第12类RD Sharma解决方案–第1章关系–练习1.1 |套装2

问题11.对称和可传递的每个关系是否也是反身的，这是真的吗?说明原因。

解决方案：

We will verify this by taking an example.

Consider a set A = {1, 2, 3} and a relation R on A such that R = { (1, 2), (2,1), (2,3), (1,3) }

The relation R over the set A is symmetric and transitive.

But, it is not reflexive.

(1,1),(2, 2) and (3,3) ∉ R.

Therefore, R is not a reflexive relation.

Hence, it not true that every relation which is symmetric and transitive is also reflexive because it is possible that all pairs of type (x, x) is not present in the relation.

问题12.如果m是n的倍数，则整数m被认为与另一个整数n有关。检查该关系是否对称，自反和可传递。

解决方案：

Let us define a relation such that 

R = {m, n : m, n ∈ Z, m = k×n} where, k ∈ N (natural number)

First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀  x ∈ A.

Let m be an element of R.

Then, m = k×m is true for k=1

(m, m) ∈ R.

So, R is reflexive. 

Check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

Let (m, n) ∈ R 

⇒ m = k×n for some k ∈ N

and according to transitivity, n = (1/k)×m for some k ∈ N but, 1/k ∉ N.

So, R is not a symmetric relation. 

Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈  R  ∀  x, y, z ∈ A.

Let m, n, o be any elements of R then, (m, n) and (n, o) ∈ R

⇒ m = k1×n and n = k2×o for some k1, k2 ∈ N

⇒ m = (k1×k2)×o 

⇒ (m, o) ∈ R.

So, R is a transitive relation. 

问题13：证明所有实数的集合R上的关系“≥ ”是自反的，可传递的，但不是对称的。

解决方案：

Let us define a relation R as

R = { (a, b) a, b ∈  R ; a ≥ b }

First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A.

Let a be an element of R.

⇒ a ∈ R

⇒ a ≥ a , which is always true.

⇒ (a, a) ∈ R 

Hence, R is a reflexive relation.

Check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

Let (a, b) ∈ R.

⇒ a ≥ b 

⇒ b ≥ a [according to transitivity] 

but it is not always true except when a=b.

⇒ (b, a) ∉ R 

Hence, R is not a symmetric relation.

Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R  ∀  x, y, z ∈ A.

Let (a, b) and (b, c) ∈ R

⇒ a ≥ b and b ≥ c

⇒ a ≥ b ≥ c 

⇒ a ≥ c

⇒ (a, c) ∈ R

Hence, R is a transitive relation.

问题14.举例说明一种关系。哪一个

(i)反身和对称，但不及物动词。
(ii)自反和及物但不对称。
(iii)对称和传递但不反身。
(iv)对称但既不反身也不及物动词。
(v)具有传递性，但既非反身性也不对称。

解决方案：

Reflexive Relation:
A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀  x ∈ A.

Symmetric Relation:
A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

Transitive Relation:
A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R  ∀  x, y, z ∈ A.

Let A be a set as,
A = {1, 2, 3J

(i) Let R be the relation on A such that

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3}

Thus, R is reflexive and symmetric, but not transitive.

(ii) Let R be the relation on A such that AR= { (1,1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3) }

Clearly, the relation R on A is reflexive and transitive, but not symmetric.

(iii) Let R be the relation on A such that R = { (1, 2), (2, 1), (1, 3), (3,1), (2, 3) }

We see that the relation R on A is symmetric and transitive, but not reflexive.

(iv) Let R be the relation on A such that R = { (1, 2), (2, 1), (1, 3), (3, 1) }

The relation R on A is symmetric, but neither reflexive nor transitive.

(v) Let R be the relation on A such that R = { (1, 2), (2, 3) ,(1, 3) }

The relation R on A is transitive, but neither symmetric nor reflexive.

问题15。给定关系A = {1,2,3}上的关系R = {((1，2)，(2，3)}，请添加最小数量的有序对，以便扩大的关系是对称的，自反的和。

解决方案：

We have given the relation,

R = {(1, 2), (2,3)}

For R to be reflexive it must have (1,1), (2, 2), (3,3).

For R to be a symmetric relation, all the ordered pairs upon interchanging the elements must be present in the relation R. Therefore, R must contain (2,1 ) and (3, 2), (3,1), (1,3).

And for to be a transitive relation, it must contain (1,3).

Hence, the number of ordered pairs to be added to R is 7, i.e. {(1,1), (2, 2), (3,3), (1,3), (3,1), (2,1), (3, 2)}.

问题16。令A = {1、2、3}和R = {(1、2)，(1、1)，(2、3)}是与A的关系。要添加的最小对数为多少在R中，以便它可以成为A的传递关系。

解决方案：

The relation R on A is given such that R = {(1, 2), (1,1), (2,3)}

For the relation R to be transitive, we must have (1, 2) ∈ R, since (2, 3) ∈ R 

⇒ (1,3) ∈ R

Therefore, the minimum number of ordered pairs need to be added to relation R is 1, i.e. (1, 3) to make it a transitive relation on A. 

问题17.令A = {a，b，c}，并在A上定义关系R如下
R = {(a，a)，(b，c)，(a，b)}。然后在R中写下要添加的最小数量的有序对，使其具有自反性和可传递性。

解决方案：

We have given a set A = {a, b, c} and a relation R={(a, a), (b, c), (a, b)}.

For relation to be reflexive, it should contain (b, b) and (c, c).

And for relation to be transitive R should contain (a, c) since (a, b) ∈ R and (b, c) ∈ R

Therefore, the minimum number of ordered pair to be added to relation R is (b, b) , (c, c) and (a, c) i.e. 3. 

问题18.以下各项均定义了与N的关系

(i)x> y，x，y∈N
(ii)x + y = 10，x，y∈N
(iii)xy是一个整数平方， x，y∈N
(iv)x + 4y = 10，x，y∈N

确定上述关系中的哪些是对称的，自反的和可传递的。

解决方案：

(i) We have given the relation defined as 
R = {(x > y),   x, y ∈ N}

First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀  x ∈ A.

if (x, x) ∈ R then, x > x, which is not true.
⇒ (x, x) ∉ R

So, the relation is not a reflexive relation.

Now check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

Let (x, y) ∈ R, then x R y 

⇒ x > y

and according to symmetric property, (y, x) ∈ R

⇒ y > x, but it is not true since x > y

⇒ (x, y) ∈ R but (y, x) ∉ R

So, the relation is not a symmetric relation.

Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R  ∀  x, y, z ∈ A.

Let (x, y) ∈ R and (y, z) ∈ R

⇒ x > y and y > z

⇒ x > z

⇒ (x, z) ∈ R

So, R is a transitive relation as well.

(ii) We have given the relation defined as
     R = { x + y =10,   x, y ∈ N }

Clearly, the relation will be R = { (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1) }

First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀  x ∈ A.

We can see that, (1, 1) ∉ R.

So, R is not a reflexive relation.

Now check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

By observing the above relation, we can say that ∀ (x, y) ∈ R, (y, x) ∈ R.

So, R is a symmetric relation.

Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R  ∀  x, y, z ∈ A.

In the relation, (1, 9) ∈ R and (9, 1) ∈ R but (1, 1) ∉ R

So, R is not a transitive relation.

(iii) We have given the relation as
      R = { xy is a square of integer,   x, y ∈ N }

First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀  x ∈ A.

Clearly, (x, x) ∈ R ∀ x ∈ N

since, x² is square of an integer for any x ∈ N.

Hence, R is a reflexive relation.

Now check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

Let (x, y) ∈ R

⇒ xy is a square of an integer

⇒ yx is also a square of the same integer since, xy = yx

⇒ (y, x) ∈ R

So, the relation is a symmetric relation.

Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R  ∀  x, y, z ∈ A.

Let (x, y) ∈ R and (y, z) ∈ R

⇒ xy is an square of an integer and yz is an square of integer

Then let xy = m² and yz = n² for some m, n ∈ Z

⇒ x = m²/y and z = n²/y

⇒ xz = (m²n²)/y² , which is also a square of an integer

⇒ (x, z) ∈ R

So, R is a transitive relation.

(iv) We have given the relation as
      R = { x + 4y =10,   x, y ∈ N }

First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀  x ∈ A.

Clearly, the relation will be R = {(2, 2), (6, 1)}      [since  x, y ∈ N]

(1, 1) ∉ R

So, the relation R is not reflexive.

Now check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

We can see that, (1, 6) R but (6, 1) R

So, R is not a symmetric relation.

Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R  ∀  x, y, z ∈ A.

From the definition, we can see that R is a transitive relation.