Mathematical Induction COM1022 Functional Programming Techniques - PowerPoint PPT Presentation

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 Proving statements about functions 1 The principle of induction 2 Examples 3 Take and drop 4 Exercises (in place of lab) 5 Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 2 / 13

Proving statements about functions Proving statements about functions oddsum 0 = 0 oddsum n = oddsum (n-1) + (2*n - 1) oddsum 0 = ? oddsum 1 = ? oddsum 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 = n 2 ” ∀ 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 ) oddsum 0 = 0 2 P ( 0 ) is true: 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. oddsum ( n + 1 ) = oddsum n + ( 2 n + 1 ) n 2 = + ( 2 n + 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 ( 5 n − 1 ) We can prove that 5 n − 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: 2 0 + 2 1 + ... + 2 n = 2 ( n + 1 ) − 1 for all n 1 2 + 2 2 + ... + n 2 = ( n ( n + 1 )( 2 n + 1 )) for all n 6 4 n + 6 n − 1 is divisible by 9 for all n Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 13 / 13