Friday 18 January 2013

relations and functions

Exercise 1.1



Question 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 < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y): y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
R = {(x, y): x  y is as integer}
(v) Relation R in the set A of human beings in a town at a particular time given by
(a) R = {(x, y): x and y work at the same place}
(b) R = {(x, y): x and y live in the same locality}
(c) R = {(x, y): x is exactly 7 cm taller than y}
(d) R = {(x, y): x is wife of y}
(e) R = {(x, y): x is father of y}
solutions =
(i) A = {1, 2, 3 … 13, 14}
R = {(x, y): 3x  y = 0}
∴R = {(1, 3), (2, 6), (3, 9), (4, 12)}
R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R.
Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0]
Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.
[3(1) − 9 ≠ 0]
Hence, R is neither reflexive, nor symmetric, nor transitive.
(ii) R = {(x, y): y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)}
It is seen that (1, 1) ∉ R.
∴R is not reflexive.
(1, 6) ∈R
But,
(6, 1) ∉ R.
∴R is not symmetric.
Now, since there is no pair in R such that (x, y) and (y, z) ∈R, then (x, z) cannot belong to R.
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(iii) A = {1, 2, 3, 4, 5, 6}
R = {(x, y): y is divisible by x}
We know that any number (x) is divisible by itself.
 (x, x) ∈R
∴R is reflexive.
Now,
(2, 4) ∈R [as 4 is divisible by 2]
But,
(4, 2) ∉ R. [as 2 is not divisible by 4]
∴R is not symmetric.
Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y.
z is divisible by x.
⇒ (x, z) ∈R
∴R is transitive.
Hence, R is reflexive and transitive but not symmetric.
(iv) R = {(x, y): x  y is an integer}
Now, for every x  Z, (x, x) ∈R as x  x = 0 is an integer.
∴R is reflexive.
Now, for every x, y  Z if (x, y) ∈ R, then x  y is an integer.
⇒ −(x  y) is also an integer.
⇒ (y  x) is an integer.
∴ (y, x) ∈ R
∴R is symmetric.
Now,
Let (x, y) and (y, z) ∈R, where x, y, z  Z.
⇒ (x  y) and (y  z) are integers.
 x  z = (x  y) + (y  z) is an integer.
∴ (x, z) ∈R
∴R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(v) (a) R = {(x, y): x and y work at the same place}
 (x, x) ∈ R
∴ R is reflexive.
If (x, y) ∈ R, then x and y work at the same place.
 y and x work at the same place.
⇒ (y, x) ∈ R.
∴R is symmetric.
Now, let (x, y), (y, z) ∈ R
 x and y work at the same place and y and z work at the same place.
 x and z work at the same place.
⇒ (x, z) ∈R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(b) R = {(x, y): x and y live in the same locality}
Clearly (x, x) ∈ R as x and x is the same human being.
∴ R is reflexive.
If (x, y) ∈R, then x and y live in the same locality.
 y and x live in the same locality.
⇒ (y, x) ∈ R
∴R is symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R.
 x and y live in the same locality and y and z live in the same locality.
 x and z live in the same locality.
⇒ (x, z) ∈ R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(c) R = {(x, y): x is exactly 7 cm taller than y}
Now,
(x, x) ∉ R
Since human being x cannot be taller than himself.
∴R is not reflexive.
Now, let (x, y) ∈R.
 x is exactly 7 cm taller than y.
Then, y is not taller than x.
∴ (y, x) ∉R
Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.
∴R is not symmetric.
Now,
Let (x, y), (y, z) ∈ R.
 x is exactly 7 cm taller thany and y is exactly 7 cm taller than z.
 x is exactly 14 cm taller than z .
∴(x, z) ∉R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(d) R = {(x, y): x is the wife of y}
Now,
(x, x) ∉ R
Since x cannot be the wife of herself.
∴R is not reflexive.
Now, let (x, y) ∈ R
 x is the wife of y.
Clearly y is not the wife of x.
∴(y, x) ∉ R
Indeed if x is the wife of y, then y is the husband of x.
∴ R is not transitive.
Let (x, y), (y, z) ∈ R
 x is the wife of y and y is the wife of z.
This case is not possible. Also, this does not imply that x is the wife of z.
∴(x, z) ∉ R
∴R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(e) R = {(x, y): x is the father of y}
(x, x) ∉ R
As x cannot be the father of himself.
∴R is not reflexive.
Now, let (x, y) ∈R.
 x is the father of y.
 y cannot be the father of y.
Indeed, y is the son or the daughter of y.
∴(y, x) ∉ R
∴ R is not symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R.
 x is the father of y and y is the father of z.
 x is not the father of z.
Indeed x is the grandfather of z.
∴ (x, z) ∉ R
∴R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

1 comment:

  1. I think there is a mistake in the answer to the question (d). The relation should be a transitive one. If (x,y) belongs to the relation R i.e. 'x' is the wife of 'y' , then obviously there is no person 'z' for which (y,z) belongs to R. So, we can't prove that R is not transitive. Therefore , R is transitive.
    For another example, the relation R = { (1,2) } on set { 1,2,3 } is transitive.

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