1. Proof by Induction and contradiction Leo Cheung

2. The TAs Our office is in SHB117, feel free to come if you get problems about the course Or ask your questions in the newsgroup news://news.erg.cuhk.edu.hk/cuhk.cse.csc 2110news://news.erg.cuhk.edu.hk/cuhk.cse.csc 2110

3. The Plan Some basic mathematical background Mathematical Induction Proof by contradiction

4. Background - Summation

5. Background - Product

6. Background - Factorial

7. Mathematical Induction - Idea You want to prove something (call it f(n)) to be true for all natural number n>=1 Step 1: Base case –Prove n=1 (that is f(1)) is true Step 2: “Induction” –Prove that if f(k) is true then f(k+1) is also true, for all k>=1 Done!

8. Why it’s done? All we know (and proved) ▪f(1) is true ▪if f(k) is true than f(k+1) is true for all k >=1

9. Example Base Case (n=0): L.H.S. = 1, R.H.S. = 1(1+1)/2 = 1 Assume f(k) is true, i.e. Then So f(k+1) is also true. By the principal of induction f(n) is true for all n >= 1

10. And lets try Base Case (n=0): L.H.S. = 2 0 = 1, R.H.S. = 2 1 – 1 =1 Assume f(k) is true, i.e. Then So f(k+1) is also true. By the principal of induction f(n) is true for all n >= 0

11. Contradiction-Idea You want to prove f is true then Assume f is false By the assumption, you derive some false statements So the assumption is wrong That means f is true

12. Examples Prove: For all integers n, if n 2 is odd, then n is odd. Assume there is a n such that n 2 is odd and n is even ▪n = 2k for some k ▪n 2 = (2k)x(2k) = 4k 2, which is even ▪Contradiction! So the assumption must be wrong.

13. Examples Prove: There are infinitely many prime numbers. Assume not, there are only n primes p 1, p 2, …, p n ▪Consider k = p 1 p 2 … p n + 1 ▪k is not divisible by any prime p i ▪So k is a prime (other than p 1, p 2,..., p n ) ▪Contradiction!

14. END Exercise will be posted on the course webpage