Conjectures and proof. Book page 24-30

Size: px

Start display at page:

Download "Conjectures and proof. Book page 24-30"

Erick McKinney
5 years ago
Views:

1 Conjectures and proof Book page 24-30

2 What is a conjecture? A conjecture is used to describe a pattern in mathematical terms When a conjecture has been proved, it becomes a theorem There are many types of proofs, some of which are - logical deductive reasoning - direct proof - induction

3 Alan Turing English Mathematician who broke the enigma code in WW2 Alan Turing - Celebrating the life of a genius

4 Direct proof A direct proof can be checked by an established method Using our knowledge of odd and even numbers we can proof that the sum of two odd numbers is always even Odd number: 2k + 1, Even number: 2k Let a = 2k + 1 and b = 2l + 1 2k l + 1 = 2k + 2l + 2 = 2(k + l + 1) Let k + l + 1 = s, s Z a + b = 2s, which is an even number

5 Example 2 Consider the pattern: 1 = = = = 4 2 The pattern shows that the sum of the first two positive odd integers is a perfect square, the sum of the first three positive odd integers is a perfect square and the sum of the first four positive odd integers is a perfect square. Can we then say, based on the first few lines of this pattern, that: (2n 1) = n 2 i.e., the sum of the first n odd positive integers is a perfect square,?

6 Solution This can be checked, as the positive odd integers form an arithmetic progression with a = 1 and d = 2. The sum of the first n terms is given by In this example we had a method to verify our guess. This will not always be the case.

7 Exercise 1G page 24 all

8 Example

9 Example

11 Proof by induction

12 Problem: direct proof is not useful Pascal s triangle is named after the French Mathematician Blaise Pascal in 1653 The first rows are shown below

13 Pascal s Triangle

14 Things to think about a) Find the sum of the numbers in I row 0 ii row 1 iii row 2 iv row 3 v row 4 b) What do you suspect is the sum of the numbers in row n? c) Suppose your suggestion in (b) is true for the nth row of Pascal s triangle. i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true?

Proof by induction Proof by induction is a mathematical way

Imagine n dominos arranged in a row in the same way If the

added at end, will also fall 3 rd step We can add as many

15 Proof by induction Proof by induction is a mathematical way to prove statements about sequences Mathematical Induction Imagine n dominos arranged in a row in the same way If the first falls, the second one will fall 2 nd step N dominoes added at end, will also fall 3 rd step We can add as many dominos as we like, they will fall if the 1 st is tilted towards the 2 nd

16 This process can be summarized Step 1: the first expression must be true (the first domino falls) Step 2: assuming that a general expression is true (assume that some domino in the series falls) Step 3: prove that the next expression is true (prove that the next domino in the series falls) Step 4: if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall).

17 Proof by induction The domino analogy can be applied to mathematics

18 Strong and weak induction Weak induction uses "if p(k) is true then p(k+1) is true" while in strong induction you use "if p(i) is true for all i less than or equal to k then p(k+1) is true", where p(k) is some statement depending on the positive integer k They are NOT "identical" but they are equivalent. It is easy to see that if weak induction is true then strong induction is true: if you know that statement p(i) is true for all i less than or equal to k, then you know that it is true, in particular, for i=k and can use simple induction.

19 It is harder to prove, but still true, that if strong induction is true, then weak induction is true. That is what we mean by "equivalent". The reason this distinction remains is that it serves a pedagogical purpose. The first proofs by induction that we teach are usually sequences The proofs of these naturally suggest "weak" induction, which students learn as a pattern to mimic. Later, we teach more difficult proofs where that pattern no longer works. To give a name to the difference, we call the new pattern "strong induction" so that we can distinguish between the methods when presenting a proof in lecture. Then we can tell a student "try using strong induction", which is more helpful than just "try using induction".

20 Example 1 Proof by induction that the sum of the first n odd numbers is n 2 That is n 1 = n 2, n 1 Solution First we need to state what our proposition is. We do this as follows: Let P(n) be the proposition that n 1 = n 2, n 1

21 Using the 3 steps

22 Example 2

24 Example 3

25 Answer

26 Plan A didn t work, what is YOUR plan B?

27 What do you know? Three steps: show true for base case Assume true for n = k RTS true for n = k+1 Two different types 1. Use sequences and add the next term to show n = k + 1 is true

Type 1 proof - easy Prove that 1 + 5 + 9 + + (4n 3) = n(2n 1) Show true for n = 1 Assume true for n = k 4k 3 = k (2k 1) RTS using the assumption true for n = k + 1 4

28 Type 1 proof - easy Prove that (4n 3) = n(2n 1) Show true for n = 1 Assume true for n = k 4k 3 = k (2k 1) RTS using the assumption true for n = k (k + 1) 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1) Rewrite left side adding the last two terms of the sequence (4k 3) + (4(k + 1) 3) and replace (4k 3) with assumption

29 Type 2 proof more difficult Manipulate the expression Rewrite sums to get the sum of the assumption Insert bracket to get common factor Regroup to factor Add and subtract the same number Keep in mind the expression you want to prove

30 Yesterday s example Prove n 2 > 7n + 1, n 8 Easier solution: Rewrite the equation n 2 7n 1 > 0 Flight Of The Phoenix

31 Example Prove by induction that3 n > 1 + 2n, n > 1

32 Another way to prove read at home You chose which method you prefer 3 n 1 + 2n, n Z, n 0

33 Try this on your own Prove by induction that 2 n 1 + n, n 1

34 Another example Prove by induction that 2 n n, n 1

35 Can you do this? 8 n n 3, n 1

36 An easy example n = 5 2 n(n + 1)

37 Example 4

39 More on sequence proofs

40 Example 5

42 Practice Book page 29 exercise 1H # 2

43 Example 6 Prove by induction that 9 n 1 is divisible by 8 for all n 1, where n is an integer Solution Let P(n) be the proposition that 9 n 1 is divisible by 8 for all n 1, where n is an integer Step 1: This is true when n = 1 since = 8, which is divisible by 8 Step 2: Assume P(n) is true for n = k. Assume true for 9 k 1 = 8m, where m is an integer

45 Example 7

47 Practice Book page 30 exercise 1I all

48 continued

Discrete Mathematics. Spring 2017

Discrete Mathematics. Spring 2017 Discrete Mathematics Spring 2017 Previous Lecture Principle of Mathematical Induction Mathematical Induction: rule of inference Mathematical Induction: Conjecturing and Proving Climbing an Infinite Ladder