# Check whether the relation $$R$$ in $$R$$ defined by $R = {(a, b) : a \leq b^3}$ is reflexive, symmetric or transitive.

Toolbox:
• A relation R in a set A is called reflexive. if $(a,a) \in R\;for\; all\; a\in A$
• A relation R in a set A is called symmetric. if $(a_1,a_2) \in R\;\Rightarrow \; (a_2,a_1)\in R \;$ for $\;a_1,a_2 \in A$
• A relation R in a set A is called transitive. if $(a_1,a_2) \in\; R$ and $(a_2,a_3)\in R \Rightarrow \;(a_1,a_3)\in R\;$for all $\; a_1,a_2,a_3 \in A$
Given the relation $R=\{(a,b):a\leq b^3\}$:
Let $a = b$. For any $a \in R, a \leq a^3$ is false unless $a=1$. Since its not true in the general case, $R$ is not reflexive.
We can verify this w/ a simple substitution:
If $a=\frac{1}{2}, \frac{1}{2}.>(\frac{1}{2})^3=\frac{1}{8} \rightarrow \;(a,a) \notin R$
Let $(a,b) \in R$ such that $a\leq b^3$
Now, if $(b,a) \in R \rightarrow b \leq a^3$ must be true $\rightarrow b^3 \leq (a^3)^3 \leq a^9$
Therefore, $a\leq b^3 \rightarrow a\leq a^9$, which is not true unles $a = 1$.
Hence $R$ is not symmetric.
We can verify this w/ a simple substitution:
Let a=1 and b=2 a,b $\in R \rightarrow$ $1 < 2^3=8\;but\;2^3>1$. Hence $R$ is not symmetric.
Let $(a,b) \in R$ such that $a\leq b^3$
Now, if $(b,c) \in R \rightarrow b \leq c^3$ must be true.
Now, if $(a,c) \in R \rightarrow a \leq c^3$ must be true.
We can disprove this through simple substiution:
Let $a=3, b =\frac{3}{2}, c = \frac{4}{3}$
$\Rightarrow a\leq b^3 \rightarrow 1 \leq (\frac{3}{2})^3 \rightarrow 1 \leq \frac{27}{8}$.
$\Rightarrow b \leq c^3 \rightarrow \frac{3}{2} \leq (\frac{4}{3})^3 \rightarrow \frac{3}{2} \leq 64/27$.
However, $a \leq c^3 \rightarrow 3 \leq (\frac{4}{3})^3 \rightarrow 3 \leq 64/27$ which is not true. Hence $R$ is not transitive.
edited Mar 8, 2013