Inequality; when we switch signs? From basic to complicated

Let's say I have

$$a > b$$

Now we know that the inequality sign will switch if:

1. We take the reciprocal of both sides

2. 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?

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Look up the definitions of increasing and decreasing functions. When you apply an increasing function to both sides of an inequality, the order of numbers stays the same. When you apply a decreasing function, the order is reversed. Care needs to be taken, when the function is increasing/decreasin only on an interval. – Jyrki Lahtonen Jan 1 '12 at 20:15

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 $0$.

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Remember that the parser sometimes gets confused when you use < and >; if that happens, try \lt and \gt instead. – Arturo Magidin Jan 1 '12 at 20:21
Woah! that's really helpful! I learned the definition of "strictly increasing/increasing" etc from sequences so I am familiar with them now! – jak Jan 1 '12 at 22:48
Actually what happens if we have f(a) < g(b)? What if I apply h(x) to it? That is does the idea still hold if I do H(f(a)) < H(g(b))? – jak Jan 1 '12 at 23:06
1. Multiplication by negative numbers flips inequalities. Multiplication by positive numbers respects inequalities.

If $a\lt b$ and $c\lt 0$, then $ac\gt bc$; if $d\gt 0$, then $ad\lt bd$.

2. You can deduce the case of reciprocals (which is more complicated than what you wrote) from this.

If $0\lt a\lt b$, then multiplying by $\frac{1}{ab}$ (which is positive) gives $\frac{a}{ab} \lt \frac{b}{ab}$, which yields $\frac{1}{b}\lt \frac{1}{a}$; i.e., a "flipped inequality".

If $a\lt b \lt 0$, then multiplying by $\frac{1}{ab}$ (which is also positive, since $a$ and $b$ are both negative) again yields $\frac{1}{b}\lt \frac{1}{a}$.

If $a\lt 0 \lt b$, then multiplying by $\frac{1}{ab}$ (which is negative) yields $$\frac{a}{ab}\gt \frac{b}{ab}$$ or $\frac{1}{b}\gt \frac{1}{a}$; that is, the inequality is not "flipped."

3. A function $f$ is strictly increasing if $a\lt b$, both in the domain of $f$, implies $f(a)\lt f(b)$. Examples of increasing functions are $f(x)=x^3$, $f(x)=\arctan(x)$, $f(x)=e^x$, $f(x)=\log(x)$. A function $f$ is increasing if $a\lt b$, both in the domain of $f$, implies $f(a)\leq f(b)$.

Symmetrically, a function $f$ is strictly decreasing if $a\lt b$ and both in the domain imply $f(a)\gt f(b)$; and decreasing if $a\lt b$ and both in the domain implies $f(a)\geq f(b)$.

You can also deduce most of the case of reciprocals from this: $f(x)=\frac{1}{x}$ is strictly decreasing on the positive reals, so if $0\lt a \lt b$, then $f(a)\gt f(b)$; $g(x)=\frac{1}{x}$ is strictly decreasing on the negative numbers, so if $a\lt b\lt 0$, then $g(a)\gt g(b)$. But the case $a\lt 0\lt b$ has to be dealt with separately anyway.

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