第12类RD Sharma解决方案–第1章关系–练习1.1 |套装1

问题1.假设A是特定时间某个镇上所有人类的集合。确定以下每个关系是否是自反的，对称的和可传递的：

(i)R = {(x。y)x和y在同一位置工作}
(ii)R = {(x。y)x和y居住在同一地点}
(iii)R = {(x。y)x是y的妻子}
(iv)R = {(x。y)x是y的父亲}  

解决方案：

(i) Given the relation R = {(x, y): x and y work at the same place}

Now we need to check whether the relation is reflexive or not. 

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 x be any element of R.

Then, x ∈ R  

⇒ x and x work at the same place is true since they are same.        

⇒ (x, x) ∈ R [condition for reflexive relation]

So, R is a reflexive relation.

Now we have to 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

⇒ x and y work at the same place                                                   [since it is given in the question]

⇒ y and x work at the same place                                                  [same as “x and y work at the same place”]

⇒ (y, x) ∈ R

So, R is a symmetric relation also.

Now we have to check whether the given relation is Transitive relation or not. 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.

Then, x and y work at the same place.                                                   [Given]

y and z also work at the same place.                                                    [(y, z) ∈ R]

⇒ x, y and z all work at the same place.

⇒ x and z work at the same place.

⇒ (x, z) ∈ R

Therefore, R is a transitive relation also.

So the relation R = {(x, y): x and y work at the same place} is a reflexive relation, symmetric relation and transitive relation as well. 

(ii) Given the relation R = {(x, y): x and y live in the same locality}

Now we have to check whether the relation R is reflexive, symmetric and transitive or not.

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 x be any element of relation R.

Then, x ∈ R

It is given that x and x live in the same locality is true since they are the same.

So, R is a reflexive relation.

Now we have to 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

⇒ x and y live in the same locality                                                         [ it is given in the question ]

⇒ y and x live in the same locality             [if x and y live in the same locality, then y and x also live in the same locality]

⇒ (y, x) ∈ R

So, R is a symmetric relation as well.

Now we have to check whether the given relation is Transitive relation or not. 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 and z be any elements of R and (x, y) ∈ R and (y, z) ∈ R.

Then,

x and y live in the same locality and y and z live in the same locality

⇒ x, y and z all live in the same locality

⇒ x and z live in the same locality

⇒ (x, z) ∈ R

So, R is a transitive relation also.

So the relation R = {(x, y): x and y live in the same locality} is a reflexive relation, symmetric relation and transitive relation as well. 

(iii) Given R = {(x, y): x is wife of y}

Now we have to check whether the relation R is reflexive, symmetric and transitive relation or not.

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 x be an element of R.

Then, x is wife of x cannot be true.           [since the same person cannot be the wife of herself]

⇒ (x, x) ∉ R

So, R is not 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 (x, y) ∈ R

⇒ x is wife of y

⇒ x is female and y is male

⇒ y cannot be wife of x as y is husband of x

⇒ (y, x) ∉ R  

So, R is not a symmetric relation.

Check whether the given relation is Transitive relation or not. 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, but (y, z) ∉ R

Since x is wife of y, but y cannot be the wife of z, since y is husband of x.

⇒ x is not the wife of z.

⇒(x, z)∈ R

So, R is a transitive relation.

Hence the given relation R = {(x, y): x is wife of y} is a transitive relation but not a reflexive and symmetric relation. 

(iv) Given the relation R = {(x, y): x is father of y}

Now we have to check whether the relation R is reflexive, symmetric and transitive or not.

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 x be an arbitrary element of R.

Then, x is father of x cannot be true.                                     [since no one can be father of himself]

So, R is not 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 (x, y)∈ R

⇒ x is the father of y.

⇒ y is son/daughter of x.

⇒ (y, x) ∉ R

So, R is not a symmetric relation.

Now, check whether the given relation is Transitive relation or not. 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.

Then, x is father of y and y is father of z

⇒ x is grandfather of z

⇒ (x, z)∉ R

So, R is not a transitive relation.

Hence, the given relation R = {(x, y): x is father of y} is not a reflexive relation, not a symmetric relation and not a transitive relation as well.
 

问题2.在集合A = {a，b，c}上定义了三个关系R1，R2和R3，如下所示：

R1 = {((a，a)，(a，b)，(a，c)，(b，b)，(b，c)，(c，a)，(c，b)，(c，c) }
R2 = {(a，a)}
R3 = {(b，c)}
R4 ＝ {(a，b)，(b，c)，(c，a)}。

找出A上的每个关系R1，R2，R3，R4是否(i)自反(ii)对称和(iii)可传递。

解决方案：

i) Considering the relation R1, we have

R1 = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}

Now we have check R1 is reflexive, symmetric and transitive or not.

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.

Given (a, a), (b, b) and (c, c) ∈ R1         [since each element maps to itself]

So, R1 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.

We see that for every ordered pair (x, y), there is a pair (y, x) present in the relation R1.

So, R1 is symmetric.

Transitive: 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, (a, b) ∈ R1, (b, c) ∈ R1 and also (a, c) ∈ R1

So, R1 is transitive.

Therefore, R1 is reflexive relation, symmetric relation and transitive relation as well.

(ii) Considering the relation R2, we have

R2 = {(a, a)}

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 (a, a) ∈ R2.                                              [since each element maps to itself]

So, R2 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.

We can see that (a, a) ∈ R

⇒ (a, a) ∈ R.

So, R2 is symmetric.

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.

R2 is clearly a transitive relation.                                 [since there is only one element in it]

Therefore, R2 is reflexive relation, symmetric relation and transitive relation as well.

(iii) Considering the relation R3, we have 

R3 = {(b, c)}

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.

In the relation, (a, a)∉ R3, (b, b)∉ R3 neither (c, c) ∉ R3. 

So, R3 is not reflexive.                                [since all pairs of type (x, x) should be present in the 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.

In the relation, (b, c) ∈ R3, but (c, b) ∉ R3

So, R3 is not symmetric.

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.

R3 has only two elements.

Hence, R3 is transitive.

Therefore, R2 is transitive relation but not a reflexive relation and not a symmetric relation also.

(iv) Considering the relation R4, we have

R4 = {(a, b), (b, c), (c, a)}

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.

In the relation, (a, a) ∉ R4, (b, b) ∉ R4 (c, c) ∉ R4

So, R4 is not 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.

Here, (a, b) ∈ R4, but (b, a) ∉ R4.

So, R4 is not symmetric

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, (a, b) ∈ R4, (b, c) ∈ R4, but (a, c) ∉ R4

So, R4 is not a transitive relation.

Therefore, R2 is not a reflexive relation, not a symmetric relation and neither a transitive relation as well.

问题3.测试以下关系R ₁ ，R ₂和R ₃是否为(i)自反的(ii)对称的和(iii)传递的：

(i)对由(A，B)∈R _1⇔一个= 1 / B所定义Q0的R _1。
(ⅱ)R ₂上通过(A，B)∈R ₂定义ž⇔|一个- B | ≤5
(ⅲ)在R R ₃所定义的(a，b)中∈R _3⇔A2 – 4AB + 3B2 = 0。

解决方案：

i) Given R₁ on Q₀ defined as (a, b) ∈ R₁ ⇔ a = 1/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 R1.

Then, a ∈ R1

⇒ a ≠1/a ∀  a ∈ Q₀

So, R1 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.

Let (a, b) ∈ R1  

Then, (a, b) ∈ R1

Therefore, we can write ‘a’ as a =1/b

⇒ b = 1/a

⇒ (b, a) ∈ R1

So, R1 is symmetric.

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.

Here, (a, b) ∈ R1 and (b, c) ∈ R2

⇒ a = 1/b and b = 1/c

⇒ a = 1/ (1/c) = c

⇒ a ≠ 1/c

⇒ (a, c) ∉ R1

So, R1 is not a transitive relation.

(ii) Given R2 on Z defined as (a, b) ∈ R2 ⇔ |a – b| ≤ 5

Now we have to check whether R2 is reflexive, symmetric and transitive or not.

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 R2.

Then, a ∈ R2

On applying the given condition we will get,

⇒ | a−a | = 0 ≤ 5

So, R1 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 (a, b) ∈ R2

⇒ |a−b| ≤ 5                    [Since, |a−b| = |b−a|]

⇒ |b−a| ≤ 5

⇒ (b, a) ∈ R2

So, R2 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 (1, 3) ∈ R2 and (3, 7) ∈ R2

⇒|1−3|≤5 and |3−7|≤5

But |1−7| ≰ 5  

⇒ (1, 7) ∉ R2

So, R2 is not a transitive relation.

(iii) Given R3 on R defined as (a, b) ∈ R3 ⇔ a² – 4ab + 3b² = 0.

Now we have to check whether R2 is reflexive, symmetric and transitive or not.

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 R3.

Then, a ∈ R3.

⇒ a² − 4a × a+ 3a²= 0  

So, R3 is 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.

Let (a, b) ∈ R3

⇒ a²−4ab+3b²=0

But b²−4ba+3a² ≠ 0 ∀ a, b ∈ R

So, R3 is not symmetric.

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 (1, 2) ∈ R3 and (2, 3) ∈ R3

⇒ 1 − 8 + 6 = 0 and 4 – 24 + 27 = 0

But 1 – 12 + 9 ≠ 0

So, R3 is not a transitive relation.

问题4.让A = {1,2,3}，让R1 = {(1，1)，(1，3)，(3，1)，(2，2)，(2，1)，( 3，3)}，R2 = {(2，2)，(3，1)，(1，3)}，R3 = {(1，3)，(3，3)}。确定A上的每个关系R1，R2，R3是否(i)自反(ii)对称(iii)可传递。

解决方案：

Considering the relation R1, we have 

R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}

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.

Here, (1, 1) ∈ R, (2, 2) ∈ R, (3, 3) ∈ R

So, R1 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.

In the given relation, (2, 1) ∈ R1 but (1, 2) ∉ R1

So, R1 is not symmetric.

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, (2, 1) ∈ R1 and (1, 3) ∈ R1 but (2, 3) ∉ R1  

So, R1 is not transitive.

Therefore, the relation R1 is reflexive but not symmetric and transitive relation.

Now considering the relation R2, we have

R2 = {(2, 2), (3, 1), (1, 3)}

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, (1, 1) and (3, 3) ∉ R2  

So, R2 is not 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.

In the relation, (1, 3) ∈ R2 and (3, 1) ∈ R2

So, R2 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, 3) ∈ R2 and (3, 1) ∈ R2  but (3, 3) ∉ R2

So, R2 is not a transitive relation.

Therefore, the relation R2 is symmetric but not a reflexive and transitive relation.

Considering the relation R3, we have

R3 = {(1, 3), (3, 3)}

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.

In the relation, (1, 1) ∉ R3

So, R3 is not 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.

In the relation, (1, 3) ∈ R3, but (3, 1) ∉ R3

So, R3 is not symmetric.

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.

Here, (1, 3) ∈ R3 and (3, 3) ∈ R3  

Also, (1, 3) ∈ R3

So, R3 is transitive.

Therefore, the relation R3 is transitive but not a reflexive and symmetric relation.

问题5.在实数集上定义以下关系。

(i)如果a – b> 0，则为aRb
(ii)aRb iff 1 + ab> 0
(iii)如果| a |，则为aRb ≤b。

查找关系是自反的，对称的还是可传递的。

解决方案：

(i) Consider the relation defined as aRb if a – b > 0

Now for this relation we have to check whether it is reflexive, transitive and symmetric or not.

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.

Then, a ∈ R

But a − a = 0 ≯  0

So, this 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 (a, b) ∈ R

⇒ a − b > 0

⇒ − (b − a) > 0

⇒ b − a < 0

So, the given 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 (a, b) ∈ R and (b, c) ∈ R.  

Then, a − b > 0 and b − c > 0

Adding the two, we will get

a – b + b − c > 0

⇒ a – c > 0  

⇒ (a, c) ∈ R.

So, the given relation is a transitive relation.

(ii) Consider the relation defined as aRb iff (read as “if and only if”) 1 + a b > 0

Now for this relation we have to check whether it is reflexive, transitive and symmetric or not.

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.

Then, a ∈ R

⇒ 1 + a × a > 0

i.e. 1 + a² > 0             [since, square of any number is positive]

So, the given relation 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 (a, b) ∈ R

⇒ 1 + a b > 0

⇒ 1 + b a > 0

⇒ (b, a) ∈ R

So, the given relation is symmetric.

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) ∈ R and (b, c) ∈ R

⇒1 + a b > 0 and 1 + b c >0

But 1+ ac ≯  0

⇒ (a, c) ∉ R

So, the given relation is not a transitive relation.

(iii) Consider the relation defined as aRb if |a| ≤ b.

Now for this relation we have to check whether it is reflexive, transitive and symmetric or not.

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 relation R.

Then, a ∈ R                  [Since, |a|=a]

⇒ |a| ≮  a

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.

Let (a, b) ∈ R

⇒ |a| ≤ b  

⇒ |b| ≰  a ∀ a, b ∈ R

⇒ (b, a) ∉ 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.

Let (a, b) ∈ R and (b, c) ∈ R

⇒ |a| ≤ b and |b| ≤ c

Multiplying the corresponding sides, we will get

|a| × |b| ≤ b c

⇒ |a| ≤ c

⇒ (a, c) ∈ R

Thus, R is a transitive relation. 

问题6.检查集合{1、2、3、4、5、6}中定义为R = {(a，b)：b = a + 1}的关系R是自反的，对称的还是可传递的。

解决方案：

Given R = {(a, b): b = a + 1}

Now, for this relation we have to check whether it is reflexive, transitive and symmetric or not. 

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.

Then, a = a + 1 cannot be true for all a ∈ A.

⇒ (a, a) ∉ R  

So, R is not a reflexive relation over the given set.

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 (a, b) ∈ R

⇒ b = a + 1

⇒ −a = −b + 1

⇒ a = b − 1

So, (b, a) ∉ R

Thus, R is not a symmetric relation over the given set.

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 (1, 2) and (2, 3) ∈ R

⇒ 2 = 1 + 1 and 3  

2 + 1  is true.

But 3 ≠ 1+1

⇒ (1, 3) ∉  R

So, R is not a transitive relation over the given set.

问题7.检查定义为R = {(a，b)：a≤b ³ }的R上的关系R是自反的，对称的或可传递的。

解决方案：

We have given the relation R = {(a, b): a ≤ b³}

First let us check whether the given relation is reflexive or not.

It can be observed that (1/2, 1/2) in R as 1/2 > (1/2)³ = 1/8

So, R is not a reflexive relation.

Now, check for whether the relation is symmetric or not

(1, 2) ∈ R (as 1 < 2³ = 8)

But,

(2, 1) ∉ R (as 2 > 1³ = 1)

So, R is not a symmetric relation.

We have (3, 3/2), (3/2, 6/5) in “R as” 3 < (3/2)³ and 3/2 < (6/5)³

But (3, 6/5) ∉ R as 3 > (6/5)³

So, R is not a transitive relation.

Hence, R is neither reflexive, nor symmetric, nor transitive.

问题8：证明集合上的每个身份关系都是自反的，但相反不一定是正确的。

解决方案：

We will verify this by taking example.

Let A be a set.

Then, Identity relation IA=I_A is reflexive, since (a, a) ∈ A  ∀ a ∈ A.

The converse of this need not be necessarily true.

Now, consider the set A = {1, 2, 3}

Here, relation R = {(1, 1), (2, 2) , (3, 3), (2, 1), (1, 3)} is reflexive on A.

But, R is not an identity relation.

Hence proved, that every identity relation on a set is reflexive but the converse is not necessarily true.

问题9.如果A = {1,2,3,4}，则定义A上具有以下性质的关系：

(i)反身，及物但不对称
(ii)对称但既不反身也不及物动词。
(iii)反身，对称和及物。

解决方案：

(i) We have given the set A = {1, 2, 3, 4}

The relation on A having properties of being reflexive, transitive, but not symmetric is

R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}

Relation R satisfies reflexivity and transitivity.

⇒ (1, 1), (2, 2), (3, 3) ∈ R                                       [satisfies the reflexivity property]

and (1, 1), (2, 1) ∈ R ⇒ (1, 1) ∈ R                          [satisfies the transitivity property] 

However, (2, 1) ∈ R, but (1, 2) ∉ R                        [does not satisfies the symmetric property]

(ii) We have given the set A = {1, 2, 3, 4}

The relation on A having properties of being reflexive, transitive, but not symmetric is

R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}

Relation R satisfies reflexivity and transitivity.

⇒ (1, 1), (2, 2), (3, 3) ∈ R                                       [satisfies the reflexivity property]

And (1, 1), (2, 1) ∈ R ⇒ (1, 1) ∈ R                          [satisfies the transitivity property]

However, (2, 1) ∈ R, but (1, 2) ∉ R                        [does not satisfies the symmetric property]

(iii) We have given the set A = {1, 2, 3, 4}

The relation on A having properties of being symmetric, reflexive and transitive is

R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)}

Relation R satisfies reflexivity, symmetricity and transitivity.

⇒ (1, 1), (2, 2), (3, 3) ∈ R                                       [satisfies the reflexivity property]

⇒ (1, 1)  ∈ R and (2, 1) ∈ R                                   [satisfies the symmetric property]

⇒ (1, 1), (2, 1) ∈ R ⇒ (1, 1) ∈ R                             [satisfies the transitivity property]

问题10：令R是在自然数N集合上定义的关系，因为R = {(x，y)：x，y∈N，2x + y = 41}。找到R的域和范围。还要验证R是否(i)自反(ii)对称(iii)可传递。

解决方案：

We have given,

{(x, y) : x, y ∈ N, 2x + y = 41}

Now,

2x + y = 41

⇒ y = 41 – 2x

Put the value of x one by one to form the relation R.

The relation we will after putting x = 1, 2, 3, ……. ,20 is:
[we can’t put x=21 since y = 41 – 2(2) < 0, which is not a natural number]  

R = {(1, 39), (2, 37), (3, 35)………….., (20, 1)}

So the domain of R is
Domain(R) = {1, 2, 3, ……… ,20}

And the range of R is
Range(R) = {39, 37, 33, ……. ,1} and can be rearranged as {1, 3, 5, ……….. ,39} 

Now for this relation we have to check whether it is reflexive, transitive and symmetric or not.

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 x be an any element of relation R.

Since, (2, 2) ∉ 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.

Since, (1, 39) ∈ R but (39, 1) ∉ R.

So, R is not symmetric.

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.

Since, (15,11) ∈ R and (11,19) ∈ R but (15,19) ∉ R.

 Thus, R is not transitive.