A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form , where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by 24 or 54? Give reasons.

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KnowledgeBoat · Accepted Answer

A rational number whose terminating decimal expansion has its last non-zero digit at the 4th decimal place can be expressed as : ⇒ where p is an integer. If p were divisible by 10, then the last non-zero digit would have occurred at the 3rd decimal place (or earlier), contradicting our assumption. Hence, p must not be divisible by 10. Now, 104 = 24 × 54. Since p is not divisible by 10 = 2 × 5, p does not contain both 2 and 5 as factors. So, p contains at most powers of 2 only, or powers of 5 only (or neither), but not both. When we reduce to lowest form : Case 1 : If p is odd (no factor of 2), then no 2's in numerator can cancel with 2's in denominator. So, the lowest-form denominator still contains 24. Case 2 : If p is not divisible by 5 (no factor of 5), then no 5's can cancel. So, the lowest-form denominator still contains 54. Examples : ⇒ 0.0625 = . Here, p = 625 = 54, and lowest-form denominator is 24. ⇒ 0.0008 = . Here, p = 8 = 23, and lowest-form denominator is divisible by 54. So, the denominator in lowest form must be divisible by 24 or 54 (at least one of them), but not necessarily both. Hence, such a number can be written as where p is not divisible by 10. The denominator in lowest form must be divisible by 24 or 54 (or both), but it is not necessary for it to be divisible by both.

Mathematics

Whole Numbers

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Mathematics

Whole Numbers

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A rational number in its lowest form has denominator 23 × 5. How many decimal places will its decimal expansion have? Explain your answer.

A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.