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

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 $$0>-R \iff 0. Thus we end up with $$R>0>x≥-R$$, or, if turned around and omitting the zero, $$-R≤x.

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 $$-R\leq x\leq R$$.

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

• Thanks a bunch. In your Case 1, how do you justify $-R \le x < 0 \implies -R \le x \le R$? – schn Feb 6 at 22:10
• 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. – fleablood Feb 6 at 22:18
• 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 $0 \le x \le R$. – fleablood Feb 6 at 22:24
• 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. – fleablood Feb 6 at 22:26
• 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? – fleablood Feb 6 at 22:27