Let A be the set of all functions f: mathbb{Z | vedclass

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(B) $R = \{(f, g) : f(0) = g(1) 	ext{ and } f(1) = g(0)\}$$1.$ Reflexive: For $R$ to be reflexive,$(f, f) \in R$ must hold for all $f \in A$. This im

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Symmetric but neither reflexive nor transitive

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(B) $R = \{(f, g) : f(0) = g(1) 	ext{ and } f(1) = g(0)\}$$1.$ Reflexive: For $R$ to be reflexive,$(f, f) \in R$ must hold for all $f \in A$. This implies $f(0) = f(1)$ and $f(1) = f(0)$. Since this is not true for all functions $f: \mathbb{Z} ightarrow \mathbb{Z}$ (e.g.,consider $f(x) = x$),$R$ is not reflexive.$2.$ Symmetric: If $(f, g) \in R$,then $f(0) = g(1)$ and $f(1) = g(0)$. We need to check if $(g, f) \in R$. This requires $g(0) = f(1)$ and $g(1) = f(0)$. These are exactly the same conditions as the definition of $(f, g) \in R$. Thus,$R$ is symmetric.$3.$ Transitive: If $(f, g) \in R$ and $(g, h) \in R$,then $f(0) = g(1)$,$f(1) = g(0)$,$g(0) = h(1)$,and $g(1) = h(0)$. For $R$ to be transitive,we need $(f, h) \in R$,which implies $f(0) = h(1)$ and $f(1) = h(0)$. From the given conditions,$f(0) = g(1) = h(0)$ and $f(1) = g(0) = h(1)$. This does not necessarily imply $f(0) = h(1)$ and $f(1) = h(0)$. For example,if $f(0)=1, f(1)=2, g(0)=2, g(1)=1, h(0)=1, h(1)=2$,then $(f, g) \in R$ and $(g, h) \in R$ hold,but $(f, h) \in R$ requires $f(0)=h(1) \Rightarrow 1=2$,which is false. Thus,$R$ is not transitive.

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

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