Key Concepts and Formulas

• Reflexive Relation: A relation $R$ on a set $S$ is reflexive if for every element $x \in S$, $x R x$.
• Symmetric Relation: A relation $R$ on a set $S$ is symmetric if for every $x, y \in S$, whenever $x R y$, it implies $y R x$.
• Transitive Relation: A relation $R$ on a set $S$ is transitive if for every $x, y, z \in S$, whenever $x R y$ and $y R z$, it implies $x R z$.
• The set of natural numbers is $\mathbb{N} = \{1, 2, 3, \ldots\}$.

Step-by-Step Solution

The relation $R$ is defined on $\mathbb{N} \times \mathbb{N}$ by $(a, b) ~\mathrm{R}~(c, d)$ if and only if $a d(b-c)=b c(a-d)$.

Step 1: Simplify the relation definition. We are given the condition $a d(b-c)=b c(a-d)$. Expanding both sides, we get: $adb - adc = abc - bcd$ Rearranging the terms to group similar products: $adb - abc = adc - bcd$ Factoring out common terms: $ab(d-c) = cd(a-b)$ Since $a, b, c, d \in \mathbb{N}$, they are non-zero. We can divide both sides by $abcd$: $\frac{ab(d-c)}{abcd} = \frac{cd(a-b)}{abcd}$ $\frac{d-c}{cd} = \frac{a-b}{ab}$ Separating the terms on each side: $\frac{d}{cd} - \frac{c}{cd} = \frac{a}{ab} - \frac{b}{ab}$ $\frac{1}{c} - \frac{1}{d} = \frac{1}{b} - \frac{1}{a}$ Rearranging the terms to match the order of the pairs $(a,b)$ and $(c,d)$: $\frac{1}{a} - \frac{1}{b} = \frac{1}{c} - \frac{1}{d}$ Let $f(x, y) = \frac{1}{x} - \frac{1}{y}$. Then the relation $(a, b) ~\mathrm{R}~(c, d)$ is equivalent to $f(a, b) = f(c, d)$. This simplified form will be useful for checking symmetry and transitivity.

Step 2: Check for Reflexivity. For $R$ to be reflexive, $(a, b) ~\mathrm{R}~(a, b)$ must hold for all $(a, b) \in \mathbb{N} \times \mathbb{N}$. Using the original definition of the relation, we substitute $c=a$ and $d=b$: $a \cdot b (b-a) = b \cdot a (a-b)$ $ab(b-a) = ab(-(b-a))$ $ab(b-a) = -ab(b-a)$ Moving all terms to one side: $ab(b-a) + ab(b-a) = 0$ $2ab(b-a) = 0$ Since $a, b \in \mathbb{N}$, $a > 0$ and $b > 0$, so $2ab \neq 0$. For the equation to hold, we must have $b-a = 0$, which implies $b=a$. This means $(a, b) ~\mathrm{R}~(a, b)$ is true if and only if $a=b$. However, reflexivity requires this to be true for all pairs $(a, b) \in \mathbb{N} \times \mathbb{N}$. For example, for the pair $(1, 2)$, $(1, 2) ~\mathrm{R}~(1, 2)$ would require $1=2$, which is false. Therefore, $R$ is not reflexive.

Step 3: Check for Symmetry. For $R$ to be symmetric, if $(a, b) ~\mathrm{R}~(c, d)$, then $(c, d) ~\mathrm{R}~(a, b)$ must hold. Assume $(a, b) ~\mathrm{R}~(c, d)$. Using the simplified form, this means: $\frac{1}{a} - \frac{1}{b} = \frac{1}{c} - \frac{1}{d}$ Now, we need to check if $(c, d) ~\mathrm{R}~(a, b)$ holds. This means checking if: $\frac{1}{c} - \frac{1}{d} = \frac{1}{a} - \frac{1}{b}$ This equation is identical to our assumption. If $X=Y$, then $Y=X$. Thus, if $(a, b) ~\mathrm{R}~(c, d)$, then $(c, d) ~\mathrm{R}~(a, b)$. Therefore, $R$ is symmetric.

Step 4: Check for Transitivity. For $R$ to be transitive, if $(a, b) ~\mathrm{R}~(c, d)$ and $(c, d) ~\mathrm{R}~(e, f)$, then $(a, b) ~\mathrm{R}~(e, f)$ must hold. Assume $(a, b) ~\mathrm{R}~(c, d)$. This implies: $\frac{1}{a} - \frac{1}{b} = \frac{1}{c} - \frac{1}{d} \quad (*)$ Assume $(c, d) ~\mathrm{R}~(e, f)$. This implies: $\frac{1}{c} - \frac{1}{d} = \frac{1}{e} - \frac{1}{f} \quad (**)$ From equations $(*)$ and $(**)$, by the transitivity of equality, we can conclude that: $\frac{1}{a} - \frac{1}{b} = \frac{1}{e} - \frac{1}{f}$ This is precisely the condition for $(a, b) ~\mathrm{R}~(e, f)$. Therefore, $R$ is transitive.

Common Mistakes & Tips

• Reflexivity Check: Always use the original definition of the relation for reflexivity. Simplifying the relation can sometimes hide conditions that are necessary for reflexivity. In this case, the simplified form $f(a,b) = f(a,b)$ is always true, but the original definition $2ab(b-a)=0$ implies $a=b$, which is not universally true.
• Simplification: Simplifying the relation to a form like $f(x)=f(y)$ is very powerful for checking symmetry and transitivity, as it directly leverages the properties of equality.
• Domain: Remember that the relation is defined on $\mathbb{N} \times \mathbb{N}$, so all variables $a, b, c, d$ are positive integers. This justifies division by terms like $a, b, c, d,$ or $abcd$.

Summary

We analyzed the given relation $R$ on $\mathbb{N} \times \mathbb{N}$. By simplifying the relation's definition to $\frac{1}{a} - \frac{1}{b} = \frac{1}{c} - \frac{1}{d}$, we found that the relation is symmetric and transitive. However, upon checking reflexivity using the original definition, we discovered that $(a, b) ~\mathrm{R}~(a, b)$ only holds when $a=b$, meaning the relation is not reflexive. Therefore, $R$ is symmetric and transitive but not reflexive.

The final answer is $\boxed{A}$.