Question

Let R be a relation $\mathbb{N}\to \mathbb{N}$ defined by $R=\left\{ \left( a,b \right):a,b\in \mathbb{N};a={{b}^{2}} \right\}$.
A. $\left( a,a \right)\in R;\forall a\in \mathbb{N}$
B. $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R;\forall a,b\in \mathbb{N}$
C. \[\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R;\forall a,b,c\in \mathbb{N}\]
D. None of these

Answer 1

Hint: We first explain the condition of relations and its different characteristics like reflexivity, symmetric and transitivity. The contraction with one example can show that the characteristics are not possible. We also do the algebra part to understand it better.

Complete step-by-step solution:
It is given that $R=\left\{ \left( a,b \right):a,b\in \mathbb{N};a={{b}^{2}} \right\}$ where R be a relation $\mathbb{N}\to \mathbb{N}$.
We have to check the reflexivity, symmetric and transitivity.
If we can show one example of contradiction for any characteristics then we can say that the characteristics are not possible.
Now suppose $\left( a,a \right)\in R;\forall a\in \mathbb{N}$. We can see that if that happens then $a={{a}^{2}}$ which gives $a(a-1)=0\Rightarrow a=0, 1$ as $a\in \mathbb{N}$. For example, $\left( 2,2 \right)\notin R$ but $2\in \mathbb{N}$.
The relation is not reflexive.
Now suppose $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R;\forall a,b\in \mathbb{N}$. We can see that if that happens then $a={{b}^{2}}$ and $b={{a}^{2}}$ which gives $a={{a}^{4}}$. The simplified form is $a=0, 1$ as $a\in \mathbb{N}$. For example, for $2,4\in \mathbb{N}$, $\left( 4,2 \right)\notin R$ but $\left( 2,4 \right)\notin R$.
The relation is not symmetric.
Now suppose \[\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R;\forall a,b,c\in \mathbb{N}\]. We can see that if that happens then $a={{b}^{2}}$, $b={{c}^{2}}$ and $a={{c}^{2}}$ which gives ${{c}^{2}}={{c}^{4}}\Rightarrow c=0, 1$ as $c\in \mathbb{N}$. For example, for $2,4,16\in \mathbb{N}$, $\left( 2,4 \right)\notin R$ and $\left( 4,16 \right)\notin R$ but $\left( 2,16 \right)\notin R$.
The relation is not transitive.
The correct option is D.

Note: If a relation satisfies all three relations, then the relation is called Equivalence relation. Equivalence relations are massively used outside of mathematics, as they are simply a means of breaking some set of objects into separate subgroups.