To what extent can I square both sides of an absolute equation?

Question

I am working on some absolute equation problems like the following:

$$\begin{align} & {|x-4|} \lt 1 \\ & 1 \le |x| \le 4 \\ & |x+3| = |2x+1| \end{align}$$

Now, for both of these equations, I simply squared both sides to get rid of the absolute and then continued solving from there. Now my question is: when can I not do this and what is the alternative if I can't?

Thanks a bunch!

Answer 1

Order of a relation is preserved, when you apply a strictly monotonously increasing function (for $\leq, \geq$ you can drop the strictly). $$f: x \mapsto x^2$$ is strictly monotonously increasing on $[0, \infty)$. so you can square whenever all expressions are guaranteed to be $\geq 0$.

Answer 2

The intuition here is to notice that the squaring function $f(x) = x^2$ is monotonically increasingly for non-negative real numbers, namely if $x<y$ and if both $x$ and $y$ are non-negative, then $x^2 < y^2$. To understand what I mean, think of the graph of $f$ over the positive real axis. It starts at $0$ and gets bigger and bigger.

On the other hand, suppose that $x<y$ but that $x$ is negative, then it is not necessarily true that $x^2 < y^2$. Consider, for example, $x=-2$ and $y=1$. Similarly, if both $x$ and $y$ are negative, then $x<y$ implies that $x^2>y^2$. Think of $x=-2$ and $y=-1$ for example. These facts follow because the squaring function is a decreasing function for negative real numbers.

How does this affect inequalities with absolute values? Well, consider, for example, $|x|<|y|$. Since both $|x|$ and $|y|$ are non-negative, we can square both sides of this inequality to obtain $|x|^2<|y|^2$.

In all of your examples, all sides of all inequalities are non-negative, so you can square without reversing inequalities. You would just have to be careful if some side of an inequality is negative.

Answer 3

The following facts hold because integers, rationals, algebraic numbers, and real numbers are all "totally ordered rings", which basically means that arithmetic plays well with order:

Fact 1: If $p\le q$ and $r\ge0$, then $pr\le qr$ and $rp\le rq$.

Fact 2: If $p<q$ and $r>0$, then $pr<qr$ and $rp<rq$.

(Note that these conclusions are really the same for numbers because of the commutative law, but there are situations where they are not).

Suppose $a\ge0$ and $b\ge0$.

If $a\le b$, then $a\cdot a\le a\cdot b$ and $a\cdot b\le b\cdot b$, so $a^2=a\cdot a\le b\cdot b$.

If $a<b$, then $b>0$, so $a^2=a\cdot a \le a \cdot b < b \cdot b=b^2$.

It turns out that we can reverse these implications using the fact that $a\le b \iff \lnot (b < a)$ and the other direction.