What are the rules for multiplying inequalities by unequal quantities please?

For example, by what property can we be certain that, given


it is true to say that



  • $\begingroup$ Is there a proof, or is it "just obvious" please? What about if one or both of the multipliers is negative? $\endgroup$ Dec 28, 2021 at 9:54
  • 1
    $\begingroup$ Proof of this would use $x<y\implies f(x)<f(y)$ when $f$ is increasing. Here, for $0<a<b$ and $0<c<d$, we get $0<ac<bc$ (because, as $c>0$, the map $x\mapsto cx$ is increasing), and also $bc<bd$ (because $x\mapsto bx$ is increasing). Hence $0<ac<bd$. With negative slopes you need to pay more attention. $\endgroup$
    – nejimban
    Dec 28, 2021 at 10:14

2 Answers 2


The way to look at this isn't as "one step" that multiplies each side by a different constant, but rather as a comparison of a series of inequalities such that you know one is bigger than the next is bigger than the next.

For example, lets say we're starting with the inequality: $$ x < y $$ and we want to know when the following inequality holds true, $$ ax \stackrel{?}{<} by $$ for any pair of constants $a$ and $b$. To make things easier for now, lets define $a\geq0$ and $a<b$ (we can look into more general cases later).

So, if we multiply both sides of the inequality by the first constant we have a new expression: $$ ax < ay $$ and from $a\geq0$ and $a<b$ we also know that the following expression must be true: $$ ay < by $$ which means that we know the following chain of inequalities holds true: $$ ax < ay < by $$ So, because we know that $by$ is larger than $ay$ is larger then $ax$... we know (for $a\geq0$ and $a<b$): $$ ax < by $$ Or, if we want to be slightly more concrete in terms of your example (where $x=2/10$ and $y=1/2$): $$ \frac{1}{10} x < \frac{1}{2} y $$

If you want to be more general with $a$ and $b$ such that $a>b$ or negative numbers are involved then the expressions above won't work anymore. Mostly because they break the following steps: $$ ax < ay $$ Is false (for $x<y$) if $a<0$. And: $$ ay < by $$ is false (for $x<y$) if $a>b$.

That isn't to say that we can't "chain inequalities" anymore... is just means that we have to chain a different set of expressions and that we have to be very careful with sign changes.

  • $\begingroup$ I think this is indeed the right way to understand what's going on. $\endgroup$ Dec 28, 2021 at 20:02

One could make several rules that work of this kind. One rule, which your example is a case of:

If $0<a<b$ and $0<c<d$ then $0<ac<bd.$

Can prove in two steps by first showing $ac<ad$ then showing $ad<bd$. both using usual rules.


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