Question

Which of the following is not correct for relation $R$ on the set of real numbers?

• A. $(x,y)\in R\iff |x|–|y|\le 1$ is reflexive but not symmetric
• B. $(x,y)\in R\iff 0<|x|–|y|\le 1$ is neither transitive nor symmetric
• C. $(x,y)\in R\iff 0<|x–y|\le 1$ is symmetric and transitive
• D. $(x,y)\in R\iff |x–y|\le 1$ is reflexive and symmetric

Answer 1

The correct option is C $(x,y)\in R\iff 0<|x–y|\le 1$ is symmetric and transitive

Checking options
$(x,y)\in R\iff |x–y|\le 1$ is reflexive and symmetric
Putting $y=x$
$\Rightarrow |x-x|\le 1$ is true
So, it is reflexive.
Also, $|x–y|=|y-x|$, so it is symmetric.

$(x,y)\in R\iff 0<|x|–|y|\le 1$ is neither transitive nor symmetric
Taking $x=3,y=2,z=1$
$|x|–|y|=1\in (0,1]$
$|y|–|z|=1\in (0,1]$
$|x|–|z|=2\notin (0,1]$
As, $(x,y),(y,z)$ lies in the relation but $(x,z)$ does not, so it is not transitive.
Also, $|x|–|y|\ne |y|–|x|$, so it is not symmetric.

$(x,y)\in R\iff 0<|x–y|\le 1$ is symmetric and transitive
$|x–y|=|y-x|$, so it is symmetric.
Taking $x=3,y=2,z=1$
$|x-y|=1|y-z|=1|x-z|=2$
As, $(x,y),(y,z)$ lies in the relation but $(x,z)$ does not, so it is not transitive.

$(x,y)\in R\iff |x|–|y|\le 1$ is reflexive but not symmetric
Putting $y=x$
$\Rightarrow |x|–|x|\le 1$ is true
So, it is reflexive.
Also, $|x|–|y|\ne |y|–|x|$, so it is not symmetric.