-1
$\begingroup$

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 $x=0$.

Thanks!

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.