Let A = 0, 1, 2, 3, 4, 5.

Question

Let A = {0, 1, 2, 3, 4, 5}. Let R be a relation on A defined by (x, y) \in  R if and only if max{x, y} \in  {3, 4}. Then among the statements (S1): The number of elements in R is 18, and (S2): The relation R is symmetric but neither reflexive nor...

Accepted Answer

To evaluate the relation $ R $ on the set $ A = \{0, 1, 2, 3, 4, 5\} $, we first need to understand the conditions for an element $(x, y)$ to be in $ R $. Specifically, $(x, y) \in R$ if and only if $\max\{x, y\} \in \{3, 4\}$.

Considering this, let's list the pairs:

For $\max\{x, y\} = 3$, the possible pairs are:

$(0, 3), (3, 0), (1, 3), (3, 1), (2, 3), (3, 2), (3, 3)$

For $\max\{x, y\} = 4$, the possible pairs are:

$(0, 4), (4, 0), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4)$

Combining these, the set $ R $ consists of the following elements:

$  R = \{(0, 3), (3, 0), (1, 3), (3, 1), (2, 3), (3, 2), (3, 3), (0, 4), (4, 0), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4)\}  $

This gives us a total of 16 elements in $ R $, not 18 as initially claimed in statement $ S_1 $.

Next, we analyze the properties of the relation $ R $:

Reflexivity: A relation is reflexive if $(x, x) \in R$ for all $ x \in A$. For example, $(0, 0), (1, 1), (2, 2)$ are not in $ R $, so $ R $ is not reflexive.

Symmetry: A relation is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$ as well. For all pairs $(x, y)$ listed, both $(x, y)$ and $(y, x)$ are present. Thus, $ R $ is symmetric.

Transitivity: A relation is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. An example where transitivity fails is $(0, 3)$ and $(3, 1)$ are in $ R $ but $(0, 1)$ is not in $ R$. Therefore, $ R $ is not transitive.

In conclusion, statement $ S_2 $ is correct as $ R $ is symmetric but neither reflexive nor transitive.

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