Aspire Faculty ID #18784 · Topic: UGC NET Computer Science Sep 2022 (Paper II) · Just now
UGC NET Computer Science Sep 2022 (Paper II)

Let $\epsilon=0.0005$ and Let $R_e$ be the relation $\{(x,y)\in R^2:|x-y|<\epsilon\}$, $R_e$ could be interpreted as the relation approximately equal. $R_e$ is
(A) Reflexive
(B) Symmetric
(C) transitive
Choose the correct answer from the options given below:

Solution


Solution:

Given relation is:

$R_e=\{(x,y)\in R^2:|x-y|<\epsilon\}$

where,

$\epsilon=0.0005$

Now check each property.

Check reflexive:

A relation is reflexive if every element is related to itself.

For any real number $x$,

$|x-x|=0$

Since,

$0<\epsilon$

So,

$|x-x|<\epsilon$

Therefore, $(x,x)\in R_e$.

Hence, $R_e$ is reflexive.

So, statement (A) is true.

Check symmetric:

A relation is symmetric if $(x,y)\in R_e$ implies $(y,x)\in R_e$.

If,

$(x,y)\in R_e$

then,

$|x-y|<\epsilon$

But,

$|x-y|=|y-x|$

So,

$|y-x|<\epsilon$

Therefore, $(y,x)\in R_e$.

Hence, $R_e$ is symmetric.

So, statement (B) is true.

Check transitive:

A relation is transitive if $(x,y)\in R_e$ and $(y,z)\in R_e$ imply $(x,z)\in R_e$.

Let,

$x=0,\ y=0.0004,\ z=0.0008$

Now,

$|x-y|=|0-0.0004|=0.0004<0.0005$

So, $(x,y)\in R_e$.

Also,

$|y-z|=|0.0004-0.0008|=0.0004<0.0005$

So, $(y,z)\in R_e$.

But,

$|x-z|=|0-0.0008|=0.0008$

Since,

$0.0008>0.0005$

So, $(x,z)\notin R_e$.

Therefore, $R_e$ is not transitive.

So, statement (C) is false.

Hence, only (A) and (B) are true.

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