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Show that the following relations are neither reflexive nor symmetric nor
transitive.
a) R= {(a,b): b= a + 1} on the set A = {1,2,3,4,5,6)
b) R= {(x,y):x – 2y—3=0} on the set of all real numbers R
Answers
Answer:
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 symmetric.
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.
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