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

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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 $\; a_1,a_2,a_3 \in A$
  • In order to prove the result, we must always solve for the general case. In order to disprove some result, it is sufficient to prove it for one specific ordered pair that does not satisfy the relation.
Given in a set of real numbers, the relation $R=\{(a,b):a\leq b^2\}$:
Let $a = \frac{1}{3} \; \rightarrow \ a^2 = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$
Clearly in this case, since $\frac{1}{3} \not \leq (\frac{1}{3})^2$, i.e., $\frac{1}{3} \not \leq \frac{1}{9}$,
Therefore since $(\frac{1}{3}.\frac{1}{3}) \not \in R$, $R$ is not reflexive.
Let $a = 1$ and $b = 3 \rightarrow \; a \leq b^2 \rightarrow 1 \leq 3 \times 3 \rightarrow 1 \leq 9$.
However, $a^2 = 1 \times 1 = 1 \rightarrow 3 \not \leq 1$ i.e., $ b \not \leq a^2 $.
Therefore while $(x,y) \in R$, $(y,x) \not \in R$. Hence is $R$ is not symmetric.
Let $a = 2$ and $b = -3 \; \rightarrow \ b^2 = -3 \times -3 = 9 \rightarrow 2 \leq 9 \rightarrow a \leq b^2$.
Let $b = -3$ and $c =1 \; \rightarrow \ c^2 = 1 \times 1 = 1 \rightarrow -3 \leq 1 \rightarrow b \leq c^2$.
However $a =2$ and $c =1 \; \rightarrow \ c^2 = 1 \times 1 = 1 \rightarrow 2 \not \leq 1 \rightarrow a \not \leq c^2$.
Hence $R$ is not transitive.
answered Feb 20, 2013 by meena.p
edited Mar 8, 2013 by balaji.thirumalai
 

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