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If we multiply or divide an inequality by a negative number, the inequality symbol is reversed. Why is this true?

Example: Given $1<2,$ multiplying both sides by $-1$ we get $-2<-1.$

Blue
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Adik001
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    You know $2 < 3$. Why does that tell you $-3 < -2$? Think about what happens on the number line when you multiply by $-1$ or another negative number. – Ethan Bolker Apr 18 '17 at 18:56
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    The function $f\colon \mathbb{R}\to\mathbb{R}$ defined by $f(x)=-x$ is an orientation reversing homeomorphism. Therefore, it reverses the order $<$. – D Wiggles Apr 18 '17 at 19:57
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    I'm sure there's a duplicate out there, but I don't think that's a good duplicate -- the question there asks if dividing by something negative reverses inequalities; the question here asks why this happens (and the answer in the dupe target merely confirms that this is the case, without any details). – pjs36 Apr 19 '17 at 02:54

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Say you have two numbers $a,b \in \Bbb{R}$ with $a < b$. Then subtracting $b$ from both sides gives $a-b<0$ and subtracting $a$ from both sides gives $-b<-a.$ Thus we know that $(-1)*b < (-1) * a$, or that multiplying by $-1$ flips the inequality. Now, this isn't a very detailed proof or anything, but it does give some intuition behind why we reverse the inequality.

Logan Clark
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Multiplying by $-1$ reverses the inequality.

In general, when applying a strictly increasing function the inequality is preserved. Applying a strictly decreasing function on an inequality reverses it.

The function $f(x) = {-x}$ is strictly decreasing, hence $$a<b\quad\implies\quad f(a)>f(b) \quad$$ and in your example $$1<2\quad\implies\quad f(1)>f(2)\quad\implies\quad -1>-2.$$

On the other hand $g(x) = e^x$ is strictly increasing, hence $$1<2\quad\implies\quad e^1<e^2. $$

Eff
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$3\lt5$ but when comparing $-3$ and $-5$ then it's clear the inequality must change direction as $-3 \gt -5$.

Multiplying by $-1$ reflects numbers through the origin so e.g. positive numbers further to the right (e.g 5 vs. 3), after the multiplication, are negative numbers further to the left (i.e.$ -5$ vs. $-3$), showing the inequality must change direction.

PM.
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$a<b$ says $a-b<0$.

$a>b$ says $a-b>0$.

Dividing of positive number by negative number gives a negative number.

Let $a>b$.

Thus, if $c<0$ so $$\frac{a-b}{c}<0$$ or $$\frac{a}{c}-\frac{b}{c}<0,$$ which says $$\frac{a}{c}<\frac{b}{c}$$

In our case, $1<2$ says $1-2<0$

$-1$ and $1-2$ are negative numbers, which says that $$-1(1-2)>0$$ or $$-1-(-2)>0$$ or $$-1>-2.$$

Michael Rozenberg
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