Decimal expansion of rational numbers

Step 1: At most (4) decimal places means the number can be written with denominator (10^4). Step 2: In lowest form, the denominator must be a divisor of (10^4). Step 3: The reduced denominator of a terminating decimal is always linked to powers of (2) and (5).

Step 1: The numerator is (1), so (3^c) cannot cancel. Step 2: The reduced denominator contains (3), a prime other than (2) and (5). Hence the decimal is non-terminating recurring. Step 3: When the numerator is (1), the denominator test is direct.

Step 1: (0.00\overline{3}=0.003333\ldots). Step 2: This equals (\frac{1}{300}). So the reduced denominator is (300). Step 3: Two zeros before the recurring digit introduce the effect of (100) in the denominator.

Step 1: In (x=2.37555\ldots), (37) is the non-repeating part and (5) is repeating. Step 2: Taking (100x) and (1000x) aligns the repeating parts. Subtracting then gives the fraction. Step 3: First note the length of the non-repeating part and then the repeating part.

Step 1: (720=2^4\cdot 3^2\cdot 5). Step 2: (45=3^2\cdot 5), so (\frac{45}{720}) reduces to (\frac{1}{16}). The denominator is (16=2^4). Step 3: To get a target denominator, check which factors the numerator cancels.

Step 1: (2310=2\cdot 3\cdot 5\cdot 7\cdot 11). Step 2: For a terminating decimal, (3), (7), and (11) must cancel from the denominator. So the minimum factor is (3\cdot 7\cdot 11=231). Step 3: (2) and (5) may remain, but other prime factors must not.

Step 1: (640=64\cdot 10=2^6\cdot 2\cdot 5=2^7\cdot 5). Step 2: The larger exponent is (7), so the decimal terminates after (7) places. Step 3: Convert the denominator directly into powers of (2) and (5).

Step 1: (20=2^2\cdot 5). Step 2: The powers of (2) and (5) are (2) and (1), so the larger exponent is (2). The decimal terminates after (2) places. Step 3: If the reduced denominator is given, check its exponents directly.

Step 1: (\frac{7}{90}=\frac{7}{9\cdot 10}). Step 2: (\frac{7}{9}=0.\overline{7}), so dividing by (10) gives (0.0\overline{7}). Step 3: A factor (10) in the denominator shifts the decimal one place.

Step 1: For exactly (3) places, the larger exponent of (2) or (5) in the reduced denominator must be (3). Step 2: (8=2^3), (40=2^3\cdot 5), and (125=5^3) satisfy this. (25=5^2) gives only (2) places. Step 3: Understand the difference between exactly and at most.

Step 1: We need (10^6=2^6\cdot 5^6). Step 2: The denominator (2^2\cdot 5^6) lacks (2^4), so multiply numerator and denominator by (16). Thus (N=17\cdot 16=272). Step 3: Multiply by the missing prime power.

Step 1: (0.0999\ldots=0.1). Step 2: (0.1=\frac{1}{10}), so it is rational and equal to a terminating decimal. Step 3: When (9)'s continue at the end, check for an equivalent terminating decimal.

Step 1: In (0.\overline{12}), the block (12) repeats and the decimal does not end. Step 2: The other decimals have only zeros after some point, so they are equal to terminating decimals. Step 3: Distinguish trailing zeros from repeating non-zero digits.

Step 1: The denominator has power (4) of (2) and power (2) of (5). Step 2: Decimal places in a terminating decimal equal the larger exponent, which is (4). Step 3: If the denominator is already reduced, do not assume further cancellation.

Step 1: The numerator (13) cancels only one factor (13) from (13^2). Step 2: The reduced denominator is (2^2\cdot 5^2\cdot 13). Since (13) remains, the decimal is non-terminating recurring. Step 3: Understand the difference between complete and partial cancellation.

Step 1: (10^6=2^6\cdot 5^6). Step 2: The denominator (2^4\cdot 5^6) lacks (2^2). So multiply numerator and denominator by (2^2). Step 3: Complete the deficiency of the prime with the smaller exponent.

Step 1: The rule applies to the denominator in lowest form. Step 2: If (q=2^m5^n), the denominator has only (2) and (5), so the decimal terminates. The other statements are incomplete because (q) may contain other prime factors. Step 3: Be careful with words like always and never.

Step 1: (6250=2\cdot 5^5). Step 2: (\frac{3}{6250}) is in lowest form and the larger exponent is (5), so it terminates exactly after (5) places. The other denominators have larger exponent (4) or (3). Step 3: For exact places, match the larger exponent.

Step 1: This decimal does not terminate. Step 2: No fixed repeating block appears because natural numbers are being joined in order. So it is treated as non-terminating non-recurring. Step 3: In a long digit pattern, check for a fixed repeating block.

Step 1: For a terminating decimal, the reduced denominator must contain only (2) and (5). Step 2: The denominator has extra prime factors (3^2) and (7), so (n) must contain (3^2\cdot 7=63). Step 3: When the smallest value is asked, cancel only the unwanted prime factors.