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First of all, I will show two questions of absolute value inequality to compare the method in which a formal logic is used:

1) Find all numbers $\:x\:$ for which $$\lvert x - 1 \rvert + \lvert x - 2 \rvert > 1$$

a) When $x-1 \geq 0$ and $x-2 \geq 0$, $\: \lvert x - 1 \rvert + \lvert x - 2 \rvert = x - 1 + x - 2 = 2x - 3 > 1 \iff x>2$.

b) When $x-1 \leq 0$ and $x-2 \geq 0$, $\: \lvert x - 1 \rvert + \lvert x - 2 \rvert = -x+1+x-2 = -1 \ngtr 1$. Therefore, Contradiction. (Sub-question: can I just say there is no $\: x \:$ that satisfies $\: x \leq 1 \:$ and $\: x \geq 2 \:$, therefore, we can disregard this case? Or, is this the 'correct' approach?)

c) When $x-1 \geq 0$ and $x-2 \leq 0$, $\: \lvert x - 1 \rvert + \lvert x - 2 \rvert = x-1-x+2 = 1 \ngtr 1$. Therefore, also a contradiction.

d) When $x-1 \leq 0$ and $x-2 \leq 0$, $\: \lvert x - 1 \rvert + \lvert x - 2 \rvert = -x+1-x+2 = -2x + 3 > 1 \iff x < 1$

Therefore, I can conclude that $\: \lvert x - 1 \rvert + \lvert x - 2 \rvert > 1 \iff x > 2\: $ or $\: x < 1. \qquad \blacksquare$

2) Find all numbers $\: x \:$ for which $$\lvert x - 1 \rvert + \lvert x + 1 \rvert < 2$$

a) When $x-1 \geq 0$ and $x+1 \geq 0$, $\: \lvert x - 1 \rvert + \lvert x + 1 \rvert = x - 1 + x + 1 = 2x < 2 \iff x<1$.

b) When $x-1 \leq 0$ and $x+1 \geq 0$, $\: \lvert x - 1 \rvert + \lvert x + 1 \rvert = -x+1+x+2 = 3 \nless 2$.

c) When $x-1 \geq 0$ and $x+1 \leq 0$, $\: \lvert x - 1 \rvert + \lvert x + 1 \rvert = x-1-x-2 = -3 < 2$ (But here also, there is no $\: x \:$ that satisfies $\: x-1 \geq 0\:$ and $\: x+1 \leq 0$.)

d) When $x-1 \leq 0$ and $x+1 \leq 0$, $\: \lvert x - 1 \rvert + \lvert x + 1 \rvert = -x+1-x-2 = -2x-1 < 2 \iff x > -\frac{3}{2}$

Therefore, I can conclude that $\lvert x - 1 \rvert + \lvert x + 1 \rvert < 2 \iff -\frac{3}{2} < x < 1 \qquad \qquad \blacksquare$

My question is, why 2) entails $\quad x > -\frac{3}{2} \quad \textbf{AND} \quad x < 1 \quad$ while 1) entails $\quad x > 2 \quad \textbf{OR} \quad x < 1$?

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When you have an inequality of the form $X+Y >c$, this forces AT LEAST ONE of $X$ and $Y$ to be large. When you have $X+Y<c$, where $X,Y>0$, then it forces BOTH of $X$ and $Y$ to be small.

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  • $\begingroup$ Thank you but can you explain in terms of formal logic based on the cases I have covered above? $\endgroup$ – Sungjun Hwang Aug 31 '16 at 17:28

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