Lecture Videos Lecture 4 – Mathematical Induction & the Euclidean Algorithm

Lecture 4 – Mathematical Induction & the Euclidean Algorithm

You may want to download the Lecture 4 slides (PDF).

1. The Principle of Mathematical Induction

Here, we give a formal definition of the Principle of Mathematical Induction.  Then, we look at a connection to programs with loops and recurrences.  (12:22)

2. Example of Mathematical Induction

In this video we apply induction to show that the sum of the first n odd integers is n2.  Then, we return to discussing how to be precise with summations involving “…” (9:38)

3. More Examples of Mathematical Induction

Here, we use induction to find an equality for the sum of the first n squares.  Then, we use induction to show an expression is divisible by 9 for all n.  (9:24)

4. An Exercise in Math Induction

See if you can use induction to show that, for n ≥ 2, that 1-1/2+2-1/2+ … + n-1/2 > n1/2. (1:08)

5. The Base Case for an Exercise in Math Induction

In this video we solve the base case for the inequality 1-1/2+2-1/2+ … + n-1/2 > n1/2. The last video of this lecture solves the rest of this problem. (2:35)

6. Basis for Long Division & Greatest Common Divisors

Here we revisit long division, and prove a statement about long division by using induction.  Then, we introduce greatest common divisors (GCDs), and look at a naïve way of computing them. (9:01)

7. The Euclidean Algorithm

This video defines the Euclidean Algorithm and implements it by using a while loop. (2:52)

8. A Concrete Example and Back-Tracking

This video explores a concrete example of calculating the GCD using the Euclidean Algorithm.  Then, we look at how to find integers a and b such that an + bm = gcd(n, m), by using back-tracking.  (10:44)

9. Avoiding Recursive Calls

A practical method for calculating the gcd of two integers by using a loop instead of recursive calls is described in this video. The code that is mentioned is here. (3:08)

10. The Inequality 1-1/2+2-1/2+ … + n-1/2 > n1/2

We complete our proof that, for n ≥ 2, that 1-1/2+2-1/2+ … + n-1/2 > n1/2. (9:20)