Determining a relation if reflexive, symmetric, and transitive

Question

I just get stuck in this relation and need to find if this relation is

• Reflexive/ Irreflexive or Neither,
• Symmetric/ Antisymmetric or Neither,
• Transitive or Not.

$$W_1 = \{(a , b) \in \mathbb N \times\mathbb N : 2a \geqslant b\}$$

What I have now is reflexive, antisymmetric and transitive. But I just get confused by the the inequality.

Answer 1

Assuming $a\sim b$ when $2a\geq b$ let us first check reflexivity.

A relation is reflexive if $(a,a)\in R$ for every $a$

So, is it true that $2a\geq a$ for all $a\in\mathbb{N}$?

Is $2a\geq a$ for all $a\in\mathbb{N}$? Divide both sides by $a$ to see that $2a\geq a$ simplifies to $2\geq 1$, which is always true, so yes, it is reflexive.

As for symmetry:

A relation is symmetric if whenever $(a,b)\in R$, then $(b,a)$ is also in $R$.

If $2a\geq b$, is it automatically true that $2b\geq a$?

Lets try to find a counterexample: Simple examples are always helpful. Consider $(100,1)$. You have $2\cdot 100 \geq 1$ is true, but $2\cdot 1\geq 100$ is not true, so it is not symmetric.

Antisymmetric?

A relation is antisymmetric if whenever $(a,b)\in R$ and $a\neq b$, then $(b,a)$ is not in $R$.

Is it always true that $2a\geq b$ means that $2b<a$?

Before we tried numbers far apart to find our counterexample. This time, it looks like numbers close together can cause problems. So, let us inspect $(3,4)$. We have $2\cdot 3=6\geq 4$, so $(3,4)\in R$. How about the reverse, $(4,3)$. We have $2\cdot 4=8\geq 3$ so $(4,3)$ is also in $R$. Therefore, it is not antisymmetric.

Finally, transitivity.

A relation is transitive if whenever $(a,b)\in R$ and $(b,c)\in R$, then you necessarily have $(a,c)\in R$.

So, if $2a\geq b$ and $2b\geq c$, is $2a\geq c$?

Well, $2a\geq b$ means the same as $4a\geq 2b$, and so since $2b\geq c$, we know that $4a\geq 2b\geq c$ and that $4a\geq c$, but it might not necessarily be that $2a\geq c$. Let us search for a counterexample. Bad things might happen when the $2a$ is very close in size to $b$, so let us look at pairs of numbers like those. $(1,2)\in R$ since $2\cdot 1\geq 2$, and $(2,4)\in R$ since $2\cdot 2\geq 4$. But look at $(1,4)$. You have $2\cdot 1=2\geq 4$ is not true, so $(1,4)\notin R$ even though both $(1,2)$ and $(2,4)$ are in $R$.