Mathematical Proofs, Equivalence Relation Question I'm new so this will be my first question!
The question is this:
Define a Relation R on $\mathbb Z$ (integer set) as follows: $ (x, y) \in R $ iff $x = |y|$.
Is $R$ symmetric, reflexive, and transitive?
My work:
$xRx$ -> $x = |x|$ but $x \in \mathbb Z$, therefore, R is not reflexive as $-x \neq |-x|$
$xRy, yRx$
WTS: $y= |x|$
$x=|y|$
$x^2 = |y|^2 = y^2$
$-y^2 = -x^2$
$y^2 = x^2$
$\sqrt{y^2} = \sqrt{x^2}$
$|y| = |x|$ which does not equal $y = |x|$
Therefore, R is not symmetric
$xRy, yRh$, WTS: $xRh$
$y=|h| $
From work in previous step:
$|y| = |h|$
$x = |y|$ -> $x = |h|$
Therefore, R is transitive.
Is my thinking correct? Did I make a mistake?
 A: When proving things, it is generally easier to build on things like you did with the transitive property.
However, when you're DISproving things, you might want to just give a counterexample.
Assume R is reflexive.
$\Rightarrow \forall x \in \mathbb{Z} xRx$
$\Rightarrow$ Since $-1 \in \mathbb{Z}$, then $-1R-1 \Rightarrow -1 = |-1|$
but we know this to be false, so our assumption must be false.
Similarly for the symmetric property
Assume R is symmetric.
$xRy \Rightarrow yRx$
Since $(1,-1)\in R, \Rightarrow (-1,1)\in R \Rightarrow -1 = |1|$
but we know this to be false, so our assumption must be false.
As far as transitive, I think your proof works. Could be prettier, but it works.
I would assume you're starting some sort of Math/CS course in late HS or early college and this is getting graded, so I would add $x = |y|$ and $y = |h|$, then we know $|y| = ||h|| = |h|$ so $x = |h|$ so $xRh$
A: You should give counter-examples when it's false, instead of giving general reasons why it shouldn't be true.
For example, your first argument saying that $-x\neq |-x|$ is in fact only true for $x$ positive. So you haven't really prove it's false, since your proof is wrong (sorry).
For the second point, it's even worse. If $y=|x|$ then $y$ is positive and then $y=|y|$. So if $|y|=|x|$ then you must have $y= |x|$. So your conclusive argument is certainly false.
At least, the last one seems correct. :-)
