For $$\alpha \in \mathbb{N}$$, consider a relation R on $$\mathbb{N}$$ given by $$R = \{(x, y) : 3x + \alpha y$$ is a multiple of 7$$\}$$. The relation R is an equivalence relation if and only if

We need to find the condition on $$\alpha \in \mathbb{N}$$ for the relation $$R = \{(x, y) : 3x + \alpha y \text{ is a multiple of } 7\}$$ on $$\mathbb{N}$$ to be an equivalence relation.

Step 1: Check Reflexivity.

$$(x, x) \in R \Leftrightarrow 3x + \alpha x \equiv 0 \pmod{7}$$ for all $$x \in \mathbb{N}$$

$$\Leftrightarrow (3 + \alpha)x \equiv 0 \pmod{7}$$ for all $$x$$

This requires $$3 + \alpha \equiv 0 \pmod{7}$$, i.e., $$\alpha \equiv 4 \pmod{7}$$.

Step 2: Check Symmetry (with $$\alpha \equiv 4 \pmod 7$$).

If $$(x, y) \in R$$, then $$3x + 4y \equiv 0 \pmod{7}$$ (using $$\alpha \equiv 4 \pmod 7$$).

We need to show $$3y + 4x \equiv 0 \pmod{7}$$.

From $$3x + 4y \equiv 0 \pmod{7}$$: $$3x \equiv -4y \pmod{7}$$.

Since $$3^{-1} \equiv 5 \pmod{7}$$ (as $$3 \times 5 = 15 \equiv 1$$), we get $$x \equiv -20y \equiv -6y \equiv y \pmod{7}$$.

Now: $$3y + 4x \equiv 3y + 4y \equiv 7y \equiv 0 \pmod{7}$$. Symmetric!

Step 3: Check Transitivity.

If $$(x, y) \in R$$ and $$(y, z) \in R$$:

$$3x + 4y \equiv 0 \pmod{7}$$ and $$3y + 4z \equiv 0 \pmod{7}$$

From above, $$x \equiv y \pmod{7}$$ and $$y \equiv z \pmod{7}$$, so $$x \equiv z \pmod{7}$$.

Then $$3x + 4z \equiv 3z + 4z = 7z \equiv 0 \pmod{7}$$. Transitive!

Step 4: Conclusion.

R is an equivalence relation if and only if $$\alpha \equiv 4 \pmod{7}$$, i.e., 4 is the remainder when $$\alpha$$ is divided by 7.

The correct answer is Option D: 4 is the remainder when $$\alpha$$ is divided by 7