For real numbers x and y, define xRy if and only if x – y + √2 is an irrational number. Then the relation R is
Answers
Answer:
Given, for real numbers x and y,
xRy => x - y + √2 is an irrational number.
Let R is a binary relation on real numbers x and y.
Now, R is transitive iff for all (x, y) ∈ R and (y, z) ∈ R implies (x, z) ∈ R
Given, xRy => x - y + √2 is irrational ............1
and yRz => y - z + √2 is irrational ............2
Add equation 1 and 2, we get
(x - y + √2) + (y - z + √2) is irrational
= x - z + √2 is irrational
= xRz is irrational
So, the relation R is transitive.
Answer:
Reflexsive only
Step-by-step explanation:
In Symmetric to get yRx take minus so -(y-x- root2) your const is now minus root2 but we needed plus root 2. also try example x=1 and y=root2
In transitive take eq x-y+ root 2, y-z +root2 adding them you'll get x-z +2root2
const is changed again we needed only root2