Q. 1 C3.9( 9 Votes )

# Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:

R = {(x, y) : x is wife of y}

Answer :

We have been given that,

A is the set of all human beings in a town at a particular time.

Here, R is the binary relation on set A.

So, recall that

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Using these criteria, we can solve these.

We have,

R = {(x, y): x is wife of y}

**Check for Reflexivity:**

Since x and x are the same people.

Then, x cannot be the wife of itself.

A person cannot be a wife of itself.

Wendy is the wife of Sam; Wendy can’t be the wife of herself.

So, ∀ x ∈ A, then (x, x) ∉ R.

**∴** **R is not reflexive.**

**Check for Symmetry:**

If x is the wife of y.Then, y cannot be the wife of x.

If Wendy is the wife of Sam, then Sam is the husband of Wendy.

Sam cannot be the wife of Wendy.

So, if (x, y) ∈ R, then (y, x) ∉ R.

∀ x, y ∈ A

**∴** **R is not symmetric.**

**Check for Transitivity:**

If x is the wife of y and y is the wife of z, which is not logically possible.

Then, x is not the wife of z.

It’s easy, take Wendy, Sam, and Mac.

If Wendy is the wife of Sam, Sam can’t be the wife of Mac.

Thus, the possibility of Wendy being the wife of Mac also eliminates.

So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∉ R.

∀ x, y, z ∈ A

**∴** **R is not transitive.**

**Hence, R is neither reflexive, nor symmetric, nor transitive.**

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