第 12 课 NCERT 解决方案 - 数学第一部分 - 第 1 章关系和函数 - 练习 1.4 |设置 1
问题 1：确定下面给出的每个 ∗ 的定义是否给出二元运算。如果 * 不是二元运算，请给出理由。
(i) 在 Z+ 上，用 a ∗ b = a – b 定义 ∗
解决方案：
If a, b belongs to Z+
a * b = a – b which may not belong to Z+
For eg: 1 – 3 = -2 which doesn’t belongs to Z+
Therefore, * is not a Binary Operation on Z+
(ii) 在 Z+ 上，用 a * b = ab 定义 *
解决方案：
If a, b belongs to Z+
a * b = ab which belongs to Z+
Therefore, * is Binary Operation on Z+
(iii) 在 R 上，用 a * b = ab² 定义 *
解决方案：
If a, b belongs to R
a * b = ab2 which belongs to R
Therefore, * is Binary Operation on R
(iv) 在 Z+ 上，用 a * b = |a – b| 定义 *
解决方案：
If a, b belongs to Z+
a * b = |a – b| which belongs to Z+
Therefore, * is Binary Operation on Z+
(v) 在 Z+ 上，用 a * b = a 定义 *
解决方案：
If a, b belongs to Z+
a * b = a which belongs to Z+
Therefore, * is Binary Operation on Z+
问题 2：对于下面定义的每个二元运算 *，确定 * 是二元、交换还是结合。
(i) 在 Z 上，定义 a * b = a – b
解决方案：
a) Binary:
If a, b belongs to Z
a * b = a – b which belongs to Z
Therefore, * is Binary Operation on Z
b) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a – b
RHS = b * a = b – a
Since, LHS is not equal to RHS
Therefore, * is not Commutative
c) Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a – b + c
RHS = (a – b) * c = a – b- c
Since, LHS is not equal to RHS
Therefore, * is not Associative
(ii) 在 Q 上，定义 a * b = ab + 1
解决方案：
a) Binary:
If a, b belongs to Q, a * b = ab + 1 which belongs to Q
Therefore, * is Binary Operation on Q
b) Commutative:
If a, b belongs to Q, a * b = b * a
LHS = a * b = ab + 1
RHS = b * a = ba + 1 = ab + 1
Since, LHS is equal to RHS
Therefore, * is Commutative
c) Associative:
If a, b, c belongs to Q, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (bc + 1) = abc + a + 1
RHS = (a * b) * c = abc + c + 1
Since, LHS is not equal to RHS
Therefore, * is not Associative
(iii) 在 Q 上，定义 a ∗ b = ab/2
解决方案 ：
a) Binary:
If a, b belongs to Q, a * b = ab/2 which belongs to Q
Therefore, * is Binary Operation on Q
b) Commutative:
If a, b belongs to Q, a * b = b * a
LHS = a * b = ab/2
RHS = b * a = ba/2
Since, LHS is equal to RHS
Therefore, * is Commutative
c) Associative:
If a, b, c belongs to Q, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (bc/2) = (abc)/2
RHS = (a * b) * c = (ab/2) * c = (abc)/2
Since, LHS is equal to RHS
Therefore, * is Associative
(iv) 在 Z+ 上，定义 a * b = 2 ab
解决方案：
a) Binary:
If a, b belongs to Z+, a * b = 2ab which belongs to Z+
Therefore, * is Binary Operation on Z+
b) Commutative:
If a, b belongs to Z+, a * b = b * a
LHS = a * b = 2ab
RHS = b * a = 2ba = 2ab
Since, LHS is equal to RHS
Therefore, * is Commutative
c) Associative:
If a, b, c belongs to Z+, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * 2bc = 2a * 2^(bc)
RHS = (a * b) * c = 2ab * c = 22abc
Since, LHS is not equal to RHS
Therefore, * is not Associative
(v) 在 Z+ 上，定义 a * b = a b
解决方案：
a) Binary:
If a, b belongs to Z+, a * b = ab which belongs to Z+
Therefore, * is Binary Operation on Z+
b) Commutative:
If a, b belongs to Z+, a * b = b * a
LHS = a * b = ab
RHS = b * a = ba
Since, LHS is not equal to RHS
Therefore, * is not Commutative
c) Associative:
If a, b, c belongs to Z+, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * bc = ab^c
RHS = (a * b) * c = ab * c = abc
Since, LHS is not equal to RHS
Therefore, * is not Associative
(vi) 在 R – {– 1} 上，定义 a ∗ b = a / (b + 1)
解决方案：
a) Binary:
If a, b belongs to R, a * b = a / (b+1) which belongs to R
Therefore, * is Binary Operation on R
b) Commutative:
If a, b belongs to R, a * b = b * a
LHS = a * b = a / (b + 1)
RHS = b * a = b / (a + 1)
Since, LHS is not equal to RHS
Therefore, * is not Commutative
c) Associative:
If a, b, c belongs to A, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * b / (c+1) = a(c+1) / b+c+1
RHS = (a * b) * c = (a / (b+1)) * c = a / (b+1)(c+1)
Since, LHS is not equal to RHS
Therefore, * is not Associative
问题 3. 考虑由 a ∧ b = min {a, b} 定义的集合 {1, 2, 3, 4, 5} 上的二元运算 ∧。写出操作∧的操作表。
解决方案：
问题 4：考虑由下面的乘法表给出的集合 {1, 2, 3, 4, 5} 上的二元运算 *。
（提示：使用下表）^ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 2 2 2 3 1 2 3 3 3 4 1 2 3 4 4 5 1 2 3 4 5
(i) 计算 (2 * 3) * 4 和 2 * (3 * 4)
解决方案：
Here, (2 * 3) * 4 = 1 * 4 = 1
2 * (3 * 4) = 2 * 1 = 1
(ii) ∗ 是可交换的吗？
解决方案：
The given composition table is symmetrical about the main diagonal of table. Thus, binary operation ‘*’ is commutative.
(iii) 计算 (2 * 3) * (4 * 5)。
解决方案：
(2 * 3) * (4 * 5) = 1 * 1 = 1
问题 5：设 ∗′ 为 a ∗′ b = a 和 b 的 HCF 定义的集合 {1, 2, 3, 4, 5} 上的二元运算。运算 ∗′ 是否与上面练习 4 中定义的运算 ∗ 相同？证明你的答案。
解决方案：
Let A = {1, 2, 3, 4, 5} and a ∗′ b = HCF of a and b.*’ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 1 1 3 1 1 4 1 2 1 4 1 5 1 1 1 1 5
We see that the operation *’ is the same as the operation * in Exercise 4 above.
问题 6：设 * 是由 a * b = a 和 b 的 LCM 给出的 N 上的二元运算。找
(i) 5 * 7、20 * 16
解决方案：
If a, b belongs to N
a * b = LCM of a and b
5 * 7 = 35
20 * 16 = 80
(ii) ∗ 是可交换的吗？
解决方案：
If a, b belongs to N
LCM of a * b = ab
LCM of b * a = ba = ab
a*b = b*a
Thus, * binary operation is commutative.
(iii) ∗ 是关联的吗？
解决方案：
a * (b * c) = LCM of a, b, c
(a * b) * c = LCM of a, b, c
Since, a * (b * c) = (a * b) * c
Thus, * binary operation is associative.
(iv) 在 N 中找到 * 的恒等式
解决方案：
Let ‘e’ is an identity
a * e = e * a, for a belonging to N
LCM of a * e = a, for a belonging to N
LCM of e * a = a, for a belonging to N
e divides a
e divides 1
Thus, e = 1
Hence, 1 is an identity element
(v) N 的哪些元素对于操作 * 是可逆的？
解决方案：
a * b = b * a = identity element
LCM of a and b = 1
a = b = 1
only ‘1’ is invertible element in N.