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 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. June 2nd, 2016, 01:02 PM #2 Senior Member   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 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 June 4th, 2016, 04:56 AM   #4
Newbie

Joined: Mar 2016
From: Israel

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Quote:
 Originally Posted by skaa 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 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. June 6th, 2016, 06:56 AM #5 Senior Member   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 $\displaystyle n-1$. Tags exercise, quick Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post bao Pre-Calculus 3 March 24th, 2015 03:52 PM ManosG Real Analysis 3 March 26th, 2013 02:11 PM Touya Akira Abstract Algebra 8 May 10th, 2011 07:57 AM Touya Akira Abstract Algebra 2 February 9th, 2011 08:59 PM 100dollars Algebra 0 June 9th, 2009 01:32 PM

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