Multiplying Inequalities

Multiplying Two Inequalities

The previous tutorial showed how to combine inequalities by addition. Multiplication requires closer attention to signs: the order theorem tells us how an inequality changes when both sides are multiplied by a number of known sign, but multiplying two inequalities involves factors that may themselves vary. In particular, one cannot multiply corresponding sides without first checking that the factors are nonnegative.

Throughout this tutorial, \(F\) denotes an ordered field. The arguments therefore apply in \(\mathbb{R}\). We will first establish a rule for two pairs of nonnegative quantities. Then we will use signs to find bounds when one or both intervals lie below zero. The multiplication-and-order theorem from “Properties of Inequalities” will be used when one factor is fixed and its sign is known.

Proof. Since \(c\geq0\), the multiplication-and-order theorem gives \(ac\leq bc\) from \(a\leq b\). Since \(b\geq0\), the same theorem gives \(bc\leq bd\) from \(c\leq d\). Transitivity of non-strict inequalities therefore gives \(ac\leq bd\).

For the equality statement, suppose first that \(ac=bd\). The chain \(ac\leq bc\leq bd\) then forces \(ac=bc\) and \(bc=bd\): if either comparison were strict, transitivity would give \(ac<bd\). Now \(bc-ac=(b-a)c\), so \(ac=bc\) exactly when \((b-a)c=0\). The zero-product property in a field gives \(b-a=0\) or \(c=0\), that is, \(a=b\) or \(c=0\). Similarly, \(bd-bc=b(d-c)\), so \(bc=bd\) exactly when \(b=0\) or \(c=d\). Thus equality implies both stated conditions. Conversely, if both conditions hold, then \(ac=bc\) and \(bc=bd\), so \(ac=bd\). \(\square\)

The two comparisons in the proof also explain strictness. If \(a<b\) and \(c>0\), multiplying \(a<b\) by \(c\) gives \(ac<bc\), and hence \(ac<bd\). If \(b>0\) and \(c<d\), multiplying \(c<d\) by \(b\) gives \(bc<bd\), again yielding \(ac<bd\). These are useful sufficient conditions for a strict result; when neither holds, the equality conditions in the theorem help determine what happens.

One Nonnegative Interval and One Nonpositive Interval

When one factor is nonnegative and the other nonpositive, products are nonpositive. The bounds are paired differently than they are when all quantities are nonnegative: the largest nonnegative factor combined with the smallest (most negative) factor gives a lower bound, while the smallest nonnegative factor combined with the largest (least negative) factor gives an upper bound.

Proof. For the lower bound, \(c\leq0\), so multiplying \(x\leq b\) by \(c\) reverses the comparison and gives \(bc\leq xc\). Also \(x\geq0\) and \(c\leq y\), so multiplying \(c\leq y\) by \(x\) gives \(xc\leq xy\). Transitivity yields \(bc\leq xy\).

For the upper bound, \(y\leq0\), so multiplying \(a\leq x\) by \(y\) reverses the comparison and gives \(xy\leq ay\). Since \(a\geq0\) and \(y\leq d\), multiplying \(y\leq d\) by \(a\) gives \(ay\leq ad\). Therefore \(xy\leq ad\), as required. Both conclusions include the cases in which a factor is zero. \(\square\)

When Both Factors Are Nonpositive

If both factors are nonpositive, their product is nonnegative. A convenient way to see the correct bounds is to negate both factors. Negation reverses order, as established in “Properties of Inequalities,” and the product of the two negatives is unchanged because \((-x)(-y)=xy\).

Proof. Negating \(a\leq x\leq b\leq0\) gives \(0\leq -b\leq -x\leq -a\), and negating \(c\leq y\leq d\leq0\) gives \(0\leq -d\leq -y\leq -c\). Apply the theorem for nonnegative inequalities to the lower endpoints and to the upper endpoints. It gives

Using \((-u)(-v)=uv\), this becomes \(bd\leq xy\leq ac\). The inequalities include zero endpoints, so no separate exception is needed when one of the original factors can equal zero. \(\square\)

Why Signs Cannot Be Ignored

The rule for multiplying by a fixed number, from “Properties of Inequalities,” has three cases: a positive multiplier preserves order, a negative multiplier reverses order, and a zero multiplier makes both sides equal. In a product of two inequalities, the factors used as multipliers may vary, so their signs must be established before the rule is applied.

For example, \(1<2\) and \(-3<-2\) are both true, but multiplying their left sides and right sides in parallel would suggest \(1(-3)<2(-2)\), or \(-3<-4\), which is false. The issue is that the second interval is negative. Indeed, multiplying \(1<2\) by the negative number \(-2\) reverses the order, giving \(-4<-2\), not \(-2<-4\). This is why the opposite-sign theorem pairs endpoints differently from the nonnegative theorem.

A reliable method is to locate each interval relative to zero, then decide whether multiplication preserves or reverses each comparison used in the argument. If both intervals cross zero, none of the three product-bound theorems above applies directly to the whole intervals. Such a case requires a separate analysis; assuming that the lower endpoints multiply to a lower bound and the upper endpoints to an upper bound can give an incorrect result.