Let $$R = \{(x, y) : x, y \in N$$ and $$x^2 - 4xy + 3y^2 = 0\}$$, where N is the set of all natural numbers. Then the relation R is :

We can factor this quadratic expression by splitting the middle term:

$$x^2 - 3xy - xy + 3y^2 = 0$$  $$x(x - 3y) - y(x - 3y) = 0$$  $$(x - y)(x - 3y) = 0$$

This gives two possible conditions for any ordered pair $$(x, y) \in R$$:

1. $$x = y$$
2. $$x = 3y$$

Now, let's test this relation for reflexivity, symmetry, and transitivity over the set of natural numbers $$\mathbb{N}$$.

1. Reflexivity

A relation is reflexive if $$(x, x) \in R$$ for every $$x \in \mathbb{N}$$.

• For $$(x, x)$$, substitute $$y = x$$ into the factored equation:$$(x - x)(x - 3x) = 0 \cdot (-2x) = 0$$
• Since the equation holds true for all $$x \in \mathbb{N}$$, the relation is reflexive.

2. Symmetry

A relation is symmetric if whenever $$(x, y) \in R$$, then $$(y, x) \in R$$.

• Let's test with a counterexample. Take the pair $$(3, 1)$$.
• Here, $$x = 3$$ and $$y = 1$$. Since $$x = 3y$$ ($$3 = 3 \cdot 1$$), the pair $$(3, 1) \in R$$.
• For symmetry, its flipped counterpart $$(1, 3)$$ must also be in $R$. Let's check $$(1, 3)$$:
  • Does $$1 = 3$$? No.
  • Does $$1 = 3 \cdot 3$$? No.
• Since $$(1, 3) \notin R$$, the relation is not symmetric.

3. Transitivity

A relation is transitive if whenever $$(x, y) \in R$$ and $$(y, z) \in R$$, then $$(x, z) \in R$$.

• Let's test with a counterexample using our $$x = 3y$$ condition.
• Pick $$z = 1 \in \mathbb{N}$$.
  • If $$y = 3z \implies y = 3(1) = 3$$. So, $$(3, 1) \in R$$.
  • If $$x = 3y \implies x = 3(3) = 9$$. So, $$(9, 3) \in R$$.
• For transitivity, if $$(9, 3) \in R$$ and $$(3, 1) \in R$$, then $$(9, 1)$$ must be in $$R$$. Let's check $$(9, 1)$$:
  • Does $$9 = 1$$? No.
  • Does $$9 = 3 \cdot 1$$? No.
• Since $$(9, 1) \notin R$$, the relation is not transitive.

Conclusion

The relation $$R$$ is reflexive but neither symmetric nor transitive.