Why can't $|x-2|$ be less than zero? Just confused with the concepts in Absolute Value.
So we know that to solve absolute value equations such,
$$|x-2| = 5 \tag1$$
In this case we have, $x-2 = 5$ and $x-2 = -5$. then solve for $x$.
However,
$$|x-2| = -5 \tag2$$
Here is no solution.

Why can't the absolute value be less than zero?

Is it from a graph that we cannot get negative values from the $y$-axis? Are there any other explanations?
A proof is highly appreciated.
 A: Saying " the absolute value of number $N$ is equal to $5$" means " I do not know what number is $N$ , but I know for sure that the distance from $N$ to $0$ is $5$ units".

Defining intuitively the absolute value as a distance explains why the
absolute value cannot be negative .

In your example, it is the number $x-2$ that plays the role of $N$.
Also, when $N$ is positive the distance from $N$ to $0$ is $N-0$ and when $N$ is negative, the distance from $N$ to $0$ is $0-N $.
Consequently, the general formula is as follows, and you simply have to apply it mechanically  , by replacing $A$ by your specific number $5$, and   $N$ by your specific number : $x-2$ :

$|N| = A $
$\iff  [ N-0 = A \space $ OR $\space  0-N =A ] $
$\iff [ N=A \space $ OR $\space 0-N =A]$
$\iff [ N=A \space \mathbb {OR}\space  -N = A ]$

So, your equation means:

(1) $(x-2)- 0 = 5 \implies x-2= 5 \implies x=7$
OR
(2)$ 0 - (x-2) = 5 \implies -x +2 =5 \implies -x = 3 \implies x=-3$
The solution set of the equation is : $\{ -3, 7\}$.
A: What's called absolute value of  a number is just the numbers's distance from zero.  As such it's always non-negative. Furthermore $|a-b|$ is the distance from $a$ to $b$.
Your first equation, $|x-2|=5$ reads: "The distance from $x$ to $2$ equals $5$.", hence $x=-3$ or $x=7$.  Now the second equation, $|x-2|=-5$, reads: "The distance from $x$ to $2$ equals $-5$.", which is impossible.
A: By definition we have,$$|x|=\begin{cases}x&\text{ if }x\geqslant0\\-x&\text{ if }x<0,\end{cases}$$and therefore we always have $|x|\geqslant0$.
In order to solve the equation $|x-2|=5$ you can consider two possibilities:

*

*$x\geqslant2$. Then $|x-2|=5$ is the same thing as $x-2=5$. So, $x=7$.

*$x<2$. Then $|x-2|=5$ is the same thing as $-(x-2)=5$. So, $x=-3$.

A: The absolute value of a number returns the non-negative part of it, even if the number is negative.
So you need to consider two cases, one where the term inside the absolute value is negative and one where it is positive:
Case 1:
$|x-2| = x-2$
Case 2:
$|x-2| = 2-x$
Your second equation is wrong, remember that you already removed the absolute value when you stated that $x-2 = -5$ (considered the second possibility), and not $|x-2|$ = $-5$, because the absolute value only returns the magnitude(length), not direction, hence there exists no such thing as a negative length.
A: The definition of the absolute value function is defined as
$$|x|=\begin{cases}x&\text{ if }x\geq0\\-x&\text{ if }x<0,\end{cases}.$$
So if $|x-2|=-5$ has no solutions because the function $|x-2|$ is defined to be greater than $0$ no matter what.
However, $0<-5$, so it has no solutions.

I hope this helped.
A: To expand on @michael.hoppe's answer: $|a|$ is in fact the distance from $a$ to $0$ because you can see that $$|a|=|a-0|$$
To get $a$'s distance to any other number $b$, just substitute the $0$ with $b$. So the distance on a line from $a$ to $b$ is $|a-b|$.
The intuition behind $|a-b|\geq0$ is that a distance is always positive, as others have said. Of course, the real reason for this is the way $|x|$ is defined, so
$$|a-b|=\begin{cases}
a-b, \quad &\text{if} \;\;a-b\geq 0\\
-(a-b)=b-a \quad &\text{if}\;\; a-b<0
\end{cases}$$
