# How to solve equations with absolute value and using the Archimedean property

I'm trying to learn Real Analysis on my own, but I found that i'm a bit rusty with the elementary stuff.

How do I solve equations like $|x| + |x+1| = 1$ and $|x-1| + |x+1| = 2$? I don't want the whole solution, because I have a feeling that what I'm really looking for is a property of the absolute value which I can't remember.

Also, there is another exercise, that given the Archimedean property (for every $x \in \mathbb{R}$ there is a number $[x] \in \mathbb{Z}$ and you know the rest) prove that for $x, y \in \mathbb{R}$, x greater than 0, than there $\exists n \in \mathbb{N} : nx > y$.

Edit: Given the fact the floor function exists and knowing that $x \in R_+$ and $y \in R$ prove that there $\exists n \in N : nx > y$.

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For the first set: Usually, it's best to consider a different case for $x < 0$ and $x > 0$, or when you have something like $|x+1|$, $x < -1$ and $x > 1$ (since then the absolute value reduces to a simple expression in x, since you know the sign of its argument). For the second: What properties would the number $\lfloor x/y \rfloor$ have, with regards to $x/y$? How about $\lfloor x/y \rfloor + 1$? – Calvin McPhail-Snyder Jan 10 '12 at 16:37
The absolute value equations I still don't get how to do. $[x/y] \leqslant x/y$ whilst $[x/y] + 1 > x/y$. Also, if I have other questions should I post them in another thread or not? – andreas.vitikan Jan 10 '12 at 16:44
@andreas.vitikan: By definition $\lfloor\frac{y}{x}\rfloor\le \frac{y}{x}<\lfloor\frac{y}{x}\rfloor+1$. (I am using $y/x$ to be consistent with your $nx>y$.) Or do you want a proof that the floor function exists? For that we need to use least upper bound principle, or something similar. What you want to know is not absolutely clear. To make *small changes in this question, edit. For any new question, please make a new post. P.S. I like the whilst! Welcome to the site. – André Nicolas Jan 10 '12 at 17:16
Thanks for the comment. I have edited the questions, and hope it is more clear now. – andreas.vitikan Jan 10 '12 at 17:26

your absolute value questions read: the sum of the distances from $x$ to $0$ and $-1$ is $1$ (so $x=-1/2$ is the obvious geometric choice) and similarly the sum of the distances from $x$ to $1$ and $-1$ is $2$ (with $x=0$). or you could write down all combinations (for the first) $$x+(x+1)=1, x+1\geq0, x\geq0$$ $$x-(x+1)=1, x+1\leq0, x\geq0$$ $$-x+(x+1)=1, x+1\geq0, x\leq0$$ $$-x-(x+1)=1, x+1\leq0, x\leq0$$ and solve them, looking for solutions in the appropriate domain.
So, apply the basic principle that $|x| = a$ then $x = a$ or $x = -a$, but applied to more, right? – andreas.vitikan Jan 10 '12 at 17:22