CSci 2011 Discrete Mathematics Lecture 15. Yongdae Kim. Admin. Due dates and quiz Groupwork 7 is due on Oct 26 th . Homework 4 is due on Oct 28 th . Quiz 4: Nov 4 th . 1 page cheat sheet is allowed. Email: [email protected] Put [2011] in front. Check class web site
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
n
1
ar
a
+

n
if
r
1
j
ar
r
1
=

j
0
(
n
1
)
a
if
r
1
=
+
=
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
(Proof) Proof by contradiction
Assume there are a finite number of primes
List them as follows: p1, p2 …, pn.
Consider the number q = p1p2 … pn + 1
Since we have only finite number of primes and q is not one of them, pi should divide q for some i.
Obviously pi  p1p2 … pn.
Recall that a  b, a  c a  b + c.
Therefore, pi  (q  p1p2 … pn) = 1. (*)
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011
(Disprove)
When n = 1601,
n2  79n + 1601 = 1601 (1601  79 + 1)
2*3*5*7*11*13+1 = 30031 = 59 * 509
CSci 2011
= (11110101)2
= (F5)16
CSci 2011
CSci 2011
CSci 2011
CSci 2011
CSci 2011