Question

Let $S$ be the set of all real numbers. A relation $R$ has been defined on $S$ by $aRb\iff |a-b|\le 1$, then $R$ is.

• A. Reflexive and transitive but not symmetric
• B. An equivalence relation
• C. Symmetric and transitive but not reflective
• D. Reflexive commulative ans associative

Answer 1

Answer: (A)

Reflexive and transitive but not symmetric

A relation $R$ is said to be reflexive if $(a,a)\in R$ for all $a\in A$, that is , every element of $A$ is related to itself.

Given,

$aRb\iff |a-b|\le 1$, $R$ defined on set of real numbers.

$\Rightarrow -1\le a-b\le 1$

In given inequality if $a\in$ Real Numers then $b\in$ Real Numers which is satisfying above reflexive relation property i.e. every element satisfying $a$ also satisfies $b$.

A relation $R$ is said to be symmetric if $(a,b)\in R\Rightarrow (b,a)\in R$, that is $aRb\Rightarrow bRa$ for all$(a,b)\in R$

Given, $-1\le a-b\le 1$

$-1+b\le a\le 1+bor-1+a\le b\le 1+a$

From above inequality for one value of $a$ multiple values of $b$ are produced which lies between $[-1+a,1+a]$

For above inequality to be symmetric all these multiple values of $b$ should map to single $a$. But, for one value of $b$ multiple values of $a$ are produced which lies between $[-1+b,1+b]$. That means instead multiple values of $b$ mapping to one element, one element of $b$ mapping to multiple values of $a$.

Hence above relation is not symmetric.

A relation $R$ is said to be transitive if $(a,b)\in Rand(b,c)\in R\Rightarrow (a,c)\in R$

In the given inequality $|a-b|\le 1$ one value of $a$ given multiple values of $b\in R[-1+a,1+a]$ and all these multiple values of $b$ produce another set of multiple values say $c$.

But for every $(a,b)\in R$ and $(b,c)\in R$ $(a,c)\in R$

So, $R$ is transitive.