Each of the following defines a relation on `N :`
(i) `x > y ,\ x ,\ y in N`
(ii) `x+y=10 ,\ x ,\ y in N`
(iii) `x y` is square of an integer, `x ,\ y in N`
(iv) `x+4y=10 ,\ x ,\ y in N`
Determine which of the above relations are reflexive, symmetric and transitive.

Explanation

(i) x is greater than `y,x,y in N`
`(x,x) in R`
For `xRx " " x gt x ` is not true for any `x in N`.
Therefore, R is not reflexive.
Let ` (x,y) in R implies xRy`
`x gt Y`
but `y gt x ` is not true for any `x, y in N`
Thus, R is not symmetric.
Let `xRY and yRz`
`x gt y and y gt z implies x gt z`
`implies xRz`
So, R is transitive.
(ii) `x+y =10, x, y in N`
` R = {(x,y), x+y=10, x,y in N }`
`R ={(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1)} (1,1) notin R `
So, R is not reflexive,
`(x,y) in R implies (y,x) in R`
Therefore, R is symmetric,
`(1,9) in R, (9,1) in R implies (1,1) notin R`
Hence, R is not transitive.
(iii) Given xy, is square of an integer `x, y in N. `
`implies R={(x,y):xy` is square of an integer `x,y in N } `
`(x,x) in R, AA x in N`
As `x^(2)` is square of an integer for any `x in N`
Hence, R is reflexive.
If `(x,y) in R implies (y,x) in R`
Therefore, R is xymmetric.
If `(x,y) in R , (y,z) in R`
So, xy is square of an integer and yz is square of an integer.
Let `xy=m^(2)` and `yz=n^(2)` for some `m, n in Z`
`x=(m^(2))/(y)` and `z=(x^(2))/(y)`
`xz =(m^(2)n^(2))/(y^(2)),` which is square of an integer.
So, R is transitive.
(iv) `x+4y=10, x,y in N`
`R={(x,y):x+4y=10, x,y in N} `
`R={(2, 2),(6,1)}`
`(1,1),(3,3), ... notin R`
Thus, R is not reflexive.
`(6,1) in R` but `(1,6) notin R`
Hence, R is not symmetric.
`(x,y) in R implies x+4y =10` but `(y,z) in R`
`y+4z =10 implies (x,z) in R`
So, R is transitive.