My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum


Thanks Tree3Thanks
  • 1 Post By illidan5
  • 1 Post By skaa
  • 1 Post By skipjack
Reply
 
LinkBack Thread Tools Display Modes
June 2nd, 2016, 09:44 AM   #1
Newbie
 
Joined: Mar 2016
From: Israel

Posts: 5
Thanks: 1

need help with 2 quick exercise



I know I should only post 1 ex, but you don't have to solve everything; I will appreciate any help!
Thanks from manus

Last edited by skipjack; June 2nd, 2016 at 07:45 PM.
illidan5 is offline  
 
June 2nd, 2016, 01:02 PM   #2
Senior Member
 
skaa's Avatar
 
Joined: Mar 2011
From: Chicago, IL

Posts: 214
Thanks: 77

Exercise 11, 2.
Use math induction.
A(1):$\displaystyle \frac{n^1-1}{n-1}=1\in \mathbb{Z}$

A(k):$\displaystyle \frac{n^k-1}{n-1}=z\in \mathbb{Z}$

A(k+1):$\displaystyle \frac{n^{k+1}-1}{n-1}=\frac{n\cdot n^k-1}{n-1}=\frac{n\cdot n^k-n^k+n^k-1}{n-1}=\frac{n^k(n-1)+n^k-1}{n-1}=n^k+\frac{n^k-1}{n-1}=n^k+z\in \mathbb{Z}$
Thanks from manus
skaa is offline  
June 2nd, 2016, 07:24 PM   #3
Global Moderator
 
Joined: Dec 2006

Posts: 20,919
Thanks: 2203

What does the remainder theorem tell you about the polynomial $n^k - 1$?
Thanks from manus
skipjack is offline  
June 4th, 2016, 04:56 AM   #4
Newbie
 
Joined: Mar 2016
From: Israel

Posts: 5
Thanks: 1

Quote:
Originally Posted by skaa View Post
Exercise 11, 2.
Use math induction.
A(1):$\displaystyle \frac{n^1-1}{n-1}=1\in \mathbb{Z}$

A(k):$\displaystyle \frac{n^k-1}{n-1}=z\in \mathbb{Z}$

A(k+1):$\displaystyle \frac{n^{k+1}-1}{n-1}=\frac{n\cdot n^k-1}{n-1}=\frac{n\cdot n^k-n^k+n^k-1}{n-1}=\frac{n^k(n-1)+n^k-1}{n-1}=n^k+\frac{n^k-1}{n-1}=n^k+z\in \mathbb{Z}$
thank you , can you follow with the solution to 11.3 ?

Quote:
Originally Posted by skipjack View Post
What does the remainder theorem tell you about the polynomial $n^k - 1$?
well , its slightly diffrent from the exercise given , also that the induction solution is pretty straight-forward, so how I apply this on 11,3 ?

Last edited by illidan5; June 4th, 2016 at 04:59 AM.
illidan5 is offline  
June 6th, 2016, 06:56 AM   #5
Senior Member
 
skaa's Avatar
 
Joined: Mar 2011
From: Chicago, IL

Posts: 214
Thanks: 77

11.3 means that $\displaystyle P(n)=\frac{n^k-1}{n-1}-k$ can be divided by $\displaystyle n-1$.

$\displaystyle \frac{n^k-1}{n-1}=n^{k-1}+n^{k-2}+n^{k-3}+...+n^2+n+1$, so:

$\displaystyle P(n)=n^{k-1}+n^{k-2}+n^{k-3}+...+n^2+n+1-k$

$\displaystyle P(1)=1+1+1+...+1+1(total\ (k-1)\ times)+1-k=0$

It means that $\displaystyle P(n)$ can be divided by $\displaystyle n-1$.
skaa is offline  
Reply

  My Math Forum > College Math Forum > Number Theory

Tags
exercise, quick



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Exercise bao Pre-Calculus 3 March 24th, 2015 03:52 PM
Help with this exercise ManosG Real Analysis 3 March 26th, 2013 02:11 PM
Could someone help me with this exercise Touya Akira Abstract Algebra 8 May 10th, 2011 07:57 AM
Need some help with an exercise Touya Akira Abstract Algebra 2 February 9th, 2011 08:59 PM
Quick exercise in probability 100dollars Algebra 0 June 9th, 2009 01:32 PM





Copyright © 2019 My Math Forum. All rights reserved.