Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

KingGeorgeBest ResponseYou've already chosen the best response.3
Well, suppose towards a contradiction that \(n=a\cdot b\) is composite with \(a,b>1\). Then, \[2^n1=2^{ab}1=(2^a)^b1\]Now substitute \(2^a=x\) to get \(x^b1\), and use the formula I helped you prove by induction earlier. Make sense?
 one year ago

johnny0929Best ResponseYou've already chosen the best response.0
I understand all the way up to the former formula. I know that it would look like\[(x1)(x^{b1}+x^{b2}+...+x^{2}+x+1)\] but why does that make \(2^{n}1\) not prime?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
\[2^n1=(2^a)^b1=(2^a1)(2^{a(b1)}+2^{a(b2)}+...+2^a+1)\]Since \(a>1\), \(2^a1>1\), and so we have that \((2^a1)(2^n1)\).
 one year ago

johnny0929Best ResponseYou've already chosen the best response.0
thank you so much i understand it now. Proof writing is a very difficult task for me :(
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
Proof writing is definitely an acquired skill :P
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.