TITLE: Multiply each side of an inequality by a different amount
QUESTION [0 upvotes]: 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
$\frac{1}{10}.\frac{2}{10}<\frac{1}{2}.\frac{1}{2}$
?

REPLY [2 votes]: 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.
{"set_name": "stack_exchange", "score": 0, "question_id": 4343576}