第1章关系和功能–练习1.4 |套装1
问题7：是否在集合{1、2、3、4、5}上通过∗ b = a和ba二进制运算的LCM来定义∗?证明你的答案。
解决方案：
The operation * on the set {1, 2, 3, 4, 5} is defined as
a * b = L.C.M. of a and b
Let a=3, b=5
3 * 5 = 5 * 3 = L.C.M. of 3 and 5 = 15 which does not belong to the given set
Thus, * is not a Binary Operation.
问题8：设*为对a的二元运算，其中a = b = a和b的HCF。 *是可交换的吗? *是关联的吗?在N上是否存在此二进制操作的标识?
解决方案：
If a, b belongs to N
LHS = a * b = HCF of a and b
RHS = b * a = HCF of b and a
Since LHS = RHS
Therefore, * is Commutative
Now, If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = HCF of a, b and c
RHS = (a – b) * c = HCF of a, b and c
Since, LHS = RHS
Therefore, * is Associative
Now, 1 * a = a * 1 ≠ a
Thus, there doesn’t exist any identity element.
问题9：令∗是有理数集Q的二进制运算，如下所示：
(i)a * b = a – b
(ii)a * b = a 2 + b 2
(iii)a * b = a + ab
(iv)a * b =(a – b) 2
(v)a * b = ab / 4
(vi)a * b = ab 2
查找哪些二进制运算是可交换的，哪些是关联的。
解决方案：
(i) 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
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = 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) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a2 + b2
RHS = b * a = b2 + a2
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b2 + c2) = a2 + (b2 + c2)2
RHS = (a * b) * c = (a2 + b2) * c = (a2 + b2)2 + c2
Since, LHS is not equal to RHS
Therefore, * is not Associative
(iii) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a + ab
RHS = b * a = b + ba
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b + bc) = a + a(b + bc)
RHS = (a * b) * c = (a + ab) * c = a + ab + (a + ab)c
Since, LHS is not equal to RHS
Therefore, * is not Associative
(iv) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = (a – b)2
RHS = b * a = (b – a)2
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b – c)2 = [a – (b – c)2]2
RHS = (a * b) * c = (a – b)2 * c = [(a – b)2 – c]2
Since, LHS is not equal to RHS
Therefore, * is not Associative
(v) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = ab / 4
RHS = b * a = ba / 4
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * bc/4 = abc/16
RHS = (a * b) * c = ab/4 * c = abc/16
Since, LHS is equal to RHS
Therefore, * is Associative
(vi) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = ab2
RHS = b * a = ba2
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (bc)2 = a(bc2)2
RHS = (a * b) * c = (ab2) * c = ab2c2
Since, LHS is not equal to RHS
Therefore, * is not Associative
问题10：找出上面给出的哪些操作具有标识
解决方案：
An element e ∈ Q will be the identity element for the operation * if
a * e = a = e * a, for a ∈ Q
for (v) a * b = ab/4
Let e be an identity element
a * e = a = e * a
LHS : ae/4 = a
=> e = 4
RHS : ea/4 = a
=> e = 4
LHS = RHS
Thus, Identity element exists
Other operations doesn’t satisfy the required conditions.
Hence, other operations doesn’t have identity.
问题11：令A = N×N，并且∗是对A的二进制运算，其定义如下：
(a，b)∗(c，d)=(a + c，b + d)
证明*是可交换的和关联的。找到A上的∗的标识元素(如果有)。
解决方案：
Given (a, b) * (c, d) = (a+c, b+d) on A
Let (a, b), (c, d), (e,f) be 3 pairs ∈ A
Commutative :
LHS = (a, b) * (c, d) = (a+c, b+d)
RHS = (c, d) * (a, b) = (c+a, d+b) = (a+c, b+d)
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = (a, b) * [(c, d) * (e, f)] = (a, b) * (c+e, d+f) = (a+c+e, b+d+f)
RHS = [(a, b) * (c, d)] * (e, f) = (a+c, b+d) * (e, f) = (a+c+e, b+d+f)
Since, LHS is equal to RHS
Therefore, * is Associative
Existence of Identity element:
For a, e ∈ A, a * e = a
(a, b) * (e, e) = (a, b)
(a+e, b+e) = (a, b)
a + e = a
=> e = 0
b + e = b
=> e = 0
As 0 is not a part of set of natural numbers. So, identity function does not exist.
问题12：陈述以下陈述是对还是错。证明合法。
(i)对于集合N上的任意二进制运算∗，a ∗ a = a∀a∈N.
(ii)如果∗是对N的交换二元运算，则a ∗(b ∗ c)=(c ∗ b)∗ a
解决方案：
(i) Let * be an operation on N, defined as:
a * b = a + b ∀ a, b ∈ N
Let us consider b = a = 6, we have:
6 * 6 = 6 + 6 = 12 ≠ 6
Therefore, this statement is false.
(ii) Since, * is commutative
LHS = a ∗ (b ∗ c) = a * (c * b) = (c * b) * a = RHS
Therefore, this statement is true.
问题13：考虑对N的二元运算*，定义为a * b = a 3 + b 3 。选择正确的答案。
(A)∗是否既具有关联性又具有可交换性?
(B)∗是可交换的而不是关联的吗?
(C)∗是关联的而不是可交换的吗?
(D)∗既不是可交换的也不是相关的?
解决方案：
On N, * is defined as a * b = a3 + b3
Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a3 + b3
RHS = b * a = b3 + a3
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b3 + c3) = a3 + (b3 + c3)3
RHS = (a * b) * c = (a3 + b3) * c = (a3 + b3)3 + c3
Since, LHS is not equal to RHS
Therefore, * is not Associative
Thus, Option (B) is correct.