Find solutions to $|x|<x$

Question

Find the solution set of

$$|x|<x.$$

I know that the solution set is $\emptyset$. But I am stuck in the case when $x<0$.

Shall I intersect the condition with the obtained result? I mean the following.

case) if $x<0$, then $|x|=-x$,

We have $$-x<x\implies 2x>0\implies x>0$$

Therefore, $$x<0 \wedge x>0=\emptyset$$

Thus the solution set of this case is $\emptyset$. Is my solution correct?

Answer 1

or alternatively we have $$0\le |x|<x$$ and we can square the inequality then we get $$x^2<x^2$$ a contradiction

Answer 2

Very simple: since $x<0$ and $|x| \geq 0$, then $x<0 \leq |x|$.

Answer 3

Your solution is correct. For $x\geqslant0$:

$$x<x$$

it's false. For $x<0$: $$-x<x\implies 0<2x$$

but since $x<0$, it's a contradiction. Hence, $x\in\emptyset$.

Answer 4

The following inequality is true for any $x$: $$ x\le |x| $$ If we also add $|x|<x$ we get a contradiction: a solution would satisfy $$x\le |x|<x$$

Answer 5

The inequality $|x| < x$ is obviously false for $x = 0$. So, let us assume that $x \neq 0$. In this case, we can divide both sides of the inequality by $|x|$ to get $$ 1< \dfrac{x}{|x|} = \pm 1. $$ This, too, is false, proving that $|x| < x$ is false for all $x$.

Answer 6

If $x$ is a solution, then $x$ is strictly positive because $$0\leq |x|<x$$

So, it's not needed to consider the case $x<0$.