Say we have the inequality $x^2 \leq 4$. This should imply, unless I'm mistaken, $-2 \leq x \leq 2$.
This sounds rather trivial, but I'm having some trouble understanding why exactly it's the case or what the general rule may be in this case. (A proof would actually be extremely helpful, if anyone has ideas on how to put one together.)
Thanks.
On
$$x^2 \le 4$$
$$x^2-4 \le 0$$
$$(x-2)(x+2) \le 0$$
The sign changes when $(x-2)(x+2)=0$, by testing the $x < -2, -2 < x < 2, x>2$, you can find the corresponding region easily.
Sketching of graph might help. Look for the region that is below the $x$-axis.
On
It's not that trivial.
We have that if $0 < a$ and $b < c$ than $ab < ac$. (an axiom)
So if $0 < a < b$ then we must have $a^2 = a*a < a*b < b*b = b^2$ and if $0 < b < a$ the reverse.
So for $0< a; 0 < b$ then $a < b \iff a^2 < b^2$.
And we have if $a < 0$ and $b < c$ then $ab > ac$. (an easy proposition)
So if $b < a < 0$ then $b^2 = b*b > b*a > a*a = a^2$ (and the converse if $a < b < 0$.
So if $b < 0; a < 0$ then $a < b \iff a^2 > b^2$.
And therefore if we don't know the signs of $a$ or $b we have:
$a^2 < b^2$ if and only if $-|b| < |a| < |b|$.
Thus we can assume if $x^2 \le 4$ then $-2 \le x \le 2$ and if $x^2 \ge 4$ then either $x \le -2$ or $x \ge 2$.
It is treated as something obvious and in one sense it is obvious, but... it's not as trivial in my opinion as people treat it.
However once you prove it, you've proven it forever and never have to worry about it again.
On
$x^2 -4\le 0.$
$\star$: $(x-2)(x+2) \le 0.$
One of the factors is $\ge 0$, the other factor is $\le 0$.
1) Assume $x \ge 0: $
The second factor in $\star$ is positive.
Hence $x-2 \le 0$, or $x \le 2:$
Putting together : $0 \le x \le 2$.
2) Assume $x \lt 0.$
The first factor in $\star$ is negative.
Hence $x+2 \ge 0$, or $x\ge -2.$
Putting together: $-2 \le x \lt 0.$
Finally:
$-2 \le x \le 2.$
It's $$\sqrt{x^2}\leq2$$ or $$|x|\leq2,$$ which is $$-2\leq x\leq2.$$