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Mathematical Induction

COM1022 Functional Programming Techniques Professor Steve Schneider

University of Surrey

Semester 2, 2010 – Week 09

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 1 / 13

Outline

1

Proving statements about functions

2

The principle of induction

3

Examples

4

Take and drop

5

Exercises (in place of lab)

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 2 / 13

Proving statements about functions

Proving statements about functions

• ddsum 0 = 0
• ddsum n = oddsum (n-1) + (2*n - 1)
• ddsum 0 = ?
• ddsum 1 = ?
• ddsum 2 = ?

... Can we spot a pattern? For all n: oddsum n = ? Can we prove it?

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 3 / 13

Proving statements about functions

The general form

Let P(n) be a statement (predicate) about n. e.g. P(n) ≡ ”oddsum n = n2” ∀n.P(n) asserts that P(n) is true for all n. How can we prove this? We can’t check each n in turn, because there are infinitely many of them.

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 4 / 13

The principle of induction

The principle of induction

If P(0) and P(n) ⇒ P(n + 1) for all n Then we can conclude ∀n.P(n) An infinite sequence of dominoes...

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 5 / 13

The principle of induction

Base case: P(0)

P(0) is true:

• ddsum 0 = 02

The initial case, that is proved directly, is called the base case. It often corresponds to the base case of the functional definition. It is typically established by examining the statement P(0) directly. [The base case can be some number other than 0. The result will hold for all numbers from the base case.]

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 6 / 13

The principle of induction

Inductive step: P(n) ⇒ P(n + 1)

We have to show, for any n, that if P(n) is true, then P(n + 1) is also true. So assume, for some arbitrary n, that P(n) is true. Let’s see if we can prove that P(n + 1) is true.

• ddsum (n + 1)

=

• ddsum n + (2n + 1)

= n2 + (2n + 1) = (n + 1)2 So if P(n) is true, then P(n + 1) is also true Hence by induction, ∀n.P(n)

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 7 / 13

Examples

Example: 4 divides (5n − 1)

We can prove that 5n − 1 is always divisible by 4. Base case: n = 0 Inductive step: assume true for n, prove for n + 1

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 8 / 13

Examples

Example: sigma n

sigma 0 = 0 sigma n = n + sigma (n-1) Q(n) : sigma n = n(n+1)

2

Base case: show Q(0) Inductive step: show Q(n) ⇒ Q(n + 1) Then conclude ∀n.Q(n)

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 9 / 13

Take and drop

take and drop

take 0 xs = [] take n [] = [] take n (x:xs) = x:(take (n-1) xs) drop 0 xs = xs drop n [] = [] drop n (x:xs) = drop (n-1) xs Is it true that (take n xs) ++ (drop n xs) = xs? What if n is greater than the length of xs?

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 10 / 13

Take and drop

a property of take and drop

Want to prove that take n xs ++ drop n xs = xs for all n. We can prove this by induction: Base case: (take 0 xs) ++ (drop 0 xs) = [] ++ xs = xs Inductive step: need to consider [] and x:xs:

(take (n+1) []) ++ (drop (n+1) []) = [] ++ [] = [] take (n+1) (x:xs) ++ drop (n+1) (x:xs) = x:(take n xs) ++ (drop n xs) = x:((take n xs) ++ (drop n xs)) = x:(xs) = x:xs

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 11 / 13

Take and drop

Summary

The principle of (mathematical) induction To prove some property P(n) is true for all n: Show the base case: P(0) Show the inductive step: P(n) ⇒ P(n + 1)

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 12 / 13

Exercises (in place of lab)

Exercises

These exercises are instead of a lab for this week. Please try them in advance of next week. Prove each of the following by induction: 20 + 21 + ... + 2n = 2(n+1) − 1 for all n 12 + 22 + ... + n2 = (n(n+1)(2n+1))

6

for all n 4n + 6n − 1 is divisible by 9 for all n

Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 13 / 13