lect_18 - Strong induction Margaret M Fleck 3 March 2010...

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Strong induction Margaret M. Fleck 3 March 2010 This lecture presents proofs by “strong” induction, a slight variant on normal mathematical induction. 1 Announcements We aren’t quite ±nished grading the midterm. We’re hoping to have them done late Thursday. 2 A geometrical example As a warm-up, let’s see another example of the basic induction outline, this time on a geometrical application. Tiling some area of space with a certain type of puzzle piece means that you ±t the puzzle pieces onto that area of space exactly, with no overlaps or missing areas. A right triomino is a 2-by-2 square minus one of the four squares. (See pictures in Rosen pp. 277-278.) I then claim that Claim 1 For any positive integer n , a 2 n × 2 n checkerboard with any one square removed can be tiled using right triominoes. Proof: by induction on n . 1
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Base: Suppose n = 1. Then our 2 n × 2 n checkerboard with one square remove is exactly one right triomino. Induction: Suppose that the claim is true for some integer k . That is a 2 k × 2 k checkerboard with any one square removed can be tiled using right triominoes. Suppose we have a 2 k +1 × 2 k +1 checkerboard C with any one square removed. We can divide C into four 2 k × 2 k sub-checkerboards P , Q , R , and S . One of these sub-checkerboards is already miss- ing a square. Suppose without loss of generality that this one is S . Place a single right triomino in the middle of C so it covers one square on each of P , Q , and R . Now look at the areas remaining to be covered. In each of the sub-checkerboards, exactly one square is missing ( S ) or already covered ( P , Q , and R ). So, by our inductive hypothesis, each of these sub-checkerboards minus one square can be tiled with right triominoes. Combining these four tilings with the triomino we put in the middle, we get a tiling for the whole of the larger checkerboard C . This is what we needed to construct. 3 Strong induction The inductive proofs you’ve seen so far have had the following outline: Proof: We will show P ( n ) is true for all n , using induction on n . Base
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lect_18 - Strong induction Margaret M Fleck 3 March 2010...

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