第12类RD Sharma解决方案–第1章关系–练习1.2 |套装2

问题11.让O为原点。如果OP = OQ，我们定义平面中两个点P和Q之间的关系。证明这样定义的关系是等价关系。

解决方案：

Let A be the set of points on plane

and let R = {(P, Q): OP = OQ} be a relation on A where O is the origin.

now (i) reflexibility:

let P ∈ A

since, OP = OP

:. P, P ∈ R

so, relation R is reflexive

(ii) symmetry:

let (P, Q) ∈ R for P, Q ∈ A

=>OQ = OP

:. (Q, P) ∈ R

therefore, relation R is symmetric .

lly (iii) transitive:

let (P,Q) ∈ R and (Q,S) ∈ R

OP = OS

(P, S) ∈ R

hence, relation R is transitive.

so, relation R is an equivalence relation on A .

问题12。设R为在集合A = {1,2,3,4,5,6,7}上通过R = {(a，b)定义的关系：a和b均为奇数或偶数}。证明R是一个等价关系。此外，表明子集{1,3,5,7}的所有元素都相互关联，并且子集{2,4,6}的所有元素都相互关联，但子集的元素不相关{1,3,5,7}与子集{2,4,6}中的任何元素有关

解决方案：

Given, A={1,2,3,4,5,6,7}

R={(a,b): both a and b are either odd or even}

clearly, (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7) ∈ R

so, relation R is reflexive.

Also, for symmetry,

a, b ∈ A such that (a, b) ∈ R

both a and b are either odd or even

both b and a are either odd or even

=>(b, a) ∈ R

so, relation R is also symmetry

lly, for transitivity,

let a, b, c ∈ Z such that (a,b) ∈ R, (b,c) ∈ R

both a and b are either odd or even

both b and c are either odd or even

if both a and b are even then,

(b,c) ∈ R => both b and c are even

:. both a and c are even

and if both a and b are odd, then

(b,c) ∈ R =>both b and c are odd

both a and c are odd

Thus, both a and c are even and odd

therefore, (a,c) ∈ R

so, (a,b) ∈ R and (b,c) ∈ R => (a,c) ∈ R

so, relation R is transitive

it proves that R is an equivalence relation.

We observe that {1,3,5,7} are related with each other only and {2,4,6} are related with each other.

问题13.设集合S是定义为S = {(a，b)∈R x R：a ² + b ² = 1}的所有实数集的关系

解决方案：

Let us observe the following properties of S

(i) Reflexibility: let a be an arbitrary element of R

a ∈ R

=> a² + a² ≠ 1 for all a ∈ R

so, S is not reflexive on R .

(ii) symmetry:

let (a, b) ∈ R

a² + b² = 1

b² + a² = 1

=> (b, a) ∈ S for all a, b ∈ R

So, S is symmetric on R

(iii) transitivity:

Let (a,b) and (b,c) ∈ S

a² + b² =1 and b² + c² = 1

Adding both the equations we will get,

a² + c² = 2 – 2b² ≠ 1 for all a, b, c ∈ R

So, S is not transitive on R

Hence, S is not an equivalence relation on R

问题14.令Z为所有整数的集合，Z ₀为所有非零整数的集合。令Z x Z ₀上的关系R定义为(a，b)R(c，d)<=> ad = bc对于所有(a，b)，(c，d)∈Z x Z ₀ 。证明R是Z x Z ₀上的等价关系。

解决方案：

Let us observe the properties of R

(i) reflexibility:

let (a,b) be an arbitrary element of Z x Z₀

(a,b) ∈ ZxZ₀

a,b ∈ Z,Z₀

ab = ba

(a,b) ∈ R for all (a,b) ∈ ZxZ₀

R is reflexive

(ii) symmetry:

Let (a,b), (c,d) ∈ ZxZ₀ such that (a,b) R (c,d)

=> ad = bc

=> cb = da

(c,d) R (a,b)

Thus, (a,b) R (c,d) => (c,d) R (a,b) for all (a,b), (c,d) ∈ ZxZ₀

so, R is symmetric.

(iii) transitivity:

Let (a,b), (c,d), (e,f) ∈ NxN₀ such that (a,b) R (c,d) and (c,d) R (e,f)

(a,b) R (c,d) => ad = bc

(c,d) R (e,f) => cf = de

from this, we get (ad) (cf) = (bc) (de)

=> af = be

(a,b) R (e,f)

so, R is transitive.

Hence it is proved that relation R is an equivalence relation.

问题15：如果R和S是集合A上的关系，则证明

(i)R和S是对称的= R∩S和R∪S是对称的

(ii)R是自反的，并且S是任意关系=>R∪S是自反的。

解决方案：

(i) R and S are symmetric relations on the set A

=> R ⊂ A x A and S ⊂ A x A

=> R ∩ S ⊂ A x A

thus, R ∩ S is a relation on A

let a, b ∈ A auch that (a,b) ∈ R ∩ S

(a,b) ∈ R ∩ S

=> (a,b) ∈ R and (a,b) ∈ S

=> (b,a) ∈ R and (b,a) ∈ S

thus, (a,b) ∈ R ∩ S

=> (b,a) ∈ R ∩ S for all a, b ∈ A

so, R ∩ S is symmetric on A

Also, let a, b ∈ R such that (a,b) ∈ R ∪ S

=> (a,b) ∈ R or (a,b) ∈ S

= (b,a) ∈ R or (b,a) ∈ S [since R and S are symmetric]

=> (b,a) ∈ R ∪ S

hence, R ∪ S is symmetric on A

(ii) R is reflexive and S is any relation:

Suppose a ∈ A

then, (a,a) ∈ R [since R is reflexive]

=> (a,a) ∈ R ∪ S

=> R ∪ S is reflexive on A .

问题16。如果R和S是集合A上的传递关系，则证明R∪S可能不是A上的传递关系。

解决方案：

Let A = {a,b,c} and R and S be two relations on A, given by

R = {(a,a), (a,b), (b,a), (b,b)} and

S = {(b,b), (b,c), (c,b), (c,c)}

Here, the relations R and S are transitive on A

(a,b) ∈ R ∪ S and (b,c) ∈ R ∪ S

but (a,c) ∉ R ∪ S

Hence, R ∪ S is not a transitive relation on A.

问题17。令C为所有复数的集合，C ₀为所有非零复数的集合。让关系R和C ₀被定义为Z ₁ RZ ₂ <=> Z ₁ – Z ₂ / Z ₁ + Z ₂是实为所有的Z _{1，Z 2∈C 0。}显示的是，R是一个等价关系。

解决方案：

(i) reflexibility:

since, z₁ – z₂ / z₁ + z₂ = 0 which is a real number

so, (z₁, z₁) ∈ R so,

relation R is reflexive.

(ii)symmetry;

z₁ – z₂ / z₁ + z₂ = x, where x is real number

=> – (z₁ – z₂ / z₁ + z₂ ) = -x

so, (z₂, z₁) ∈ R

Hence, R is symmetric.

(iii) transitivity:

Let (z₁,z₂) ∈ R and (z₂,z₃) ∈ R

then,

z₁-z₂ / z₁+z₂ = x where x is a real number

=> z₁ – z₂ = xz₁ + xz₂

=> z₁ – xz₁ = z₂ + xz₂

=> z₁ (1 – x) = z₂ (1+x)

=> z₁/z₂ = (1+x)/(1-x) …. eqn. (1)

also, z₂-z₃/z₂+z₃ = y where y is a real number

=> z₂ – z₃ = yz₂ + yz₃

=> z₂ – yz₂ = z₃ + yz₃

=> z₂(1-y) = z₃ (1+y)

=> z₂/z₃ = (1+y)/(1-y)….eqn.(2)

dividing (1) and (2) we will get

z₁/z₃ = (1+x / 1-x) X (1-y / 1+y) = z where z is a real number

=> z₁-z₃ / z₁+z₃ = z-1/z+1 which is real

=> (z₁, z₃) ∈ R

Hence, R is transitive

Therefore, it is proved that relation R is an equivalence relation.