1 to a differentheap. 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 mathe-matical induction P(n) is true for all natural numbers n.

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4 MONEY

MONEY

Suppose you have an infinite supply of 2p and 5p coins. Prove that you can obtain n pence using only 2p and 5pcoins for all n ≥ 4.

Base case When n = 4 we use two 2p 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 2pand 5p coins.

Induction step We must show that P(k +1) is true; that is, k +1 pence can be obtained using only 2p and 5pcoins.

Firstly suppose there is a 5p coin used to create the k pence.

Remove this 5p coin and replace it with three 2p coins.

This will give k +1 pence as required.

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Now suppose there are no 5p coins used to create the k pence (for example if k = 4).

In this case we remove two 2p 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 Mathe-matical Induction, the proposition P(n) is true for all value of the natural number n ≥ 4.

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5 COINS — REARRANGING

COINS

You have a collection of coins. You want to arrange them in continuous rows so each coin touches 2 coins below.How many different ways are there to arrange the coins when you have 2 coins, 3 coins, 4 coins, and so on.

Some initial arrangements are:

The sequence goes:1,2,3,5,8,12,18,26,38,53, . . .

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 2 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.

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6 GOODSTEIN’S THEOREM

The hereditary representation of:

◁ 266 in base 2 is 266 = 28+23+2 = 222+1+22+1+2.

◁ 35 in base 2 is 35 = 222+1+2+1.

◁ Take a number in hereditary base two notation and increase the base from 2 to 3:

– 222+1+22+1+2 goes to 333+1+33+1+3.

– 222+2+1 goes to 333

+3+1.

◁ Subtract one.

– 333+1+33+1+3 goes to 333+1+33+1+2.

– 333+3+1 goes to 333

+3.

◁ Now increase the base by 1 and then subtract 1.

– 333+1+33+1+2 goes to 444+1+44+1+1.

– 333+3 goes to 444

+3.

◁ Amazingly this sequence will always converge to zero!

– Starting with 4 we reach a maximum of 3× 10402653210 − 1. Stay there for 3× 10402653209 steps, thendecrease to zero.

– Starting with 19, after 7 steps we have a number of the order 10369693099.

◁ ’Amazingly to the power of 11’ it is mathematical impossible to prove this result by induction. (It is indepen-dent of Peano arthimetic.)

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PART AINDUCTION — FURTHER DETAILS

1 INDUCTION AT A-LEVEL

TYPICAL INDUCTION QUESTION

Prove ∑nr=1 r = 1

2 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, weonly want half of them so the final total is 10×11

2 . This method clearly generalises to n.

ARGUMENT 2

Let s =n

∑r=1

r . Then

s = 1+ 2+ 3+ . . . +98 +99 +100s = 100+ 99+ 98+ . . . +3 +2 +1+

2s = 101 +101 +101 . . . +101 +101 +101

Simplifying gives 2s = 100×101 since there are 100 lots of 101 on the right-hand side. Thus

s =100×101

2.

This clearly generalises to n.

FORMAL ARGUMENT — PROOF BY INDUCTION

At A-level, after introducting these motivational arguments, we then tell them to prove the result by mathematicalinduction. 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 1/2 Rank 1/22 Rank 2 Rank 1/2 Rank 1/23 Rank 1 Rank 3 Rank 3

A. Induction — Further details Page 12

QUESTION 2

Prove by induction that

12−22

+32−42

+ . . .+(−1)(n−1)n2= (−1)(n−1)n(n+1)

2.

for all natural numbers n.

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?


2 WHAT IS MATHEMATICAL INDUCTION?


Prove by induction that

12−22

+32−42

+ . . .+(−1)(n−1)n2= (−1)(n−1)n(n+1)

2.


What are we REALLY being asked to do?

◁ Statement 1:

12= (−1)(1−1) 1(1+1)

2.

◁ Statement 2:

12−22

= (−1)(2−1) 2(2+1)

2.

◁ Statement 3:

12−22

+32= (−1)(3−1) 3(3+1)

2.

◁ The proposition is asking us to verify an ’infinite number of statements’.


Prove by induction that2n

> n


◁ Statement 1: 21 > 1.




◁ ...


Prove the following by induction:d

d x(xn

) = nxn−1


◁ Statement 1:d

d x(x) = 1.

◁ Statement 2:d

d x(x2) = 2x.

◁ Statement 3:d

d x(x3) = 3x2.

◁ Statement 4:d

d x(x4) = 4x3.

◁ ...



Prove by induction that7n−2n

is divisible by 5, for all natural numbers n.

◁ Statement 1: 71−21 is divisible by 5.




◁ ...


Prove by induction that you can obtain n pence using only 2p and 5p coins for all n ≥ 4.

◁ 4 pence can be obtained using only 2p and 5p coins.




◁ ...


For n ≥ 3, the sum of the angles in a convex n-gon is 180(n−2) degrees.

◁ A convex 3-gon has ‘angle sum’ 180 degrees.




◁ ...


Every natural number is the sum of distinct, non-consecutive Fibonacci numbers.

◁ The number 1 is the sum of distinct, non-consecutive Fibonacci numbers.




◁ ...



If x1, x2, . . . , xn are real numbers then

∣x1∣+ ∣x2∣+⋅ ⋅ ⋅+ ∣xn ∣ ≥ ∣x1+ x2+ . . .+ xn ∣

◁ ∣x1∣ ≥ ∣x1∣.

◁ ∣x1∣+ ∣x2∣ ≥ ∣x1+ x2∣.

◁ ∣x1∣+ ∣x2∣+ ∣x3∣ ≥ ∣x1+ x2+ x3∣.

◁ ...


(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.



◁ ...


Suppose there are n different sized discs which can be placed in three heaps 1, 2 and 3. A disc may be movedlegally 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. Theother two piles are empty. Prove there exists a sequence of legal moves which will transfer all the discs to heap2.

◁ Given two integers m and n, with n ≠ 0, there exist unique integers q and r such that m = bn + r and0 ≤ r < ∣n∣.

◁ Prime factorisation.

◁ Remainder and Factor theorem.

◁ For x1, . . . , xn ∈R and positive,x1+ x2+ . . .+ xn

n≥ n√

x1x2 . . . xn .

◁ and so on ...

PROPOSTIONS

Let P(n) be the proposition that

12−22

+32−42

+ . . .+(−1)n−1n2= (−1)n−1 n(n+1)

2.

We want to show:


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 toshow 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 neverwrite 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. }.


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 +2 < k +1.

k +2 = (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 < nfor all values of n. Since there is no base case the proof fails.

QUESTION 3

◁ Suppose P(n) is the statement n2+5n+1 is even. Prove that if P(k) is true then P(k +1) is true.

◁ Suppose P(n) is the statement n2+5n+1 is odd. Prove that if P(k) is true then P(k +1) is true.

QUESTION 4

In fact, n2+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.


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.


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) Ô⇒ P(k +1) for k ≥ 2.

The mistake here is that P(k) does not imply P(k +1) for any k ≥ 1. In particular, it fails when k = 2, indeed theimplication fails only for this case.

FIBONACCI NUMBERS

The Fibonacci number are defined by F1 = 0, F2 = 1 and Fn+1 = Fn +Fn−1 for all n ≥ 2.

Let r be the positive real solution of r 2 = r +1; that is, r =1+

√5

2.

Let P(n) be the proposition that Fn = r n−2 for each Fibonacci number smaller than n.Suppose P(k) is true. Then Fk = r k−2.We need to show that P(k +1) is true; that is Fk+1 = r k−1.We have

Fk+1 = Fk +Fk−1

= r k−2+ r k−3

= r k−3(r +1)

= r k−3r 2 since r 2= r +1

= r k−1.

Therefore, P(k +1) is true.But we didn’t check P(1) which is false, F1 = 1, which is not r 1−2 = r−1.

QUESTION 5

Where is the mistake in the following proof by induction?Let F1 = 0 and F2 = 1 and define Fn+1 = Fn +Fn−1 for all n ≥ 1. Let P(n) be the proposition that Fr is even 0 ≤ r ≤ nfor all natural numbers n.Clearly P(1) is true, since F1 = 0 is even.


Suppose P(k) is true. That is, all Fr are even for 0 ≤ r ≤ k.Now Fk+1 = Fk +Fk−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 inductionP(n) is true for all natural numbers n.

QUESTION 6

What is wrong with the following proof?Let P(n) be the proposition that n2−n+41 is prime for all natural numbers n.We have

12−1+41 = 41,

22−2+41 = 43,

32−3+41 = 47,

42−4+41 = 53,

52−5+41 = 61.

Each of these numbers is prime and this pattern clearly continues for each n, so n2−n+41 is prime for all naturalnumbers 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 thesame point, then they divide the disc into 2n−1 parts.

Counting the number of regions gives:

Points Number of regions

1 12 23 44 85 16

Clearly the pattern for the number of regions continues and we have 2n−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.


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 2n×2n square grid with one squareremoved.

For any of these infinitely many boards we are going to show how if precisely one square is removed at randomfrom such a board, you can always tile what remains using the below tiles!With a 2 by 2 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 onepower 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 hada square removed. We now just have the problem of dealing with 4 by 4 boards each with one squareremoved!

◁ So, using the same idea as before, break each of these 4 by 4 boards up into four 2 by 2 boards.

◁ In each of the 4 by 4 boards, three of the 2 by 2 boards have no square removed. So place a tile coveringthe ‘middle corner’ square of each 2 by 2 board which does not yet have a square removed.

◁ We are left with sixteen 2 by 2 boards, all with a square missing, so slot in one L-shaped tile for each 2 by2 board!

The logic behind proving the statement for an 8×8 board is

◁ We can prove the statement for any 2×2 board;

◁ Hence we can prove the statement for any 4×4 board;


◁ 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 2n ×2n board.

Written out properly Let P(n) be the proposition that for any n a 2n × 2n with one square removed can becovered using only the four tiles

Suppose n = 1 then we have a 2×2 board. Removing one square will leave exactly one of the tiles given in thelist. 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 2k ×2k board with one square removed can be tiled using onlythe four given tiles.Consider a 2k+1 ×2k+1 board. This board consists of 4 copies of a 2k ×2k board. We remove one tile, this mustbe contained in one of the four 2k ×2k boards. At the meeting point of the remaining three boards we can useone of the four permitted tiles to cover the three squares at the meeting point. Hence, we now have four 2k ×2k

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 2k+1×2k+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 numbern.

TOWER OF HANOI

Suppose there are n different sized discs which can be placed in three heaps 1, 2 and 3. A disc may be movedlegally 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. Theother two piles are empty. Prove there exists a sequence of legal moves which will transfer all the discs to heap3.

A solution to this problem when there are three discs is given by:


The proposition we wish to prove is, P(n): The minimum number of moves required to move n disc from theheap 1 to heap 3 is 2n −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 henceone move is required to move this disc from heap 1 to heap 3. In this case P(1) is the statement thatthe minimum number of moves to move the roof from the left tower to the right tower is 21 −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 movesto required to move k discs from the heap 1 to heap 3 is 2k −1.

Induction step We now need to show that this supposition allows us to deduce that the minimum number ofmoves with k +1 discs is 2k+1−1.

Suppose there are k+1 discs on heap 1. The induction hypothesis allows us to move the top k disc to heap3 in a minimum of 2k −1 moves. But it actually tells us more than this, it tells us we can move the top k discfrom any heap to any different heap in a minimum of 2k −1 moves. Therefore, by the induction hypothesiswe can move the k discs on the heap 1 to heap 2 in 2k −1 moves.

We can now move the k + 1th disc (the largest disc) to heap 3. Once we have done this the inductionhypothesis tells us that we can move the k disc on heap 2 to heap 3 in a minimum of 2k −1 moves.

To complete the induction step we just need to add up the number of moves and check it is what we expectfrom P(k +1). Indeed, the total number of moves is:

2k−1

²Move k discs to heap 2

+ 1®

Move large disc to heap 3

+ 2k−1

²Move k discs from heap 2 to heap 3

= 2×2k−1 = 2k+1

−1.

Conclusion Therefore P(1) is true and if P(k) is true, then P(k +1) is true. Hence by the Principle of Mathe-matical Induction, the proposition P(n) is true for all value of the natural number n.

INTEGRATION

The number of subsets of a set with n elements is 2n .

Let P(n) be the proposition that a set with n elements has 2n subsets.

Base case (n = 0 or n = 1 both work) Suppose n = 1. Since any 1-element set has 2 subsets, namely the emptyset and the set itself, and 21 = 2, 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 2k

subsets.

Induction step We need to show that P(k +1) is true; that is, any set with k +1 elements has 2k+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 2k 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 2k subsets of A′ and thus 2k subsets of A that contain a.

Therefore, there are 2k +2k = 2k+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 Mathe-matical Induction the proposition P(n) is true for all natural numbers n.

DIFFERENTIATION

Prove the following by induction:d

d x(xn

) = nxn−1



Let P(n) be the proposition thatd

d x(xn

) = nxn−1.

Base case When n = 1, we have dd x (x1) = 1. Moreover, 1×x1−1 = 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

d x(xk

) = kxk−1.

Induction step We need to show that P(k +1) is true; that is,

d

d x(xk+1

) = (k +1)xk .

Using the product rule we have

d

d x(xk+1

) =d

d x(xk

× x)

=d

d x(xk

)× x + xk d

d x(x)

= k × xk−1× x + xk

×1

= xk(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 Mathe-matical Induction the proposition is true for all natural numbers n.

INEQUALITIES

Prove the following by induction: 2n+1 < 2n for all natural numbers n ≥ 3.

Let P(n) be the proposition that2n+1 < 2n .

Base case When n = 3, we have 2×3+1 = 7. Moreover, 23 = 8. Therefore, the proposition is true when n = 3.

Induction hypothesis Suppose the P(k) is true for some k ≥ 3; that is,

2k +1 < 2k

Induction step When n = k +1 we need to show that

2k +3 = 2(k +1)+1 < 2k+1

When n = k +1.

2(k +1)+1 = 2k +2+1

= 2k +1+2

< 2k+2

< 2k+2k

= 2×2k

= 2k+1

Here the induction hypothesis is used for the first inequality. The second inequality follows since 2 < 2k fork ≥ 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 Mathe-matical Induction the proposition is true for all natural numbers n ≥ 3.


DIVISIBILITY

Prove by mathematical induction that f (n) = 7n −2n is divisible by 5 for all natural numbers n.

Let P(n) be the proposition thatf (n) = 7n

−2n

is divisible by 5.

Base case Suppose n = 1, then f (1) = 71−21 = 5. This is clearly divisible by 5. Therefore, the proposition is truewhen n = 1.

Induction hypothesis Suppose the proposition P(k) is true for some k ≥ 1; that is,

f (k) = 7k−2k

is divisible by 5. Thus f (k) = 7k −2k = 5M for some natural number M .

Induction step Consider the case when n = k +1, we need to show that

f (k +1) = 7k+1−2k+1

is divisible by 5.

When n = k +1 it follows that

f (k +1)− f (k) = 7k+1−2k+1

−(7k−2k)

= 7k+1−7k

−2k+1+2k

= 7k×(7−1)−2k

×(2−1)

= 6×7k−2k

= (5+1)×7k−2k

= 5×7k+7k

−2k

= 5×7k+ f (k).

Hence f (k + 1)− f (k) = 5× 7k + f (k). By the induction hypothesis f (k) is divisible by 5, which impliesf (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. HenceP(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 Mathe-matical Induction the proposition P(n) is true for all natural numbers n.

SEQUENCES

A sequence (un) is defined by u1 = 7 and un+1 = 7un −3. Prove the following using mathematical induction:

un =(13×7n−1)+1

2



un =(13×7n−1)+1

2.

Base case Suppose n = 1, then u1 =(13×70)+1

2 = 7. This is just the value of u1 as defined. Therefore, theproposition is true when n = 1.


uk =(13×7k−1)+1

2


Induction step Consider the case when n = k +1, we need to show that

uk+1 =(13×7k)+1

2.

The definition of the sequence (un) implies that:

uk+1 = 7uk −3

= 7((13×7k−1)+1

2)−3

=13×7k +7

2−3

=13×7k +7−6

2

=13×7k +1

2.

The second equality follows from the induction hypothesis. Hence, P(k +1) is true.


INTEGRATION

Show that for all n ≥ 0

∫

∞0

xne−x d x = n!.


∫

∞0

xne−x d x = n!

Base case Suppose n = 0. We have

∫

∞0

e−x d x = [−e−x]∞0 = 0+1 = 1 = 0!.

Therefore, P(1) is true.


∫

∞0

xk e−x d x = k !

Induction step We need to show that P(k +1) is true; that is,

∫

∞0

xk+1e−x d x = (k +1)!

The induction step is performed using integration by parts (which is exactly how you would derive the resultin the first place)!

∫

∞0

xk+1e−x d x = [xk+1×(−e−x

)]∞0+∫

∞0

e−x(k +1)xk d x

= 0+(k +1)∫∞

0e−x xk d x

= (k +1)k ! By the induction hypothesis.

= (k +1)!

Therefore, P(k +1) is true.



CONVERGENCE OF SEQUENCES

1. Prove that the cube root of an irrational number is irrational.

2. You are given that 3√

5 is irrational. Let an = 51

3n . Prove that the sequence (an) is irrational for all n ∈N.

3. Deduce that given m ∈ Z there exists a sequence of irrational numbers converging to m. (You may assumethat as n tend to infinity that 1

3n → 0.)

1. Let y be irrational. Suppose for a contradiction that y13 =

pq where p, q ∈Z and q ≠ 0. Then y = p3

q3 is a rational

number since p3 and q3 ≠ 0 are integers. This contradiction establishes the result.

2. Let P(n) be the proposition that an = 51

3n is irrational.

Base case When n = 1 we have a1 =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 ak = 51

3k is irrational.

Induction step We need to prove that ak+1 = 51

3k+1 is irrational. Now

51

3k+1 = (51

3k )

13

by the indices laws. By the induction hypothesis 51

3k is irrational and we have proved that the cube root

of an irrational number is irrational. Therefore, 51

3k+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 ofMathematical Induction P(n) is true for all n ∈N.

3. Let m ∈Z. Define un =man . Then as n→∞ it follows that an → 1 and so un →m.

This proof has combined several things we’ve seen before: direct proof, proof by contradiction, proof by inductionand then combining results to prove a new result.The fact we have assumed 1

3n → 0 as n →∞. This is used at A-level when we sum an infinite geometric seriesand deduce that

a(1− xn+1)

1− x→

a

1− x

when −1 < x < 1.FIBONACCI NUMBERS

Every positive integer can be represented as the sum of one or more distinct Fibonacci numbers in such a waythat the sum does not include any two consecutive Fibonacci numbers. (In fact, it is unique, but we don’t provethis, although it can be done by induction.)

An example and non-example of such expressions are:

100 = 89+8+3, yes100 = 89+8+2+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 moredistinct 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 with1 ≤ r ≤ k can be represented as the sum of distinct Fibonacci numbers and the sum does not include anytwo consecutive Fibonacci numbers.


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 berepresented as the sum of distinct Fibonacci numbers and the sum does not include any two consecutiveFibonacci numbers. Clearly, we just need to check the case r = k +1 (the induction hypothesis deals withthe 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 Fibonaccinumber, then there exists i such that Fi < k +1 < Fi+1.

Now k+1 = Fi +n for some positive integer n. If we can show that n is the sum of distinct Fibonacci numbersthat do not include Fi−1 then we are done.

Now n = k+1−Fi . The largest that n can be is k, so 0 < n ≤ k. (If n = 0, then k+1 = Fi and if n = k, then Fi = 1and so k +1 = 2 and the proposition is clearly two.) So we may now assume n = k +1−Fi < k. But by theinduction hypothesis we know that n is the sum of distinct Fibonacci numbers with the sum not includingtwo consecutive Fibonacci numbers.

We finally need to check that the sum n does not include Fi−1 for otherwise k +1 = Fi +n would contain theconsecutive Fibonacci numbers Fi and Fi−1.

Suppose Fi−1 was contained in the representation of n, then since Fi+1 = Fi +Fi−1 we have

k +1 = Fi +n = Fi +Fi−1+m = Fi+1+m > Fi+1.

This is a contraction, therefore, the representation of n does not contain Fi−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 mathematicalinduction implies that P(n) holds for all natural numbers n.

INFINITELY MANY PRIMES

The nth Fermat number, Fn , is defined byFn = 22n

+1

for n ∈N.

1. It is claimed thatF0F1F2 . . .Fn−1 = Fn −2 (A.1)

for all n ∈N.

Prove by induction that Equation (A.1) holds for all natural numbers.

2. 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−2 is 1.]

1. Let P(n) be the proposition thatF0F1F2 . . .Fn−1 = Fn −2.

Base case When n = 1 we need to check thatF0 = F1−2.

We have F0 = 3 and F1 = 5. Hence, since 3 = 5−2, P(1) is true.

Induction hypothesis Suppose P(k) is true for some k ≥ 1; that is,

F0F1F2 . . .Fk−1 = Fk −2.

Induction step We need to show thatF0F1F2 . . .Fk−1Fk = Fk+1−2


By the induction hypothesis

F0F1F2 . . .Fk−1Fk = (Fk −2)Fk

= (22k+1−2)(22k

+1)

= (22k−1)(22k

+1)

= 22k×22k

−1

= 22k+1−1

= 22k+1+1−2

= Fk+1−2.

Thus if P(k) is true it follows 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 it follows by the Principle

of Mathematical Induction that P(n) is true for all natural numbers n.

2. Suppose Fa and Fb have a common factor m. We may assume that a < b (otherwise just flip the two numbersround). By our proved result

F0F1F2 . . .Fa . . .Fb−1 = Fb −2.

So m divides F0F1F2 . . .Fa . . .Fb−1 and thus m divides Fb −2. This implies that Fb and Fb −2 have a commonfactor m. Since Fb is odd, it follows that the only common factor of Fb and Fb −2 is 1. Therefore, m = 1.

3. Any Fermat number must either be prime or can be factorised into prime factors. Since no two Fermat numbershave a common factor greater than 1, each Fermat number must consist of primes or prime factors different toall other Fermat numbers. Clearly there are an infinite number of Fermat numbers so there must be an infinitenumber of primes.

To prove the fact given in the question, suppose n and n −2 are divisible by p. Then n = pr for some integer rand n −2 = ps for some integer s. Subtracting gives 2 = pr −ps = p(r − s). Since p and r − s are integers eitherp = 1 and r − s = 2 or p = 2 and r − s = 1. Suppose p = 2, then an even number divides into the odd number n. Thiscontradiction shows that p = 1 giving the result.

PIRATES GOLD

Some pirates have got hold of some gold bars. The gold bars are 2×1 units in size. The pirates have boxes whichare 2×n in size.The pirates wonder how many different ways there are to arrange the gold in: 2×2, 2×3, 2, 2×5 and 2×n boxes.

One arrangement in a 2×8 box is:

It shoud be clear that that this actually can be proved by induction:

So, we get the relationPn = Pn−1+Pn−2

This relation can be proved by induction.


5 PROPOSITIONS THAT LOOK LIKE THEY CAN BE PROVED BY INDUCTION

GOLDBACH CONJECTURE

Every even integer greater than 2 can be expressed as the sum of two primes.

Comments:

◁ Examples are: 4 = 2+2, 10 = 3+7 = 5+5, 100 = 47+53.

◁ Not asserting uniqueness or distinct primes.

◁ Proved by hand up to 105 by Nils Pipping in 1938.

◁ Extended by computer up to 1018.

◁ Unproved generally as of when I wrote this sentence.


PART BSOLUTIONS AND FURTHER DETAILS

SOLUTION TO QUESTION 1:

Coins Final total

1 02 13 34 65 106 157 218 289 36

10 45

We might conjecture that: the final total is always 12 n(n−1). How do we prove this?

◁ There are 12 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 splitthe final total is always

f (n) =1

2n(n−1).

When n = 1, we have one coin, there is no second coin so the final total will be 0. Moreover, 12 ×1×0 = 0. Therefore, P(1) is true.

Suppose P(k) is true for all 2 ≤ i ≤ k with k ≥ 1. That is,

f (i) =1

2i(i −1).

We need to show that P(k +1) is true; that is,

f (k +1) =1

2k(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

f (i) =1

2i(i −1) f (k +1− i) =

1

2(k +1− i)(k − i).

So the final product will be

i(k +1− i)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

Product for current pile

+ f (i)+ f (k +1− i) = i(k +1− i)+1

2i(i −1)+

1

2(k +1− i)(k − i)

=1

2[2i k +2i −2i 2

+ i 2− i +k2

−ki +k − i − i k + i 2]

=1

2(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 naturalnumbers n.In a very real sense, the formal proof by induction is far less convincing that the argument using the graph!



12−22

+32−42

+ . . .+(−1)n−1n2= (−1)n−1 n(n+1)

2.

B. Solutions and further details Page 34

When n = 1, then

(−1)(1−1) 1(1+1)

2=

1×2

2= 1.

Also 12= 1. Therefore, P(1) is true.

Assume P(k) is true for some k ≥ 1; that is,

12−22

+32−42

+ . . .+(−1)k−1k2= (−1)k−1 k(k +1)

2.

We need to prove that P(k +1) is true; that is,

12−22

+32−42

+ . . .+(−1)k−1k2+(−1)k

(k +1)2= (−1)k (k +1)(k +2)

2.

We have

12−22

+32−42

+ . . .+(−1)k−1k2

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

First k terms

+(−1)k(k +1)2

= (−1)k−1 k(k +1)

2+(−1)k

(k +1)2 (by the induction hypothesis)

= (−1)k (k +1)

2[(−1)−1k +2(k +1)]

= (−1)k (k +1)

2[−k +2k +2]

= (−1)k (k +1)(k +2)

2.

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 numbersn.


◁ Suppose P(k) is true; that is k2+5k +1 is even. We need to show that P(k +1) is true; that is,

(k +1)2+5(k +1)+1 = k2

+2k +1+5k +5+1 = k2+7k +7

is even.Since

k2+7k +7 = k2

+5k +1+(2k +6) = k2+5k +1

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

Assumed to be even

+2(k +3)´¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶

even

it follows that k2+7k +7 is even. Indeed, since we’ve assumed P(k) is true, k2

+5k +1 is even and 2(k +3) is even and the sum oftwo even numbers is even. Therefore, P(k +1) is true.

◁ Suppose P(k) is true; that is k2+5k +1 is odd. We need to show that P(k +1) is true; that is,

(k +1)2+5(k +1)+1 = k2

+2k +1+5k +5+1 = k2+7k +7

is odd.Since

k2+7k +7 = k2

+5k +1+(2k +6) = k2+5k +1

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

Assumed to be odd

+2(k +3)´¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶

even

it follows that k2+7k +7 is odd. Indeed, by the induction hypothesis k2

+5k +1 is odd and 2(k +3) is even and the sum of an evennumber and an odd number is odd. Therefore, P(k +1) is true.


Suppose n is even. Then n = 2m for some m ∈Z. Thus

n2+5n+1 = (2m)

2+5(2m)+1 = 4m2

+10m+1 = 2(2m2+5m)+1

Since 2m2+5m is an integer, 2(2m2

+5m)+1 is odd and it follows that n2+5n+1 is odd.

Suppose that n is odd. Then n = 2m+1 for some m ∈Z. Then

n2+5n+1 = (2m+1)2

+5(2m+1)+1 = 4m2+14m+7 = 2(2m2

+7m+3)+1.

Since 2m2+7m+3 is an integer, 2(2m2

+7m+3)+1 is odd. Therefore, n2+5n+1 is odd.


The statement Fk+1 = Fk +Fk−1 is false when k = 1. Since Fk+1 = F2 = 1.


We have proved P(1); P(1) and P(2) implies P(3); P(2) and P(3) implies P(4).Since P(2) is not true and hence we have not verified that P(k) implies P(k +1).


Except, 412−41+41 = 412 clearly is not prime!

Let P(n) be the proposition that n2−n+41 is prime for all natural numbers n.

When n = 1, 12−1+41 = 41 is prime. Therefore, P(1) is true.

Suppose P(k) is true; that is, k2−k +41 is prime.

We need to prove that P(k +1) is true; that is, (k +1)2−(k +1)+41 is prime.

(k +1)2−(k +1)+41 = k2

+2k +1−k −1+41

= (k2−k +41)+2k.

Now (k2−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.


In fact we actually have the following:

Points Number of regions

1 12 23 44 85 166 317 57

In fact it gets worse, the actual formula is

(n

4)+(

n

2)+1.

You need to know that for a planar graph with V vertices, E edges and F faces, then V −E +F = 2. (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 anedge we need to select 2 of the n points. So to create two edges we need to select 4 of the n points. There are (

n4) ways to do

this.

– Therefore, there are n+(n4) vertices.

◁ We now need to determine E .

– Firstly there are n edges which are the circular arcs.

– Next, at each of the (n4) interior vertices there are 4 edges. This gives 4(n

4) edges.

– Next there are (n2) chords. (Each chord is determined by selecting 2 of the n points.)

– We have counted 2(n2)+4(n

4) edges. But each has been counted twice. So there are n+(n2)+2(n

4) edges.

◁ Euler’s formula now gives

F = 2+E −V = 2+n+(n

2)+2(

n

4)−(n+(

n

4)) = 2+(

n

2)+(

n

4).

◁ But remember we have counted the face exterior to the circle, so the number of regions is F −1.


martyn parker - meimei.org.uk/files/conference13/k2_2013.pdfthe sum of the interior angles in any...

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