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Ch. 8 - Sequences, Induction, and Probability
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 9, Problema 49

Express each repeating decimal as a fraction in lowest terms. 0.257‾0.\(\overline{257}\)

Guida verificata passo dopo passo
1
Let \( x = 0.257257257\ldots \), where the digits 257 repeat indefinitely.
Since the repeating block has 3 digits, multiply \( x \) by \( 10^3 = 1000 \) to shift the decimal point three places to the right: \( 1000x = 257.257257257\ldots \).
Set up the equation by subtracting the original \( x \) from this new expression to eliminate the repeating decimal part: \( 1000x - x = 257.257257257\ldots - 0.257257257\ldots \).
Simplify the subtraction to get \( 999x = 257 \), since the repeating decimals cancel out.
Solve for \( x \) by dividing both sides by 999: \( x = \frac{257}{999} \). Then, simplify the fraction to its lowest terms by finding the greatest common divisor (GCD) of 257 and 999 and dividing numerator and denominator by it.

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