Algebra Pre-Algebra and Basic Algebra Math Forum

 April 28th, 2014, 10:43 PM #1 Senior Member   Joined: Apr 2013 Posts: 425 Thanks: 24 A fraction Hello! Show that the fraction $\displaystyle \frac{6n-1}{n^3-n}$ is a periodic decimal fraction mixed for all the natural numbers $\displaystyle n>1$. Thanks from agentredlum April 29th, 2014, 02:37 AM #2 Senior Member   Joined: Feb 2010 Posts: 708 Thanks: 142 I'm not sure I understand the question. What is special about $\displaystyle \frac{6n-1}{n^3-n}$? Isn't any quotient of integers (with non-zero 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:
 Show that the fraction $\displaystyle \frac{6n-1}{n^3-n}$ is a periodic decimal fraction for all the natural numbers $\displaystyle n>1$.

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 $\displaystyle n^3-n$ 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:
 Originally Posted by soroban Hello, Dacu! 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 $\displaystyle n^3-n$ is not of the form $\displaystyle 2^a5^b.$
Hello! 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:
 Originally Posted by soroban 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 $\displaystyle n^3-n$ is not of the form $\displaystyle 2^a5^b.$
Provided $\displaystyle \frac ND$ is a fraction in its lowest terms.

You must also show that $n^3-n$ and $6n-1$ 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: $n^3-n$ and $6n-1$ are not always coprime – e.g. $n=6$ $\implies$ $n^3-n=210$ and $6n-1=35$. Thus $\displaystyle \frac{6n-1}{n^3-n}$ is not always a fraction in the lowest terms. Still, $D\ne2^a5^b$ is not a sufficient guarantee that $\dfrac ND$ is non-terminating; 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 , $n^3 - n = (n-1)n(n+1)$ 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 Thanks from Olinguito May 1st, 2014, 02:17 AM   #8
Senior Member

Joined: Apr 2013

Posts: 425
Thanks: 24

Quote:
 Originally Posted by agentredlum 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 , $n^3 - n = (n-1)n(n+1)$ 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 $$\frac{6n -1}{n^3 -n}$$
Correctly! 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.  Tags fraction 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 Chikis Elementary Math 13 February 25th, 2013 05:23 AM ThePope Elementary Math 3 November 4th, 2012 06:05 AM daigo Algebra 3 July 15th, 2012 01:05 AM Tommy_Gun Algebra 5 June 3rd, 2012 06:14 PM K Sengupta Elementary Math 1 February 9th, 2010 03:34 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      