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

We all learned in our early years that when dividing both sides by a negative number, we reverse the inequality sign.

Take $-3x < 9$

To solve for x, we divide both sides by -3 and get

$x > -3$

Why is the reversal of inequality? What is going in terms of number line that will help me understand the concept better?

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I won't be able to vote on answers until the "SE day" ends, so: the right idea is to reduce your inequality to the either of the forms $\text{stuff} > 0$ or $\text{stuff} \geq 0$. Multiplication by $-1$ could be treated as reflection about the number line. –  Ｊ. Ｍ. Dec 28 '11 at 23:34
I already found the answer which I wanted to rehash as answer to my own question but I must wait for 5 more hours. Instead I give link on Inequalities by Lawrence Spector which answers it in thorough detail introducing theorem of inequalities and their proofs. –  Sniper Clown Dec 28 '11 at 23:53

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 $-a> -b$.

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Note: In particular, intuitively, dividing/multiplying by a negative number is like "flipping" the whole number line over itself. You should be able to see why that reverses > and <. –  Lopsy Dec 28 '11 at 21:45
-1 I don't find the above "explanation" to be helpful. Either it is circular or so incomplete to be of little help. Alas, the OP doesn't know what he's missing by accepting it. Perhaps you may wish to do him a favor by improving the answer. –  Bill Dubuque Dec 28 '11 at 22:31
@BillDubuque I did fear it did not explain why the signs must be switched but I accepted the answer based on the fact after inputting values in (1) it made an intuitive sense to reverse the inequality sign. However, after reading your answer it appears the short (and straight) answer is because of law of signs which I think is what I was looking for. –  Sniper Clown Dec 28 '11 at 23:21
@Mahmud You might find it helpful to look at the concept of an ordered ring. This is a ring whose nonzero elements can be partitioned into disjoint sets $\rm\:P\:$ ("positives") and $\rm\:N\:$ (negatives") such that the sum and product of positives are positive, i.e. $\rm\ P+P \subset P,\ \ P\cdot P\subset P\:.\:$ One then defines $\rm\ a > b\$ to mean $\rm\ a-b \in P\$ and one may then easily deduce all of the usual inequality laws from these basic hypotheses combined with the ring axioms. –  Bill Dubuque Dec 28 '11 at 23:59
@Mahmud Thus all of the inequality laws may be deduced ("generated") from the basic axioms $\rm\ x,\:y > 0\ \Rightarrow x+y,\: x\cdot y > 0\:,\:$ combined with the ring axioms. –  Bill Dubuque Dec 29 '11 at 0:02

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 $-2x > 3$...

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For my money, this is the best of the answers (and I'm up-voting it), because it needs just a touch to give the complete picture of what's happening, namely, that the relation is ALWAYS reversed. In other words, it's also reversed for EQUALITY. The reason that the reversal for equality escapes notice is that the symbol for equality is symmetric. –  Hexagon Tiling Jan 7 '12 at 11:10

Consider $-2<4$ Dividing by $-2$ gives $1>-2$. If not you would have $1<-2$ which is false.

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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).

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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.

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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.

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It's because negative numbers appear bigger but are actually smaller. Take this example: Which is bigger, $-7$ or $-9999$? First instinct, and absolute value tells us that the latter is the larger, but as per human definition of negatives, $-7$ is to the right of $-9999$ on the number line, therefore making it larger.

Consider the following: $$7<9999$$ $$7(-1)<9999(-1)$$ $$-7 < -9999$$ Wait, the last statement is not true! We need to flip the sign to make it true. $-7$ appears smaller than $-9999$ but it really isn't. For negative numbers, things are flip-flopped.

NOTE: Multiplying by a negative is multiplying by a positive (signs still same) then multiplying by -1 (Signs now flipped). Dividing is multiplying by a fraction, and dividing by a negative number is multiplying by a negative fraction. Also, if you think about it, this also holds when you reciprocate both sides.

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That "Also, if you think about it, this also holds when you reciprocate both sides" part might be confusing, because it only works when both sides have the same sign; otherwise, if one is positive and one is negative, you don't have to flip the >. –  Lopsy Dec 28 '11 at 21:48