Absolute values don't work

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I don't understand, how absolute valued could possibly be considered well defined.

As shown here,

$|a| = |-a| , ||a|| = |-|a||$

So lets take $a=-2, |a| = -2 = |-a|,$ but $|-a| = |2| = 2$

But it doesn't just stop here:$ ||-a|| = |-|-a|| = -|-a| = --|a|=---a=-a$

It doesn't even matter how many times you do it, it always inverts, when it shouldn't. Hence $|a| \ne |-a|$

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The absolute value is never negative, by definition. Hence, for any $a>0$, $|-a|=|a|=a$.

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Just to add that $|x|$ is never negative, take a look at the graph of $|x|$.

enter image description here

As you can see, if you have $x<0$, then $|-x|=x$ and for $x\ge 0$, $|x|=x$.

This is because by definition of absolute value, we have $$|x|=\begin{cases}x&x\ge 0\\-x&x<0\end{cases}$$

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Let's look at the standard definition of $|x|$ $$|x|=\begin{cases}x&x\ge 0\\-x&x<0\end{cases}$$ That is: - $|x|=x$ when $x \ge 0$ - $|x|=-x$ when $x < 0$

For any real number $a$ it is either greater than zero, equal to zero, or less than zero.

So lets take $a=-2, |a| = -2 = |-a|, but |-a| = |2| = 2$

Since $a=-2$ is less than zero, $|a|=-(-2)=2$

this also gives $-a=2$ is greater than zero so $|-a|=2$.

Looks consistent to me. We can also do this in general.

  • Whenever $a \ge 0$ we have $-a \le 0$ hence $|a|=a$ and $|-a|=-(-a)=a$

  • Whenever $a < 0$ we have $-a > 0$ hence $|a|=-a$ and $|-a|=-a$

The only possible point that could be not well-defined is zero but since $\pm 0 = 0$ it's not an issue.

Now let's look at the double absolute value problem (work like an onion one layer at a time):

  • $||a||$ since $|a|>0$ $||a||=|a|$
  • $|-|a||$ since $|a|>0$ we have $-|a|<0$. Then $|-|a||=-(-|a|)=|a|$