MACM 101 Lecture 18: Lecture 18 Part 3_ Mathematical Induction

24 views2 pages

Document Summary

Show that every 2n x 2n checkerboard with one square removed can be tiled using triominos. Basis step: p(1) is true, as 2 2 checkerboards with one square removed have one of the following shapes. Inductive step: suppose that p(k) is true that is every 2k x 2k checkerboard with one square removed can be tiled with triominos. We have to prove p(k + 1), that is, every 2k+1 x 2k+1 checkerboard without one square can be tiled. Split the big checkerboard into 4 half-size checkerboards. Put one triomino as shown in the picture. We have 4 2k x 2k checkerboards, each without one square. By the induction hypothesis, they can be tiled. We can use the principle of strong induction. To prove that p(n) is true for all positive integers n, we complete two steps: Basis step: verify that p(1) is true. [ p(1) p(2) p(k) ] p(k + 1) is true for all positive integers k.

Get access

Grade+20% off
$8 USD/m$10 USD/m
Billed $96 USD annually
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
40 Verified Answers
Class+
$8 USD/m
Billed $96 USD annually
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
30 Verified Answers

Related textbook solutions

Related Documents

Related Questions