Q.Let $R$ be a relation from $\mathbb{N}$ to $\mathbb{N}$ defined by $R = \{(a, b) : a, b \in \mathbb{N}\ \text{and}\ a = b^2\}$. Are the following true?

Concept: Relation Arrow Diagram — a relation is a set of ordered pairs; each property (reflexive, symmetric, transitive) must hold for all elements that satisfy the condition.

Given $R = \{(a, b) : a, b \in \mathbb{N}, a = b^2\}$.

(i) $(a, a) \in R$ for all $a \in \mathbb{N}$?

For $(a, a)$ to be in $R$, we need $a = a^2$, i.e. $a^2 - a = 0 \Rightarrow a(a-1)=0$. In $\mathbb{N}$, only $a=1$ satisfies this. So false for $a=2,3,\dots$.

(ii) $(a, b) \in R \implies (b, a) \in R$?

If $(a,b) \in R$, then $a = b^2$. For $(b,a)$ to be in $R$, we need $b = a^2 = (b^2)^2 = b^4$, so $b^4 - b = 0 \Rightarrow b(b^3-1)=0$, giving $b=0$ or $b=1$. Only $b=1$ works in $\mathbb{N}$. So false for $b=2$ (e.g. $(4,2) \in R$ but $(2,4) \notin R$).

(iii) $(a,b) \in R$ and $(b,c) \in R \implies (a,c) \in R$? …

Why the Arrow Diagram Tells the Story

Think of $R$ as a machine: you feed in a natural number $b$, and it spits out its square $a = b^2$. So the pairs in $R$ look like $(1,1)$, $(4,2)$, $(9,3)$, $(16,4)$, and so on. The first element is always a perfect square, and the second element is its square root.

If you draw an arrow diagram with $\mathbb{N}$ on both sides, an arrow goes from $b$ (left) to $a = b^2$ (right). This immediately shows the relation is very one-directional and sparse — most numbers don't appear as the first element at all.

Checking Each Property

1. Is $(a,a) \in R$ for every $a \in \mathbb{N}$?

For $(a,a)$ to be in $R$, we need $a = a^2$. That means $a^2 - a = 0$, so $a(a-1)=0$. The only natural numbers satisfying this are $a=0$ (not in $\mathbb{N}$ as defined here — $\mathbb{N}$ usually starts from 1) and $a=1$.

So only $(1,1)$ is in $R$. For $a=2$, we'd need $2 = 2^2 = 4$, which is false. For $a=3$, $3 = 9$? No.

Conclusion: $(a,a) \in R$ is false for all $a \neq 1$. So statement (i) is false.

2. Does $(a,b) \in R$ imply $(b,a) \in R$?

If $(a,b) \in R$, then $a = b^2$. For symmetry, we'd need $(b,a) \in R$, which means $b = a^2$.

Substituting $a = b^2$ into $b = a^2$ gives $b = (b^2)^2 = b^4$, so $b^4 - b = 0$, i.e. $b(b^3 - 1)=0$. The only natural solution is $b=1$, giving $a=1$.

Take a concrete counterexample: $(4,2) \in R$ because $4 = 2^2$. But $(2,4) \in R$ would require $2 = 4^2 = 16$, which is false.

Conclusion: Statement (ii) is false. …