第1章关系和功能–练习1.1 |套装1
问题11：证明在由R = {(P，Q)给出的平面中的点集A中的关系R：点P与原点的距离与点Q与原点的距离相同是等价关系。此外，表明与点P≠(0，0)相关的所有点的集合是通过P的圆，其原点为中心。
解决方案：
We can see (P, P) ∈ R since, the distance of point P from the origin is always the same as the distance of the same point P from the origin. Therefore, R is reflexive.
Let (P,Q)∈ R.
⇒The distance of point P from the origin is the same as the distance of point Q from the origin.
⇒The distance of point Q from the origin is the same as the distance of point P from the origin.
So, (Q,P) ∈ R. Therefore, R is symmetric.
Let (P,Q), (Q,S) ∈ R.
⇒The distance of point P from the origin is the same as the distance of point Q from the origin and also, the distance of point P from the origin is the same.
⇒ The distance of points P and S from the origin is the same.
⇒(P, S) ∈ R. Therefore, R is transitive.
Therefore, R is an equivalence relation.
The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.
In other words, if O(0 0) is the origin and OP = k, then the set of all points related to P is at the same as k from the origin. Hence, this set of points forms a circle with the centre as the origin and the circle passes through point P.
问题12：证明所有三角形的集合A中定义为R = {(T 1 ，T 2 )：T 1类似于T 2 }的关系R是等价关系。考虑三个直角三角形T 1与侧面3，4，5，T 2与侧面5，12，13和T 3与侧部6，8，10，T 1，T 2和T 3之间哪个三角形有关系吗?
解决方案：
R is reflexive since, every triangle is similar to itself.
If (T1, T2) ∈ R, then T1 is similar to T2. So, (T2, T1) ∈ R. Therefore, R is symmetric.
Let (T1, T2), (T2, T3) ∈ R, then T1 is similar to T2 and T2 is similar to T3. So, T1 is also similar to T3. Therefore, (T1, T3) ∈ R so, R is transitive.
Thus, R is an equivalence relation.
We observe,
(3/6)=(4/8)=(5/10)=1/2
Therefore, the corresponding sides of triangles T1 and T3 are in the same ratio. Then, triangle T1 is similar to triangle T3.
Hence, T1 is related to T3.
问题13：证明在所有多边形的集合A中定义为R = {(P 1 ，P 2 )：P 1和P 2具有相同边数}的关系R是等价关系。 A中与边3、4和5的直角三角形T相关的所有元素的集合是什么?
解决方案：
R is reflexive since(P1, P2) ∈ R as the same polygon has the same number of sides with itself.
Let (P1, P2) ∈ R, then P1 and P2 have the same number of sides. So, (P2, P1) ∈ R. Therefore, R is symmetric.
Let (P1, P2), (P2, P3) ∈ R, then P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides. So, P1 and P3 have the same number of sides ,i.e., (P1, P3) ∈ R. Therefore, R is transitive.
Hence, R is an equivalence relation.
The elements in A related to the right-angled triangle (T) with sides 3, 4 and 5 are those polygons which have 3 sides(since, T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.
问题14.设L为XY平面上所有线的集合，R为L的关系，定义为R = {(L 1 ，L 2 )：L 1平行于L 2 }。证明R是一个等价关系。查找与线y = 2x + 4相关的所有线的集合。
解决方案：
R is reflexive as any line L1 is parallel to itself i.e., (L1, L2) ∈ R.
Let (L1, L2) ∈ R, then L1 is parallel to L2. So, (L2, L1) ∈ R. Therefore, R is symmetric.
Let (L1, L2), (L2, L3) ∈ R, then L1 is parallel to L2 and L2 is parallel to L3. So, L1 is parallel to L3. Therefore, R is transitive.
Hence, R is an equivalence relation.
The set of all lines related to the line y=2x +4 is the set of all lines that are parallel to the line y=2x+4.Slope of the line is m=2.
It is known that parallel lines have the same slopes. The line parallel to the given line is of the form y=2x +c, where c ∈ R.
Hence, the set of all lines related to the given line is given by y=2x +c, where c ∈ R.
问题15。设R为R = {(1,2,2，(2,2)，(1,1)，(4,4)，( 1,3)，(3,3)，(3,2)}。选择正确的答案。
(A)R是反身和对称的，但不及物动词。
(B)R是自反的和可传递的，但不是对称的。
(C)R是对称的和可传递的，而不是自反的。
(D)R是等价关系。
解决方案：
It is seen that (a,a) ∈ R, for every a ∈ {1, 2, 3, 4}. Therefore, R is reflexive.
It is seen that (1,2) ∈ R but (2,1) ∉ R. Therefore, R is not symmetric.
Also, it is observed that (a,b), (b,c) ∈ R⇒ (a,c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}. Therefore, R is transitive.
Hence, R is reflexive and transitive but not symmetric. The correct answer is B.
问题16。令R为R = {(a，b)：a = b-2;给出的集合N中的关系。 b> 6}。选择正确的答案。
(A)(2,4)∈R
(B)(3,8)∈R
(C)(6,8)
(D)(8,7)∈R
解决方案：
Since b>a, (2,4) ∉ R also, as 3≠8-2, (3,8)∉R and as 8≠7-2. Therefore, (8,7)∉R
Consider (6,8). We have 8>6 and also, 6=8-2. Therefore, (6,8) ∈ R.
The correct answer is C.