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Terminating decimals

  • Thread starter Natasha1
  • Start date
  • #1
467
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Homework Statement



Which of the following produce terminating decimals for all integers n?

n/10

n/7

n/12

2. The attempt at a solution

I read on the internet that reduced fraction a/b (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only if b (denominator) is of the form 2^n5^m, where m and n are non-negative integers.

That would mean only n/10 and n/12 are the answers. But I don't really understand why only denominators in the form 2^n5^m work?

Can someone explain please?
 
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Answers and Replies

  • #2
tnich
Homework Helper
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Homework Statement



Which of the following produce terminating decimals for all integers n?

n/10

n/7

n/12

2. The attempt at a solution

I read on the internet that reduced fraction a/b (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only if b (denominator) is of the form 2^n5^m, where m and n are non-negative integers.

That would mean only n/10 and n/12 are the answers. But I don't really understand why only denominators in the form 2^n5^m work?

Can someone explain please?
First of all, can you express 12 as ##2^n5^m## for some non-negative integers ##m## and ##n##?
 
  • #3
467
7
No
 
  • #4
467
7
But it is 2^2*3
 
  • #5
33,273
4,982
But I don't really understand why only denominators in the form 2^n5^m work?

Can someone explain please?
The denominators would be powers of 2 (2, 4, 8, 16, ...), powers of 5 (5, 25, 125, ...), or products of 2 to some power times 5 to some power. All such fractions with any of these denominators have terminating decimal expansions If the denominator is 2, 4, 8, and so on, the fraction will be some multiple of .5, .25, or .125, and so on. If the denominator is 5, 25, 125, and so on, the fraction will be a multiple of .2, .04, .008, and so on. All of these are artifacts of writing fractions as decimal (base-10) fractions, with 10 being divisible by 2 and 5.

But it is 2^2*3
Which is not of the form ##2^m5^n##. The problem did not mention factors of 3 in the denominator.
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
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Homework Statement



Which of the following produce terminating decimals for all integers n?

n/10

n/7

n/12

2. The attempt at a solution

I read on the internet that reduced fraction a/b (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only if b (denominator) is of the form 2^n5^m, where m and n are non-negative integers.

That would mean only n/10 and n/12 are the answers. But I don't really understand why only denominators in the form 2^n5^m work?

Can someone explain please?
Any fraction of the form
$$ \frac{1}{2^n 5^m}$$
can be made into either
$$\frac{2^k}{10^m} \hspace{2em} \text{or} \hspace{2em} \frac{5^j}{10^n},$$
depending on whether ##n < m## or ##n > m##. For example, ##1/(2^3 5^2) = 5/ (2^3 5^3) = 5/10^3##, and ##1/(2^2 5^3) = 2/(2^3 5^3) = 2/10^3.##

Anyway: why are you asking us if you understand why denominators of the form ##2^n 5^m## work? Telling us you do not understand it is different from asking us if you understand it, and when you end with a question mark "?" you are asking, not telling.
 
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  • #7
467
7
Thanks all for your help. I understand a little better now :)
 

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