Symmetry, transitivity and reflexivity

Question

I need some help on how to approach this problem. I can't seem to find any examples that help me understand this, so if anyone has an approach example to post I would be very grateful:

"Consider a relation $R$ defined on the set of integers. Determine for the following if the relation is reflexive, symmetric, and transitive: $R = \{(x, y)|x = 2y \}.$"

Answer 1

I assume you have a clear concept about the definition of Relations.

As given, $R=\{(x,y) : x=2y\}$ i.e. in the set of integers the relation $R$ is defined as,

for $x,y\in\mathbb{Z}$, $x R y \iff x=2y$.

Now $R$ is not reflexive, as $1\ne2$ so, $1\not R 1$ i.e. $(1,1)\notin R$.

$R$ is not symmetric, as $2R1\iff 2=2\times1$ but $1\not R 2$ as $\exists$ no integer $n$ s.t. $1=2n$.

$R$ is not transitive, as $4R2$ and $8R4$ but $8\not R 2$.

Answer 2

1. is it true that for all $x \in \mathbb Z$ we have $x=2x$ ? If yes, then $R$ is reflexive, if no, then $R$ is not reflexive.

2. suppose that $x,y \in \mathbb Z$ and $x=2y$. Does it always follow that $y=2x$ ? If yes, then $R$ is symmetric, if no, then $R$ is not symmetric.

3. suppose that $x,y,z \in \mathbb Z$ and $x=2y$ and $y=2z.$ Does it always follow that $x=2z$ ? If yes, then $R$ is transitive, if no, then $R$ is not transitive.

Answer 3

You need to think about the basic definitions. Any relation, x~ y, is "reflexive" if and only if x~ x is true for any x. It is "symmetric" if and only if x~ y implies y~ x. It is "transitive" if and only if x~ y and y~ z implies x~ z.

Reflexive: is it true that x~ x, that is, x= 2x, for any real number x?

Symmetric: Suppose x~ y. That is, x= 2y. Does it the follow that y~ x, that y= 2x?

Transitive: Suppose x~ y and y~ z. That is, x= 2y and y= 2z. Does it follow that x~ z, that x= 2z?

Answer 4

Let's ask if $R$ is symmetric. For $R$ to be symmetric, we would have to have $$y\ R\ x\qquad\text{whenever}\qquad x\ R\ y.$$

According to the definition of $R$ that you gave, $x\ R \ y$ just means $x=2y$. So what we are really asking about is whether it's true that $$y=2x\qquad\text{whenever}\qquad x=2y.$$

But simple algebra tells us this only happens for $x=0, y=0$. In particular, when $x=2, y=1$ we have $x\ R\ y$ but not $y\ R\ x$. So $R$ is not symmetric.

Now you try the other two.