The split \tfrac{a + b}{c} = \tfrac{a}{c} + \tfrac{b}{c} gets used so often in school algebra that it starts to feel like an inviolable law. Fine — as long as the sum lives on top. Every year, students apply the same move when the sum lives on the bottom — \tfrac{c}{a + b} \stackrel{?}{=} \tfrac{c}{a} + \tfrac{c}{b} — and lose half their marks on a simplification they thought was automatic. This satellite is the picture that shows why the top split works and the bottom split does not.

The rule, stated carefully

If c \neq 0, then

\frac{a + b}{c} \;=\; \frac{a}{c} + \frac{b}{c}.

In words: you may distribute division over an addition in the numerator. The denominator stays put.

The analogous-looking statement with the sum in the denominator is not a rule:

\frac{c}{a + b} \;\neq\; \frac{c}{a} + \frac{c}{b} \text{ in general}.

Try numbers. a = 2, b = 3, c = 12.

Not even close. One answer is about one-fourth of the other.

The numerator split is just the distributive law

Division by c is really multiplication by 1/c. Rewrite the top split:

\frac{a + b}{c} \;=\; (a + b) \cdot \tfrac{1}{c} \;=\; a \cdot \tfrac{1}{c} + b \cdot \tfrac{1}{c} \;=\; \tfrac{a}{c} + \tfrac{b}{c}.

Why: the middle step is the plain distributive law for multiplication over addition — the same rule as (a + b) \times k = ak + bk, with k = 1/c. Nothing new was invented for fractions.

So the "distribute division on top" rule isn't really about division. It is about multiplication distributing over addition, dressed up in fraction notation.

Why the denominator split isn't the distributive law

Now try to mimic the derivation with the sum in the denominator. You want

\frac{c}{a + b} \stackrel{?}{=} c \cdot \frac{1}{a + b} \stackrel{?}{=} c \cdot \left(\frac{1}{a} + \frac{1}{b}\right).

The first equality is fine. The second equality says

\frac{1}{a + b} \stackrel{?}{=} \frac{1}{a} + \frac{1}{b}. \quad (\text{WRONG})

And this is the whole crux: the reciprocal of a sum is not the sum of reciprocals. Check with a = 2, b = 3: \tfrac{1}{5} = 0.2, but \tfrac{1}{2} + \tfrac{1}{3} = \tfrac{5}{6} = 0.833. They are not equal. The reciprocal operation doesn't distribute over addition, and that is the single reason the denominator split fails.

The picture: two rectangles stacked, one ladoo

Imagine c = 12 ladoos and you want to share them. The two fractions represent two different sharing problems.

Two side-by-side pictures showing why numerator splits but denominator does notTwo pictures side by side. On the left, a stack of seven candies divided by three children. Below it, the same seven candies split into groups of four and three, each still shared among the same three children — same total, same answer. On the right, seven candies divided between five children. Below it, an incorrect split that gives seven candies to two children and seven candies to three children — total candies are doubled from seven to fourteen, which is why the right-hand split is wrong. TOP split — legal (a+b) / c = a/c + b/c split the 7 candies, keep 3 kids 7/3 = 2.33… = 4/3 + 3/3 = 1.33 + 1 = 2.33 same total shared among same denom. BOTTOM split — illegal c / (a+b) ≠ c/a + c/b split the 5 kids, keep 12 candies 12/5 = 2.4 but 12/2 + 12/3 = 6 + 4 = 10 right side secretly used 24 candies
The green panel: splitting the numerator is legal because the denominator is unchanged — the pool of sharers stays the same. The red panel: splitting the denominator is illegal because the *numerator* quietly doubles — the right-hand side counts the candies twice. The asymmetry isn't a quirk of notation; it is the difference between rearranging a sum and accidentally duplicating a resource.

Test it before you use it

If you ever find yourself tempted to split a denominator, stop and test with two small numbers. Pick a = 2, b = 3, c = 1, and plug both sides in. If they agree, you might have stumbled onto a true identity; if they disagree, your move was wrong. This takes five seconds and saves marks.

The only case where \tfrac{c}{a + b} = \tfrac{c}{a} + \tfrac{c}{b} accidentally holds is — well, it doesn't, unless c = 0 (both sides are 0) or some degenerate configuration. For any real values with a, b, c nonzero, the two sides are different.

Where it bites: simplifying fractions

Students simplify \dfrac{1}{x + h} - \dfrac{1}{x} by "splitting" the denominators and get nonsense. The correct move is to find a common denominator:

\dfrac{1}{x + h} - \dfrac{1}{x} \;=\; \dfrac{x - (x + h)}{x(x + h)} \;=\; \dfrac{-h}{x(x + h)}.

This expression is at the heart of the derivative of 1/x. Splitting a denominator wrongly here is the single most common mistake in limits chapters on JEE — and it propagates silently all the way to the wrong answer.

Where it bites: resistors in parallel

This is also why resistors in parallel don't add their resistance values — they add their reciprocals. Two resistors of 4\,\Omega and 6\,\Omega in parallel give

R_{\text{total}} = \frac{1}{\tfrac{1}{4} + \tfrac{1}{6}} = \frac{1}{\tfrac{5}{12}} = \tfrac{12}{5}\,\Omega = 2.4\,\Omega.

If you had naively written "R_{\text{total}} = 4 + 6 = 10", you would be confusing series with parallel. The correct formula is "one over the sum of the one-overs," and the reason it is that ugly — rather than just 1/(R_1 + R_2) — is precisely that \tfrac{1}{R_1 + R_2} \neq \tfrac{1}{R_1} + \tfrac{1}{R_2}.

The reflex, in one line

Numerator contains a sum? Split freely. Denominator contains a sum? Find a common denominator — never split, never distribute. The rule lives on one side only, because the distributive law lives on the multiplication-over-addition side only.

Related: Operations and Properties · Distributive Property as an Area · Fractions and Decimals · Algebraic Identities