第12类RD Sharma解决方案–第3章二进制运算–练习3.2

问题1.假设“ *”是对a的二元运算，对于所有a，b∈N，a * b = 1.cm(a，b)

(i)找到2 * 4，3 * 5，1 * 6

解决方案：

We are given that a * b = L.C.M. (a, b) 

⇒ 2 * 4 = L.C.M. (2, 4) = 4

and, 3 * 5 = L.C.M. (3, 5) = 15

now, 1 * 6 = L.C.M. (1, 6) = 6

Hence, 2 * 4 = 4, 3 * 5 = 15 and 1 * 6 = 6.

(ii)检查N上’*’的可交换性和可结合性。

解决方案：

For Commutativity:

Let a, b ∈ N

a * b = L.C.M. (a, b) = L.C.M. (b, a) = b * a

Therefore, a * b = b * a ∀ a, b ∈ N

Thus * is commutative on N.

For Associativity:

Let a, b, c ∈ N

⇒ a * (b * c) = a * L.C.M. (b, c) = L.C.M. (a, (b, c)) = L.C.M. (a, b, c)

And, (a * b) * c = L.C.M. (a, b) * c = L.C.M. ((a, b), c) = L.C.M. (a, b, c)

Therefore, (a * (b * c) = (a * b) * c, ∀ a, b, c ∈ N

Thus, * is associative on N.

问题2.确定以下哪个二进制运算是关联的，哪个是可交换的：

(i)对于所有a，由a * b = 1定义的N上的b∈N

解决方案：

For commutativity:

Let a, b ∈ N

a * b = 1 and b * a = 1

Therefore, a * b = b * a, for all a, b ∈ N

Thus * is commutative on N.

For associativity:

Let a, b, c ∈ N

Then a * (b * c) = a * (1) = 1

and, (a * b) *c = (1) * c = 1

Therefore, a * (b * c) = (a * b) * c for all a, b, c ∈ N

Thus, * is associative on N.

Hence, * is both commutative and associative on N.

(ii)对于所有a，由a * b =(a + b)/ 2定义的Q，b∈Q

解决方案：

For Commutativity:

Let a, b ∈ N

a * b = (a + b)/2 = (b + a)/2 = b * a

Therefore, a * b = b * a, ∀ a, b ∈ N

Thus * is commutative on N.

For Associativity:

Let a, b, c ∈ N

⇒ a * (b * c) = a * (b + c)/2 = [a + (b + c)]/2 = (2a + b + c)/4

Now, (a * b) * c = (a + b)/2 * c = [(a + b)/2 + c] /2 = (a + b + 2c)/4

Thus, a * (b * c) ≠ (a * b) * c

If a = 1, b= 2, c = 3

1 * (2 * 3) = 1 * (2 + 3)/2 = 1 * (5/2) = [1 + (5/2)]/2 = 7/4

and, (1 * 2) * 3 = (1 + 2)/2 * 3 = 3/2 * 3 = [(3/2) + 3]/2 = 4/9

Therefore, there exist a = 1, b = 2, c = 3 ∈ N such that a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N.

Hence * is commutative on N but not associative on N.

问题3.让A为包含多个元素的任何集合。假设’*’是对a的a * b = b定义的A的二元运算，b∈A’*’在A上是可交换的还是相关的?

解决方案：

For Commutativity:

Let a, b ∈ A.

Then, a * b = b

⇒ b * a = a

Therefore, a * b ≠ b * a

Thus, * is not commutative on A.

Now we have to check associativity:

Let a, b, c ∈ A

a * (b * c) = a * c = c

Therefore, a * (b * c) = (a * b) * c, ∀ a, b, c ∈ A

Thus, * is associative on A.

问题4.检查以下每个二进制运算的可交换性和关联性：

(i)Z上的’*’由a * b = a + b + ab定义，适用于所有a，b∈Z

解决方案：

For Commutativity:

Let a, b ∈ Z

Then a * b = a + b + ab = b + a + ba = b * a

Therefore, a * b = b * a, ∀ a, b ∈ Z

Hence, * is commutative on Z.

For Associativity:

Let a, b, c ∈ Z, Then,

a * (b * c) = a * (b + c + b c)

= a + (b + c + b c) + a (b + c + b c)

= a + b + c + b c + a b + a c + a b c

Now, (a * b) * c = (a + b + a b) * c

= a + b + a b + c + (a + b + a b) c

= a + b + a b + c + a c + b c + a b c

Clearly, a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Z

Thus, * is associative on Z.

(ii)对所有a而言，^{由a * b = 2 ab}定义的N上的’*’，b∈N

解决方案：

For Commutativity:

Let a, b ∈ N

a * b = 2^ab = 2^ba = b * a

Therefore, a * b = b * a, ∀ a, b ∈ N

Thus, * is commutative on N

For Associativity:

Let a, b, c ∈ N

Then, a * (b * c) = a * (2^bc) = 2a^2bc

and, (a * b) * c = (2^ab) * c = 2ab^2c

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N.

(iii)对于所有a，b∈Q，a上的*定义为a * b = a – b

解决方案：

For Commutativity:

Let a, b ∈ Q, then

a * b = a – b

b * a = b – a

Clearly, a * b ≠ b * a

Thus, * is not commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c) = a – (b – c) = a – b + c

and, (a * b) * c = (a – b) * c = a – b – c

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(iv)Q上的’⊙’由a a b = a ² + b ²对所有a，b∈Q定义

解决方案：

For Commutativity:

Let a, b ∈ Q, then

a ⊙ b = a² + b² = b² + a² = b ⊙ a

Clearly, a ⊙ b = b ⊙ a, ∀ a, b ∈ Q

Thus, ⊙ is commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

a ⊙ (b ⊙ c) = a ⊙ (b² + c²)

= a² + (b² + c²)2

= a² + b⁴ + c⁴ + 2b²c²

(a ⊙ b) ⊙ c = (a² + b²) ⊙ c

= (a² + b²)² + c²

= a⁴ + b⁴ + 2a²b² + c²

Clearly, (a ⊙ b) ⊙ c ≠ a ⊙ (b ⊙ c)

Thus, ⊙ is not associative on Q.

(v)对所有a，b∈Q由aob =(ab / 2)定义的Q上的’o’

解决方案：

For Commutativity:

Let a, b ∈ Q, then

a o b = (ab/2) = (b a/2) = b o a

Clearly, a o b = b o a, ∀ a, b ∈ Q

Thus, o is commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

a o (b o c) = a o (b c/2) = [a (b c/2)]/2

= [a (b c/2)]/2 = (a b c)/4

and, (a o b) o c = (ab/2) o c = [(ab/2) c] /2 = (a b c)/4

Clearly, a o (b o c) = (a o b) o c, ∀ a, b, c ∈ Q

Thus, o is associative on Q.

(vi)对所有a，b∈Q^{由a * b = ab 2}定义的Q上的’*’

解决方案：

For Commutativity:

Let a, b ∈ Q, then

a * b = ab²

b * a = ba²

Clearly, * b ≠ b * a

Thus, * is not commutative on Q.

Now we have to check associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (bc²)

= a (bc²)²

= ab² c⁴

(a * b) * c = (ab²) * c

= ab²c²

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(vii)Q上的’*’由a * b = a + ab定义，适用于所有a，b∈Q

解决方案：

For commutative:

Let a, b ∈ Q, then

a * b = a + ab

b * a = b + ba = b + ab

Clearly, a * b ≠ b * a

Thus, * is not commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

a * (b * c) = a * (b + bc)

= a + a (b + bc)

= a + ab + abc

(a * b) * c = (a + ab) * c

= (a + ab) + (a + ab)c

= a + ab + ac + abc

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(viii)R上的’*’由a * b = a + b -7定义，对于所有a，b∈R

解决方案：

For Commutativity:

Let a, b ∈ R, then

a * b = a + b – 7

= b + a – 7 = b * a

Clearly, a * b = b * a, for all a, b ∈ R

Thus, * is commutative on R.

For Associativity:

Let a, b, c ∈ R, then

a * (b * c) = a * (b + c – 7)

= a + b + c -7 -7

= a + b + c – 14

and, (a * b) * c = (a + b – 7) * c

= a + b – 7 + c – 7

= a + b + c – 14

Clearly, a * (b * c ) = (a * b) * c, for all a, b, c ∈ R

Thus, * is associative on R.

(ix)对于所有a，b上的* *由a * b =(a – b) ^{2定义，b∈Q}

解决方案：

For Commutativity:

Let a, b ∈ Q, then

a * b = (a – b)²

= (b – a)²

= b * a

Clearly, a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c)²

= a * (b² + c² – 2bc)

= (a – b² – c² + 2bc)²

(a * b) * c = (a – b)² * c

= (a² + b² – 2ab) * c

= (a² + b² – 2ab – c)²

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(x)对所有a，b∈Q由a * b = ab + 1定义的Q上的’*’

解决方案：

For Commutativity:

Let a, b ∈ Q, then

a * b = ab + 1

= ba + 1

= b * a

Clearly, a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

a * (b * c) = a * (bc + 1)

= a (bc + 1) + 1

= abc + a + 1

(a * b) * c = (ab + 1) * c

= (ab + 1) c + 1

= abc + c + 1

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(xi)对于所有a，^{由a * b = a b}定义的N上的’*’，b∈N

解决方案：

For Commutativity:

Let a, b ∈ N, then

a * b = a^b

b * a = b^a

Clearly, a * b ≠ b * a

Thus, * is not commutative on N.

For Associativity:

a * (b * c) = a * (bc) = ${a^b}^c$

and, (a * b) * c = (ab) * c = (a^b)^c = a^bc

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N.

(xii)Z上的“ *”由a * b = a – b定义，所有a，b∈Z

解决方案：

Let a, b ∈ Z, then

a * b = a – b

b * a = b – a

Clearly, a * b ≠ b * a

Thus, * is not commutative on Z.

For Associativity:

Let a, b, c ∈ Z, then

a * (b * c) = a * (b – c)

= a – (b – c)

= a – (b + c)

(a * b) * c = (a – b) – c

= a – b – c

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Z.

(xiii)对所有a，b∈Q由a * b =(ab / 4)定义的Q上的’*’

解决方案：

For Commutativity:

Let a, b ∈ Q, then

a * b = (ab/4)

= (ba/4)

= b * a

Therefore, a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

For Associativity:

Let a, b, c ∈ Q, then

a * (b * c) = a * (bc/4)

= [a (b c/4)]/4

= (a b c/16)

(a * b) * c = (ab/4) * c

= [(ab/4) c]/4

= abc/16

Clearly a * (b * c) = (a * b) * c for all a, b, c ∈ Q

Thus, * is associative on Q.

(xiv)Z上的“ *”由a * b = a + b – ab对于所有a定义，b∈Z

解决方案：

For Commutativity:

Let a, b ∈ Z, then

a * b = a + b – ab

= b + a – ba

= b * a

Clearly, a * b = b * a, for all a, b ∈ Z

Thus, * is commutative on Z.

For Associativity:

Let a, b, c ∈ Z

a * (b * c) = a * (b + c – bc)

= a + b + c- b c – ab – ac + abc

(a * b) * c = (a + b – ab) c

= a + b – ab + c – (a + b – ab) 

= a + b + c – ab – ac – bc + a b c

Clearly, a * (b * c) = (a * b) * c, for all a, b, c ∈ Z

Thus, * is associative on Z.

(xv)Q上的’*’由a * b = gcd(a，b)定义，对于所有a，b∈Q

解决方案：

For Commutativity:

Let a, b ∈ N, then

a * b = gcd (a, b)

= gcd (b, a)

= b * a

Therefore, a * b = b * a, for all a, b ∈ N

Thus, * is commutative on N.

Now we have to check associativity of *

Let a, b, c ∈ N

a * (b * c) = a * [gcd (a, b)]

= gcd (a, b, c)

(a * b) * c = [gcd (a, b)] * c

= gcd (a, b, c)

Clearly, a * (b * c) = (a * b) * c, for all a, b, c ∈ N

Thus, * is associative on N.

问题5.如果二进制运算o由除1以外的所有有理数的集合Q – {-1}上的a0b = a + b – ab定义，则表明o在Q – [–1]上是可交换的。

解决方案：

Let a, b ∈ Q – {-1}.

Then aob = a + b – ab

= b+ a – b = boa

Therefore,

aob = boa for all a, b ∈ Q – {-1}

Thus, o is commutative on Q – {-1}.

问题6.证明由a * b = 3a + 7b定义的Z上的二进制运算*是可交换的吗?

解决方案：

Let a, b ∈ Z

a * b = 3a + 7b

and, b * a = 3b + 7a

Clearly, a * b ≠ b * a for all a, b ∈ Z.

Example, Let a = 1 and b = 2

1 * 2 = 3 × 1 + 7 × 2 = 3 + 14 = 17

2 * 1 = 3 × 2 + 7 × 1 = 6 + 7 = 13

Therefore, there exist a = 1, b = 2 ∈ Z such that a * b ≠ b * a

Thus, * is not commutative on Z.

问题7.在整数Z上，对于所有a，b∈Z，二进制运算*由a * b = ab + 1定义。证明*与Z无关。

解决方案：

Let a, b, c ∈ Z

a * (b * c) = a * (bc + 1)

= a (bc + 1) + 1

= a b c + a + 1

(a * b) * c = (ab+ 1) * c

= (ab + 1) c + 1

= a b c + c + 1

Clearly, a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Z

Thus, * is not associative on Z.

问题8。令S为除-1以外的所有实数的和，并令*为所有a，b∈S的a * b = a + b + ab定义的运算。确定*是否为S上的二进制运算。如果是，请检查其可交换性和关联性。

解决方案：

Given: a * b = a + b + ab, a, b ∈ S = R − {−1}

Let a, b ∈ S.

Thus, ab ∈ S and hence, a + b − ab ∈ S or a * b ∈ S

Hence, a * b S is a binary operation.

For Commutativity:

a * b = a + b + ab = b +a + ba = b * a

Hence, * is commutative.

For Associativity:

Let a, b, c ∈ Z, Then,

(a * b) * c = (a + b + ab) * c

= a + b + ab + c + (a + b + ab)c

= a + b + c + ab + ac + bc + abc      …..(a)

Now, a * (b * c) = a * (b + c + bc)

= a + b + c + bc + ac +ab +abc        …..(b)

From (a) and (b), it is clear that a * (b * c) = (a * b) * c for all a, b, c ∈ Q

Hence, * is associative on Q.

问题9.在Q上，有理数*的集合由a * b =(a – b)/ 2定义，表明*不是关联的。

解决方案：

Let a, b, c ∈ Q. Then,

(a * b) * c = $\frac{a − b}{2}$* c = $\frac{\frac{a-b}{2}-c}{2}$= $\frac{a-b-2c}{4}$      …….(a)

Now, a * (b * c) = a * $\frac{b - c}{2}$= $\frac{2a − b + c}{4}$                      ……….(b)

From (a) and (b), it is clear that a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Q

Hence, * is not associative on Q.

问题10.让二元运算*：R×R⇥R定义为a * b = 2a + b。查找(2 * 3)* 4

解决方案：

Given, a * b = 2a + b

⇒ (2 * 3) * 4 = (2 × 2 + 3) * 4 = 7 * 4 = (2 × 7 + 4) = 18

Hence, (2 * 3) * 4 = 18.

问题11。在整数集Z上，将二进制运算*定义为* * b = a + 3b −4。证明*在Z上既不是可交换的也不是相关的。

解决方案：

For Commutativity:

a * b = a + 3b − 4 ≠ b + 3a − 4 = b * a

⇒ a * b ≠ b * a

Hence * is not commutative on Z.

For Associativity:

Let a, b, c ∈ Z, Then,

(a * b) * c = (a + 3b − 4) * c 

= a + 3b − 4 + 3c − 4 

= a + 3b + 3c − 8     …….(a)

Now, a * (b * c) = a + 3(b + 3c − 4) − 4

= a + 3b + 9c − 16    ……(b)

From (a) and (b),  we get a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Q

Hence, * is not associative on Q.

问题12.如果将二进制运算*定义为，则在所有有理数的集合Q上

a * b = ab / 5，证明*与Q相关。

解决方案：

Let a, b, c ∈ Z, then,

(a * b) * c = ab/5 * c = abc/25      …..(a)

and, a * (b * c) = a * bc/5 = abc/25    ….(b)

From eq (a) and (b), we have 

a * (b * c) = (a * b) * c for all a, b, c ∈ Q

Hence, * is associative on Q.

问题13.在有理数集Q上，二进制运算*被定义为a * b = ab / 7。证明*与Q相关。

解决方案：

Let a, b, c ∈ Z, then,

(a * b) * c = ab/7 * c = abc/49      …..(a)

and, a * (b * c) = a * bc/7 = abc/49   ….(b)

From eq(a) and (b), we have 

a * (b * c) = (a * b) * c for all a, b, c ∈ Q

Hence, * is associative on Q.

问题14.在Q(所有有理数的集合)上，将二进制运算*定义为(a + b)/ 2。证明*与Q无关。

解决方案：

Let a, b, c ∈ Z, then,

(a * b) * c = $\frac{a + b}2$ * c = $\frac{\frac{a-b}{2}-c}{2}$= $\frac{a-b-2c}{4}$      …(a)

a * (b * c) = a * $\frac{b - c}{2}$= $\frac{a+\frac{b+c}{2}}{2}$= $\frac{2a+b+c}4$     …(b)

From eq(a) and (b), we have, 

a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Q

Hence, * is not associative on Q.

问题15。设S为除1以外的所有实数的和，并设*为所有a，b∈S的a * b = a + b-ab定义的运算。证明：

(i)*是对S的二进制运算。

解决方案：

Let a, b ∈ S

Thus, ab ∈ S and hence, 

a + b − ab ∈ S or a * b ∈ S

Hence, a * b S is a binary operation.

(ii)是可交换的和关联的。

解决方案：

For Commutativity:

a * b = a + b − ab = b + a − ba = b * a

Hence, * is commutative.

For Associativity:

Let a, b, c ∈ Z, Then,

(a * b) * c = (a + b − ab) * c

= a + b − ab + c + (a + b − ab)c

= a + b + c − ab − ac − bc + abc      …..(a)

Now, a * (b * c) = a * (b + c − bc)

= a + b + c − bc − ac − ab +abc        …..(b)

From eq(a) and (b), it is clear that 

a * (b * c) = (a * b) * c for all a, b, c ∈ Q

Hence, * is associative on Q.