Question

Let $\mathrm{W}$ denote the words in the English dictionary. Define the relation $\mathrm{R}$ by $R=\left\{\left(x,y\right)\in W\times W\mathrm{the}\mathrm{words}\mathrm{x},\mathrm{y}\mathrm{have}\mathrm{atleast}\mathrm{one}\mathrm{letter}\mathrm{in}\mathrm{common}\right\}$ then $\mathrm{R}$ is.

• A. reflexive, symmetric, and not transitive
• B. reflexive, symmetric, and transitive
• C. reflexive, not symmetric, and transitive
• D. not reflexive, symmetric, and transitive

Answer 1

The correct option is A

reflexive, symmetric, and not transitive

Explanation for the correct answer:

Finding the characteristics of $\mathrm{R}$:

Let $y$ be a word, $y$ has every letter in common.

So,

$\left(\mathrm{y},\mathrm{y}\right)\in \mathrm{R}$

Thus $R$ is reflexive.

Let, $\left(\mathrm{x},\mathrm{y}\right)\in \mathrm{R}$ $x,y$ have at least one letter is common, similarly $y,x$ have at least one letter is common.

Thus $R$ is symmetric.

Let us assume, $x=apple$, $y=tap$, $z=thing$

Here,

$\left(\mathrm{x},\mathrm{y}\right)\in \mathrm{R},\left(\mathrm{y},\mathrm{z}\right)\in \mathrm{R}\mathrm{but}\left(\mathrm{x},\mathrm{z}\right)\notin \mathrm{R}$

Thus $R$ is not transitive.

Hence, the correct option is A.