Define a Relation R on the set of real numbers as follows:

Question

(x,y) ∈ R if and only if |x+y| = |x|+|y|. Is this reflective? Symmetric? Transitive? Is it an equivalence relation?

My attempt:

Reflective: Yes, it is reflective. |x+x| = |x|+|x| => 2|x|=2|x| which is true.

Symmetric: Yes, it is symmetric(?). Suppose (x,y) ∈ R if and only if |x+y| = |x|+|y|. We want to show (y,x) ∈ R iff |y+x| = |y|+|x|. Indeed |y+x| = |y|+|x|.

Transitive: Suppose (x,y) ∈ R if and only if |x+y| = |x|+|y| and (y,z) ∈ R if and only if |y+z| = |y|+|z|. We want to show (x,z) ∈ R if and only if |x+z| = |x|+|z|. Not sure where to begin for this one.

I don't really know how to prove the symmetric and the transitive one.

Answer 1

Let $x=1$, $y=0$, and $z=-1$. Then $x\sim y$ and $y\sim z$, but $x\not\sim z$.

Answer 2

Note that $|x+y|=|x|+|y|$ if and only if $|x+y|^2=(|x|+|y|)^2$, which simplifies to $|xy|=xy$.

Thus your relation can be rewritten as $$ (x,y)\in R\text{ if and only if } \begin{aligned}[t] &(x>0\text{ and }y>0)\text{ or} \\ &(x<0\text{ and }y<0)\text{ or} \\ &(x=0\text{ or }y=0) \end{aligned} $$ Now you can prove more easily that the relation is reflexive and symmetric, but not transitive, essentially because $(0,z)\in R$, for every $z$.

If transitive, it would be an equivalence relation, but the class of $0$ would be the whole set and the relation is clearly not the trivial relation containing every pair.