Conjectures and proof - WordPress.com

Conjectures and proof Book page 24 - 30

copycgrahamphysicscom 2016

What is a conjecture bull A conjecture is used to describe a pattern in

mathematical terms

bull When a conjecture has been proved it becomes a theorem

bull There are many types of proofs some of which are - logical deductive reasoning - direct proof - induction


Alan Turing

bull English Mathematician who broke the enigma code in WW2

bull Alan Turing - Celebrating the life of a genius


Direct proof bull A direct proof can be checked by an established method

bull Using our knowledge of odd and even numbers we can proof that the sum of two odd numbers is always even

bull Odd number 2k + 1 Even number 2k

bull Let 119886 = 2119896 + 1 119886119899119889 119887 = 2119897 + 1

bull 2119896 + 1 + 2119897 + 1 = 2119896 + 2119897 + 2 = 2(119896 + 119897 + 1)

bull Let 119896 + 119897 + 1 = 119904 119904 isin 119937 119886 + 119887 = 2119904 which is an even number


Example 2 bull Consider the pattern 1 = 12 1 + 3 = 22 1 + 3 + 5 = 32 1 + 3 + 5 + 7 = 42

bull 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

bull Can we then say based on the first few lines of this pattern that

bull 1 + 3 + 5 + 7 + hellip + (2n ndash 1) = 1198992 bull ie the sum of the first n odd positive integers is a perfect

square


Solution bull 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

bull In this example we had a method to verify our guess

bull This will not always be the case


Exercise 1G page 24 all


Example


Example



Proof by induction copycgrahamphysicscom 2016

Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician

Blaise Pascal in 1653

bull The first rows are shown below


Pascalrsquos Triangle


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 Pascalrsquos 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 bull Proof by induction is a mathematical way to prove

statements about sequences

bull Mathematical Induction

bull Imagine n ndash dominos arranged in a row in the same way

If the first falls the second one will fall ndash 2nd step

N dominoes added at end will also fall ndash 3rd step

We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd


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)


Proof by induction

The domino analogy can be applied to mathematics


Strong and weak induction bull 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

bull They are NOT identical but they are equivalent

bull 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


bull 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

bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences

bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic

bull 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


Example 1 bull Proof by induction that the sum of the first n odd numbers

is 1198992

bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1

Solution

bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1


Using the 3 steps


Example 2



Example 3


Answer


Plan A didnrsquot work what is YOUR plan B


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 copycgrahamphysicscom 2016

Type 1 proof - easy

bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)

bull Show true for n = 1

bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)

bull RTS using the assumption true for n = k + 1

bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)

bull Rewrite left side adding the last two terms of the sequence

bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016

Type 2 proof ndash more difficult

bull Manipulate the expression

bull Rewrite sums to get the sum of the assumption

bull Insert bracket to get common factor

bull Regroup to factor

bull Add and subtract the same number

bull Keep in mind the expression you want to prove


Yesterdayrsquos example

bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8

bull Easier solution Rewrite the equation

bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix


Example

bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1


Another way to prove ndash read at home

bull You chose which method you prefer

bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0


Try this on your own

bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1


Another example

bull Prove by induction that 2119899 ge 119899 forall119899 ge 1


Can you do this bull 8119899 ge 1198993 forall119899 ge 1


An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4



More on sequence proofs


Example 5



Practice

Book page 29 exercise 1H 2


Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all

n ge 1 where n is an integer

bull Solution

bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer

bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8

bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer



Example 7



Practice

Book page 30 exercise 1I all


hellip continued


What is a conjecture bull A conjecture is used to describe a pattern in

mathematical terms

bull When a conjecture has been proved it becomes a theorem

bull There are many types of proofs some of which are - logical deductive reasoning - direct proof - induction


Alan Turing







bull Let 119886 = 2119896 + 1 119886119899119889 119887 = 2119897 + 1

bull 2119896 + 1 + 2119897 + 1 = 2119896 + 2119897 + 2 = 2(119896 + 119897 + 1)







square









Example


Example

























Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Alan Turing







bull Let 119886 = 2119896 + 1 119886119899119889 119887 = 2119897 + 1

bull 2119896 + 1 + 2119897 + 1 = 2119896 + 2119897 + 2 = 2(119896 + 119897 + 1)







square









Example


Example

























Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued





bull Let 119886 = 2119896 + 1 119886119899119889 119887 = 2119897 + 1

bull 2119896 + 1 + 2119897 + 1 = 2119896 + 2119897 + 2 = 2(119896 + 119897 + 1)







square









Example


Example

























Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued






square









Example


Example

























Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued









Example


Example

























Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued




Example


Example

























Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Example


Example

























Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Example

























Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued

























Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued
























Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued























Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued



















Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued

















Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued












Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued




Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Proof by induction














is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued













is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued








is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued



is 1198992


Solution



Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Using the 3 steps


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Example 2



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued



Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Example 3


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Answer




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued




What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


What do you know



Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Type 1 proof - easy





bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued















Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued







Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Example










Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued









Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued





Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Another example





An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued




An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


An easy example

bull 5 + 10 + 15 + ⋯ 5119899 =5

2119899(119899 + 1)


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Example 4





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued





Example 5



Practice





bull Solution






Example 7



Practice



hellip continued




Example 5



Practice





bull Solution






Example 7



Practice



hellip continued


Example 5



Practice





bull Solution






Example 7



Practice



hellip continued



Practice





bull Solution






Example 7



Practice



hellip continued


Practice





bull Solution






Example 7



Practice



hellip continued




bull Solution






Example 7



Practice



hellip continued



Example 7



Practice



hellip continued


Example 7



Practice



hellip continued



Practice



hellip continued


Practice



hellip continued


hellip continued


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Conjectures and proof - WordPress.com