问题1.确定以下每个关系是否是自反的，对称的和可传递的：
(i)集合A中的关系R = {1，2,3，。。 。 。，13，14}定义为R = {(x，y)：3x-y = 0}
解决方案：
A={1 ,2 ,3,…,13, 14}
R={(x,y): 3x-y=0}
Therefore R={(1,3),(2,6),(3,9),(4,12)}
R is not reflexive since (1,1),(2,2),(3,3),…,(14,14)∉R
Also, R is not reflexive since (1,3)∈R, but (3,1)∉R.[since 3(3)-1≠0]
Also, R is not transitive as (1,3), (3,9)∈R, but (1,9)∉R.[since 3(1)-9≠0]
Hence, R is not reflexive, nor symmetric nor transitive.
(ii)定义为R = {(x，y)的自然数集合N中的关系R：y = x + 5和x <4}
解决方案：
R={(x, y): y=x+5 and x<4}={(1,6), (2, 7), (3,8)}
It is seen that (1, 1)∉R. Therefore, R is not reflexive.
(1, 6)∈R. But, (6, 1)∉R so, R is not symmetric.
Now, since their is no pair in R such that (x, y) and (y, z)∈R, so (x, z) can not belong to R. Therefore, R is not transitive.
So, we can conclude that R is neither reflexive, nor symmetric, nor transitive.
(iii)集合A = {1,2,3,4,5,6}中的关系R为R = {(x，y)：y可被x整除}
解决方案：
A={1, 2, 3, 4, 5, 6}
R={(x, y): y is divisible by x}
We know that any number is always divisible by itself.⇒ (x, x)∈ R. Therefore, R is reflexive.
We see, (2, 4)∈R [4 is divisible by 2]. But, (4, 2)∉R [2 is not divisible by 4]. Therefore, R is not symmetric.
Let’s assume (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y. Therefore, z is divisible by x.⇒ (x, z) ∈ R. Therefore, R is transitive.
(iv)所有Z的集合Z中的关系R定义为R = {(x，y)：xy是一个整数}
解决方案：
R={(x, y): x-y is an integer}
For every x ∈ Z, (x, x) ∈ R [x-x=0 which is an integer]. Therefore, R is reflexive.
For every x, y ∈ Z if (x, y) ∈ R, then x-y is an integer. ⇒-(x-y) is also an integer. ⇒ (y-x) is also an integer. Therefore, R is symmetric.
Let’s assume, (x, y) and (y, z) ∈ R, where x, y and z ∈ Z. ⇒ (x-y) and (y-z) are integers. ⇒ x-z=(x-y)+(y-z) is an integer. ⇒ (x, z) ∈ R. Therefore, R is transitive.
(v)特定时间在一个城镇的一组人类中的关系R，由下式给出：
(a)R = {(x，y)：x和y在同一位置工作。
Solution:
We can see (x,x) ∈ R .Therefore, R is reflexive.
If (x,y) ∈ R, then x and y work at the same place. So, (y,x) ∈ R. Therefore, R is symmetric.
Let, (x,y), (y,z) ∈ R.
⇒ x and y work mat the same place and y and z work at the same place.
⇒ x and z work at the same place.
Therefore, R is transitive.
(b)R = {(x，y)：x和y居住在同一地点}。
解决方案：
We can see (x,x) ∈ R. Therefore, R is reflexive.
If (x,y) ∈ R, then x and y live in same locality. So, (y,x) ∈ R. Therefore, R is symmetric.
Let (x,y) ∈ R and (y,z) ∈ R. So, x, y and z live in the same locality. So, (x,z) ∈ R. Therefore, R is transitive.
(c)R = {(x，y)：x正好比y高7厘米}。
解决方案：
(x,x)∉R since, human being can not be taller than himself. So, R is not reflexive.
Let (x,y) ∈ R ,then x is exactly 7 cm taller than y. Then, y is not taller than x. Therefore, R is not symmetric.
Let (x,y), (y,z) ∈ R, then x is exactly 7 cm taller than y and y is exactly 7 cm taller than z which means x is 14 cm taller than z. So, (x,z)∉R . Therefore, R is not transitive.
(d)R = {(x，y)：x是y的妻子}。
解决方案：
(x,x) ∉ R. Since, x can not be the wife of herself. Therefore, R is not reflexive.
Let (x,y) ∈ R, then x is the wife of y. So, y is not the wife of x ,i.e., (y,x) ∉ R. Therefore, R is not symmetric.
Let (x,y), (y,z) ∈ R, then x is the wife of y and y is the wife of z which is not possible. So, R can not be transitive.
(e)R = {(x，y)：x是y的父亲}。
解决方案：
(x,x) ∉ R. Since, x can not be the father of himself. Therefore, R is not reflexive.
Let (x,y) ∈ R, then x is the father of y. So, y can not be the father of x. So, (y,x) ∉ R. Therefore, R is not symmetric.
Let (x,y), (y,z) ∈ R, then x is the father of y and y is the father of z which means x is the grandfather of z. So, So, (x, z) ∉ R. Therefore, R is not transitive.
问题2。证明实数集合R中的关系R(定义为R = {(a，b)：a≤b 2 })既不是自反的，也不是对称的，也不是传递的。
解决方案：
It can be observed that (½, ½) ∉ R, since ½>(½)2 =¼. Therefore, R is not reflexive.
(1,4) ∈ R as 1<42 .But, (4,1) ∉ R. Therefore, R is not symmetric.
(3,2), (2,1.5) ∈ R. But, 3> (1.5)2 =2.25 . So, (3,1.5) ∉ R. Therefore, R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
问题3.检查集合{1、2、3、4、5、6}中定义为R = {(a，b)：b = a + 1}的关系R是自反，对称还是可传递}。
解决方案：
Let the set {1, 2, 3, 4, 5, 6} be named A.
R={(1,2), (2, 3), (3, 4), (4, 5
), (5, 6)}
We can see (x, x) ∉ R. Since, x ≠ x+1. Therefore, R is not reflexive.
It is observed that (1,2) ∈ R but, (2,1) ∉ R. Therefore, R is not symmetric.
We can see, (1,2), (2, 3) ∈ R, but (1,3) ∉ R. Therefore, R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
问题4表明，在R中的关系R定义为R = {(A，B)：A≤B}，是自反的和传递的但不是对称的。
解决方案：
Clearly, (a,a) ∈ R as a=a. Therefore, R is reflexive.
(2,4) ∈ R (as 2<4) but (4,2) ∉ R as 4 is greater than 2. Therefore, R is not symmetric.
Let (a,b), (b,c) ∈ R. Then, a≤ b and b≤ c.
⇒a ≤ c.
(a, c) ∈ R. Therefore, R is transitive.
Hence, R is reflexive and transitive but not symmetric.
问题5.检查R中的关系R定义为R =是否{(A，B)：一个≤b 3分配}是自反的，对称的或可传递的。
解决方案：
It is observed that (½, ½) ∉ R as ½ > (½)3 =(1/8). Therefore, R is not reflexive.
(1,2) ∈ R(as 1<8) but, (2,1) ∉ R. Therefore, R is not symmetric.
We have, (3, 3/2), (3/2, 6/5) ∈ R but, (3, 6/5) ∉ R. Therefore, R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
问题6：证明由R = {(1,2 ,,(2,1)}}给出的集合{1，2，3}中的关系R是对称的，但既不是自反的也不是传递的。
解决方案：
Let the set {1, 2, 3} be named A.
It is seen that, (1, 1), (2,2), (3,3)∉ R. Therefore, R is not reflexive.
As (1, 2) ∈ R and (2, 1) ∈ R. Therefore, R is symmetric.
However, (1, 1)∉ R. Therefore, R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.
问题7：证明由R = {(x，y)：x和y具有相同的页数}给出的大学图书馆中所有图书的集合A中的关系R是等价关系。
解决方案：
Set A is the set of all books in the library of a college.
R={(x,y):x and y have the same number of pages}
R is reflexive since (x,x) ∈ R as x and x have the same number of pages.
Let (x,y) ∈ R
⇒x and y have the same number of pages
⇒y and x have the same number of pages.
⇒(y,x)∈ R
Therefore , R is symmetric.
Let (x,y) ∈ R and (y,z)∈ R.
⇒x and y have the same number of pages and y and z have the same number of pages.
⇒x and z have the same number of pages.
⇒(x,z) ∈ R
Therefore, R is transitive.
Hence, R is an equivalence relation.
问题8.证明集合A = {1,2,3,4,5}中的关系R由R = {((a，b)：| ab |)给出。等于}，是等价关系。证明{1,3,5}的所有元素相互关联，并且{2，4}的所有元素相互关联。但是{1,3,5}的元素都与{2,4}的任何元素都不相关。
解决方案：
A={ 1,2,3,4,5}
R={(a,b):|a-b| is even }
It is clear that for any element a∈ A, we have |a-a|=0 (which is even).
Therefore, R is reflexive.
Let (a,b) ∈ R.
⇒|a-b| is even.
⇒|-(a-b)|=|b-a| is also even.
⇒(b,a)∈ R
Therefore, R is symmetric.
Now , let (a,b)∈ R and (b,c)∈ R.
⇒|a-b| is even and |b-c| is even.
⇒(a-b) is even and (b-c) is even.
⇒(a-c)=(a-b)+(b-c) is even. [Sum of two even integers is even]
⇒|a-c| is even.
Therefore, R is transitive.
Hence, R is an equivalence relation.
All elements of the set {1,2,3} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.
Similarly, all elements of the set {2,4} are related to each other as all the elements of this subset are even.
Also, no element of the subset {1,3,5} can be related to any element of {2,4} as all elements of {1,3,5} are odd and all elements of {2,4} are even . Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.
问题9：证明集合A = {x∈Z：0 <= x <= 12}中的每个关系R由下式给出：
(i)R = {((a，b)：| ab |是4的倍数}
(ii)R = {((a，b)：a = b}
是等价关系。找出每种情况下与1相关的所有元素的集合。
解决方案：
A={x∈ Z:0<=x<=12}={0,1,2,3,4,5,6,7,8,9,10,11,12}
(i) R={(a,b):|a-b| is a multiple of 4}
For any element a ∈A , we have (a,a)∈R as |a-a|=0 is a multiple of 4.
Therefore, R is reflexive.
Now , let (a,b)∈R ⇒|a-b| is a multiple of 4.
⇒|-(a-b)|=|b-a| is a multiple of 4.
⇒(b,a)∈R
Therefore, R is symmetric.
Let (a,b) ,(b,c) ∈ R.
⇒|a-b| is a multiple of 4 and |b-c| is a multiple of 4.
⇒(a-b) is a multiple of 4 and (b-c) is a multiple of 4.
⇒(a-c)=(a-b)+(b-c) is a multiple of 4.
⇒|a-c| is a multiple of 4.
⇒(a,c)∈R
Therefore, R is transitive.
Hence, R is an equivalence relation .
The set of elements related to 1 is {1,5,9} since |1-1|=0 is a multiple of 4,
|5-1|=4 is a multiple of 4,and
|9-1|=8 is a multiple of 4.
(ii) R={(a,b):a=b}
For any element a∈A, we have (a,a) ∈ R , since a=a.
Therefore, R is reflexive.
Now , let (a,b)∈R.
⇒a=b
⇒b=a
⇒(b,a)∈R
Therefore, R is symmetric.
Now, let (a,b)∈R and (b,c)∈R.
⇒a=b and b=c
⇒a=c
⇒(a,c)∈R
Therefore, R is transitive.
Hence, R is an equivalence relation.
The elements in R that are related to 1 will be those elements from set A which are equal to 1.
Hence, the set of elements related to 1 is {1].
问题10：举一个关系的例子。哪一个
(i)对称但既不反身也不及物动词。
(ii)具有传递性，但既非反身性也不对称。
(iii)反身和对称，但不及物。
(iv)自反和及物但不对称。
(v)对称和传递但不反身。
解决方案：
(i) Let A ={5,6,7}.
Define a relation R on A as R ={(5,6),(6,5)}.
Relation R is not reflexive as (5,5) , (6,6),(7,7)∉R.
Now, as (5,6)∈R and also (6,5)∈R, R is symmetric.
⇒(5,6),(6,5)∈R , but (5,5)∉R
Therefore , R is not transitive.
Hence, relation R is symmetric but not reflexive or transitive.
(ii) Consider a relation R in R defined as:
R= {(a,b): a For any a∈R, we have (a,a)∉R since a cannot be strictly less than itself . In fact a=a. Therefore, R is not reflexive. Now, (1,2)∈R (as 1<2) But, 2 is not less than 1. Therefore, (2,1)∉R Therefore, R is not symmetric. Now, let (a,b),(b,c)∈R. ⇒a
⇒a ⇒ (a,c)∈R Therefore, R is transitive. Hence, relation R is transitive but not reflexive and symmetric. (iii) Let A={4,6,8}. Define a relation R on A as: A={(4,4),(6,6),(8,8),(4,6),(6,4),(6,8),(8,6)} Relation R is reflexive since for every a∈A , (a,a)∈R i.e.,(4,4),(6,6),(8,8)∈R. Relation R is symmetric since (a,b)∈R⇒(b,a)∈R for all a,b∈R. Relation R is not transitive since (4,6),(6,8)∈R , but (4,8)∉R. Hence, relation R is reflexive and symmetric but not transitive. (iv) Define a relation R in R as: R={ (a,b): a3 ≥ b3 } Clearly (a,a)∈R as a^3=a^3 Therefore, R is reflexive. Now, (2,1)∈R(as 23>=13) But, (1,2)∉ R (as 13< 23) Therefore, R is not symmetric Let (a,b), (b,c) ∈ R. ⇒a3 >= b3 and b3 >=c3 ⇒a3>=c3 ⇒(a, c)∈R Therefore, R is transitive. Hence, relation R is reflexive and transitive but not symmetric. (v) Let A={-5,-6}, Define a relation R on A as: R={(-5,-6), (-6,-5), (-5,-5)} Relation R is not reflexive as (-6,-6)∉ R. Relation R is symmetric as (-5,-6)∈ R and (-6,-5)∈ R. It is seen that (-5,-6),(-6,-5)∈R . Also , (-5,-5)∈R. Therefore, the relation R is transitive . Hence, relation R is symmetric and transitive but not reflexive.