How does Solving Inequations Work?¶

Recall from solving equations that we solve equations by rewriting them with $⟺$ at each step, and if $f$ is a function that has an inverse function, then we can apply it on both sides.

For solving inequations, we do a similar thing, but we can't just apply any function with some inverse function on both sides. For example, the function $f(x)=-2x$ has an inverse function, but applying it on both sides of the inequation $1<2$ gives $-2<-4$, which is false.

Next we will prove that if $f$ is a strictly increasing function, then we can apply it on both sides of an inequation with a $<$ sign, and it will work correctly (we will consider other kinds of inequations soon): $$x<y⟺f(x)<f(y)$$

The $\Rightarrow$ direction is easy; if $x<y$, then $f(x)<f(y)$, by the definition of strictly increasing. We still need the $\Leftarrow$ direction: we need to show that if $f(x)<f(y)$, then $x<y$. We have proved that the inverse of a strictly increasing function is also strictly increasing, so if $f(x)<f(y)$, then we have $$x=f^{-1}(f(x))<f^{-1}(f(y))=y,$$ where in the middle, we used the definition of strictly increasingness for the function $f^{-1}$, and the first and last equality came from definition of inverse function. This shows that $x<y$, which is what we wanted.

Of course, this works similarly for inequations with $>$ instead of $<$. It even works for inequations having $\le$ or $\ge$, because $$\begin{array}{rl}x\ge y& ⟺\text{we do NOT have }\underset{\begin{array}{c}\text{equivalent}\\ \text{with:}\\ f(x)<f(y)\end{array}}{\underbrace{x<y}}\\ & ⟺\text{we do NOT have }f(x)<f(y)\\ & ⟺f(x)\ge f(y).\end{array}$$ We assumed that the function $f$ is strictly increasing, so by the previous result, $x<y$ and $f(x)<f(y)$ are equivalent. It means that they have the same truth value (either both true, or both false), so we can replace one with the other.

Again, a similar calculation for strictly decreasing functions shows that we get similar results, but this time, the sign must be flipped; this comes from the definition of strictly decreasing, because there we had different signs in $a<b$ and $f(a)>f(b)$.

As special cases of these results, we let $a$ be any number, and we consider the functions $f(x)=x+a$, $g(x)=x-a$, $h(x)=ax$ and $i(x)=\frac{x}{a}$:

• The functions $f$ and $g$ are strictly increasing (graph is a line with slope $1$, which is positive). This means that they can be applied on both sides of any inequation, without flipping sign. Applying $f$ on both sides of an inequation is adding $a$ on both sides, and applying $g$ on both sides of an inequation is subtracting $a$ on both sides.
• The graph of $h$ is a line with slope $a$, and
  • if $a>0$, then $h$ is strictly increasing;
  • if $a<0$, then $h$ is strictly decreasing;
  • if $a=0$, then $h$ is neither strictly nor strictly decreasing.
  Applying $h$ on both sides of an inequation is multiplying both sides by $a$.
• If $a=0$, then $i(x)$ is undefined for any $x$. Suppose that $a$ is nonzero. Now $i$ is the inverse function of $h$, so it is strictly increasing whenever $h$ is strictly increasing, and strictly decreasing whenever $h$ is strictly decreasing. Applying $i$ on both sides of an inequation is dividing both sides $a$.

Combining all this with the previous results, we get the following.