Prove by cases that $|x|≤R \iff -R≤x≤R$

Question

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?

Answer 1

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 $-R\leq x\leq R$.

Answer 2

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 $|x| \le R$.