April 28th, 2014, 10:43 PM  #1 
Senior Member Joined: Apr 2013 Posts: 425 Thanks: 24  A fraction
Hello! Show that the fraction $\displaystyle \frac{6n1}{n^3n}$ is a periodic decimal fraction mixed for all the natural numbers $\displaystyle n>1$. 
April 29th, 2014, 02:37 AM  #2 
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140 
I'm not sure I understand the question. What is special about $\displaystyle \frac{6n1}{n^3n}$? Isn't any quotient of integers (with nonzero denominator) by definition a rational number and thus can be expressed as periodic decimal?

April 29th, 2014, 07:30 AM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, Dacu! Quote:
A fraction $\displaystyle \frac{N}{D}$ is a terminating decimal if $\displaystyle D$ is of the form: $\displaystyle \,2^a5^b$ $\displaystyle \quad$for nonnegative integers $\displaystyle a$ and $\displaystyle b.$ You must show that $\displaystyle n^3n$ is not of the form $\displaystyle 2^a5^b.$  
April 30th, 2014, 06:18 AM  #4  
Senior Member Joined: Apr 2013 Posts: 425 Thanks: 24  Quote:
I do not understand!Details please!  
April 30th, 2014, 07:42 AM  #5  
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  Quote:
You must also show that $n^3n$ and $6n1$ have no common divisor greater than $1$.  
April 30th, 2014, 08:48 AM  #6 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 
I beg your pardon: $n^3n$ and $6n1$ are not always coprime – e.g. $n=6$ $\implies$ $n^3n=210$ and $6n1=35$. Thus $\displaystyle \frac{6n1}{n^3n}$ is not always a fraction in the lowest terms. Still, $D\ne2^a5^b$ is not a sufficient guarantee that $\dfrac ND$ is nonterminating; e.g. $30\ne2^a5^b$ but $\dfrac3{30}=0.1$ terminates. 
April 30th, 2014, 09:55 AM  #7 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
The decimal value of equivalent fractions does not change. To put it another way , common factors in the numerator and denominator do not affect the decimal representation. $$\frac{3}{30} = \frac{1}{10} = \frac{20}{200} = 0.1$$ If the fraction IS in lowest terms then the decimal will terminate IFF the denominator is of the form $2^a5^b$ Now , $n^3  n = (n1)n(n+1)$ is the product of 3 consecutive integers and therefore a multiple of $6 = 2 \cdot 3$ $6n  1$ is never a multiple of $6 = 2 \cdot 3$ Therefore there will always be an irreducible factor of 6 in the denominator of $$ \frac{6n 1}{n^3 n}$$ The 2 in 6 is no problem BUT the 3 in 6 gaurantees the denominator of the REDUCED fraction will never be of the form $2^a5^b$ This approach solves the OP's question 
May 1st, 2014, 02:17 AM  #8  
Senior Member Joined: Apr 2013 Posts: 425 Thanks: 24  Quote:
 
May 1st, 2014, 04:28 AM  #9 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
For the numerator you can also use $$6n \pm 1$$ $$6n \pm 2$$ $$6n \pm 4$$ $$6n \pm 5$$ All are gauranteed to be non terminating because the factor 3 (inside 6) in the denominator is irreducible. 

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