# Taking the absolute value in inequalities.

If I've a expression: $$-4<3$$ and take the absolute value $$|-4|<|3|\implies 4<3$$ which is false. So I though that maybe the inequality sign would change. But $$|-2|<|3| \implies 2>3$$ which is also false.

My problem is that I have the inequality $$-x where $$x,y>0$$. What will happen if I take the absolute value of it: $$|-x|<|y|.$$ I though it would be $$0<|x|<|y|\implies 0.

But from the examples above it seems that this ain't true.

EDIT:

I have a increasing sequence $$(x_1>x_2>x_3...$$ etc.), $$x_1,x_2,...,x_n$$ where all $$x's$$ are positive and a constant $$m=1/2$$. The inequality: $$-x_n<1/2$$ is always valid because the $$x's$$ are positive. But $$x_3=1$$. So for all $$n \geq 3$$, $$|-x_n|>1/2$$ .

Given this context helps in solving my problem in taking the absolute value?

Taking the absolute value on both side we have for all $$n>3, |-x_n|>|1/2|>0 \implies x_n>1/2>0$$.

Is this correct?

• Absolute value is not a monotonic function – J. W. Tanner Apr 2 '19 at 2:42
• You dont know which number has a greater distance from 0. Talking about x (or-x) and y. – randomgirl Apr 2 '19 at 2:47
• As you have demonstrated, this is nonsensical. $x < y$ in no way implies $|x| < |y|$. "Taking the absolute value" of both sides of an inequality is a misleading phrase since you are not justified in doing so without additional context. It may be helpful to mentally replace the absolute value function with an arbitrary function $f$ and consider what it would be to apply $f$ to both sides of an inequality. – Brian Apr 2 '19 at 2:47
• So if I assumed that $x>y>0$, for example, then I would know that $x$ is far from $0$ than $y$ and I would be able to do somthing about it? There must be more information some sort of relation between this numbers? – Pinteco Apr 2 '19 at 2:52
• I'll give some context and put in the post. – Pinteco Apr 2 '19 at 2:55