Question

Show that the relation R in the set R of real numbers, defined as $R=\left\{\right(a,b):a\le b^2\}$ is neither reflexive nor symmetric nor transitive.

Here, the result is disproved using some speicific examples. In order to prove a result. we must prove it in generlity and in order to disprove a result we can just provide one example. where the condition is false. It is important to pick up the examles suitably. Since there are certain ordered pairs like (1,1) for which the relation is reflexive.

Answer 1

We have $R=\left\{\right(a,b):a\le b^2\}$ where $a,bϵR$
For reflexivity, we observe that $\frac{1}{2}\le (\frac{1}{2}{)}^2$ is not true.
So, R is not reflexive as $(\frac{1}{2},\frac{1}{2})\text{/}ϵR$
For symmetry, we observe that $-1\le 3^{2}but3\text{/}\le (-1{)}^2$
$\therefore (-1,3)ϵRbut(3-1)\text{/}ϵR.$
So, R is not symmetric.
For transitivity, we observe that $2\le (-3{)}^{2}and-3\le (1{)}^2$ but $2\text{/}\le (1{)}^2$
$\therefore (2,-3)ϵR$ and $(-3,1)ϵR$ but $(2,1)\text{/}ϵR$. So, R is not transitive.
Hence, R is neither reflexive, nor symmetric and nor transitive.