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Each of the following defines a relation on `N :``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 arereflexive, symmetric and transitive. |
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Answer» (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. |
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