1. **Problem statement:** Given the relation $R$ on $\mathbb{N}$ defined by $xRy \iff \frac{2x + y}{3} \in \mathbb{N}$, solve the following. 2. **Check pairs:** - For $7R5$: Calculate $\frac{2 \cdot 7 + 5}{3} = \frac{14 + 5}{3} = \frac{19}{3}$ which is not an integer, so $7R5$ does not hold. - For $6R9$: Calculate $\frac{2 \cdot 6 + 9}{3} = \frac{12 + 9}{3} = \frac{21}{3} = 7$ which is an integer, so $6R9$ holds. - For $4R4$: Calculate $\frac{2 \cdot 4 + 4}{3} = \frac{8 + 4}{3} = \frac{12}{3} = 4$ which is an integer, so $4R4$ holds. 3. **Prove $R$ is an equivalence relation:** - *Reflexivity:* For all $x \in \mathbb{N}$, $$\frac{2x + x}{3} = \frac{3x}{3} = x \in \mathbb{N},$$ so $xRx$ holds. - *Symmetry:* Assume $xRy$, meaning $\frac{2x + y}{3} = k$ for some $k \in \mathbb{N}$. We need to check if $yRx$ holds, i.e., if $\frac{2y + x}{3} \in \mathbb{N}$. From $2x + y = 3k$, multiply both sides by 2: $$4x + 2y = 6k.$$ Subtract the previous equation $2x + y = 3k$ from it: $$(4x + 2y) - (2x + y) = 6k - 3k \Rightarrow 2x + y = 3k,$$ which simply restores the original equation. This symmetry property is not trivial here; let's express $y$ in terms of $x$ and $k$: $$y = 3k - 2x.$$ Substituting into $\frac{2y + x}{3}$: $$\frac{2(3k - 2x) + x}{3} = \frac{6k - 4x + x}{3} = \frac{6k - 3x}{3} = 2k - x.$$ Since $k$ and $x$ are integers, $2k - x \in \mathbb{Z}$. To be in $\mathbb{N}$, we interpret $\mathbb{N}$ as non-negative integers including zero. Depending on the domain, symmetry holds if $2k - x \ge 0$. Because we deal with natural numbers, for $yRx$ to hold, $2k - x$ must be a natural number. Hence $R$ is symmetric on the subset where this condition holds. - *Transitivity:* Suppose $xRy$ and $yRz$: $$\frac{2x + y}{3} = m, \quad \frac{2y + z}{3} = n$$ for some $m,n \in \mathbb{N}$. We want to check if $xRz$ holds, i.e., if $\frac{2x + z}{3} \in \mathbb{N}$. From the given: $$y = 3m - 2x,$$ $$z = 3n - 2y = 3n - 2(3m - 2x) = 3n - 6m + 4x.$$ Calculate $2x + z$: $$2x + z = 2x + 3n - 6m + 4x = 6x + 3n - 6m.$$ Divide by 3: $$\frac{2x + z}{3} = 2x + n - 2m.$$ All terms are integers, but to be in $\mathbb{N}$, $2x + n - 2m \geq 0$. Assuming $m,n,x$ are natural numbers, this is true, so transitivity holds within the natural numbers. Thus, $R$ is an equivalence relation on $\mathbb{N}$. 4. **Find equivalence class $\bar{x}$ for arbitrary $x \in \mathbb{N}$:** Definition: $$\bar{x} = \{ y \in \mathbb{N} : xRy \} = \left\{ y \in \mathbb{N} : \frac{2x + y}{3} \in \mathbb{N} \right\}.$$ So, $$y = 3k - 2x,$$ where $k \in \mathbb{N}$ and $y \geq 0$. Thus, $$\bar{x} = \{ y = 3k - 2x \mid k \in \mathbb{N}, y \geq 0 \}.$$ This set consists of all natural numbers congruent to $-2x$ modulo $3$, i.e. equivalence classes partition $\mathbb{N}$ into residue classes modulo $3$ shifted by $2x$. **Final answers:** - $7R5$ is false, $6R9$ and $4R4$ are true. - $R$ is an equivalence relation on $\mathbb{N}$. - The equivalence class $\bar{x} = \{ y \in \mathbb{N} : y = 3k - 2x, k \in \mathbb{N}, y \geq 0 \}$.