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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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    Thanks for the comment. I have edited the questions, and hope it is more clear now.2012-01-10

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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.

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    @andreas.vitikan yes, although it can seem tedious, it is thorough.2012-01-10