Common errors: Induction

Marking time

Proofs Workshop

How many marks (out of 10) would you give this student's effort? And what would need to change in order to get full marks.

Problem:

Prove or disprove the following proposition:

1 1 1 1 + +… + n = 2 ! n for all positive integers n.

2 4 2 2 Solution:

Let P(n) be the statement: 1 1 1 1 + +… + n = 2 ! n 2 4 2 2

Assume the statement P(k) is true.

P(k) is the statement: P(k+1) is the statement:

Subtraction gives

1 1 1 1 + +… + k = 2 ! k 2 4 2 2

1 1 1 1 1 + +… + k + k+1 = 2 ! k+1 2 4 2 2 2 1 1 " 1% = 2 ! !

2 ! $ ' 2k+1 2k+1 # 2k & =

1 1 1 ! k+1 = k+1 k 2 2 2

which is true.

Hence, by induction, P(n) is true for all positive integers n.

Shirleen Stibbe

http://www.shirleenstibbe.co.uk

Marking time – tutors' view

Proofs Workshop

Mark: 1/10.

What's wrong with it: Method:

a) No initial step - so we have no 'starting off' point for n = 1. (lose 4 marks) b) For the inductive step, he assumed that both P(k) and P(k+1) are true, and proved a true (lose 4 marks) statement. The correct method is to assume P(k) is true, and prove P(k) ⇒ P(k+1). eg "Assume the statement P(k) is true, so that

1 1 1 1 + +… + k = 2 ! k 2 4 2 2

Then, for n = k + 1, we have

1 1 1 1 1 1 + +… + k + k+1 = 2 ! k + k+1 2 4 2 2 2 2 = 2!

by assumption

2 1 1 + = 2 ! k+1 2k+1 2k+1 2

This is the statement P(k+1) and completes the inductive step, P(k) ⇒ P(k+1)."

Conclusion:

Wrong! The statement is actually false. (lose 1 mark)

The Fix: Problem: Prove or disprove the following proposition:

1 1 1 1 + +… + n = 2 ! n for all positive integers n.

2 4 2 2 Solution:

1 1 3 , RHS = 2 ! = " LHS. 2 2 2 The counter-example shows that the proposition is false.

When n = 1, LHS =

Comment:

1 1 1 1 + +… + n = 2 ! n .

2 4 2 2 We have disproved the proposition by finding a particular n for which the statement is false; ie we have proved !n ¬( P(n)) which is equivalent to ¬(!n P(n)) . (See Logic and Proof Structures, p2, Counter-examples) The proposition is of the form !n P(n) , where P(n) is the statement

We could have gone down a different route: We proved (in the first session) that 1+ a + a 2 + ... + a n!1 =

a n !1 for all positive integers n, when a ≠1. a !1

k

!1$ 1 By substituting # & = k 2 2 " %

for a k , and summing up to n instead of n −1, we have

1 !1 " 1 % n+1 1 1 1 1 2 1+ + 2 +… + n = = ! 2 $ n+1 !1' = 2 ! n 1 2 2 2 2 #2 & !1 2 Then

! 1 1 !

1 1 1 1$ 1$ 1 + +… + n = #1+ + 2 +… + n & '1 = # 2 ' n & '1 = 1' n for all positive integers n 2 4 2 2 % 2 " 2 2 " 2 %

This shows that the proposition is false.

(

Note: What we have proved here is !n ¬ P(n)

)

(

)

which is equivalent to ¬ !n P(n) .

(

)

This is sufficient but not necessary to prove ¬ !n P(n) .

Shirleen Stibbe

http://www.shirleenstibbe.co.uk