Why does the sign get reversed in an inequality?
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 $-2<-1.$
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$\begingroup$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 $-1$ changes the sign. Now, this isn't a very detailed proof or anything, but it does give some intuition behind why we change the sign.
$\endgroup$ $\begingroup$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 $$1<2\quad\implies\quad e^1<e^2. $$
$\endgroup$ $\begingroup$$3\lt5$ but when comparing $-3$ and $-5$ then it's clear the inequality must change direction as $-3 \gt -5$.
Multiplying by $-1$ reflects numbers through the origin so e.g. positive numbers further to the right (e.g 5 vs. 3), after the multiplication, are negative numbers further to the left (i.e.$ -5$ vs. $-3$), showing the inequality must change direction.
$\endgroup$ $\begingroup$$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 $$-1>-2.$$
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