Aspire's Library
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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:
(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.
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
