Analyzing the Relation R on Functions
We are given a set A of all functions $f: \mathbb{Z} \to \mathbb{Z}$ and a relation R defined as:
$R = \{(f, g) \in A \times A \mid f(0) = g(1) \text{ and } f(1) = g(0)\}$
We need to determine if R is reflexive, symmetric, and transitive.
Relation Properties Check
Reflexivity
A relation R is reflexive if $(f, f) \in R$ for every function $f \in A$. This requires $f(0) = f(1)$ and $f(1) = f(0)$ for all $f$.
Consider the function $f(x) = x$. Here, $f(0) = 0$ and $f(1) = 1$. Since $f(0) \neq f(1)$, the condition $f(0) = f(1)$ is not met.
Therefore, R is not reflexive.
Symmetry
A relation R is symmetric if $(f, g) \in R$ implies $(g, f) \in R$.
Assume $(f, g) \in R$. This means:
- $f(0) = g(1)$
- $f(1) = g(0)$
Now, let's check the conditions for $(g, f) \in R$:
- We need $g(0) = f(1)$. From our assumption, $f(1) = g(0)$, which is true.
- We need $g(1) = f(0)$. From our assumption, $f(0) = g(1)$, which is true.
Since both conditions are met, if $(f, g) \in R$, then $(g, f) \in R$.
Therefore, R is symmetric.
Transitivity
A relation R is transitive if $(f, g) \in R$ and $(g, h) \in R$ implies $(f, h) \in R$.
Assume $(f, g) \in R$ and $(g, h) \in R$. This means:
- $f(0) = g(1)$
- $f(1) = g(0)$
- $g(0) = h(1)$
- $g(1) = h(0)$
We need to check if $(f, h) \in R$, which requires $f(0) = h(1)$ and $f(1) = h(0)$.
From the assumptions:
- Combining (1) and (4): $f(0) = g(1) = h(0)$. This gives $f(0) = h(0)$.
- Combining (2) and (3): $f(1) = g(0) = h(1)$. This gives $f(1) = h(1)$.
The conditions derived ($f(0) = h(0)$ and $f(1) = h(1)$) are not the required conditions for $(f, h) \in R$ (which are $f(0) = h(1)$ and $f(1) = h(0)$).
Let's use a counterexample:
- Let $f(x) = x$. Then $f(0)=0, f(1)=1$.
- Let $g(x) = 1-x$. Then $g(0)=1, g(1)=0$. $(f, g) \in R$ because $f(0)=0=g(1)$ and $f(1)=1=g(0)$.
- Let $h(x) = x$. Then $h(0)=0, h(1)=1$. $(g, h) \in R$ because $g(0)=1=h(1)$ and $g(1)=0=h(0)$.
- Now check $(f, h)$: We need $f(0) = h(1)$ and $f(1) = h(0)$.
- Here, $f(0) = 0$ and $h(1) = 1$. Since $0 \neq 1$, $f(0) \neq h(1)$.
- Thus, $(f, h) \notin R$.
Therefore, R is not transitive.
Conclusion
Based on the analysis:
- R is not reflexive.
- R is symmetric.
- R is not transitive.
The relation R is symmetric but neither reflexive nor transitive.
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