How do you algebraically derive "x <= 0" from "-x = | x |"

Question

A = "-x = | x |"
B = "x <= 0"
If A, then B.

By plugging in numbers or testing ranges less than zero, greater than zero, and equal to zero, I can verify that A implies B. But how do I derive "x <= 0" from "-x = | x |" algebraically i.e. without testing points and ranges?

For example, I tried:

-x = | x |
-1 = ( | x | / x ) // cannot divide by x if x = 0
0 = ( | x | / x ) + 1 // [range of right side is [0,2] ]
At this point, we can safely square both sides because the ranges of both sides are non-negative
0 ^ 2 = (| x | / x ) + 1) * (| x | / x ) + 1)
0 = (|x| / x )^2 + ( 2*(|x| / x ) ) + 1
Hold it. (|x| / x )^2 = 1
0 = ( 1 ) + ( 2*(|x| / x ) ) + 1 // assuming x != 0
0 >= ( 1 ) + ( 2*(|x| / x ) ) + 1 // without loss of generality
0 >= ( 2*(|x| / x ) ) + 2
0 >= 2*( (|x| / x ) ) + 1 )
0 >= (|x| / x ) ) + 1
-1 >= (|x| / x ) )
(|x| / x ) ) =< -1
|x| =< -x

But I ended right back where I started. I want to get to x <= 0, but the only way I can see how to verify it is like this:

0 = |x| + x
assume x > 0
0 = (1 + 1)*x // in this case | x | is equal to x
0 = x // False because positive numbers cannot be less than or equal to zero.
False. x > 0 is invalid.
assume x < 0
0 = (-1 + 1) x
0 = 0*x
True. x < 0  is valid.
assume x = 0
0 = |0| + 0
True. x = 0  is valid.

Since there were only three possibilities [x >0, x < 0, and x = 0], we have proven that for all possible cases, -x = | x | implies x <= 0. But I want to do it algebraically, without any plug-and-chug of points. How can I prove it algebraically?

Answer 1

It's always true that $|x|\geq0$. So if $-x=|x|$, then this means $-x\geq0$. That is equivalent to $x\leq0$.

Answer 2

You can approach it by using the definition of absolute value as the square root of $x$ squared.

Given $-x = |x|$, we can then say

$-x = \sqrt{x^2}$.

But for all $x \in\mathbb{R}$, $\sqrt{x^2}\geq 0$. Then,

$-x \geq 0$

$x \leq 0$

Answer 3

The key, I think, is to appeal to the definition of the absolute value function: $|x| = x$ if $x > 0$, and $|x| = -x$ if $x \le 0$. Your desired implication then follows immediately.

Answer 4

Yet another approach: if $¬ B → ¬ A$ then $A → B$. Since by definition of the absolute value of real numbers, $x>0 → x=|x|$, it must be that $-x=|x| → x \leq 0$, and conversely.