Inequalities and equations look similar, but they obey a different multiplication rule. An equation lets you multiply both sides by anything non-zero and the statement still holds. An inequality only lets you multiply by something whose sign you know — because multiplying by a negative flips the arrow, and multiplying by a zero kills the inequality entirely. When the thing you want to multiply by is an expression involving a variable, you do not know its sign. So the reflex cannot be "cross-multiply." The reflex has to be "split into cases."

This is the single most-common bug in rational inequality problems. Students see \dfrac{1}{x} > 2, cross-multiply to get 1 > 2x, divide by 2, write x < \tfrac{1}{2}, and hand in a wrong answer. The wrong answer includes every negative number — none of which satisfy the original. Here is the reflex that prevents it, every time.

The rule you cannot break

If you want to multiply or divide both sides of an inequality by an expression whose sign you do not know, stop. Split into cases first.

The rule has exactly two escape hatches, and only these two:

  1. The expression is a specific number whose sign you can see (e.g., 7, -3, \pi). Then you multiply and keep or flip the arrow based on that sign.
  2. The expression is provably positive (e.g., x^2 + 1 for real x, or e^x, or a known positive quantity from context). Then you multiply and the arrow stays.

Anything else — x, x - 3, 2x + 1, 1 - x — has unknown sign until you pin it down. Until then, you cannot multiply both sides by it without destroying the logic.

The worked bug: \dfrac{1}{x} > 2

Watch what happens when you ignore the rule. The naive student sees a fraction, wants it gone, and cross-multiplies:

\frac{1}{x} > 2 \quad \overset{\text{?}}{\Longrightarrow} \quad 1 > 2x \quad \Longrightarrow \quad x < \frac{1}{2}.

According to this, every number less than \tfrac{1}{2} works. Try x = -1: the original inequality becomes \tfrac{1}{-1} > 2, i.e., -1 > 2. That is plainly false. Try x = -100: the original becomes -0.01 > 2. Also false. Every negative value fails the original, yet the naive answer endorses them all.

The sickness is right at the cross-multiply step. Multiplying both sides by x assumes x > 0 (so the arrow stays). But we never checked — and if x < 0, the multiplication should flip the arrow. One branch of logic was ignored.

The fix: split into cases

The correct move is to enumerate the sign cases of the expression you want to multiply by and solve each case separately. Here x can be positive or negative (zero is not allowed because \tfrac{1}{x} is undefined there), so there are exactly two cases.

Case 1: x > 0. Multiply both sides by x; since x > 0, the arrow stays.

1 > 2x \quad \Longrightarrow \quad x < \tfrac{1}{2}.

Now intersect with the case assumption x > 0. The numbers satisfying both x > 0 and x < \tfrac{1}{2} form the interval (0,\ \tfrac{1}{2}).

Case 2: x < 0. Multiply both sides by x; since x < 0, the arrow flips.

1 < 2x \quad \Longrightarrow \quad x > \tfrac{1}{2}.

Now intersect with the case assumption x < 0. No number is both less than 0 and greater than \tfrac{1}{2}. This case contributes nothing — the empty set.

Combine. The solution set is (0, \tfrac{1}{2}) \cup \varnothing = (0, \tfrac{1}{2}).

Spot check: x = \tfrac{1}{4} gives \tfrac{1}{1/4} = 4 > 2. Correct. x = 1 gives 1 > 2. False — and indeed 1 \notin (0, \tfrac{1}{2}). x = -1 gives -1 > 2. False — and -1 \notin (0, \tfrac{1}{2}). The answer stands.

The two-branch decision tree

Every case split has the same shape. You pick the expression whose sign you need, list the cases, solve inside each case, then intersect with the case's own assumption. Picture it:

Two-case decision tree for multiplying an inequality by a variable expressionA tree diagram. At the top sits the inequality one over x greater than two with a box labelled multiply by x — sign unknown. The tree branches into two children: case 1 x greater than zero on the left, case 2 x less than zero on the right. Under case 1 the result is x less than one half, intersected with x greater than zero gives the open interval from zero to one half. Under case 2 the result is x greater than one half flipped, intersected with x less than zero gives empty. At the bottom the two contributions are unioned into the final answer, the open interval from zero to one half. 1/x > 2 multiply by x — sign unknown Case 1: x > 0 (no flip) Case 2: x < 0 (flip) Solve: x < 1/2 Intersect with x > 0: (0, 1/2) Solve: x > 1/2 Intersect with x < 0: empty Union of the two cases: x ∈ (0, 1/2)
Every "multiply by a variable expression" step in an inequality is a branch point. Do not try to collapse both branches into one line — that is where the wrong answer comes from. Solve each branch inside its own sign assumption, then union the intersected results.

Why intersecting with the case assumption matters

The easiest-to-miss step is the intersection with the case's own assumption at the bottom of each branch. Inside Case 1 you assumed x > 0, so any solution you derive in that case is only valid where that assumption holds. If the case-branch gave x < \tfrac{1}{2} but you forgot to intersect with x > 0, you would report "x < \tfrac{1}{2}" — and again include all the negative numbers your case had specifically excluded. The intersection is not a technicality; it is the reason splitting works.

When this reflex fires

Any one of these patterns should trigger "case split":

In every case, the trigger is not "fraction" — it is "the thing I want to multiply by has an unknown sign." Denominators just happen to be the most common source of such things.

Aside: the other safe move

For rational inequalities specifically, there is a tidy alternative that avoids the case split: move everything to one side, combine over a common denominator, and do a sign chart. For \tfrac{1}{x} > 2:

\frac{1}{x} - 2 > 0 \quad \Longrightarrow \quad \frac{1 - 2x}{x} > 0.

Now ask where the fraction \tfrac{1 - 2x}{x} is positive. The numerator changes sign at x = \tfrac{1}{2}, the denominator at x = 0. A sign chart on the three intervals (-\infty, 0), (0, \tfrac{1}{2}), (\tfrac{1}{2}, \infty) shows the fraction is positive only on (0, \tfrac{1}{2}). Same answer, no case split. See Quadratic Inequalities for the sign-chart technique in full.

Both methods are correct. Pick whichever feels cleaner for the problem in front of you — but never the naive cross-multiply, which is correct for neither.

The compressed version

See a variable in a divisor or on a side you want to multiply by? Its sign is unknown. Do not cross-multiply. Split into two cases (positive and negative), solve inside each with the correct arrow direction, intersect each case-answer with the case's own sign assumption, and union the results. The naive single-step is the universal inequality bug; the two-case split is the universal fix.

Related: Intervals and Inequalities Preview · Forgetting to Flip the Inequality After Multiplying by a Negative · Why You Can't Just "Square Both Sides" of an Inequality · Quadratic Inequalities