Mathematical Proofs, Equivalence Relation Question

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?

Answer 1

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. :-)

Answer 2

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 $xRh$