第 11 类 RD Sharma 解决方案 - 第 2 章关系 - 练习 2.3 |设置 1
问题 1. 如果 A = {1, 2, 3}, B = {4, 5, 6}，下列哪项是从 A 到 B 的关系？给出支持你答案的理由。
(i) {(1, 6), (3, 4), (5, 2)}
(ii) {(1, 5), (2, 6), (3, 4), (3, 6)}
(iii) {(4, 2), (4, 3), (5, 1)}
(iv) A × B
解决方案：
Given:
A = {1, 2, 3}, B = {4, 5, 6}
A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(i) {(1, 6), (3, 4), (5, 2)}
No, it is not a relation from A to B. The given set is not a subset of A × B
because (5, 2) is not a part of the relation from A to B.
(ii) {(1, 5), (2, 6), (3, 4), (3, 6)}
Yes, it is a relation from A to B. Hence, the given set is a subset of A × B.
(iii) {(4, 2), (4, 3), (5, 1)}
No, it is not a relation from A to B. The given set is not a subset of A × B
because (4, 2), (4, 3), (5, 1) are not a part of the relation from A to B.
(iv) A × B
A × B is a relation from A to B:
{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
问题 2. 从集合 A = {2, 3, 4, 5} 到集合 B = {3, 6, 7, 10} 定义关系 R 如下： (x, y) R x 与是的。将 R 表示为一组有序对并确定其域和范围。
解决方案：
Given: (x, y) ∈ R = x is relatively prime to y -(Relatively prime numbers are also known as co-prime numbers)
Here,
2 is co-prime to 3 and 7.
3 is co-prime to 7 and 10.
4 is co-prime to 3 and 7.
5 is co-prime to 3, 6 and 7.
∴ R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}
So, Domain(R) = {2, 3, 4, 5} and Range(R) = {3, 6, 7, 10}
问题 3. 设 A 是前五个自然的集合，设 R 是 A 上的一个关系，定义如下：(x, y) R x ≤ y。将 R 和 R -1表示为有序对的集合。也确定
(i) R -1的域
(ii) R 的范围。
解决方案：
Given: A= {1, 2, 3, 4, 5} -(A is set of first five natural numbers)
(x, y) R x ≤ y
1 is less than 2, 3, 4 and 5.
2 is less than 3, 4 and 5.
3 is less than 4 and 5.
4 is less than 5.
5 is not less than any number A
So, R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 5)}
∴ R-1 = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (2, 2), (3, 2), (4, 2), (5, 2), (3, 3), (4, 3), (5, 3), (4, 4), (5, 4) (5, 5)}
(i) Domain(R-1) = {2, 3, 4, 5}
(ii) Range(R) = {2, 3, 4, 5}
Note: You can see that Domain of R-1 is same as Range of R. Similarly, Domain of R is same as Range of R-1
问题 4. 找出下列每种情况下的逆关系 R -1 ：
(i) R= {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}
(ii) R= {(x, y) : x, y ∈ N; x + 2y = 8}
(iii) R 是从 {11, 12, 13} 到 (8, 10, 12} 的关系，由 y = x – 3 定义
解决方案：
(i) Given: R= {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}
So the inverse relation R-1 = {(2, 1), (3, 1), (3, 2), (2, 3), (6, 5)}
(ii) Given: R= {(x, y) : x, y ∈ N; x + 2y = 8}
Here, x + 2y = 8
x = 8 – 2y
As y ∈ N, Put the values of y = 1, 2, 3,…… till x ∈ N
On putting y = 1, x = 8 – 2(1) = 8 – 2 = 6
On putting y = 2, x = 8 – 2(2) = 8 – 4 = 4
On putting y = 3, x = 8 – 2(3) = 8 – 6 = 2
On putting y = 4, x = 8 – 2(4) = 8 – 8 = 0
Now, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.
Therefore, R = {(2, 3), (4, 2), (6, 1)}
R-1 = {(3, 2), (2, 4), (1, 6)}
(iii) Given: R is a relation from {11, 12, 13} to (8, 10, 12} defined by y = x – 3
So, x = {11, 12, 13} and y = (8, 10, 12}
y = x – 3
On putting x = 11, y = 11 – 3 = 8 ∈ (8, 10, 12}
On putting x = 12, y = 12 – 3 = 9 ∉ (8, 10, 12}
On putting x = 13, y = 13 – 3 = 10 ∈ (8, 10, 12}
Therefore, R = {(11, 8), (13, 10)}
R-1 = {(8, 11), (10, 13)}
问题 5. 将下列关系写成有序对的集合：
(i) 从集合 {2, 3, 4, 5, 6} 到由 x = 2y 定义的集合 {1, 2, 3} 的关系 R。
解决方案：
Let A = {2, 3, 4, 5, 6} and B = {1, 2, 3}
Given: x = 2y where x = {2, 3, 4, 5, 6} and y = {1, 2, 3}
On putting y = 1, x = 2(1) = 2 A
On putting y = 2, x = 2(2) = 4 A
On putting y = 3, x = 2(3) = 6 A
∴ R = {(2, 1), (4, 2), (6, 3)}
(ii) 由 (x, y)∈ R <=>x 定义的集合 {1,2,3,4,5,6,7} 上的关系 R 与 y 互质。
解决方案：
Given: (x, y) R x is relatively prime to y
Here, 2 is co-prime to 3, 5 and 7.
3 is co-prime to 2, 4, 5 and 7.
4 is co-prime to 3, 5 and 7.
5 is co-prime to 2, 3, 4, 6 and 7.
6 is co-prime to 5 and 7.
7 is co-prime to 2, 3, 4, 5 and 6.
∴ R = {(2,3), (2,5), (2,7), (3,2), (3,4), (3,5), (3,7), (4,3), (4.5), (4,7), (5,2), (5,3), (5,4), (5,6), (5,7), (6,5), (6,7), (7,2), (7,3), (7,4), (7,5), (7,6)}
(iii) 由 2x + 3y = 12 定义的集合 {0,1,2,…,10} 上的关系 R。
解决方案：
Given: (x, y) R 2x + 3y = 12
Where x and y {0,1,2,…,10}
2x + 3y = 12
⇒ 2x = 12 – 3y
⇒ x = (12-3y)/2
On putting y=0, we get
⇒ x = ( 12 – 3(0) )/2 = 6
On putting y=2,
⇒ x = ( 12 – 3(2) )/2 = 3
On putting y=4, we get
⇒ x= ( 12 – 3(4) )/2 = 0
∴ R = {(0, 4), (3, 2), (6, 0)}
(iv) 关系 R 形成集合 A = {5, 6, 7, 8} 到由 (x, y) R x 除以 y 定义的集合 B = {10, 12, 15, 16, 18}。
解决方案：
Given: (x, y) R x divides y
Where x = {5, 6, 7, 8} and y = {10, 12, 15, 16, 18}
Here,
5 divides 10 and 15.
6 divides 12 and 18.
7 divides none of the value of set B.
8 divides 16.
∴ R = {(5, 10), (5, 15), (6, 12), (6, 18), (8, 16)}
问题 6. 设 R 是由 (x, y) R x + 2y = 8 定义的 N 中的一个关系。将 R 和 R -1表示为有序对的集合。
解决方案：
Given: (x, y) R x + 2y = 8 where x N and y N
x = 8 – 2y
As y N, Put the values of y = 1, 2, 3,…… till x N
On putting y = 1, x = 8 – 2(1) = 8 – 2 = 6
On putting y = 2, x = 8 – 2(2) = 8 – 4 = 4
On putting y = 3, x = 8 – 2(3) = 8 – 6 = 2
On putting y = 4, x = 8 – 2(4) = 8 – 8 = 0
Now, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.
Therefore, R = {(2, 3), (4, 2), (6, 1)}
R-1 = {(3, 2), (2, 4), (1, 6)}
问题 7：设 A = {3, 5} 和 B = {7, 11}。令 R = {(a, b): a ∈ A, b ∈ B, ab 为奇数}。证明 R 是从到 B 的空关系。
解决方案：
Given: A = {3, 5} and B = {7, 11}
R = {(a, b): a ∈ A, b ∈ B, a-b is odd}
When a = 3 and b = 7
a – b = 3 – 7 = -4 which is not odd
When a = 3 and b = 11
⇒ a – b = 3 – 11 = -8 which is not odd
When a = 5 and b = 7
⇒ a – b = 5 – 7 = -2 which is not odd
When a = 5 and b = 11
⇒ a – b = 5 – 11 = -6 which is not odd
∴ R = { } = Φ
⇒ R is an empty relation from into B
问题 8：设 A = {1, 2} 且 B={3, 4}。求从 A 到 B 的关系总数。
解决方案：
Given: A= {1, 2}, B= {3, 4}
n(A) = 2 -(Number of elements in set A).
n(B) = 2 -(Number of elements in set B).
We know,
n(A × B) = n(A) × n(B) = 2 × 2 = 4
Therefore, the number of relations from A to B are 24 = 16
问题 9. 确定由下式定义的关系 R 的域和范围
(i) R = {(x, x+5): x ∈ {0, 1, 2, 3, 4, 5}
解决方案：
Given: R = {(x, x+5): x ∈ {0, 1, 2, 3, 4, 5}
Therefore, R = {(0, 0+5), (1, 1+5), (2, 2+5), (3, 3+5), (4, 4+5), (5, 5+5)}
R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
So,
Domain(R) = {0, 1, 2, 3, 4, 5}
Range(R) = {5, 6, 7, 8, 9, 10}
(ii) R= {(x, x 3 ): x 是小于 10 的素数}
解决方案：
Given: R = {(x, x3): x is a prime number less than 10}
Prime numbers less than 10 are 2, 3, 5 and 7
Therefore, R = {(2, 23), (3, 33), (5, 53), (7, 73)}
R = {(2, 8), (3, 27), (5, 125), (7, 343)}
So,
Domain(R) = {2, 3, 5, 7}
Range(R) = {8, 27, 125, 343}
问题 10. 确定下列关系的域和范围：
(i) R= {a, b): a ∈ N, a < 5, b = 4}
解决方案：
Given: R= {a, b): a ∈ N, a < 5, b = 4}
Natural numbers less than 5 are 1, 2, 3 and 4
Therefore, a = {1, 2, 3, 4} and b = {4}
R = {(1, 4), (2, 4), (3, 4), (4, 4)}
So,
Domain(R) = {1, 2, 3, 4}
Range(R) = {4}
(ii) S= {a, b): b = |a-1|, a ∈ Z 和 |a| ≤ 3}
解决方案：
Given: S= {a, b): b = |a-1|, a Z and |a| ≤ 3}
Z denotes integer which can be positive as well as negative
Now, |a| ≤ 3 and b = |a-1|
∴ a {-3, -2, -1, 0, 1, 2, 3}
S = {a, b): b = |a-1|, a Z and |a| ≤ 3}
S = {a, |a-1|): b = |a-1|, a Z and |a| ≤ 3}
S = {(-3, |-3 – 1|), (-2, |-2 – 1|), (-1, |-1 – 1|), (0, |0 – 1|), (1, |1 – 1|), (2, |2 – 1|), (3, |3 – 1|)}
S = {(-3, |-4|), (-2, |-3|), (-1, |-2|), (0, |-1|), (1, |0|), (2, |1|), (3, |2|)}
S = {(-3, 4), (-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1), (3, 2)}
So,
Domain(S) = {-3, -2, -1, 0, 1, 2, 3}
Range(S) = {0, 1, 2, 3, 4}
问题 11. 让 A = {a, b}。列出 A 上的所有关系并找到它们的编号。
解决方案：
As we know that the total number of relations that can be defined from a set A to a set B is the number of possible subsets of A × B. If n(A) = p and n(B) = q, then n(A × B) = pq. So, the total number of relations is 2pq.
Now,
A × A = {(a, a), (a, b), (b, a), (b, b)}
Total number of relations are all possible subsets of A × A:
{Φ, {(a, a)}, {(a, b)}, {(b, a)}, {(b, b)}, {(a, a), (a, b)}, {(a, a), (b, a)}, {(a, a), (b, b)}, {(a, b), (b, a)}, {(a, b), (b, b)}, {(b, a), (b, b)}, {(a, a), (a, b), (b, a)}, {(a, b), (b, a), (b, b)}, {(a, a), (b, a), (b, b)}, {(a, a), (a, b), (b, b)}, {(a, a), (a, b), (b, a), (b, b)}}
n(A) = 2 ⇒ n(A × A) = 2 × 2 = 4
Hence, the total number of relations = 24 = 16