FP1 Chapter 6 – Proof By Induction Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 31st March 2015

What is Proof by Induction? We tend to use proof by induction whenever we want to show some property holds for all integers (usually positive). Example Show that 𝑖=1 𝑛 𝑖 = 1 2 𝑛 𝑛+1 for all 𝑛∈ℕ. Showing it’s true for a few values of 𝑛 is not a proof. We need to show it’s true for ALL values of 𝑛. If it’s true for 𝑛=1, it must be true for 𝑛=2. If it’s true for 𝑛=3, it must be true for 𝑛=4. 𝒏=𝟏 𝒏=𝟐 𝒏=𝟑 𝒏=𝟒 𝒏=𝟓 … If it’s true for 𝑛=2, it must be true for 𝑛=3. Suppose we could prove the above is true for the first value of 𝑛, i.e. 𝑛=1. And so on. Hence the statement must be true for all 𝑛. Suppose that if we assumed it’s true for 𝑛=𝑘, then we could prove it’s true for 𝑛=𝑘+1. Then:

Chapter Overview Summation Proofs Divisibility Proofs We will use Proof of Induction for 4 different types of proof: 1 Summation Proofs Show that 𝑖=1 𝑛 𝑖 = 1 2 𝑛 𝑛+1 for all 𝑛∈ℕ. 2 Divisibility Proofs Prove that 𝑛 3 −7𝑛+9 is divisible by 3 for all 𝑛∈ ℤ + . 3 Recurrence Relation Proofs Given that 𝑢 𝑛+1 =3 𝑢 𝑛 +4, 𝑢 1 =1, prove that 𝑢 𝑛 = 3 𝑛 −2 Prove that 1 −1 0 3 𝑛 = 1 1− 2 𝑛 0 2 𝑛 for all 𝑛∈ ℤ + . 4 Matrix Proofs Bro Tip: Recall that ℤ is the set of all integers, and ℤ + is the set of all positive integers. Thus ℕ= ℤ + .

The Four Steps of Induction For all these types the process of proof is the same: ! Proof by induction: Step 1: Basis: Prove the general statement is true for 𝑛=1. Step 2: Assumption: Assume the general statement is true for 𝑛=𝑘. Step 3: Inductive: Show that the general statement is then true for 𝑛=𝑘+1. Step 4: Conclusion: The general statement is then true for all positive integers 𝑛. (Step 1 is commonly known as the ‘base case’ and Step 3 as the ‘inductive case’)

Type 1: Summation Proofs Step 1: Basis: Prove the general statement is true for 𝑛=1. Step 2: Assumption: Assume the general statement is true for 𝑛=𝑘. Step 3: Inductive: Show that the general statement is then true for 𝑛=𝑘+1. Step 4: Conclusion: The general statement is then true for all positive integers 𝑛. Show that 𝑖=1 𝑛 (2𝑖−1) = 𝑛 2 for all 𝑛∈ℕ. 1 Basis Step ? When 𝑛=1, 𝐿𝐻𝑆= 𝑖=1 1 2𝑖−1 =2 1 −1 =1 𝑅𝐻𝑆= 1 2 =1. 𝐿𝐻𝑆=𝑅𝐻𝑆 so summation true for 𝑛=1. 2 Assumption ? Assume summation true for 𝑛=𝑘. So 𝑖=1 𝑘 (2𝑖−1) = 𝑘 2 3 Inductive ? When 𝑛=𝑘+1: 𝑖=1 𝑘+1 2𝑖−1 = 𝑖=1 𝑘 2𝑖−1 + 2 𝑘+1 −1 = 𝑘 2 +2𝑘+1 = 𝑘+1 2 Hence true when 𝑛=𝑘+1. 4 Conclusion ? Since true for 𝑛=1 and if true for 𝑛=𝑘, true for 𝑛=𝑘+1, ∴ true for all 𝑛.

More on the ‘Conclusion Step’ “As result is true for 𝑛=1, this implies true for all positive integers and hence true by induction.” I lifted this straight from a mark scheme, hence use this exact wording! The mark scheme specifically says: (For method mark) Any 3 of these seen anywhere in the proof: “true for 𝒏=𝟏” “assume true for 𝒏=𝒌” “true for 𝒏=𝒌+𝟏” “true for all 𝒏/positive integers”

Test Your Understanding Edexcel FP1 Jan 2010 ?

Exercise 6A All questions except Q2.

Type 2: Divisibility Proofs Step 1: Basis: Prove the general statement is true for 𝑛=1. Step 2: Assumption: Assume the general statement is true for 𝑛=𝑘. Step 3: Inductive: Show that the general statement is then true for 𝑛=𝑘+1. Step 4: Conclusion: The general statement is then true for all positive integers 𝑛. Prove by induction that 3 2𝑛 +11 is divisible by 4 for all positive integers 𝑛. 1 Basis Step ? Let 𝑓 𝑛 = 3 2𝑛 +11, where 𝑛∈ ℤ + . 𝑓 1 = 3 2 +11=20 which is divisible by 4. 2 Assumption ? Assume that for 𝑛=𝑘, 𝑓 𝑘 = 3 2𝑘 +11 is divisible by 4 for 𝑘∈ ℤ + . …

Type 2: Divisibility Proofs Step 1: Basis: Prove the general statement is true for 𝑛=1. Step 2: Assumption: Assume the general statement is true for 𝑛=𝑘. Step 3: Inductive: Show that the general statement is then true for 𝑛=𝑘+1. Step 4: Conclusion: The general statement is then true for all positive integers 𝑛. Prove by induction that 3 2𝑛 +11 is divisible by 4 for all positive integers 𝑛. 3 Inductive ? Method 1 (the textbook’s method, which I don’t like) Find the difference between successive terms and show this is divisible by the number of interest. 𝑓 𝑘+1 = 3 2 𝑘+1 +11 = 3 2𝑘 ⋅ 3 2 +11 =9 3 2𝑘 +11 ∴𝑓 𝑘+1 −𝑓 𝑘 = 9 3 2𝑘 +11 − 3 2𝑘 +11 =8 3 2𝑘 =4 2 3 2𝑘 ∴𝑓 𝑘+1 =𝑓 𝑘 +4 2 3 2𝑘 Therefore 𝑓(𝑛) is divisible by 4 when 𝑛=𝑘+1. Bro Tip: Putting in terms of 3 2𝑘 allows us to better compare with the assumption case since the power is now the same. Method: You need to explicitly factor out 4 to show divisibility by 4. Method: Then write 𝑓 𝑘+1 back in term of 𝑓(𝑘) and the difference. We know both the terms on the RHS are divisible by 4. 4 Conclusion ? Since true for 𝑛=1 and if true for 𝑛=𝑘, true for 𝑛=𝑘+1, ∴ true for all 𝑛.

Type 2: Divisibility Proofs Prove by induction that 3 2𝑛 +11 is divisible by 4 for all positive integers 𝑛. 3 Inductive Method 2 (presented as ‘Alternative Method’ in mark schemes) Directly write 𝑓(𝑘+1) in form 𝑓 𝑘 +𝑑(…) by ‘splitting off’ 𝑓(𝑘) or some multiple of it, where 𝑑 is the divisibility number. 𝑓 𝑘+1 = 3 2 𝑘+1 +11 = 3 2𝑘 ⋅ 3 2 +11 =9 3 2𝑘 +11 = 3 2𝑘 +11+8 3 2𝑘 =𝑓 𝑘 +4 2 3 2𝑘 Therefore 𝑓(𝑛) is divisible by 4 when 𝑛=𝑘+1. ? Method: We have separated 𝟗 𝟑 𝟐𝒌 into 3 2𝑘 +8 3 2𝑘 because it allowed us to then replace 3 2𝑘 +11 with 𝑓(𝑘) This method is far superior for two reasons: The idea of ‘separating out 𝑓(𝑘)’ is much more conceptually similar to both summation proofs and (when we encounter it) matrix proofs, whereas using 𝑓 𝑘+1 −𝑓(𝑘) is specific to divisibility proofs. Thus we needn’t see divisibility proofs as a different method from the other types. We’ll encounter a question next where we need to get in the form 𝑓 𝑘+1 =𝑎 𝑓 𝑘 +𝑑(…), i.e. we in fact separate out a multiple of 𝑓(𝑘). Thus causes absolute havoc for Method 1 as our difference 𝑓 𝑘+1 −𝑓(𝑘) has to refer to 𝑓(𝑘), which is mathematically inelegant.

Type 2: Divisibility Proofs Prove by induction that 8 𝑛 − 3 𝑛 is divisible by 5. 2 Assumption ? Assume that for 𝑛=𝑘, 𝑓 𝑘 = 8 𝑘 − 3 𝑘 is divisible by 5 for 𝑘∈ ℤ + . 3 Inductive ? Using Method 2 When 𝑛=𝑘+1: 𝑓 𝑘+1 = 8 𝑘+1 − 3 𝑘+1 =8 8 𝑘 −3 3 𝑘 =3 8 𝑘 − 3 𝑘 +5 8 𝑘 =3𝑓 𝑘 +5( 8 𝑘 ) therefore true for 𝑛=𝑘+1. Method: We have 3 lots of 8 𝑘 and 3 𝑘 we can split off.

Test Your Understanding Step 1: Basis: Prove the general statement is true for 𝑛=1. Step 2: Assumption: Assume the general statement is true for 𝑛=𝑘. Step 3: Inductive: Show that the general statement is then true for 𝑛=𝑘+1. Step 4: Conclusion: The general statement is then true for all positive integers 𝑛. Prove by induction that 𝑛 3 −7𝑛+9 is divisible by 3 for all positive integers 𝑛. 1 Basis Step ? Let 𝑓 𝑛 = 𝑛 3 −7𝑛+9, where 𝑛∈ ℤ + . 𝑓 1 =3 which is divisible by 3. 2 Assumption ? Assume that for 𝑛=𝑘, 𝑓 𝑘 = 𝑘 3 −7𝑘+9 is divisible by 3 for 𝑘∈ ℤ + . 3 Inductive ? Method 1: 𝑓 𝑘+1 = 𝑘+1 3 −7 𝑘+1 +9 =… = 𝑘 3 +3 𝑘 2 −4𝑘+3 ∴𝑓 𝑘+1 −𝑓 𝑘 =…=3 𝑘 2 +𝑘−2 ∴𝑓 𝑘+1 =𝑓 𝑘 +3 𝑘 2 +𝑘−2 Therefore 𝑓(𝑛) is divisible by 3 when 𝑛=𝑘+1. Method 2: 𝑓 𝑘+1 = 𝑘+1 3 −7 𝑘+1 +9 =… = 𝑘 3 +3 𝑘 2 −4𝑘+3 = 𝑘 3 −7𝑘+9+3 𝑘 2 +3𝑘−6 =𝑓 𝑘 +3 𝑘 2 +𝑘−2 Therefore 𝑓(𝑛) is divisible by 3 when 𝑛=𝑘+1. 4 Conclusion ? Since true for 𝑛=1 and if true for 𝑛=𝑘, true for 𝑛=𝑘+1, ∴ true for all 𝑛.

Exercise 6B Page 130. All questions.

Type 3: Recurrence Relation Proofs Step 1: Basis: Prove the general statement is true for 𝑛=1. Step 2: Assumption: Assume the general statement is true for 𝑛=𝑘. Step 3: Inductive: Show that the general statement is then true for 𝑛=𝑘+1. Step 4: Conclusion: The general statement is then true for all positive integers 𝑛. Notice that we’re trying to prove a position-to-term formula from a term-to-term one. Given that 𝑢 𝑛+1 =3 𝑢 𝑛 +4 and 𝑢 1 =1, prove by induction that 𝑢 𝑛 = 3 𝑛 −2 1 Basis Step ? When 𝑛=1, 𝑢 1 = 3 1 −2=1 as given. IMPORTANT NOTE: The textbook gets this wrong (bottom of Page 131), saying that 𝒖 𝟐 is required. I checked a mark scheme – it isn’t. Bro Tip: You need to carefully reflect on what statement you are trying to prove, i.e. 𝑢 𝑛 = 3 𝑛 −2, and what statement is GIVEN (and thus already know to be true), i.e. 𝑢 𝑛+1 =3 𝑢 𝑛 +4 2 Assumption ? Assume that for 𝑛=𝑘, 𝑢 𝑘 = 3 𝑘 −2 is true for 𝑘∈ ℤ + . Then 𝑢 𝑘+1 =3 𝑢 𝑘 +4 =3 3 𝑘 −2 +4 = 3 𝑘+1 −6+4 = 3 𝑘+1 −2 3 Inductive ? Method: Sub in your assumption expression into the recurrence. Bro Tip: For harder questions, write out what you’re trying to prove in advance, i.e. that we want to get 3 𝑘+1 −2, , to help guide your manipulation. 4 Conclusion ? Since true for 𝑛=1 and if true for 𝑛=𝑘, true for 𝑛=𝑘+1, ∴ true for all 𝑛.

Test Your Understanding Edexcel FP1 Jan 2011 ?

(Warning: slightly different to before) Recurrence Relations based on two previous terms Step 1: Basis: Prove the general statement is true for 𝑛=1. Step 2: Assumption: Assume the general statement is true for 𝑛=𝑘. Step 3: Inductive: Show that the general statement is then true for 𝑛=𝑘+1. Step 4: Conclusion: The general statement is then true for all positive integers 𝑛. Given that 𝑢 𝑛+2 =5 𝑢 𝑛+1 −6 𝑢 𝑛 and 𝑢 1 =13, 𝑢 2 =35, prove by induction that 𝑢 𝑛 = 2 𝑛+1 + 3 𝑛+1 1 Basis Step ? When 𝑛=1, 𝑢 1 = 2 2 + 3 2 =13 as given. When 𝑛=2, 𝑢 2 = 2 3 + 3 3 =35 as given. We have two base cases so had to check 2 terms! 2 Assumption ? Assume that for 𝑛=𝑘 and 𝑛=𝑘+1, 𝑢 𝑘 = 2 𝑘+1 + 3 𝑘+1 and 𝑢 𝑘+1 = 2 𝑘+2 + 3 𝑘+2 is true for 𝑘∈ ℤ + . We need two previous instances of 𝑛 in our assumption case now. 3 Inductive ? Let 𝑛=𝑘+1 (Brominder: it’s worth noting in advance we’re aiming for 𝑢 𝑘+1 = 2 𝑘+1 +1 + 3 𝑘+1 +1 ) Then 𝑢 𝑘+2 =5 𝑢 𝑘+1 −6 𝑢 𝑘 =5 2 𝑘+1 + 3 𝑘+1 −6 2 𝑘+1 + 3 𝑘+1 =… = 2 𝑘+3 + 3 𝑘+3 = 2 𝑘+2+1 + 3 𝑘+2+1 4 Conclusion ? (Warning: slightly different to before) Since true for 𝑛=1 and 𝑛=2 and if true for 𝑛=𝑘 and 𝑛=𝑘+1, true for 𝑛=𝑘+2, ∴ true for all 𝑛.

Exercise 6C Page 132. Q1, 3, 5, 7

Type 4: Matrix Proofs Step 1: Basis: Prove the general statement is true for 𝑛=1. Step 2: Assumption: Assume the general statement is true for 𝑛=𝑘. Step 3: Inductive: Show that the general statement is then true for 𝑛=𝑘+1. Step 4: Conclusion: The general statement is then true for all positive integers 𝑛. Prove by induction that 1 −1 0 2 𝑛 = 1 1− 2 𝑛 0 2 𝑛 for all 𝑛∈ ℤ + . 1 Basis Step ? When 𝑛=1, 𝐿𝐻𝑆= 1 −1 0 2 1 = 1 −1 0 2 𝑅𝐻𝑆= 1 1− 2 1 0 2 1 = 1 −1 0 2 . As 𝐿𝐻𝑆=𝑅𝐻𝑆, true for 𝑛=1. Assume true 𝑛=𝑘, i.e. 1 −1 0 3 𝑘 = 1 1− 2 𝑘 0 2 𝑘 2 Assumption ? When 𝑛=𝑘+1, 1 −1 0 2 𝑘+1 = 1 −1 0 2 𝑘 1 −1 0 2 = 1 1− 2 𝑘 0 2 𝑘 1 −1 0 2 =…= 1 1− 2 𝑘+1 0 2 𝑘+1 Therefore true when 𝑛=𝑘+1. 3 Inductive ? Method: As with summation and divisibility proofs, it’s just a case of ‘separating off’ the assumption expression. 4 Conclusion ? Since true for 𝑛=1 and if true for 𝑛=𝑘, true for 𝑛=𝑘+1, ∴ true for all 𝑛.

Test Your Understanding Step 1: Basis: Prove the general statement is true for 𝑛=1. Step 2: Assumption: Assume the general statement is true for 𝑛=𝑘. Step 3: Inductive: Show that the general statement is then true for 𝑛=𝑘+1. Step 4: Conclusion: The general statement is then true for all positive integers 𝑛. Prove by induction that −2 9 −1 4 𝑛 = −3𝑛+1 9𝑛 −𝑛 3𝑛+1 for all 𝑛∈ ℤ + . 1 Basis Step ? When 𝑛=1, 𝐿𝐻𝑆= −2 9 −1 4 1 = −2 9 −1 4 𝑅𝐻𝑆= −3+1 9 −1 3+1 = −2 9 −1 4 . As 𝐿𝐻𝑆=𝑅𝐻𝑆, true for 𝑛=1. Assume true 𝑛=𝑘, i.e. −2 9 −1 4 𝑘 = −3𝑘+1 9𝑘 −𝑘 3𝑘+1 2 Assumption ? When 𝑛=𝑘+1, −2 9 −1 4 𝑘+1 = −2 9 −1 4 𝑘 −2 9 −1 4 = −3𝑘+1 9𝑘 −𝑘 3𝑘+1 −2 9 −1 4 =…= −3𝑘−2 9𝑘+9 −𝑘−1 3𝑘+4 = −3 𝑘+1 +1 9(𝑘+1) −(𝑘+1) 3 𝑘+1 +1 Therefore true when 𝑛=𝑘+1. 3 Inductive ? 4 Conclusion ? Since true for 𝑛=1 and if true for 𝑛=𝑘, true for 𝑛=𝑘+1, ∴ true for all 𝑛.

Exercise 6D Page 134. All questions.