Inequalities tell us that one expression is larger (or smaller) than another. So when we write $2 < 3$, we know that $2$ is smaller than $3$. Equivalently, there must be some positive value that we can add to $2$ to get $3$ (in this case, that value is $1$). Indeed, we can define an inequality in this way. We say $a < b$ if and only if there exists some positive $\epsilon$ such that $a + \epsilon = b$.

If variables appear in one or both of the sides of an inequality, then it is likely that only certain values for those variables make the inequality true.

When we find the largest set of values that can be used in place of these variables to make it true, we say we have found the solution to the inequality -- very similarly to what is the case when we have found the solution to an equation.

As an example, consider

$$2x-1 > 5$$Note, this bears so much similarity to the equation $2x-1=5$. We might hope that we can solve it in the same way, namely -- add $1$ to both sides, and then divide both sides by $2$ (or equivalently, multiply both sides by $1/2$.

However, before we do this, we must assure ourselves that such operations are legitimate -- that if we add some value (maybe even a negative value) to both sides of an inequality, or if we multiply both sides by some value (again, possibly even a negative value), then the inequality that results is equivalent to the one with which we started (i.e., it has the same solution set).

Let's take a look first at the legitimacy of adding a common value to both sides of an inequality to produce an equivalent inequality.

Given any real value $c$ can we show that if $a <b$, then $a + c <b + c$?

Absolutely! If $a <b$, then by our definition in the first paragraph above, there has to be some positive $\epsilon$ so that $$a + \epsilon = b$$

Since we know that we can add $c$ to both sides of an equation, we can then say

$$(a + \epsilon) + c = b + c$$or equivalently,

$$(a + c) + \epsilon = b + c$$which in turn tells us what we wanted to know, namely $$a + c <b + c$$

This process is completely reversible too -- if $a+c <b+c$, then there is some positive $\epsilon$ such that $a + c + \epsilon = b + c$. This being an equation, we are free to subtract the $c$ from both sides to obtain $a + \epsilon = b$. Noting that $\epsilon$ is positive, we have $a < b$.

But then any solution to $a < b$ is a solution to $a+c < b+c$ and vice-versa, so adding any value $c$ to both sides of an inequality results in an equivalent inequality (one with the same solution set)

Now, consider multiplying both sides by some constant $c$.

If $c$ is positive, we are in luck...

Just as before, if $a <b$, then there is some positive $\epsilon$ where $a + \epsilon = b$. Being an equation now, we can multiply both sides by $c$ to get $ac + c\epsilon = bc$. We hope that there is some positive $\epsilon_2$ where $ac + \epsilon_2 = bc$. There is of course -- just take $\epsilon_2 = c\epsilon$.

This too is reversible. Starting with $ac \lt bc$, we have $ac + \epsilon = bc$ for some positive $\epsilon$. Now, multiply both sides by $1/c$ (which is defined since $c \neq 0$, by virtue of it being positive). This gives us $a + \epsilon/c = b$. Noting that $\epsilon/c$ is positive, we have $a \lt b$.

As before, this means that when $c$ is positive, $a < b$ and $ac < bc$ are completely equivalent in terms of their solutions sets.

However, notice what happens when we try to multiply both sides by $c=-1$...

Starting again with an assumption that $a \lt b$, which implies there is some positive $\epsilon$ where $a + \epsilon = b$, we multiply both sides by $-1$ to get $-a - \epsilon = -b$. Adding $\epsilon$ to both sides, we see this is equivalent to $-b + \epsilon = -a$. But this tells us that $-a \gt -b$, not the other way around! However, the implication does work in reverse... If $-b \gt -a$, then there exists some positive $\epsilon$ where $-b + \epsilon = -a$. Multiplying both sides of this equation by $-1$, we have $b - \epsilon = a$, which tells us $a + \epsilon = b$ and consequently $a \lt b$. Thus, if we multiply both sides of an inequality by $-1$, we only produce an equivalent inequality if we also reverse the inequality symbol.

With regard to multiplying both sides of an inequality by any other negative value, note that we can split that into two operations: one where we multiply by the positive magnitude of the value in question (which produces an equivalent inequality), and the other where we multiply by $-1$ (which again produces an equivalent inequality only if we reverse the inequality symbol in what results).

Putting this all together, we can multiply both sides of any inequality by a positive value to produce an equivalent inequality, but if we multiply both sides by a negative value we only produce an equivalent inequality if we also reverse the inequality symbol as well!