Converting between Decimals and Fractions

0.75 means 75 hundredths, so \( 0.75 = \frac{75}{100} \). To simplify, find the highest common factor of 75 and 100, which is 25: \( 75 \div 25 = 3 \) and \( 100 \div 25 = 4 \). So \( \frac{75}{100} = \frac{3}{4} \).

We can verify by going the other way: \( 3 \div 4 = 0.75 \). We can also use money: 75p out of £1 is three quarters of a pound. Or think of a 10×10 grid — shading 75 squares is the same as shading 3 out of 4 equal rows of 25.

If \( \frac{1}{3} \) were exactly 0.3, then \( \frac{1}{3} \times 3 \) should equal \( 0.3 \times 3 \). But \( \frac{1}{3} \times 3 = 1 \), while \( 0.3 \times 3 = 0.9 \). Since \( 0.9 \neq 1 \), the two values cannot be equal. The decimal 0.3 is actually \( \frac{3}{10} \), not \( \frac{1}{3} \).

The exact decimal form of \( \frac{1}{3} \) is \( 0.333\ldots \) (the 3 repeats forever). Each extra 3 gets closer to \( \frac{1}{3} \) but never reaches it exactly: \( 0.3 = \frac{3}{10} \), \( 0.33 = \frac{33}{100} \), \( 0.333 = \frac{333}{1000} \) — always slightly less than \( \frac{1}{3} \). This is why mathematicians use the recurring dot notation: \( 0.\dot{3} \) to show the digit repeats infinitely.

0.5 means 5 tenths \( = \frac{5}{10} = \frac{1}{2} \). And 0.50 means 50 hundredths \( = \frac{50}{100} = \frac{1}{2} \). Both simplify to the same fraction, so they are equal. Adding a zero after the last decimal digit does not change the value — it’s like saying “I have 50 hundredths” instead of “I have 5 tenths,” which are equivalent.

Students who believe “longer decimals are bigger” may think 0.50 > 0.5 because it has more digits. But trailing zeros after the last significant decimal digit never change the value. Ask yourself: is 50cm longer than 5dm (50 centimetres vs 5 decimetres, or 50 hundredths of a metre vs 5 tenths)? Contextualising decimals as units of measurement destroys the “more digits = bigger” illusion.

0.125 means 125 thousandths, so \( 0.125 = \frac{125}{1000} \). Now simplify: 125 and 1000 share a factor of 125 (since \( 125 \times 1 = 125 \) and \( 125 \times 8 = 1000 \)). So \( \frac{125}{1000} = \frac{1}{8} \).

We can also work from the fraction side using the halving pattern: \( \frac{1}{2} = 0.5 \), and halving gives \( \frac{1}{4} = 0.25 \), and halving again gives \( \frac{1}{8} = 0.125 \). Each time we halve, we get one more decimal place. A third approach: divide \( 1 \div 8 \) using short division — 8 doesn’t go into 1, so we write 0. and proceed: \( 10 \div 8 = 1 \) remainder 2, then \( 20 \div 8 = 2 \) remainder 4, then \( 40 \div 8 = 5 \). Result: 0.125.

Example: \( \frac{1}{2} = 0.5 \)

Another: \( \frac{3}{5} = 0.6 \)

Creative: \( \frac{7}{2} = 3.5 \) — an improper fraction; students rarely consider these, but \( 7 \div 2 = 3.5 \) which has exactly one decimal place.

Trap: \( \frac{1}{3} \approx 0.3 \) — a student might stop their division after one step and write 0.3, but \( \frac{1}{3} = 0.333\ldots \) recurring. It never terminates, so it doesn’t have “exactly one decimal place.”

Example: \( 0.5 = \frac{5}{10} = \frac{1}{2} \) (denominator 2 in simplest form)

Another: \( 0.4 = \frac{4}{10} = \frac{2}{5} \) (denominator 5)

Creative: \( 0.375 = \frac{375}{1000} = \frac{3}{8} \) — requires recognising that 375 and 1000 share a factor of 125. Students rarely go beyond halving or dividing by 5 when simplifying.

Trap: \( 0.7 = \frac{7}{10} \) — the denominator IS 10 in its simplest form because 7 and 10 share no common factors. A student might assume all decimals simplify to “nicer” denominators like 2, 4, or 5, but sevenths of ten cannot be reduced further.

Example: \( \frac{1}{2} \) and \( \frac{2}{4} \) — both equal 0.5

Another: \( \frac{3}{5} \) and \( \frac{6}{10} \) — both equal 0.6

Creative: \( \frac{3}{6} \) and \( \frac{5}{10} \) — both equal 0.5, yet neither looks like the “obvious” answer of \( \frac{1}{2} \). Students focused on the digits may not notice these are all equivalent to one half.

Trap: \( \frac{3}{8} \) and \( \frac{3}{10} \) — a student might argue these are equivalent because both have a numerator of 3. But \( \frac{3}{8} = 0.375 \) and \( \frac{3}{10} = 0.3 \) — the denominator completely changes the value, even when the numerator stays the same. Having the same numerator does not make two fractions equal.

Example: \( \frac{1}{3} = 0.333\ldots \)

Another: \( \frac{1}{6} = 0.1666\ldots \)

Creative: \( \frac{1}{7} = 0.142857142857\ldots \) — a repeating block of 6 digits. Students rarely see recurring decimals with long repeating cycles.

Trap: \( \frac{1}{8} = 0.125 \) — a student might think this recurs because 8 is not a factor of 10 or 100. But \( 8 = 2^3 \), and since 2 is a prime factor of 10, the decimal terminates (after 3 decimal places). Whether a fraction recurs depends on whether the simplified denominator has only factors of 2 and 5 — not on whether it divides neatly into 10 or 100.

Example: \( \frac{16}{5} = 3.2 \)

Another: \( \frac{13}{4} = 3.25 \)

Creative: \( \frac{10}{3} = 3.333\ldots \) — an improper fraction that yields a recurring decimal right in the requested range.

Trap: \( 3 \frac{1}{5} \) — a student might write this because it equals 3.2, but it is a mixed number, not an improper fraction!

A fraction in simplest form gives a terminating decimal only when its denominator has no prime factors other than 2 and 5 (the prime factors of 10). For example, \( \frac{3}{8} = 0.375 \) terminates because \( 8 = 2^3 \), and \( \frac{3}{5} = 0.6 \) terminates because 5 is already a prime factor of 10.

But \( \frac{1}{3} = 0.333\ldots \) recurs because 3 is not a factor of any power of 10. Similarly, \( \frac{1}{7} = 0.142857\ldots \) recurs. The key test: simplify the fraction fully, then check whether the denominator’s prime factorisation contains only 2s and 5s.

Equivalent fractions represent the same number — they are just different ways of writing the same value. Multiplying the numerator and denominator by the same number doesn’t change the value of a fraction: \( \frac{1}{4} = \frac{2}{8} = \frac{3}{12} = \frac{25}{100} \), and they all equal 0.25.

Students sometimes think that different-looking fractions should give different decimals, especially when the denominators are very different (e.g. \( \frac{1}{2} \) vs \( \frac{50}{100} \)). But if two fractions are equivalent, their decimal forms are identical — since the decimal is just another representation of the same number.

The opposite is true: as the denominator increases, the unit fraction gets smaller. \( \frac{1}{2} = 0.5 \), \( \frac{1}{3} = 0.333\ldots \), \( \frac{1}{4} = 0.25 \), \( \frac{1}{5} = 0.2 \), \( \frac{1}{10} = 0.1 \), \( \frac{1}{100} = 0.01 \). Dividing 1 by a larger number always gives a smaller result.

Students may confuse this because “bigger numbers usually mean bigger answers.” But with unit fractions, the denominator tells you how many pieces you’ve divided 1 into — more pieces means each piece is smaller. The fraction wall clearly shows how increasing the denominator shrinks the pieces.

Many fractions with denominators that are multiples of 3 do produce recurring decimals: \( \frac{1}{3} = 0.333\ldots \), \( \frac{1}{6} = 0.1666\ldots \), \( \frac{5}{9} = 0.555\ldots \) all recur. However, what matters is the denominator in the fraction’s simplest form, not its original form.

For example, \( \frac{3}{6} \) has a denominator of 6 (a multiple of 3), but it simplifies to \( \frac{1}{2} = 0.5 \), which terminates. Similarly, \( \frac{9}{12} = \frac{3}{4} = 0.75 \) — the original denominator 12 is a multiple of 3, but the simplified denominator \( 4 = 2^2 \) has only factors of 2. The termination rule applies to the simplified denominator.

The student has concatenated the numerator and denominator as digits after the decimal point. This is not how fraction-to-decimal conversion works. To convert \( \frac{3}{5} \), you divide the numerator by the denominator: \( 3 \div 5 = 0.6 \). Alternatively, find an equivalent fraction with a power-of-10 denominator: \( \frac{3}{5} = \frac{6}{10} = 0.6 \).

A quick check shows the error: \( \frac{3}{5} \) is more than a half (since 3 > half of 5), so the decimal must be greater than 0.5. But 0.35 < 0.5, which should ring alarm bells. This method would also claim that \( \frac{1}{2} = 0.12 \), which is clearly absurd.

The answer is correct, but the reasoning relies on a dangerous rule: “always halve to simplify.” This only works when both the numerator and denominator happen to be even. It fails completely when they aren’t.

Consider \( 0.3 = \frac{3}{10} \). Halving gives \( \frac{1.5}{5} \), which is not a valid simplified fraction. Or \( 0.9 = \frac{9}{10} \) — halving gives \( \frac{4.5}{5} \), which is nonsensical. The correct method is to divide both numerator and denominator by their highest common factor (HCF). For \( \frac{8}{10} \), the HCF is 2, so dividing by 2 gives \( \frac{4}{5} \). The student’s “halve” method gave the right answer here only because the HCF happened to be 2.

The student has used 10 as the denominator regardless of how many decimal places the number has. This is the “all decimals are tenths” misconception. While 0.1 (one decimal place) is indeed \( \frac{1}{10} \), the denominator must match the place value of the last digit.

0.125 has three decimal places, so the last digit is in the thousandths column. The correct fraction is \( \frac{125}{1000} \), which simplifies to \( \frac{1}{8} \). The rule is: one decimal place → divide by 10, two decimal places → divide by 100, three decimal places → divide by 1000. Checking: \( \frac{125}{10} = 12.5 \), not 0.125.

The student has rounded a recurring decimal to a terminating one, losing the key information that \( \frac{2}{3} \) is a repeating decimal. 0.6 is not equal to \( \frac{2}{3} \) — it equals \( \frac{6}{10} = \frac{3}{5} \), which is a different number entirely. If we convert back: \( 0.6 = \frac{3}{5} \), not \( \frac{2}{3} \).

The correct decimal representation is \( 0.666\ldots \) or, using recurring notation, \( 0.\dot{6} \). Recurring decimals are a perfectly valid way to write numbers — they don’t need to be rounded. This matters particularly because rounding changes the value: \( 0.6 \times 3 = 1.8 \), but \( \frac{2}{3} \times 3 = 2 \). The recurring decimal preserves the exact relationship to the original fraction; the rounded version does not.

The student is confusing the digits of the fraction with place value. The whole number 2 is correct, but \( \frac{2}{5} \) is equal to \( \frac{4}{10} \), which is 0.4. The correct answer is 2.4.

The student’s answer of 2.25 is actually \( 2 \frac{25}{100} \) or \( 2 \frac{1}{4} \). You cannot simply concatenate the numerator and denominator to form the decimal part of a mixed number.