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April 28th, 2014, 10:43 PM   #1
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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 $\displaystyle n>1$.
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April 29th, 2014, 02:37 AM   #2
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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?
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April 29th, 2014, 07:30 AM   #3
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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 $\displaystyle 2^a5^b.$

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April 30th, 2014, 06:18 AM   #4
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Quote:
Originally Posted by soroban View Post
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!
I do not understand!Details please!
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April 30th, 2014, 07:42 AM   #5
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Quote:
Originally Posted by soroban View Post

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 $1$.
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April 30th, 2014, 08:48 AM   #6
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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.
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April 30th, 2014, 09:55 AM   #7
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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

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May 1st, 2014, 02:17 AM   #8
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Quote:
Originally Posted by agentredlum View Post
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!
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May 1st, 2014, 04:28 AM   #9
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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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