My Math Forum need help with 2 quick exercise

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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: 20,613 Thanks: 2071 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 $\displaystyle n-1$.

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