Decimal expansion of rational numbers

Step 1: In the reduced denominator, (3) is a prime other than (2) and (5). Step 2: So the decimal will not terminate but will repeat. Step 3: A non-terminating decimal of a rational number is recurring.

Step 1: (160=2^5\times5). Step 2: The larger exponent is (5), so the decimal terminates after (5) places. Step 3: Factorising and noting the larger exponent is the safest method.

Step 1: (625=5^4). Step 2: The denominator has only (5) with exponent (4). Step 3: Therefore the decimal terminates after (4) places.

Step 1: (40=2^3\times5). Step 2: The larger exponent is (3), so the decimal terminates after (3) places. Step 3: This is also confirmed by (\frac{1}{40}=0.025).

Step 1: (2500=2^2\times5^4). Step 2: The larger exponent is (4). Step 3: Hence the decimal terminates within (4) places.

Step 1: The number of decimal places comes from the larger exponent of (2) and (5). Step 2: The exponent of (2) is (2), so for exactly (6) places we need (x=6). Step 3: Match the larger exponent with the required places.

Step 1: The rule applies to the denominator in lowest form. Step 2: Without reducing, extra factors may appear in the denominator. Step 3: In exams, reduce the fraction using the highest common factor and then check the denominator.

Step 1: (44) and (105) are coprime. Step 2: (105=3\times5\times7), so the denominator has (3) and (7) as well. Step 3: A reduced denominator with primes other than (2) and (5) gives a non-terminating recurring decimal.

Step 1: (\frac{75}{300}=\frac{1}{4}). Step 2: The reduced denominator is (4=2^2), so the decimal terminates. Step 3: Apply the rule to the reduced denominator, not the original one.

Step 1: The terminating decimal rule applies to the denominator in lowest form. Step 2: Such a denominator has no prime factors other than (2) and (5). Step 3: This rule is very useful in direct exam questions.

Step 1: A decimal with two places is measured in hundredths. Step 2: So it can be written as (\frac{n}{100}), where (n) is an integer. Step 3: Do not forget to reduce the fraction afterward.

Step 1: (0.125=\frac{125}{1000}). Step 2: Reducing gives (\frac{125}{1000}=\frac{1}{8}). Step 3: Write a terminating decimal first with denominator (10), (100), or (1000), then reduce.

Step 1: The repeating block is (27), so (0.\overline{27}=\frac{27}{99}). Step 2: (\frac{27}{99}=\frac{3}{11}). Step 3: For recurring decimals, the number of (9)s matches the repeating digits.

Step 1: (0.2\overline{3}=0.2333\ldots). Step 2: Converting it gives (\frac{7}{30}). Step 3: For a mixed recurring decimal, separate the non-repeating and repeating parts carefully.

Step 1: (4.125=\frac{4125}{1000}=\frac{33}{8}). Step 2: The reduced denominator is (8=2^3). Step 3: The reduced denominator of a terminating decimal is made only of (2) and (5).

Step 1: (3.1416) has only four decimal places. Step 2: Therefore it is terminating and rational. Step 3: A number is not irrational just because it has a decimal point.

Step 1: A non-terminating non-recurring decimal neither ends nor has a fixed repeating pattern. Step 2: Such a number cannot be written as (\frac{p}{q}). Step 3: In exams, carefully distinguish recurring from non-recurring decimals.

Step 1: (2^5=32). Step 2: The denominator has exponent (5) of (2), while the exponent of (5) can be taken as (0). Step 3: The larger exponent is (5), so the decimal terminates after (5) places.

Step 1: (\frac{15}{48}=\frac{5}{16}). Step 2: Since (16=2^4), the decimal terminates after (4) places. Step 3: Reducing first is necessary in such fractions.

Step 1: (125\times8=1000). Step 2: (\frac{18}{125}=\frac{144}{1000}=0.144). Step 3: Converting the denominator to (10), (100), or (1000) is a quick method.