# Multiply each side of an inequality by a different amount

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

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

$$\frac{2}{10}<\frac{1}{2}$$

it is true to say that

$$\frac{1}{10}.\frac{2}{10}<\frac{1}{2}.\frac{1}{2}$$

?

• Is there a proof, or is it "just obvious" please? What about if one or both of the multipliers is negative? Dec 28, 2021 at 9:54
• 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. Dec 28, 2021 at 10:14

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 (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 $$a we also know that the following expression must be true: $$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): $$ax < by$$ 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 $$x) if $$a<0$$. And: $$ay < by$$ is false (for $$x) 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.

• I think this is indeed the right way to understand what's going on. 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 $$0 and $$0 then $$0

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