Let A$A$ be the set of all functions fcolon mathbbZ to mathbbZ$f\colon \mathbb{Z} \to \mathbb{Z}$ and R$R$ be a relation on A$A$ such that R = \(f, g): f(0) = g(1) text and f(1) = g(0)\$R = \{(f, g): f(0) = g(1) \text{ and } f(1) = g(0)\}$. Then R$R$ is:

### Related Formula Definition of properties of binary relations: * Reflexive: (f, f) in R iff f(0) = f(1)$(f, f) \in R \iff f(0) = f(1)$ * Symmetric: (f, g) in R implies (g, f) in R$(f, g) \in R \implies (g, f) \in R$ * Transitive: (f, g) in R text and (g, h) in R implies (f, h) in R$(f, g) \in R \text{ and } (g, h) \in R \implies (f, h) \in R$ ### Core Logic Evaluate reflexivity, symmetry, and transitivity sequentially by plugging standard arbitrary function values into the condition definition. ### Step 1: Reflexivity Audit For (f,f) in R$(f,f) \in R$, we require f(0) = f(1)$f(0) = f(1)$ and f(1) = f(0)$f(1) = f(0)$. This holds true only for functions whose values at 0$0$ and 1$1$ are identical. Since it does not hold true for *all* possible functions mapping mathbbZ to mathbbZ$\mathbb{Z} \to \mathbb{Z}$ (e.g., f(x)=x$f(x)=x$), R$R$ is **not reflexive**. ### Step 2: Symmetry Audit Assume (f,g) in R implies f(0) = g(1)$(f,g) \in R \implies f(0) = g(1)$ and f(1) = g(0)$f(1) = g(0)$. To check if (g,f) in R$(g,f) \in R$, check its matching constraints: g(0) = f(1)$g(0) = f(1)$ and g(1) = f(0)$g(1) = f(0)$. Both statements are perfectly identical to our assumption. Therefore, R$R$ is **symmetric**. ### Step 3: Transitivity Audit Assume (f,g) in R implies f(0) = g(1), f(1) = g(0)$(f,g) \in R \implies f(0) = g(1), f(1) = g(0)$. Assume (g,h) in R implies g(0) = h(1), g(1) = h(0)$(g,h) \in R \implies g(0) = h(1), g(1) = h(0)$. For (f,h) in R$(f,h) \in R$, we need f(0) = h(1)$f(0) = h(1)$ and f(1) = h(0)$f(1) = h(0)$. From assumptions: f(0) = g(1) = h(0)$f(0) = g(1) = h(0)$ and f(1) = g(0) = h(1)$f(1) = g(0) = h(1)$. This means f(0) = h(0)$f(0) = h(0)$ and f(1) = h(1)$f(1) = h(1)$, which does *not* necessarily satisfy f(0)=h(1)$f(0)=h(1)$. Hence, R$R$ is **not transitive**. ### Pattern Recognition The relation swaps indices 0$0$ and 1$1$. Swapping twice returns you to the original position, which visually justifies why symmetry holds trivially, while transitivity creates a cyclic dependency that fails standard property constraints. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions