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If m and n are integers, is m/n an infinite decimal? 04 May 2022, 05:00
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If m and n are integers, is m/n an infinite decimal?

(1) m is a prime factor of 100.
(2) n is a prime factor of 50.
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If m and n are integers, is m/n an infinite decimal? 04 May 2022, 05:08
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If I can prove \(\frac{m}{n}\) to be a terminating fraction then it can be proven that \(\frac{m}{n}\) is not an infinite decimal. So, for that, I need the nature of \(n\). If \(n \) has \(2/5,\) \(\frac{m}{n}\) can be proven to be terminating decimal.

(1) Talks nothing about '\(n\)'. Not sufficient.
(2) Tells us that \(n \) is either \(2 \) or \(5\). Thus, we can safely conclude that the fraction \(\frac{m}{n}\) is a terminating decimal and thus not an infinite decimal. Sufficient.
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If m and n are integers, is m/n an infinite decimal? 27 Dec 2024, 05:16
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Why can we conclude that this division results in a termnating decimal?
PyjamaScientist
If I can prove \(\frac{m}{n}\) to be a terminating fraction then it can be proven that \(\frac{m}{n}\) is not an infinite decimal. So, for that, I need the nature of \(n\). If \(n \) has \(2/5,\) \(\frac{m}{n}\) can be proven to be terminating decimal.

(1) Talks nothing about '\(n\)'. Not sufficient.
(2) Tells us that \(n \) is either \(2 \) or \(5\). Thus, we can safely conclude that the fraction \(\frac{m}{n}\) is a terminating decimal and thus not an infinite decimal. Sufficient.
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If m and n are integers, is m/n an infinite decimal? 27 Dec 2024, 05:24
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barbosacrenata
Why can we conclude that this division results in a termnating decimal?
PyjamaScientist
If m and n are integers, is m/n an infinite decimal?

(1) m is a prime factor of 100.
(2) n is a prime factor of 50.

If I can prove \(\frac{m}{n}\) to be a terminating fraction then it can be proven that \(\frac{m}{n}\) is not an infinite decimal. So, for that, I need the nature of \(n\). If \(n \) has \(2/5,\) \(\frac{m}{n}\) can be proven to be terminating decimal.

(1) Talks nothing about '\(n\)'. Not sufficient.
(2) Tells us that \(n \) is either \(2 \) or \(5\). Thus, we can safely conclude that the fraction \(\frac{m}{n}\) is a terminating decimal and thus not an infinite decimal. Sufficient.
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A reduced fraction \(\frac{a}{b}\) (meaning that the fraction is already in its simplest form, so reduced to its lowest term) can be expressed as a terminating decimal if and only if the denominator \(b\) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as the denominator \(250\) equals \(2*5^3\). The fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and the denominator \(10=2*5\).

Note that if the denominator already consists of only 2s and/or 5s, then it doesn't matter whether the fraction is reduced or not.

For example, \(\frac{x}{2^n5^m}\), (where \(x\), \(n\), and \(m\) are integers) will always be a terminating decimal.

(We need to reduce the fraction in case the denominator has a prime other than 2 or 5, to see whether it can be reduced. For example, the fraction \(\frac{6}{15}\) has 3 as a prime in the denominator, and we need to know if it can be reduced.)

(2) says that n is a prime factor of 50 = 2 * 5^2, so n either 2 or 5, thus m/n will be a terminating decimal.
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If m and n are integers, is m/n an infinite decimal? 27 Dec 2024, 06:58
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I see know. Thank you very much!
Bunuel
barbosacrenata
Why can we conclude that this division results in a termnating decimal?
PyjamaScientist
If m and n are integers, is m/n an infinite decimal?

(1) m is a prime factor of 100.
(2) n is a prime factor of 50.

If I can prove \(\frac{m}{n}\) to be a terminating fraction then it can be proven that \(\frac{m}{n}\) is not an infinite decimal. So, for that, I need the nature of \(n\). If \(n \) has \(2/5,\) \(\frac{m}{n}\) can be proven to be terminating decimal.

(1) Talks nothing about '\(n\)'. Not sufficient.
(2) Tells us that \(n \) is either \(2 \) or \(5\). Thus, we can safely conclude that the fraction \(\frac{m}{n}\) is a terminating decimal and thus not an infinite decimal. Sufficient.

A reduced fraction \(\frac{a}{b}\) (meaning that the fraction is already in its simplest form, so reduced to its lowest term) can be expressed as a terminating decimal if and only if the denominator \(b\) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as the denominator \(250\) equals \(2*5^3\). The fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and the denominator \(10=2*5\).

Note that if the denominator already consists of only 2s and/or 5s, then it doesn't matter whether the fraction is reduced or not.

For example, \(\frac{x}{2^n5^m}\), (where \(x\), \(n\), and \(m\) are integers) will always be a terminating decimal.

(We need to reduce the fraction in case the denominator has a prime other than 2 or 5, to see whether it can be reduced. For example, the fraction \(\frac{6}{15}\) has 3 as a prime in the denominator, and we need to know if it can be reduced.)

(2) says that n is a prime factor of 50 = 2 * 5^2, so n either 2 or 5, thus m/n will be a terminating decimal.
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