Why do we reverse inequality sign when dividing by negative number?

Solution 1:

Dividing by a negative number is the same as dividing by a positive number and then multiplying by $-1$. Dividing an inequality by a positive number retains the same inequality. But, multiplying by $-1$ is the same as switching the signs of the numbers on both sides of the inequality, which reverses the inequality: $$ \tag{1} a\lt b\quad\iff -a\gt -b. $$ You should be able to convince yourself why the above is true by looking at the number line and considering the various cases involved.

Seeing why (1) is true is not too hard.

Here is the hand waving approach I suggested above:

Consider, for example, in (1), the case when $a$ is negative and $b$ is positive. We have $a<b$. Then $-a$ is positive and $-b$ is negative. Thus, we have $-b<-a$.

As another case, suppose $a$ and $b$ are both negative with $a<b$. Switching the signs here makes the resulting numbers both positive with $-a>-b$ (you can see this by drawing the points on the number line and noting that with the given conditions, $b$ is closer to the origin than $a$):

).

The other cases can be handled similarly.

But, perhaps a bit of rigor is needed here.

Recall that
$$a<b\quad\text{ if and only if }\quad b-a\quad \text{ is positive.}$$ Now, $b-a$ is positive if and only if $(-a)-(-b) =-a+b=b-a$ is positive. So $a<b$ if and only if $-a> -b$.

Solution 2:

Multiplying or dividing an inequality by $-1$ is exactly the same thing as moving each term to the other side. But then if you switch side for all terms, each term faces the opposite "side" of inequality sign...

For example:

$2x < -3$

Moving them on the other side yields:

$3 < -2x$ which is the same as $-2x > 3$...

Solution 3:

Let $c$ be a negative number. In the case of multiplying both sides of an inequality by $c$, note that the function $f$ defined by $f(x) = cx$ is strictly decreasing on the entire real line. By definition, this means that if $x_{1} < x_{2},$ then $f\left(x_{1}\right) > f\left(x_{2}\right)$ (i.e. $cx_{1} > cx_{2}$). Incidentally, this is equivalent to $x_{2} > x_{1}$ implying $f\left(x_{2}\right) < f\left(x_{1}\right)$, so $f$ also reverses both types of strict inequalities. Moreover, it is not difficult to see that a strictly decreasing function reverses both types of non-strict inequalities. As for dividing both sides by a negative number, note that the function $g$ defined by $g(x) = \frac{1}{c}x$ is strictly decreasing on the entire real line. The same explanation can be used for taking the reciprocal of both sides of an inequality, when both sides are positive or when both sides are negative. In general, if a function $h$ is strictly decreasing on an interval $I$, then we can "take $h$" of both sides of an inequality as long as both sides belong to $I$ and we reverse the inequality. Similarly, strictly increasing functions preserve inequalities. This gives a sometimes useful application of the calculus task of determining on what interval(s) a function might be increasing or decreasing, by the way. For example, $\arctan(x)$ is strictly increasing on the entire real line, so you can take the arctangent of both sides of an inequality (keeping the inequality type the same).

Solution 4:

HINT $\rm\ \ \ 9\: >\: -3\:x \iff 3(x+3)\: >\: 0 \iff x+3\: >\: 0 \iff x\: >\: -3$

Therefore, by shifting it to comparison to $\:0$, we've reduced the comparison to an application of the law of signs, viz. above if $\rm\ y > 0\ $ then $\rm\ yz > 0\iff z>0\ $ where $\rm\ y = 3,\ \ z = x+3\ $ above.

Solution 5:

Consider the simplest inequality of all: $x>0$. This simply says that $x$ is a positive number. Then $ -x > -0 $ is FALSE (note that -0=0) but $-x< 0$ is true, since $-x$ is obviously negative if $x$ is positive.