Let A = {1, 2, 3, 4, 5, 6, 7}. Then the relat | vedclass

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(D) Given $A = \{1, 2, 3, 4, 5, 6, 7\}$ and $R = \{(x, y) \in A 	imes A : x + y = 7\}$.Listing the elements of $R$: $R = \{(1, 6), (2, 5), (3, 4), (4

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symmetric but neither reflexive nor transitive

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(D) Given $A = \{1, 2, 3, 4, 5, 6, 7\}$ and $R = \{(x, y) \in A 	imes A : x + y = 7\}$.Listing the elements of $R$: $R = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$.$1$. Reflexive: For $R$ to be reflexive,$(x, x) \in R$ for all $x \in A$. Since $(1, 1) 
otin R$,$R$ is not reflexive.$2$. Symmetric: For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x) \in R$. Since $(1, 6) \in R$ and $(6, 1) \in R$,$(2, 5) \in R$ and $(5, 2) \in R$,etc.,$R$ is symmetric.$3$. Transitive: For $R$ to be transitive,if $(x, y) \in R$ and $(y, z) \in R$,then $(x, z) \in R$. Here $(1, 6) \in R$ and $(6, 1) \in R$,but $(1, 1) 
otin R$. Thus,$R$ is not transitive.Therefore,$R$ is symmetric but neither reflexive nor transitive.

Let $A = \{1, 2, 3, 4, 5, 6, 7\}$. Then the relation $R = \{(x, y) \in A \times A : x + y = 7\}$ is

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