13-Mathematical Induction - EECS 210 Discrete Structures...

Info icon This preview shows pages 1–8. Sign up to view the full content.

Mathematical Induction David O. Johnson EECS 210 (Fall 2016) 1 EECS 210 Discrete Structures David O. Johnson Fall 2016
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Reminders Connect 3 due: 11:59 PM, Thursday, October 13 (today) Exam 2: Thursday, October 20 Mathematical Induction David O. Johnson EECS 210 (Fall 2016) 2
Image of page 2
Any Questions? Mathematical Induction 3 David O. Johnson EECS 210 (Fall 2016)
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Mathematical Induction (Section 5.1) Mathematical Induction Template for Proofs by Mathematical Induction Examples of Proof by Mathematical Induction Mathematical Induction David O. Johnson EECS 210 (Fall 2016) 4
Image of page 4
Mathematical Induction & Recursion In Chapter 2 we described sets by listing their elements or by giving some property that characterizes these elements. We gave formulae for the values of functions. There is another important way to define such objects, based on mathematical induction : To define functions, some initial terms are specified, and a rule is given for finding subsequent values from values already known. We briefly touched on this sort of definition in Chapter 2 when we showed how sequences can be defined using recurrence relations. Sets can be defined by listing some of their elements and giving rules for constructing elements from those already known to be in the set. Such definitions, called recursive definitions, are used throughout discrete mathematics and computer science. Mathematical Induction David O. Johnson EECS 210 (Fall 2016) 5
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Mathematical Induction & Recursion Many mathematical statements assert that a property is true for all positive integers (n): n! ≤ n n n 3 − n is divisible by 3 a set with n elements has 2 n subsets sum of the first n positive integers is n(n + 1)/2 What we will learn about in this lecture is mathematical induction , which is used to prove assertions like this. Proofs using mathematical induction have two parts. 1. They show that the statement holds for the positive integer 1 2. They show that if the statement holds for a positive integer then it must also hold for the next larger integer Mathematical induction is based on the rule of inference that tells us that if P(1) and k (P(k) → P(k + 1)) are true for the domain of positive integers, then n P(n) is true. Mathematical Induction David O. Johnson EECS 210 (Fall 2016) 6 Remember: All positive integers = {1, 2, …}
Image of page 6
Climbing an Infinite Ladder Mathematical induction is like climbing an infinite ladder … Suppose we have an infinite ladder: 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder, then we can reach the next rung. From (1), we can reach the first rung. Then by applying (2), we can reach the second rung. Applying (2) again, the third rung. And so on. We can apply (2) any number of times to reach any particular rung, no matter how high up.
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern