1.

Determine whether each of the following relations are reflexive, symmetric and transitive : (i) Relation R in the set A= {1, 2, 3, …, 13, 14} defined as "" R = { (x, y ) : 3x - y =0} (ii) Relation R in the set N of natural numbers defined as"" R = {(x, y) : y = x + 5 and x lt 4} (iii) Relation R in the set A= { 1, 2, 3, 4, 5, 6} as ""R = { (x, y) : yis divisible by x} (iv) Relation R in the set Z of all integers defined as "" R = {(x,y) : x -yis an integer } (v) Relation R in the set A of human beings in a town at a particular time given by (a) R= {(x, y) : and y work at the same place } (b) R = { (x, y) : x and ylive in the same locality } (c) R = {(x, y) : xis exactly 7 cm taller than y } (d) R = {(x, y): xis wife of y} (e) R= { (x, y) :xis father of y}

Answer»

Solution :(i) A =` { 1, 2, 3, …, 13, 14}`
and`"" R = {(x, y) : 3x -y =0}`
For reflexive `(x,x) in R AA x in A`
but ` "" 3x -y =0rArry =3x`
`therfore "" (x, x) notin R ` if ` x = 2 in A`
`rArr R `is not reflexive,
For symmetricity `(x, y) in R rArr (y, x) in R AA x, y in R `
Now, `"" (x, y) in R rArr 3 x -y =0`
`"" rArr 3y -x ne 0`
`"" rArr (y, x) notin R`
`therefore R` is not symmetric.
e.g., `(1, 3) in R and (3, 1) notin R`
For transitivity `(x, y) in R, (y, z) in R rArr (x, z) in R`
`therefore (1, 3) in R and (3, 9) in R rArr (1, 9) in R`
`rArr R` is not transitive.
(ii) `R= {(x,y) : y = x + 5 and x lt 4 }` and N is the set of natural numbers.
`rArr R = {(1, 6), (2, 7), (3, 8)}`
For reflexive, `(1,1) notin R`
`rArr R` is not reflexive.
For transitivity, `(x,y) in R (y, z) in R rArr (x, z) in R`.
No pair satisfies this condition.
`therefore R` is not transitive.
(iii) `A= {1, 2, 3, 4, 5, 6}`
and `R = {(x, y) : y` is divisible by `x}`
`rArr R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5, (6, 6)}`
For each `x in A, (x, x) in R`
`therefore R ` is reflexive.
For each `x, y in R, (x, y) in R cancel(rArr) (y, x) in R`
`therefore R` is not symmetric.
For each `x, y, z in A` if `(x, y) in R, (y, z) in R` then `(x, z) in R`
`therefore R` is transitive.
(iv) In the set of all integers Z
`"" R = {(x, y) : x -y ` is an integer. }
Which is true.
`therefore R` is reflexive.
For symmetricity,
`(x, y) in R rArr (x -y)` is an integer.
`"" rArr (y-x) in R`
` therefore R` is symmetric.
`therefore ` For transitivity.
`(x, y) in R and (y, z) in R`
`rArr (x -y) ` is an integer and `(y-z)` is an integer.
`rArr (x-y)+ (y-z)` is an integer.
`rArr (x, z) in R`
`therefore R` is transitive.
(v) (a) `R= {(x, y) : x and y` work at the same place }
This relation is reflexive, symmetric and transitive.
(b) `R= {(x, y) : x and y` live in the same locality}.
This relation is reflexive, symmetric and transitive.
(c) `R= {(x, y) : x` is EXACTLY 7 cmtaller than y }
`(x, x) notin R` because x is no exactly 7 cm taller than y
`therefore R ` is not reflexive.
`(x, y) in R rArrx ` is exactly 7 cm taller than y.
`"" cancel ( rArr) y` is exactly 7 cm taller than x.
`"" cancel (rArr) (y, x) in R`
`therefore R` is not symmetric.
`therfore (x, y) in R and (y, z) in R rArr x ` is exactly 7 cm taller than y and y is exactly7 cm taller than z.


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