Let $A = \mathbb{N}$ and let $aRb$ mean that $b|a$...

Question

Let $A = \mathbb{N}$ and let $aRb$ mean that $b\mid a$.

1. Is $R$ reflexive on $A$?
2. Is $R$ symmetric on $A$?
3. Is $R$ anti-symmetric on $A$?
4. Is $R$ transitive on $A$?
5. Is $R$ an equivalence relation on $A$?

I do not know where to start on this question, any help into guiding me into the right path would be very much appreciated!

Answer 1

A good way to start is to draw the relation $R$ for a few examples. Let's try $1, 2, 3, 4, 5,$ and $6$. Here are all the edges you have to draw:

• $1$ divides $1, 2, 3, 4, 5,$ and $6$, so draw an arrow from all the numbers to $1$ (including from $1$ to itself).

• $2$ divides $2, 4,$ and $6$, so draw three arrows (from $2$ to itself, from $4$ to $2$, and from $6$ to $2$.)

• $3$ divides $3$ and $3$ divides $6$

• $4$ divides itself

• $5$ divides itself

• $6$ divides itself

Once you draw all those edges you can check.

• "Reflexive" means every number has an edge to itself. Does this look true or false, for these examples?

• "Transitive" means that if there is an edge from $a$ to $b$ and from $b$ to $c$, then there's an edge from $a$ to $c$. Does this seem to be true in the picture?

• "Symmetric" means that if there's an edge from $a$ to $b$, then there's an edge from $b$ to $a$. Does it look true, for these exampls ($1, 2, 3, 4, 5$, and $6$)?

Once you decide which properties are true for $1, 2, 3, 4, 5,$ and $6$, you can decide if the properties that are true here will be true for all other numbers, as well. That would be a good overall way to build intuition and to approach this problem and similar problems.

Answer 2

You should start by looking at some definitions. A relation $R$ (will also be denoted by $\sim$) on a set $X$ is

• reflexive if each element of $X$ is in relation with itself that is

$$ aRa\qquad\text{or}\qquad a\sim a,\qquad \forall a \in X. $$

In your specific case this means that for $a\in \mathbb{N}$ we will check whether $a\mid a$. You could argue that $a=1\cdot a+0$ so we get $0$ as a rest when dividing $a$ with itself so $R$ is reflexive. Every natural number divides itself.

• symmetric if $$ aRb\Rightarrow bRa \qquad\text{or}\qquad a\sim b\Rightarrow b\sim a\qquad \forall a,b\in X $$

In your specific case this is not true, you could argue similarly as above but it may be easier to find a counterexample say $a=2$ and $b=4$. $$ 2\mid 4\not\Rightarrow 4\mid 2 $$

• antisymmetric if $$ aRb\qquad\text{together with}\qquad bRa \qquad \text{implies}\qquad a=b,\quad \forall a,b\in X $$

I think you can do this yourself.

• transitive if $$ a\sim b\quad\text{and}\quad b\sim c\Rightarrow a\sim c, \quad \forall a,b,c\in\ X $$

This one is left for you as well.

• is an equivalence relation if $R$ is reflexive, symmetric and transitive at the same time.

Hope this helped.