Simplifying and Equivalent Fractions

Visually, we can see that 6 of the smaller eighths take up exactly the same space as 3 of the larger quarters.

Mathematically, if you divide both the numerator and denominator of \(\frac{6}{8}\) by 2, you get \(\frac{3}{4}\). Dividing both parts by the same number is the same as grouping smaller pieces into larger chunks.

To simplify a fraction, the numerator and denominator must share a common factor greater than 1. The factors of 3 are just 1 and 3. The factors of 7 are just 1 and 7. The only shared factor is 1, so there is nothing to divide both by. \(\frac{3}{7}\) is already in its simplest form.

A fraction is in simplest form when the HCF of its numerator and denominator is 1. Since 3 and 7 are both prime (and different primes), their HCF is 1.

Simplify both: \(\frac{4}{6} = \frac{2}{3}\) (dividing by 2) and \(\frac{10}{15} = \frac{2}{3}\) (dividing by 5). Since both simplify to \(\frac{2}{3}\), they are equivalent.

Alternatively, cross-multiply: \(4 \times 15 = 60\) and \(6 \times 10 = 60\). Since the cross-products are equal, the fractions are equivalent. Two fractions that look very different can be equal — what matters is their simplified form.

Multiply numerator and denominator by 2 to get \(\frac{2}{6}\). By 3 to get \(\frac{3}{9}\). By 4 to get \(\frac{4}{12}\). By 100 to get \(\frac{100}{300}\). We can multiply by any whole number — and there are infinitely many whole numbers — so there are infinitely many equivalent fractions.

Each equivalent fraction is like cutting the same portion into smaller and smaller pieces. 1 out of 3 equal slices is the same amount as 2 out of 6, 10 out of 30, or a million out of three million.

Example: \(\frac{4}{10}\)

Another: \(\frac{6}{15}\)

Creative: \(\frac{200}{500}\) — using a very large multiplier. Or \(\frac{20}{50}\), which links to percentages (40%).

Trap: \(\frac{3}{6}\) — a student might add 1 to the numerator and 1 to the denominator, thinking “I did the same thing to both.” But \(\frac{2}{5} = 0.4\) while \(\frac{3}{6} = 0.5\). Adding the same number doesn’t preserve equivalence — you must multiply.

Example: \(\frac{6}{8}\)

Another: \(\frac{9}{12}\)

Creative: \(\frac{75}{100}\) — connects to percentages since 75% = \(\frac{3}{4}\). Or \(\frac{300}{400}\).

Trap: \(\frac{4}{5}\) — a student might think “3 + 1 = 4 and 4 + 1 = 5, so \(\frac{4}{5}\) simplifies to \(\frac{3}{4}\).” But \(\frac{4}{5}\) is already in simplest form (HCF of 4 and 5 is 1), and \(\frac{4}{5} \neq \frac{3}{4}\).

Example: \(\frac{3}{5}\)

Another: \(\frac{7}{10}\)

Creative: \(\frac{11}{13}\) — two primes that feel “large” make students less certain. Or \(\frac{1}{1}\), which equals 1 but IS a fraction in simplest form.

Trap: \(\frac{2}{4}\) — it looks simple because the numbers are small, but HCF(2, 4) = 2, so it simplifies to \(\frac{1}{2}\). Small numbers don’t mean “already simplified.” This is the “small numbers = simplest form” misconception.

Example: \(\frac{6}{3}\) (= 2)

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

Creative: \(\frac{100}{4}\) (= 25) — a surprisingly large value. Or \(\frac{7}{1}\) — technically improper and already a whole number.

Trap: \(\frac{7}{3}\) — a student might think “7 ÷ 3 = 2 remainder 1, so it equals 2.” But \(\frac{7}{3} = 2\frac{1}{3}\), not a whole number. The numerator must be an exact multiple of the denominator.

This is almost always false, but it works in one special case: when the original fraction equals 1. For example, \(\frac{3}{3} \to \frac{4}{4}\) — both equal 1.

For any other fraction, it fails dramatically. Consider \(\frac{1}{100}\). If you add 1 to the top and bottom, you get \(\frac{2}{101}\). Since \(\frac{1}{100} = 1\%\) and \(\frac{2}{101} \approx 2\%\), you have almost doubled the value of the fraction! Equivalence is completely destroyed.

This is one of the most persistent misconceptions in fractions. The key word is multiply, not add.

Simplifying a fraction doesn’t change its value — it just uses smaller numbers to express the same amount. \(\frac{3}{4}\) and \(\frac{75}{100}\) sit at exactly the same point on a number line.

Think of it like pizza: cutting a pizza into 100 tiny slices and eating 75 of them is the same amount of food as eating 3 quarters. You have fewer pieces in the simplified version, but each piece is larger.

If both the numerator and denominator are even, they share a factor of 2. Dividing both by 2 gives a simpler equivalent fraction. For example, \(\frac{4}{6} \to \frac{2}{3}\), and \(\frac{8}{10} \to \frac{4}{5}\).

This is a useful quick-check: spot two even numbers and you know immediately the fraction isn’t fully simplified yet.

True for \(\frac{3}{7}\) (HCF = 1, both prime), but false for \(\frac{9}{15}\) — both odd, yet HCF(9, 15) = 3, so \(\frac{9}{15}\) simplifies to \(\frac{3}{5}\).

The key is whether the numerator and denominator share a common factor, not whether they are odd or even. Students who rely on “divide by 2” as their only simplifying strategy get stuck here.

Note: All three fractions are equivalent to \(\frac{2}{3}\).

Note: All three are in simplest form.

The student is comparing numerators and denominators separately, as if they were whole numbers. This is the “bigger numbers mean a bigger fraction” misconception. A bigger denominator means the whole is cut into more pieces, making each piece smaller.

Rewrite with a common denominator: \(\frac{3}{4} = \frac{9}{12}\). Now compare: \(\frac{9}{12} > \frac{5}{12}\), so \(\frac{3}{4}\) is larger.

The answer happens to be correct — \(\frac{16}{64}\) does equal \(\frac{1}{4}\) — but the method is completely wrong. The student is cancelling digits, not dividing by common factors. This “works” here by pure coincidence.

Apply the same method to \(\frac{13}{39}\): “cancel the 3s” gives \(\frac{1}{9}\), but actually \(\frac{13}{39} = \frac{1}{3}\) (since \(39 = 13 \times 3\)). The correct method is to find that HCF(16, 64) = 16, then divide both by 16.

The student has found a correct equivalent fraction, but \(\frac{4}{6}\) is not fully simplified. 4 and 6 still share a common factor of 2: \(\frac{4}{6} = \frac{2}{3}\). This is the “incomplete simplification” error — dividing by a common factor that isn’t the highest common factor.

HCF(12, 18) = 6, so dividing both by 6 in one step gives the fully simplified answer \(\frac{2}{3}\). A fraction is only fully simplified when the HCF of numerator and denominator is 1.

The student has confused adding with multiplying. This is the “add the same to both” misconception. To create equivalent fractions, you must multiply (or divide) both the numerator and denominator by the same number — not add.

\(\frac{3}{5} = 0.6\) but \(\frac{4}{6} \approx 0.667\), so they are not equal. Correct equivalents: \(\frac{3}{5} \times \frac{2}{2} = \frac{6}{10}\), or \(\frac{3}{5} \times \frac{3}{3} = \frac{9}{15}\).

This is the “Pseudo-prime” trap. Numbers like 51, 57, and 91 often look prime because they are odd and don’t end in 5. However, 57 is actually divisible by 3 (since 5 + 7 = 12).

\(57 = 3 \times 19\). So the fraction simplifies to: \(\frac{3 \times 19}{19} = 3\).