# Terminating Decimals: 2^n5^m Explained

• Natasha1
In summary, the fraction 1/(2^n 5^m) can be made into either 2^k/10^m or 5^j/10^n, depending on whether n < m or n > m.
Natasha1

## 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?

Last edited:
Natasha1 said:

## 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?

First of all, can you express 12 as ##2^n5^m## for some non-negative integers ##m## and ##n##?

No

But it is 2^2*3

Natasha1 said:
But I don't really understand why only denominators in the form 2^n5^m work?

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.

Natasha1 said:
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.

Natasha1
Natasha1 said:

## 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?

Any fraction of the form
$$\frac{1}{2^n 5^m}$$
$$\frac{2^k}{10^m} \hspace{2em} \text{or} \hspace{2em} \frac{5^j}{10^n},$$
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.

Last edited by a moderator:
Thanks all for your help. I understand a little better now :)

## 1. What are terminating decimals?

Terminating decimals are decimal numbers that have a finite number of digits after the decimal point. This means that the decimal representation of the number ends, rather than continuing infinitely.

## 2. What is the significance of the numbers 2 and 5 in 2^n5^m?

The numbers 2 and 5 are prime factors of 10, and their combination in the form of 2^n5^m determines the decimal representation of a terminating decimal. In other words, the powers of 2 and 5 determine the number of digits after the decimal point in a terminating decimal.

## 3. How do you calculate 2^n5^m?

To calculate 2^n5^m, you would simply multiply 2 by itself n times and 5 by itself m times. For example, 2^3 = 2 x 2 x 2 = 8 and 5^2 = 5 x 5 = 25. Therefore, 2^3 5^2 = 8 x 25 = 200.

## 4. Can any decimal be represented in the form of 2^n5^m?

No, only terminating decimals can be represented in the form of 2^n5^m. Non-terminating decimals, such as pi (3.1415926...), cannot be expressed in this form because they have an infinite number of digits after the decimal point.

## 5. What is the practical application of understanding 2^n5^m in terminating decimals?

Understanding 2^n5^m can help in simplifying and comparing decimals. For example, if two decimals have the same powers of 2 and 5, they will have the same number of digits after the decimal point. This can also be useful in rounding numbers to a certain number of decimal places.

• Precalculus Mathematics Homework Help
Replies
7
Views
1K
• Precalculus Mathematics Homework Help
Replies
6
Views
991
• Precalculus Mathematics Homework Help
Replies
3
Views
1K
• Precalculus Mathematics Homework Help
Replies
6
Views
1K
• Precalculus Mathematics Homework Help
Replies
3
Views
785
• Precalculus Mathematics Homework Help
Replies
3
Views
1K
• Precalculus Mathematics Homework Help
Replies
8
Views
984
• Precalculus Mathematics Homework Help
Replies
5
Views
2K
• Precalculus Mathematics Homework Help
Replies
8
Views
379
• MATLAB, Maple, Mathematica, LaTeX
Replies
3
Views
1K