**Introduction**

In the inequality, $latex – 2x > 4$, the solution is $latex x < -2$. Similarly, $latex -3x < -5$ gives us $latex x > \frac{5}{3}$. In these inequalities, we need to multiply both sides by a negative number to make $latex x$ positive. Notice that after the multiplication, the inequality sign is flipped or reversed. Now why do we do this?

Multiplying two unique numbers by a negative number reverses their order on the number line. For instance, if we have two numbers $latex – 3$ and $latex 4$, clearly, $latex 4 > -3$. However, multiplying both sides by a negative number, say $latex -1$, reverses their order on the number line. The new pair of number is now $latex -4$ and $latex 3$, and $latex -4 < 3$. This also happens even if the pair of numbers are both positive or both negative. For example, the pair $latex 6$ and $latex 7$ gives us the inequality $latex 7 > 6$. Multiplying both sides by $latex -1$, we have $latex – 7 < -6$. In both cases, the inequality sign was flipped. Now it’s time to prove why this is so.

**Theorem**

If $latex a, b$ and $latex c$ are real numbers, $latex a > b$ and $latex c$ is a negative number, then $latex ac < bc$.

Proof

Since $latex a > b$, subtracting $latex b$ from both sides, we have $latex a – b> 0$. The inequalitity $latex a – b > 0$ means that $latex a -b$ is positive. If we multiply a positive number $latex a – b$ to a negative number $latex c$, the result is negative. So, since $latex c$ is negative $latex c(a – b)$ is negative. This means that

$latex c(a-b) <0$.

Simplifying,

$latex ac – bc <0$.

Adding bc to both sides results to

$latex ac < bc$

which is precisely what we want to show. $latex \blacksquare$.

This proof indicates that flipping the inequality is needed if we multiply both sides of the inequality by a negative number.

Photo Credit (Creative Commons) by Vinay Shivakumar.

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