Proving statements about functions Proving statements about functions Mathematical Induction oddsum 0 = 0 oddsum n = oddsum (n-1) + (2*n - 1) COM1022 Functional Programming Techniques oddsum 0 = ? Professor Steve Schneider oddsum 1 = ? University of Surrey oddsum 2 = ? ... Semester 2, 2010 – Week 09 Can we spot a pattern? For all n: oddsum n = ? Can we prove it? Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 1 / 13 Professor Steve Schneider Proving statements about functions Mathematical Induction Semester 2, 2010 – Week 09 3 / 13 Semester 2, 2010 – Week 09 5 / 13 The principle of induction The general form The principle of induction If P(0) and P(n) ⇒ P(n + 1) for all n Let P(n) be a statement (predicate) about n. e.g. P(n) ≡ ”oddsum n = n2 ” Then we can conclude ∀n.P(n) ∀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. An infinite sequence of dominoes... Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 4 / 13 Professor Steve Schneider Mathematical Induction The principle of induction The principle of induction Inductive step: P(n) ⇒ P(n + 1) Base case: P(0) P(0) is true: We have to show, for any n, that if P(n) is true, then P(n + 1) is also true. oddsum 0 = 02 So assume, for some arbitrary n, that P(n) is true. Let’s see if we can prove that P(n + 1) 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. oddsum (n + 1) = oddsum n + (2n + 1) It is typically established by examining the statement P(0) directly. = n2 = (n + 1) [The base case can be some number other than 0. The result will hold for all numbers from the base case.] + (2n + 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 6 / 13 Professor Steve Schneider Mathematical Induction Examples Semester 2, 2010 – Week 09 7 / 13 Semester 2, 2010 – Week 09 9 / 13 Examples Example: 4 divides (5n − 1) Example: sigma n sigma 0 = 0 sigma n = n + sigma (n-1) We can prove that 5n − 1 is always divisible by 4. Q(n) : sigma n = n(n+1) 2 Base case: n = 0 Inductive step: assume true for n, prove for n + 1 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 8 / 13 Professor Steve Schneider Mathematical Induction Take and drop Take and drop take and drop a property of take and drop take 0 xs = [] take n [] = [] take n (x:xs) = x:(take (n-1) xs) Want to prove that take n xs ++ drop n xs = xs for all n. drop 0 xs = xs drop n [] = [] drop n (x:xs) = drop (n-1) xs Base case: (take 0 xs) ++ (drop 0 xs) = [] ++ xs = xs Inductive step: need to consider [] and x:xs: We can prove this by induction: (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 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 Professor Steve Schneider Take and drop Mathematical Induction Semester 2, 2010 – Week 09 11 / 13 Exercises (in place of lab) Summary Exercises These exercises are instead of a lab for this week. Please try them in advance of next week. The principle of (mathematical) induction Prove each of the following by induction: To prove some property P(n) is true for all n: 20 + 21 + ... + 2n = 2(n+1) − 1 Show the base case: P(0) Show the inductive step: P(n) ⇒ P(n + 1) 12 + 22 + ... + n2 = (n(n+1)(2n+1)) 6 4n + 6n − 1 is divisible by 9 Professor Steve Schneider Mathematical Induction Semester 2, 2010 – Week 09 12 / 13 Professor Steve Schneider for all n for all n for all n Mathematical Induction Semester 2, 2010 – Week 09 13 / 13