Math, asked by vievekananda3320, 1 year ago

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

Answered by faizanbasha005
20

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.

Answered by AdibProbot
11

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

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