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I need help with a proof regarding the absolute value. I have the definition: $$ \lvert x \rvert= \begin{cases} x, \quad &\text{if } x\geq0\\ -x, \quad &\text{if } x<0 \end{cases} $$

I want to prove that $-\lvert x\rvert\leq x$ is true for all real $x$.

Proof:

If $x\geq0$: Then $\lvert x \rvert=x$ so $-x\leq x$ or equivalent $x\geq 0$. Multiplication with $-1$ gives $-x\leq 0$. The two inequalities $x\geq 0$ and $-x\leq 0$ is the same as $-x\leq x$.

If $x<0$: Then $\lvert x \rvert =-x$ so $-(-x)<x$ is the same as $x<x$. But subtraction of $x$ on both sides gives $0<0$. Isn't $x<0$ valid? What is wrong here?

Update

I think the problem in the second case was the strict inequality. So maybe the proof should be:

If $x<0$: Then $\lvert x \rvert =-x$ so $-(-x)\leq x$ is the same as $x\leq x$, which is $x=x$.

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    $\begingroup$ The inequality you want to prove is $-|x|\leq x$, with the equality case included! $\endgroup$ – Anna SdTC Mar 19 '17 at 19:09
  • $\begingroup$ @AnnaSdTC Can you clarify? Isn't the equality included in the first case, $x\geq 0$? $\endgroup$ – Donsert Mar 19 '17 at 19:47
  • $\begingroup$ The inequality you are trying to prove is $-|x|\leq x$. Then you have two subcases for $x$: $x\geq0$ and $x<0$. These two are cases for $x$, but in both cases you are trying to prove the same, $-|x|\leq x$. And, as you found out, when $x<0$, then $-|x|=-(-x)=x$, so for $x<0$ the inequality is actually an equality, and it is only a strict inequality for $x>0$. Again, the key is that the equality you are trying to ultimately prove is $-|x|\leq x$ for all $x$. $\endgroup$ – Anna SdTC Mar 19 '17 at 20:16
  • $\begingroup$ @AnnaSdTC Thank you! But do you mean $x<0$ gives $x \leq x$, $x<x$ or $x=x$? I used $x<x$ above, but how can we conclude an equality from this? $\endgroup$ – Donsert Mar 19 '17 at 20:29
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    $\begingroup$ $x<0$ leads to $-|x|=x$, $x=0$ also leads to $-|x|=x$, and $x>0$ leads to $-|x|<x$, so all cases lead to $-|x|\leq x$. The inequality $x<x$ is always false, so you cannot use it. $\endgroup$ – Anna SdTC Mar 19 '17 at 20:48
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Your own correction is right. Although $\lvert x \rvert= -x$ for $x\lt0$ it doesn't change the inequality from $\le$ to $\lt$. But you still made a minor mistake in the first part.

Proof:

If $x \ge 0$: $ \ -|x| \le x \iff -x \le x \iff 0 \le 2x $ which holds true since $ x \ge 0$.

If $x \lt 0$: $ \ -|x| \le x \iff -(-x) \le x \iff x \le x $ which holds true for all $x$.

Instead of proofing your statement for all $x \in \mathbb{R} $ at once, you split it up in $x \ge 0$ and $x \lt 0$. This enables you to replace $|x|$ with the definition each time. But each time you proof the same statement, so you don't change the actual statement (e.g. $\le$ to $\lt$).

In regard to solving inequalities you go about it like a normal equality. There are some additional rules which you have to follow with inequalities, but they don't apply to your example here. I encourage you to read this article on mathsisfun which explains them very well.

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