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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! 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):$\displaystyle \frac{n^11}{n1}=1\in \mathbb{Z}$ A(k):$\displaystyle \frac{n^k1}{n1}=z\in \mathbb{Z}$ A(k+1):$\displaystyle \frac{n^{k+1}1}{n1}=\frac{n\cdot n^k1}{n1}=\frac{n\cdot n^kn^k+n^k1}{n1}=\frac{n^k(n1)+n^k1}{n1}=n^k+\frac{n^k1}{n1}=n^k+z\in \mathbb{Z}$ 
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 $n^k  1$?

June 4th, 2016, 04:56 AM  #4  
Newbie Joined: Mar 2016 From: Israel Posts: 5 Thanks: 1  Quote:
Quote:
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 $\displaystyle P(n)=\frac{n^k1}{n1}k$ can be divided by $\displaystyle n1$. $\displaystyle \frac{n^k1}{n1}=n^{k1}+n^{k2}+n^{k3}+...+n^2+n+1$, so: $\displaystyle P(n)=n^{k1}+n^{k2}+n^{k3}+...+n^2+n+1k$ $\displaystyle P(1)=1+1+1+...+1+1(total\ (k1)\ times)+1k=0$ It means that $\displaystyle P(n)$ can be divided by $\displaystyle n1$. 

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