1.

If a relation R ..............

Answer»


SOLUTION :We have `R={(X, y) : x^(2)-4xy+3y^(2)=0, x, y in N}`
Let `x in N, x^(2)-4.x.x+3x^(2)=0`
`:. (x, x) in R :. R` is reflexive
we have `(3)^(2)-4(3)(1)+3(1)^(2)=9-12+3=0`
or `(3, 1) in R`
Also `1^(2)-4(1)(3)+3(3)^(2)=1-12+27 NE 0`
`:. (1, 3) in R, :. R` is not symmetric
again `(9, 3) in R` because
`9^(2)-4(9)(3)+3(3)^(2)=108-108=0`
and `(3, 1) in R` because
`(3)^(2)-4(3)(1)+3(1)^(2)=12-12=0`
and `(9, 1) in R` if `9^(2)-4(9)(1)+3(1)^(2)=0`
if `84-36=0` which is not possible
`:. (9, 3), (3, 1) in R` and `(9, 1) in R :. R` is not transitive
`:.` Relation R is reflexive but NEITHER symmetric nor transitive.


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