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The problem is to find the set $\{x \in \mathbb{R} : |x|+|x+1|<2\}$ and to find the set $\{x \in \mathbb{R} : |x+1|+|x-2|=7\}$, I somehow solved it and got the answer as $(\frac{-3}{2},\frac{1}{2})$ and $\{-3,4\}$ respectively, I want to know how to systematically solve this... How to solve this and to prove that the set builder form represents the set only... In other words, How the author would have wanted the learner to solve this problem... kindly help

Hanul Jeon
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Praveen Kumaran P
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    To systematically solve inequations with absolute values you can do it by "splitting into cases" and removing absolute values one at a time. For example, for the first one: Start by assuming $x\geq 0$, then remove the absolute value, then assume $x+1\geq 0$ and remove the second, then you get $2x<1$ which leads to $x<1/2$ but remember to intersect all sets: $x\geq 0$ with $x\geq -1$ and $x<1/2$. So, $0\leq x<1/2$. Then next: $x+1<0$ and remove $|x+1|$ by multiplying by $-1$ so $x-(x+1)<2$ which leads to $1<2$ which is true for all $x$, and so on. With two absolute values you have $4$ cases. – Martingalo Oct 15 '23 at 08:12
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    In general if you have $\sum^n_k |x-a_k|=c$ you could remove all absolute values by studying each possible case. Every $(x-a_k)$ could be $\ge 0$ or $<0$, so you have to study $2^n$ cases. – Bongo Oct 15 '23 at 08:19
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    You may find the idea of this post useful: https://math.stackexchange.com/questions/113270/the-median-minimizes-the-sum-of-absolute-deviations-the-ell-1-norm – Etemon Oct 15 '23 at 09:57
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    $MA+MB=k$ is an ellipse, $<k$ is the inside of it. Cannot be made more general than this. You just consider only the points on the x-axis. – zwim Oct 15 '23 at 10:45
  • @Martingalo on the $x+1<0$ case, $x< -1$ so $|x|=-x$ and hence the equation would be $-x-(x+1)<2$ which implies $x>\frac{-3}{2}$. isn't it? – Praveen Kumaran P Oct 16 '23 at 04:27
  • @Bongo that applies to $\sum_{k}^{n}|x-a_k| \leq c $ and $\sum_{k}^{n}|x-a_k| \geq c $ cases also... Isn't it? – Praveen Kumaran P Oct 16 '23 at 04:30

2 Answers2

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Denote $A=\{x\in\mathbb{R}\mid|x|+|x+1|<2\}$. Let $I_{1}=(-\infty,-1),$ $I_{2}=[-1,0),$ and $I_{3}=[0,\infty)$. Note that $I_{1}\cup I_{2}\cup I_{3}=\mathbb{R}$, so \begin{eqnarray*} A & = & A\cap\left(I_{1}\cup I_{2}\cup I_{3}\right)\\ & = & \cup_{i=1}^{3}\left(A\cap I_{i}\right). \end{eqnarray*} Then, we go to work out $A\cap I_{i}$, for $i=1,2,3$. Observe that if $x<-1$, then $x+1<0$, and $x<0$, so $|x|=-x$ and $|x+1|=-(x+1)$. Therefore, $|x|+|x+1|<2$ is equivalent to $-x-(x+1)<2\Rightarrow x>-\frac{3}{2}$. Putting everything formally, we assert that $A\cap I_{1}=(-\frac{3}{2},-1)$.

Proof of the claim: Let $x\in A\cap I_{1}$. Hence $x<-1$ and $|x|+|x+1|<2$. As $|x|=-x$ and $|x+1|=-(x+1)$, we have $-x-(x+1)<2\Rightarrow x>-\frac{3}{2}$. Therefore, $-\frac{3}{2}<x<-1$. This shows that $A\cap I_{1}\subseteq(-\frac{3}{2},-1)$. Conversely, we let $x\in(-\frac{3}{2},-1)$. Clearly, $x\in I_{1}$. Also, $x>-\frac{3}{2}\Rightarrow-2x<3\Rightarrow-2x-1<2$. Note that $|x|=-x$ and $|x+1|=-(x+1)$ because $x<0$ and $x+1<0$, so $|x|+|x+1|=-2x-1<2$, i.e., $x\in A$. This shows that $(-\frac{3}{2},-1)\subseteq A\cap I_{1}$.

We can work out $A\cap I_{2}$ and $A\cap I_{3}$ similarly...

Danny Pak-Keung Chan
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Look first at $$|x|+|x+1|=2$$ When $x\gt0$ and $x+1\gt0$ we have $x+x+1=2$ so $x=0.5$.

When $x\lt0$ and $x+1\lt0$ we have $-x-(x+1)=2$ so $x=-1.5$.

It follows that from $|x|+|x+1|=2$ we have that $|x|+|x+1|\lt2$ is valid for $\boxed{-1.5\lt x\lt0.5}$.

►Similarly $|x+1|+|x-2|=7$ has as only solutions $x=-3$ and $x=4$

Piquito
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