Can I Add 1/2 + 1/3 by Adding Tops and Bottoms to Get 2/5?

A student stares at \tfrac{1}{2} + \tfrac{1}{3} and makes the natural guess: add the tops, add the bottoms, get \tfrac{2}{5}. The move feels symmetric, it looks like how every other operation works, and it is completely wrong. The right answer is \tfrac{5}{6} — more than twice \tfrac{2}{5}. This is the most common fraction mistake in all of school arithmetic, and it is worth dismantling carefully.

The trap

The "add tops and bottoms" rule sounds reasonable because two other rules do work that way. For multiplication:

\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}

Yes — multiply tops, multiply bottoms. So why not for addition?

Because addition and multiplication are not the same operation, and fractions are not just pairs of numbers — they are single numbers that happen to be written as pairs. Adding them requires adding the actual numbers, not mashing the components together.

Why \tfrac{2}{5} can't possibly be right

Here's a sanity check that takes three seconds and kills the wrong answer instantly.

\tfrac{1}{2} is half of a whole — the bigger of the two fractions you are adding. Whatever \tfrac{1}{2} + \tfrac{1}{3} is, it has to be bigger than \tfrac{1}{2}, because you are adding a positive number to it. But \tfrac{2}{5} = 0.4, which is less than \tfrac{1}{2} = 0.5. The proposed answer is smaller than one of the things you are supposed to be adding. That is impossible for a sum of positives.

\frac{2}{5} = 0.4, \qquad \frac{1}{2} = 0.5, \qquad \frac{2}{5} < \frac{1}{2}

So before you even do the real arithmetic, the "add tops and bottoms" rule gives an answer that fails the most basic consistency check: a sum cannot be less than its own parts when the parts are positive.

The picture of what goes wrong

Take a chapati. Cut it in half — each piece is \tfrac{1}{2} of the whole. Now take a second chapati and cut it in thirds — each piece is \tfrac{1}{3} of the whole. Put one half-piece and one third-piece next to each other and ask: what fraction of one whole chapati is this pile?

Cut one chapati into six equal cells. Half the chapati takes up three cells; a third of it takes up two more. The total is five cells out of six, which is $\tfrac{5}{6}$ — not $\tfrac{2}{5}$. The picture demands a common cell size, and once you have it, the addition becomes counting.

The problem with "add tops and bottoms" is that a half and a third are not the same size of piece. Half is a big piece; a third is a medium piece. You cannot meaningfully "add them" without first chopping both into the same size of piece. The natural choice is sixths — the smallest size that fits both. A half is three sixths; a third is two sixths; together they are five sixths.

\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}

Once both fractions are in sixths, the denominator is one piece size, and you can just count shaded pieces.

Why "add tops and bottoms" is a wrong operation

What is \tfrac{a+c}{b+d} actually computing, if not \tfrac{a}{b} + \tfrac{c}{d}? It turns out this operation has a name — it is called the mediant — and it computes something different and less useful than the sum. For two fractions \tfrac{a}{b} and \tfrac{c}{d}, the mediant \tfrac{a+c}{b+d} is always a number between the two original fractions, not their sum.

For \tfrac{1}{2} and \tfrac{1}{3}, the mediant is \tfrac{2}{5}, which sits between \tfrac{1}{3} \approx 0.333 and \tfrac{1}{2} = 0.5. That is what \tfrac{2}{5} does represent — a number in between the two — and it is a meaningful operation in its own right. It just is not addition.

Why the mediant is between: if \tfrac{a}{b} < \tfrac{c}{d}, then multiplying out gives ad < bc. The mediant \tfrac{a+c}{b+d} then satisfies (a+c)b = ab + bc > ab + ad = a(b+d), so \tfrac{a+c}{b+d} > \tfrac{a}{b}. A symmetric argument shows the mediant is less than \tfrac{c}{d}. So the mediant is strictly between the two original fractions — which is why it cannot be the sum.

The mediant shows up in beautiful places — the Stern–Brocot tree, Farey sequences, the theory of continued fractions — but it is not addition, and anyone who uses it as addition will get systematically wrong answers.

The real rule

Addition of fractions requires a common denominator. Find a denominator both can be written over, rewrite, then add numerators only. The denominator stays the same because the piece size has not changed.

\frac{a}{b} + \frac{c}{d} = \frac{ad}{bd} + \frac{cb}{bd} = \frac{ad + cb}{bd}

The algorithm: multiply \tfrac{a}{b} top-and-bottom by d, multiply \tfrac{c}{d} top-and-bottom by b, now both have denominator bd, add the numerators ad + cb, put over bd. For \tfrac{1}{2} + \tfrac{1}{3}:

\frac{1}{2} + \frac{1}{3} = \frac{1 \cdot 3}{2 \cdot 3} + \frac{1 \cdot 2}{3 \cdot 2} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}

If the two denominators share a common factor, you can use the LCM instead of the full product to keep the numbers smaller. But the product bd always works — it is the foolproof fallback.

Why this error is so contagious

Three things make the "add tops and bottoms" trap genuinely seductive.

Multiplication works that way. Students see \tfrac{2}{3} \times \tfrac{4}{5} = \tfrac{8}{15} and generalise to the wrong operation.
It takes one less step. Real fraction addition needs LCMs, rewriting, and then adding. "Add tops and bottoms" needs… nothing. The shortcut is tempting precisely because it is a shortcut.
Nobody checks the sanity check. If students routinely checked "is my sum bigger than my biggest summand," they would catch this within seconds. But that habit has to be built.

The cure is to always build in a sanity check. If you are adding two positive fractions, the sum has to be at least as big as the larger of the two. \tfrac{1}{2} + \tfrac{1}{3} must be at least \tfrac{1}{2}. Any answer smaller than \tfrac{1}{2} is disqualified.

Sanity-check three candidate answers.

Which of these could be \tfrac{2}{5} + \tfrac{1}{3}? Choose one without calculating.

(a) \tfrac{3}{8} (b) \tfrac{2}{15} (c) \tfrac{11}{15}

Check size. The bigger of \tfrac{2}{5} = 0.4 and \tfrac{1}{3} \approx 0.333 is \tfrac{2}{5} = 0.4. The sum must be larger than 0.4.

(a) \tfrac{3}{8} = 0.375 — less than 0.4. Impossible.
(b) \tfrac{2}{15} \approx 0.133 — way less than 0.4. Impossible.
(c) \tfrac{11}{15} \approx 0.733 — bigger than 0.4. Plausible.

Verify. Using the real rule, \tfrac{2}{5} + \tfrac{1}{3} = \tfrac{6}{15} + \tfrac{5}{15} = \tfrac{11}{15}. Option (c) is correct.

Why the sanity check was decisive: options (a) and (b) were both the result of the "add tops and bottoms" trap applied in slightly different ways. The check "sum must exceed the larger summand" rules them both out without any computation. A habit this cheap should be automatic.

The takeaway

Fractions are numbers, not pairs of numbers. Addition of numbers does not care what notation they are written in — it cares about their positions on the line. When you add \tfrac{1}{2} and \tfrac{1}{3}, you are adding two lengths. The lengths only combine cleanly when you measure them with the same ruler — which is exactly what finding a common denominator does.

"Add tops and bottoms" forgets that fractions are numbers, treats the numerator and denominator as independent counts, and gives an answer that is sometimes a meaningful different quantity (the mediant) but never the sum.

If the rule ever tries to tempt you again, remember the chapati picture. Half a chapati plus a third of a chapati is visibly more than half a chapati. \tfrac{5}{6} is more than half. \tfrac{2}{5} is less. The trap identifies itself.