Prove by cases that:

$$|x|≤R \iff -R≤x≤R$$

$R$ is defined as $R≥0$.

I consider the two relevant cases to be $x≥0$ and $x<0$. However, for the latter, if $x<0$ then $|x|=-x$. This yields $-x≤R \iff x≥-R$. Moreover, we have $0>x≥-R$, which implies that $0>-R \iff 0<R$. Thus we end up with $R>0>x≥-R$, or, if turned around and omitting the zero, $-R≤x<R$.

I don't see how this expression is equivalent to the above. According to my book the equivalence should hold for $x≥0$ and $x<0$. Am I missing something?


Let's distinguish 2 cases.

  1. Let $x \geq 0$. Then, $|x|\leq R$ means that $x \leq R$. Moreover, $0\leq x \leq R$. Since $R\geq 0$, $-R\leq 0$. Then, $-R\leq 0 \leq x \leq R$. That is $-R\leq x \leq R$.

  2. Let $x <0$. Then $|x|\leq R$ means that $-x \leq R$. I.e. $-R\leq x$. Since $R\geq 0$, $0\leq R$. Then, $-R\leq x< 0 \leq R$. That is $-R\leq x\leq R$.

  • $\begingroup$ How do you conclude that if $-x \leq R$ and $-R \leq x$ implies $-R \leq x \leq R$? How can we know that $x \leq R$? $\endgroup$ – schn Feb 6 at 20:57
  • $\begingroup$ I have edited it. Maybe now is more clear. @user37043 $\endgroup$ – idriskameni Feb 6 at 21:05
  • $\begingroup$ In your second case, how do you justify $-R\leq x\leq 0 \leq R$ if $x<0$. Shouldn't it be $-R\leq x < 0 \leq R$? $\endgroup$ – schn Feb 6 at 21:10
  • $\begingroup$ Yes, you are right. But anyway, $-R\leq x <0 \leq R$ implies $-R\leq x \leq R$. $\endgroup$ – idriskameni Feb 6 at 21:11
  • $\begingroup$ Okay, thanks a lot! $\endgroup$ – schn Feb 6 at 21:13

It is true that if $x < 0$ then $x \ne R$ and those we have $-R \le x < 0$. This is a subcase of $-R \le x \le -R$ but we do have that $0\le x \le R$ are all not possible in this case where $x< 0$.

But that is not a problem. Those cases become possible if $x \ge 0$. If $x \ge 0$ then $0 \le x \le -R$ and this is a subcase of $-R \le x \le R$ but we have now have that $-R \le x < 0$ is impossible.

I think you are confusing "Prove by cases $X \iff Y$ with

Prove $Case 1 \iff Y$ and $Case 2 \iff Y$.

That is not what we have to prove. (In fact that is impossible as that would mean $Case 1 \iff Case 2$ which is oviously not the case).

To prove it by cases actually means prove

$Case 1 \implies Y$

$Case 2 \implies Y$ and $Y \implies Case 1$ or $Case 2$.

In this case:

Case 1: $x < 0$ and $|x| \le R$; then $-R \le x < 0$ so $-R \le x \le R$.

Case 2: $x \ge 0$ and $|x| \le R$; then $0 \le x < R$ so $-R \le x \le R$.

And if $-R \le x \le R$ then either $0 \le x \le R$ and $|x| \le R$ or $-R \le x < 0$ and $|x| \le R$.

  • $\begingroup$ Thanks a bunch. In your Case 1, how do you justify $-R \le x < 0 \implies -R \le x \le R$? $\endgroup$ – schn Feb 6 at 22:10
  • $\begingroup$ Because if $x < 0$ and $R \ge 0$ then $x \le R$. An inequality is an INequality. If $a < b$ then $a <$ everything bigger than $b$. Oh and a "less then or equal to" is a less than OR equal to. So if $a < b$ then $a \le b$ because if $a$ is less than $b$ then $a$ is less than $b$ or $a = b$. It is true that I am an earthling or I am from Alpha Centauri because.... I am an earthling. $\endgroup$ – fleablood Feb 6 at 22:18
  • $\begingroup$ Can you justive that if $x \in [-R, 0)$ then $x\in [-R, R]$ because $[-R,0) \subset [-R,R]$. I think you are tripping up when we say something might be true you are confusing it with there has to be some cases where it is true. I am a human being. Therefore I am a mammal. $x < 0$ therefore $x \le R$. Saying I am a mammal doesn't mean it's possible that I am a non-human animal. And saying $x \le R$ doesn't mean it's possible that $0 \le x \le R$. $\endgroup$ – fleablood Feb 6 at 22:24
  • $\begingroup$ Put another way. For $-R \le x \le R$ to be false we must have $x > R$ which is not possible because $x < 0$. Or we must have $x < -R$ which is not possible because $-R\le x$. So it is not false that $-R \le x \le R$. So it most be true. $\endgroup$ – fleablood Feb 6 at 22:26
  • $\begingroup$ Also $-R \le x \le R$ means $x$ is in the range of $-R$ to $R$. So if $-R \le x < 0$ is $x$ in the range of $-R$ to $R$? Well..... um... yes? $\endgroup$ – fleablood Feb 6 at 22:27

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