Terminating Decimal Proof

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January 15th 2010, 10:38 AM
Haven

Terminating Decimal Proof

I am having troubles with this deceptively simple problem:

Show that for any integer n that $\frac{1}{n}$ will be a terminating sexagesimal iff n's prime factors only consist of 2,3 and 5.

In other words $\frac{1}{n}$ will terminate in base 60 iff $n = 2^a*3^b*5^c$
January 15th 2010, 11:05 AM
Media_Man

Define $n = 2^a3^b5^c$ and $m=max(a,b,c)$ . Then $\frac1n=\frac{2^{m-a}3^{m-b}5^{m-c}}{60^m}$ . Therefore $\frac1n$ terminates on the $m^{th}$ decimal place. QED
January 16th 2010, 12:23 AM
Unbeatable0

Hint for the other direction:
If $\frac{1}{n} = 0.a_1 a_2 a_3 ... a_n$ then look at $60^n \frac{1}{n} = ?$
January 16th 2010, 04:51 AM
HallsofIvy

More generally, any integer, n, is equal to a product of powers of its prime divisors. Any terminating "decimal", base n, is equal to an integer times some power of n (the negative of the power of the last "decimal" place) and so is equal to that integer divided by that power of n. While some factors in denominator and numerator may cancel, that won't introduce new factors in the denominator. The only possible factors of the denominator are factors of n and so the only possible prime factors of the denominator are the prime factors of n.
January 18th 2010, 05:20 PM
Haven

Quote:

Originally Posted by Media_Man

Define $n = 2^a3^b5^c$ and $m=max(a,b,c)$ . Then $\frac1n=\frac{2^{m-a}3^{m-b}5^{m-c}}{60^m}$ . Therefore $\frac1n$ terminates on the $m^{th}$ decimal place. QED

Why does that terminate?
January 18th 2010, 05:52 PM
Media_Man

The (Decimal) Terminator

Haven:

In base $60$ , by definition...

$\frac1{60}=.1$
$\frac1{60^2}=.01$
$\frac1{60^3}=.001$
...

Etcetera. So a decimal terminates in base $60$ iff it is expressible in the form $\frac{N}{60^m}$ , with $N,m$ integers.