NTI 9-12

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Name:  Marshal Thrasher

\vspace{.5in}

NTI Proposition 9.  If $n$ is and integer, then $n|0$.
\vspace{.25in}

Proof: 
	
	\vspace{.1in}By Definition 1, for any integer $n$, $n$ divides an integer $b$ if there exists an integer $k$ such that $nk=b$.
Let $b=0$, any integer $n$ multiplied by zero equals zero. So $k=0$, 
\\ Therefore,
		$$n|0$$ 
	\hspace{2.8in}		  Q.E.D.
			  
			  
\vspace{.5in}


NTI Corollary 10. If $n$ and $a$ are integers then $n|(a-a)$.
\vspace{.25in}

Proof: 
	
	\vspace{.1in}
	Building on Proposition 9, $(a-a)$ will always equal zero. \\
	Therefore,
		$$n|(a-a)$$
		\vspace{.5in}
        \hspace{2.8in}		Q.E.D. 
		
NTI Proposition 11. Let $n,a,b$ be integers. If $n|(a-b)$, the $n|(b-a)$.

\vspace{.25in}Proof: \vspace{.1in}

	By Definition 1, $nk=(a-b)$, using the commutative property and \\multiplying $-1$ through we get:
	$n(-k)=(b-a)$
	$-k$ is still an integer \\
	Therefore, 
	
	$$n|(b-a)$$
			\hspace{2.8in}Q.E.D.


NTI Proposition 12. Let $n,a,b,c$ be integers. If $n|(a-b)$ and $n|(b-c)$, then $n|(a-c)$.
\vspace{.25in}

Proof: 
	
	\vspace{.1in}
	By Definition 1,  $nk_{1}=(a-b)$ and $nk_{2}=(b-c)$ such that $k_{1}$ and $k_{2}$ are integers. By adding $c$ to both sides on the second equation we obtain: 
    $$c+nk_{2}=b$$ 
    We can now substitute $b$ out of the first equation: 
    
    $$nk_{1}=a-c+nk_{2}$$ 
    
    After tidying up the equation we get: $$n(k_{1}-k_{2})=a-c$$
	
Through the basic properties we know the Left Hand Side will always be and integer. \\
	Therefore,
	$$n|(a-c)$$ 
				 
		\hspace{2.8in}		Q.E.D. 


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