
For real numbers $x$ and $y$, we define $xRy$ if $x - y + \sqrt 5 $ is irrational. Then, which of the following is true about R?
A. R is reflexive.
B. R is symmetric.
C. R is transitive.
D. None of these
Answer
575.4k+ views
Hint: We will first write the meanings of R being reflexive, symmetric and transitive. Then, we will try to see if any of the properties are followed by the given relation and then mark that as correct.
Complete step-by-step answer:
Let us first understand the meanings of all the options:
Let R be a binary relation on a set A. Then,
R is reflexive if for all \[x\; \in A,xRx\].
R is symmetric if for all $x,y \in A$, if \[xRy\], then \[yRx\].
R is transitive if for all \[x,y,z\; \in A\], if \[xRy\] and \[yRx\], then \[xRz\].
R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.
Now, let us come back to our question. We define $xRy$ if $x - y + \sqrt 5 $ is irrational.
Let us see if this relation is reflexive or not.
So, if this is reflexive for all \[x\; \in A,xRx\].
Consider $x - x + \sqrt 5 = \sqrt 5 $, which is irrational.
Hence, R is reflexive.
Let us see if this relation is symmetric or not.
Now, for all $x,y \in A$, if \[xRy\] we need \[yRx\].
Consider $(\sqrt 5 ,1)$, we have $x - y + \sqrt 5 = \sqrt 5 - 1 + \sqrt 5 = 2\sqrt 5 - 1$, which is irrational but $(1,\sqrt 5 )$, we get: $x - y + \sqrt 5 = 1 - \sqrt 5 + \sqrt 5 = 1$, which is rational.
Hence, R is not symmetric.
Let us see if this relation is transitive or not.
Now, for all \[x,y,z\; \in A\], if \[xRy\] and \[yRx\], then we need \[xRz\].
Consider $(\sqrt 5 ,1)$, we have: $x - y + \sqrt 5 = \sqrt 5 - 1 + \sqrt 5 = 2\sqrt 5 - 1$ , which is irrational.
And see $(1,2\sqrt 5 )$, we have $x - y + \sqrt 5 = 1 - 2\sqrt 5 + \sqrt 5 = 1 - \sqrt 5 $ , which is irrational.
But in $(\sqrt 5 ,2\sqrt 5 )$, we get: \[x - y + \sqrt 5 = \sqrt 5 - 2\sqrt 5 + \sqrt 5 = 0\] , which is rational.
Hence, it is not transitive as well.
So, the correct answer is “Option A”.
Note: The students might be tempted to leave the question after proving the A part to be correct. They may do that, if we have the same options and the same kind of question, but there is a possibility that you have an equivalence relation in the option. So, to prove that incorrect or correct, you will have to check every property.
Complete step-by-step answer:
Let us first understand the meanings of all the options:
Let R be a binary relation on a set A. Then,
R is reflexive if for all \[x\; \in A,xRx\].
R is symmetric if for all $x,y \in A$, if \[xRy\], then \[yRx\].
R is transitive if for all \[x,y,z\; \in A\], if \[xRy\] and \[yRx\], then \[xRz\].
R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.
Now, let us come back to our question. We define $xRy$ if $x - y + \sqrt 5 $ is irrational.
Let us see if this relation is reflexive or not.
So, if this is reflexive for all \[x\; \in A,xRx\].
Consider $x - x + \sqrt 5 = \sqrt 5 $, which is irrational.
Hence, R is reflexive.
Let us see if this relation is symmetric or not.
Now, for all $x,y \in A$, if \[xRy\] we need \[yRx\].
Consider $(\sqrt 5 ,1)$, we have $x - y + \sqrt 5 = \sqrt 5 - 1 + \sqrt 5 = 2\sqrt 5 - 1$, which is irrational but $(1,\sqrt 5 )$, we get: $x - y + \sqrt 5 = 1 - \sqrt 5 + \sqrt 5 = 1$, which is rational.
Hence, R is not symmetric.
Let us see if this relation is transitive or not.
Now, for all \[x,y,z\; \in A\], if \[xRy\] and \[yRx\], then we need \[xRz\].
Consider $(\sqrt 5 ,1)$, we have: $x - y + \sqrt 5 = \sqrt 5 - 1 + \sqrt 5 = 2\sqrt 5 - 1$ , which is irrational.
And see $(1,2\sqrt 5 )$, we have $x - y + \sqrt 5 = 1 - 2\sqrt 5 + \sqrt 5 = 1 - \sqrt 5 $ , which is irrational.
But in $(\sqrt 5 ,2\sqrt 5 )$, we get: \[x - y + \sqrt 5 = \sqrt 5 - 2\sqrt 5 + \sqrt 5 = 0\] , which is rational.
Hence, it is not transitive as well.
So, the correct answer is “Option A”.
Note: The students might be tempted to leave the question after proving the A part to be correct. They may do that, if we have the same options and the same kind of question, but there is a possibility that you have an equivalence relation in the option. So, to prove that incorrect or correct, you will have to check every property.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which animal has three hearts class 11 biology CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

The camels hump is made of which tissues a Skeletal class 11 biology CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

