I looking for comments and corrections on my proof. I'm also unsure if the second case is correct.

For $x\in \mathbb{R}$ we have:

$\lvert x\rvert \geq 0$, and $\lvert x \rvert=0$ if and only if $x=0$.

I have three cases:

  1. For $x>0$, the absolute value $\lvert x\rvert=x$ is greater than zero, so $x \neq 0$. This case not true according to the theorem.

  2. For $x<0$, we have $\lvert x\rvert =-x>0$, which is less than zero, so $x<0$. This case not true according to the theorem.

  3. For $x=0$, we have $\lvert x\rvert=0$, according to the theorem. Therefore the theorem is true only for $x=0$.


  • $\begingroup$ In the case of $n > 1$, I assume $0$ stands for the zero vector? Also, why is the condition $a \in \mathbb{R}$ in the statement? $a$ is not used anywhere for anything - what is $a$ supposed to be? $\endgroup$ – Zubin Mukerjee Jun 6 '17 at 20:26
  • $\begingroup$ either $x\in \mathbb{R}^n$ is a typo, or your proofs make no sense. If $x\in \mathbb{R}^n$ then you cannot say $x>0$, because $\mathbb{R}^n$ is not ordered like $\mathbb{R}$ is $\endgroup$ – user160738 Jun 6 '17 at 20:27
  • $\begingroup$ @ZubinMukerjee It was a typo. $\endgroup$ – JDoeDoe Jun 6 '17 at 20:29
  • $\begingroup$ @user160738 It was a typo, so $x\in \mathbb{R}$. $\endgroup$ – JDoeDoe Jun 6 '17 at 20:29
  • 2
    $\begingroup$ For $x \in \mathbb{R}$ all you need is $-|x| \le x \le |x|\,$. $\endgroup$ – dxiv Jun 6 '17 at 20:29

If the absolute value is defined as

$$|x|:=\begin{cases}x>0\to x\\x=0\to0\\x<0\to-x\end{cases},$$

you can say $$|x|\ge0\iff (x>0\land x\ge0)\lor(x=0\land 0\ge0)\lor(x<0\land-x\ge0)\\\iff x>0\lor x=0\lor x<0\iff\text{true}.$$ $$|x|=0\iff (x>0\land x=0)\lor(x=0\land 0=0)\lor(x<0\land-x=0)\\\iff x=0\lor x=0\lor x=0\iff x=0.$$

If it is defined as

$$|x|:=\begin{cases}x\ge0\to x\\x\le0\to-x\end{cases},$$

you can say

$$|x|\ge0\iff (x\ge0\land x\ge0)\lor(x\le0\land-x\ge0)\iff x\ge0\lor x\le 0\iff\text{true}.$$ $$|x|=0\iff (x\ge0\land x=0)\lor(x\le0\land-x=0)\\\iff x=0\lor x=0\iff x=0.$$

If it is defined as

$$|x|:=\sqrt{x^2},$$ then

$$\sqrt{x^2}\ge0$$ by definition of the square root, and by squaring, then solving the quadratic equation

$$\sqrt{x^2}=0\iff x^2=0^2=0\iff x=0.$$

  • $\begingroup$ Thanks! Suppose I instead have $|x|\ge x$, is the proof similar? I.e. with your first definition: $(x>0\land x\ge x)\lor(x=0\land 0\ge x)\lor(x<0\land-x\ge x)\\ \iff x>0\lor x=0\lor x<0\iff\text{true}.$ $\endgroup$ – JDoeDoe Jun 9 '17 at 19:05
  • $\begingroup$ @JDoeDoe: absolutely. $\endgroup$ – Yves Daoust Jun 10 '17 at 9:22

The theorem should be interpreted as

For $x\in \mathbb{R}$ we have: $(\lvert x\rvert \geq 0$) and ($\lvert x \rvert=0$ if and only if $x=0$).

With this in mind, the theorem should be straightforward to prove using your casework.


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