第12类RD Sharma解决方案–第3章二进制运算–练习3.1

问题1.确定以下操作是否在给定集合上定义了二进制操作：

(i)对所有a，b∈N，由a * b = ab定义的N上的’*’。

(ii)对所有a，b∈Z，由O b = ab定义的Z上的’O’。

(iii)N上的“ *”由a * b = a + b – 2定义，所有a，b∈N

(iv)S = {1,2,3,4,5}上的’×6’由a×6定义b = ab除以6时的余数。

(v)S = {0，1，2，3，4，5}上的’+6’由a +6 b定义

$=\begin{cases}a+b,\ if\ a+b<6\\a+b-6,\ if\ a+b\ge 6\end{cases}$

(vi)N上的’N’由对所有a的b = ab + ba定义，b∈N

(vii)Q对所有a的’*’由a * b =(a – 1)/(b + 1)定义，b∈Q

解决方案：

(i) Given ‘*’ on N defined by a * b = a^b for all a, b ∈ N.

Let a, b ∈ N. Then,

a^b ∈ N      [∵ ab≠0 and a, b is positive integer]

⇒ a * b ∈ N

Therefore,

a * b ∈ N, ∀ a, b ∈ N

Thus, * is a binary operation on N.

(ii) Given ‘O’ on Z defined by a O b = a^b for all a, b ∈ Z.

Both a = 3 and b = -1 belong to Z.

⇒ a * b = 3^-1

= $\frac{1}{3}$ ∉ Z

Thus, * is not a binary operation on Z.

(iii)  Given ‘*’ on N defined by a * b = a + b – 2 for all a, b ∈ N

If a = 1 and b = 1,

a * b = a + b – 2

= 1 + 1 – 2

= 0 ∉ N

Thus, there exist a = 1 and b = 1 such that a * b ∉ N

So, * is not a binary operation on N.

(iv) Given ‘×₆‘ on S = {1, 2, 3, 4, 5} defined by a ×₆ b = Remainder when a b is divided by 6.

Consider the composition table,

X6	1	2	3	4	5
1	1	2	3	4	5
2	2	4	0	2	4
3	3	0	3	0	3
4	4	2	0	4	2
5	5	4	3	2	1

Here all the elements of the table are not in S.

⇒ For a = 2 and b = 3,

a ×₆ b = 2 ×₆ 3 = remainder when 6 divided by 6 = 0 ≠ S

Thus, ×₆ is not a binary operation on S.

(v) Given ‘+₆’ on S = {0, 1, 2, 3, 4, 5} defined by a +₆ b

$=\begin{cases}a+b,\ if\ a+b<6\\a+b-6,\ if\ a+b\ge 6\end{cases}$

Consider the composition table,

+6	0	1	2	3	4	5
0	0	1	2	3	4	5
1	1	2	3	4	5	0
2	2	3	4	5	0	1
3	3	4	5	0	1	2
4	4	5	0	1	2	3
5	5	0	1	2	3	4

Here all the elements of the table are not in S.

⇒ For a = 2 and b = 3,

a ×₆ b = 2 ×₆ 3 = remainder when 6 divided by 6 = 0 ≠ Thus, ×₆ is not a binary operation on S.

(vi) Given ‘⊙’ on N defined by a ⊙ b= a^b + b^a for all a, b ∈ N

Let a, b ∈ N. Then,

ab, ba ∈ N

⇒ a^b + b^a ∈ N      [∵Addition is binary operation on N]

⇒ a ⊙ b ∈ N

Thus, ⊙ is a binary operation on N.

(vii) Given ‘*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q

If a = 2 and b = -1 in Q,

a * b = $\frac{(a - 1)}{ (b + 1)}$

= $\frac{(2 - 1)}{ (- 1 + 1)}$

= $\frac{1}{0}$ [which is not defined]

For a = 2 and b = -1

a * b does not belongs to Q

So, * is not a binary operation in Q.

问题2.确定下面给出的*的定义是否给出了二进制运算。如果*不是二进制运算，请给出理由。

(i)在Z ^+上，*由a * b = a – b定义

(ii)在Z ^+上，通过a * b = ab定义*

(iii)在R上，通过a * b = ab ^2定义*

(iv)在Z ^+上，通过a * b = | a − b |来定义*

(v)在Z ^+上*通过a * b = a定义

(vi)在R上，*由a * b = a + 4b ^2定义

在此，Z ⁺表示所有非负整数的集合。

解决方案：

(i) Given On Z⁺, defined * by a * b = a – b

If a = 1 and b = 2 in Z⁺, then

a * b = a – b

= 1 – 2

= -1 ∉ Z⁺ [because Z⁺ is the set of non-negative integers]

For a = 1 and b = 2,

a * b ∉ Z⁺

Thus, * is not a binary operation on Z⁺.

(ii) Given Z⁺, define * by a*b = a b

Let a, b ∈ Z⁺

⇒ a, b ∈ Z⁺

⇒ a * b ∈ Z⁺

Thus, * is a binary operation on R.

(iii) Given on R, define by a*b = ab²

Let a, b ∈ R

⇒ a, b² ∈ R

⇒ ab² ∈ R

⇒ a * b ∈ R

Thus, * is a binary operation on R.

(iv) Given on Z⁺ define * by a * b = |a − b|

Let a, b ∈ Z⁺

⇒ | a – b | ∈ Z⁺

⇒ a * b ∈ Z⁺

Therefore,

a * b ∈ Z⁺, ∀ a, b ∈ Z⁺

Thus, * is a binary operation on Z⁺.

(v) Given on Z⁺ define * by a * b = a

Let a, b ∈ Z⁺

⇒ a ∈ Z⁺

⇒ a * b ∈ Z⁺

Therefore, a * b ∈ Z⁺ ∀ a, b ∈ Z⁺

Thus, * is a binary operation on Z⁺.

(vi) Given On R, define * by a * b = a + 4b²

Let a, b ∈ R

⇒ a, 4b² ∈ R

⇒ a + 4b² ∈ R

⇒ a * b ∈ R

Therefore, a *b ∈ R, ∀ a, b ∈ R

Thus, * is a binary operation on R.

问题3.令*为整数集I的二进制运算，由a * b = 2a + b − 3定义。找到3 * 4的值。

解决方案：

Given:

a * b = 2a + b – 3

3 * 4 = 2 (3) + 4 – 3

= 6 + 4 – 3

= 7

问题4. *是否通过a和ba二进制运算的* b = LCM在集合{1,2,3,4,5}上定义*?证明你的答案。

解决方案：

LCM	1	2	3	4	5
1	1	2	3	4	5
2	2	2	6	4	10
3	3	5	3	12	15
4	4	4	12	4	20
5	5	10	15	20	5

In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.

If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 ∉ {1, 2, 3, 4, 5}.

Thus, * is not a binary operation on {1, 2, 3, 4, 5}.

问题5.令S = {a，b，c}。查找S上的二进制操作总数。

解决方案：

Number of binary operations on a set with n elements is $n^{n^2}$

Here, S = {a, b, c}

Number of elements in S = 3

Number of binary operations on a set with 3 elements is $3^{3^2}=3^9$

问题6.查找{a，b}上的二进制操作总数。

解决方案：

We have,

S = {a, b}

The total number of binary operation on S = {a, b} in $2^{2^2}=2^4=16$

问题7.证明操作*已设置

M = $\left\{\begin{bmatrix}a&0\\0&b\end{bmatrix} : a, b ∈ R=\{0\}\right\}$由A + B = AB定义的是二进制运算。

解决方案：

We have,

$\left\{\begin{bmatrix}a&0\\0&b\end{bmatrix} : a, b ∈ R=\{0\}\right\}$ and

A + B = AB for all A, B ∈ M

Let A =\$\begin{bmatrix}a&0\\0&b\end{bmatrix}∈M$ and B = $\begin{bmatrix}c&0\\0&d\end{bmatrix}∈M$

Now, AB = $\begin{bmatrix}a&0\\0&b\end{bmatrix}\begin{bmatrix}c&0\\0&d\end{bmatrix}=\begin{bmatrix}ac&0\\0&bd\end{bmatrix}$

Therefore, a ∈ R, b ∈ R, c ∈ R and d ∈ R

⇒ ac ∈ R and bd ∈ R

⇒ $\begin{bmatrix}ac&0\\0&bd\end{bmatrix}∈M$

⇒ A * B ∈ M

Hence, the operator * defines a binary operation on M

问题8.设S为形式的所有有理数的集合$\frac{m}{n}$其中m∈Z且n = 1、2、3。证明由a * b = ab定义的S上的*不是二元运算

解决方案：

S = set of rational numbers of the form$\frac{m}{n}$ where m ∈ Z and n = 1, 2, 3

Also, a * b = ab

Let a ∈ S and b ∈ S

⇒ ab = $\frac{35}{6}∈S$

Therefore, a * b ∉ S

Hence, the operator * does not defines a binary operation on S

问题9.二进制运算＆：R×R→R定义为a * b = 2a + b

解决方案：

It is given that, a*b = 2a + b

Now,

(2*3) = 2 × 2 + 3

         = 4 + 3

(2*3)*4 = 7*4 = 2 × 7 + 4

            = 14 + 4

            = 18

问题10。令*是对a的二元运算，对于所有a，b∈N，a * b = LCM(a，b)给定。找到5 * 7。

解决方案：

It is given that a*b = LCM (a, b)

Now,

5*7 = LCM (5, 7)

       = 35