Ttdfjt

I want to improve my theoretical background to make it rock solid.

Given

$(x - 2)^2 geq 1$

O learnt that taking the square root of boot sides yields

$| x -2| geq sqrt1$ or $| x - 2 | leq - sqrt1$

Here, I have two questions:

1. 
   

   Why does taking the square root transforms it from $(something)^2$ to $|something|$? Why the "module" symbols?

   
   
   

   I believe this is redundant, not really necessary. Am I right?

   
   

and

2. 
   

   Why does the sign change direction?

   
   
   

   I know this is because $x$ can be $ le 0$, so I need to go into a little bit more depth about the theory on swapping signs. As far as I can tell we only swap inequality sign when dividing or multiplying by a negative number

   
   

Thanks a lot in advance!

1. The key lies in understanding the square root function, which by definition should give a unique value. Unfortunately, there are always two candidates satisfying the equation $x^2=y,$ where $y> 0$ is given. Mathematicians have convened to choose the positive root satisfying this equation as the result of the square root operation. Thus taking the square root of a nonnegative quantity, say $x^2,$ always gives a nonnegative result. This operation is often symbolised by $sqrt,$ so that we always have $$sqrtx^2=|x|,$$ by definition. Thus, $sqrt 4 = 2,$ and not $-2.$ Hence, whenever you take the square root of an expression of the form $E(x,y,z,ldots)^2,$ you must have the result to be $|E(x,y,z,ldots)|.$




2. The sign does not change direction. If you have an inequality of the form $$x^2ge y,$$ where $yge 0,$ then you may take the square root of both sides (and the inequality is respected since the square-root function is monotonic), to get $$|x|ge sqrt y,$$ and the right hand side is uniquely defined by (1), so there's no negative value at all. Thus, applying this to your inequality, namely $$(x-2)^2ge 1,$$ we obtain $$|x-2|ge 1.$$ But if you want to solve this inequality, you don't need to take square roots at all. Just transpose and factor, to get $$(x-2)^2-1ge 0,$$ which gives $(x-2-1)(x-2+1)=(x-3)(x-1)ge 0,$ and so on.

The chain of equivalent transformations should progress first to
$$
|x-2|gesqrt1=1
$$
and then to
$$
x-2le -1text or x-2ge 1.
$$

If $|u|ge a$, then

• either $uge 0$ and $u=|u|ge a$.


• or $u<0$ and $-u=|u|ge aiff ule -a$.

You can go from $u^2ge a^2$, $a>0$, to $|u|ge a$ because the square function is strictly monotonically increasing on the positive half-axis. Or just simply because in $$0le u^2-a^2=(|u|-a)(|u|+a)$$ you can divide out the positive second factor.

The absolute value of a positive number $n$ is $n.$
The absolute value of a negative number $n$ is $-n,$ which is a positive number.
(For example, $lvert -2rvert = -(-2) = 2.$)
The absolute value of zero is zero.

No matter what you start out with, you end up with a non-negative number
(that is, zero or positive).
You might have to use a negative sign ($-n$ instead of $n$) in order to get a result that isn't negative.
(If you start with zero, the negative sign doesn't matter, since $-0=0.$)

So $lvert x-2rvert,$ the absolute value of the number $x-2,$ is either
$x - 2$ or $-(x-2),$ whichever one of those is non-negative.

From $(p - 2)^2 geq 1$ you can conclude that
$lvert x-2rvert geq sqrt1 = 1,$ full stop.
That is one of the other properties of the absolute value.
There is no case where $lvert x-2rvert leq -sqrt1 = -1,$
because that would say that there is some number whose absolute value is negative, which cannot happen.

From the fact that $lvert x-2rvert geq 1,$ since you know that $lvert x-2rvert$
is actually either $x-2$ or $-(x-2),$ you know that one of the following two statements is true:

$$ x-2 geq 1 quadtextorquad -(x-2) geq 1. $$

If you like, you can rewrite $-(x-2) geq 1$ as $x-2 leq -1$;
reversing the signs on both sides of an inequality also reverses the direction of the inequality. Or you could write $-x + 2 geq 1$ and proceed from there.

$|x-2|le-sqrt1$ is redundant, since absolute values are never negative.

As for why the module, look at the graph of $x^2$. What values of $x$ give $x^2>1$, for example?

As for how you get a sign change, notice that $|x|>1$ means $x>1$ or $-x>1$, but the latter becomes $x<-1$ when you multiply both sides by $-1$, resulting in a sign change..