Let's say I have
$$a > b$$
Now we know that the inequality sign will switch if:
-
We take the reciprocal of both sides
-
Multiply $-1$ to both sides.
But, are there more? What if we do more "complicated" things to it?
Will it change if I apply inverse trig on it? For instance is it still true then if
$$\arccos(a) > \arccos(b)$$
Note that $a,b \in\mathbb R$.
So what I really want to ask is, are there any other operations that would make you change the inequality signs?
Best Answer
It's not entirely true that when you take the reciprocals of both sides, you change "$\lt$" to "$\gt$" and vice-versa. If $a$ and $b$ are both positive or if they are both negative, and $a<b$, then $1/a>1/b$. But if one is positive and the other negative, and $a<b$ (which means $a$ must be the one that's negative) then $1/a<1/b$; the direction doesn't get reversed.
Generally if $a<b$ and $g$ is a strictly decreasing function, then $g(a)>g(b)$. That's actually the definition of the concept of "strictly decreasing function". So the fact that when you multiply by a negative number, you invert the inequality relation, is the same as saying that multiplication by a negative number is a strictly decreasing function. For example, if $g(x) = -5x$ for all values of $x$, then $g$ is a strictly decreasing function. If $g(x)=1/x$, then the restriction of $g$ to the positive numbers is a strictly decreasing function, and the restriction of $g$ to the negative numbers is a strictly decreasing function, but $g$, over its whole domain, is not a strictly decreasing function.
$\arccos$ is a strictly decreasing function: as a number increases from $-1$ to $1$, its arccosine decreases, i.e. if $a<b$, then $\arccos a>\arccos b$.
One thing that will tell you that a function is strictly decreasing is that it's derivative is everywhere negative and its domain has no gaps. The reciprocal function has an everywhere negative derivative, but its domain has a gap at $0$.