Functions Question 553
Question: For real numbers x and y, we define x R y iff $ x-y+\sqrt{5} $ is an irrational number. The relation R is
Options:
A) Reflexive
B) Symmetric
C) Transitive
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ x\in R\Rightarrow x-x+\sqrt{5}=\sqrt{5} $ is an irrational number.
$ \therefore (x,x)\in R $
$ \therefore $ R is reflexive. $ (\sqrt{5},1)\in R $ because $ \sqrt{5}-1+\sqrt{5} $ $ =2\sqrt{5}-1 $ Which is an irrational number.
$ \therefore (1,\sqrt{5})\notin R.\therefore R $ is not symmetric. We have, $ (\sqrt{5},1),(1,2\sqrt{5})\in R $ because $ \sqrt{5}-1+\sqrt{5}=2\sqrt{5}-1 $ if $ 1-2\sqrt{5}+\sqrt{5}=1-\sqrt{5} $ are irrational numbers. Also, $ (\sqrt{5},2\sqrt{5})\in R $ and $ \sqrt{5}-2\sqrt{5}+\sqrt{5}=0. $ which is not an irrational number.
$ \therefore (\sqrt{5},2\sqrt{5})\notin R $
$ \therefore $ R is not transitive.
BETA
