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Step-by-Step Explanation
Step 1: Understand the Relation
The relation $R$ on $\mathbb{N} \times \mathbb{N}$ is defined by:
$$(a, b) \;R\; (c, d) \quad \text{if and only if} \quad a\,d\,(b - c) \;=\; b\,c\,(a - d).$$
We need to check three properties of this relation: reflexivity, symmetry, and transitivity.
Step 2: Check Reflexivity
A relation $R$ on a set is called reflexive if every element is related to itself. In other words,
$$(a, b) \;R\; (a, b) \quad \text{for all } (a,b).$$
Applying the given condition to $(a, b) \;R\; (a, b)$, we get the left-hand side as:
$$a \times b \times (b - a)$$
and the right-hand side as:
$$b \times a \times (a - b).$$
Notice that $(b-a)$ is the negative of $(a-b)$ except in special cases where $a = b$. Generally,
$$a b (b - a) = -\,a b (a - b).$$
Hence, these two sides are not equal unless $a = b$. Since we need this to hold for all $(a,b)$ in $\mathbb{N} \times \mathbb{N}$, it fails whenever $a \neq b.$
Therefore, the relation is not reflexive.
Step 3: Check Symmetry
A relation $R$ on a set is called symmetric if for every pair $((a,b),(c,d))$ in $R$, the reversed pair $((c,d),(a,b))$ is also in $R$. Formally,
$$(a, b) \;R\; (c, d) \quad \implies \quad (c, d) \;R\; (a, b).$$
From the condition $a\,d\,(b-c) = b\,c\,(a-d)$, if we flip $(a,b)$ and $(c,d)$, we check:
$$(c, d) \;R\; (a, b) \quad \Longleftrightarrow \quad c\,b\,(d-a) = d\,a\,(c-b).$$
We can rearrange each side of $a\,d\,(b-c) = b\,c\,(a-d)$ to see that it implies
$$c\,b\,(d-a) = d\,a\,(c-b),$$
which is the same condition for $(c,d) \;R\; (a,b)$. Thus, whenever $(a,b) \;R\; (c,d)$ holds, so does $(c,d) \;R\;(a,b)$.
Therefore, the relation is symmetric.
Step 4: Check Transitivity
A relation $R$ on a set is called transitive if whenever $(a, b) \;R\; (c, d)$ and $(c, d) \;R\; (e, f)$, then $(a, b) \;R\; (e, f)$ must also hold.
We are given a counterexample in the provided solution steps:
$(2,3) \;R\; (3,2)$
$(3,2) \;R\; (5,30)$
but it turns out that
$$(2,3) \;R\; (5,30) \quad \text{does not hold}.$$
Hence, there exists at least one case where two relation links exist, but the third link does not. This example is sufficient to prove that the relation is not transitive.
Conclusion
We have shown that $R$ is symmetric but not reflexive and not transitive.
Therefore, the correct characterization is:
“symmetric but neither reflexive nor transitive.”
