Reflexive and anti reflexive, symmetric and antisymmetric problems in discrete mathematics! A = {1,2,3} let R1 = {,}; R2 = {} Why is R1 neither reflexive nor anti reflexive Why is R2 symmetric or antisymmetric

In other words, if the relation R contains but does not contain all, it is neither reflexive nor reflexive. The symmetry and antisymmetry of relation R mainly consider whether and appear at the same time when x ≠ y. if it appears at the same time, it is symmetric; if only one, it is antisymmetric

In discrete mathematics, what is the meaning of empty set? Why is it anti reflexive

An empty set X is still a set. We use a function to express the characteristics of a set, such as the number of elements of a set. Then an empty set is only f (x) = 0, and a nonempty set is only f (x) ≠ 0. The inverse of an empty set is the complete set Y (including all things in the universe) f (y) = ∞. Then, the infinite inverse is of course not, back to the empty set itself

The definition of anti reflexivity in Discrete Mathematics
The definition in the book is: let R be a binary relation on the set X. if for any x ∈ x, there must be "R is not a binary relation of X" (x / Rx), then the relation R is anti reflexive on X
According to this definition, any relation R on X can't be reflexive
For example: X ∈ x, take x = x, because R is a binary relation (XRX) on set X, so relation R is not reflexive on X
Take a different example
Suppose x = {1,2,3,4,5}, take x = {1,2,3}, r = {,}, where x ∈ X
Because R {,} is a binary relation on X, by definition R is not reflexive on X
I found some courseware on the Internet. For the definition of anti reflexivity, it is simply: if a is the element of a, then it is not the element of R
According to this definition, the relation R {,} in the second example is anti reflexive in X, because there are
Is the definition in the book wrong? Or do I understand the definition in the book wrong? New number, how many points does wood have? Please help me
In the above question, (x / Rx) means "R is not a binary relation of X", and (R) means a crossed out R

You're wrong
(x / Rx) means that it does not belong to relation R. no relation R can be reflexive
It does not define the relationship of other numbers
Relation matrix is the main diagonal of 0, other arbitrary

I want to ask about the symmetry and antisymmetry of discrete mathematics and reflexivity
First of all, I know the definition of three relationships
If you have the following sets
R1{(1.1)(2.2)(3.3)}
R2{(1.1)(1.2)(2.1)(2.2)}
R3{(1.2)(2.3)(31)}
I know R1 is reflexive
R3 is antisymmetric
According to the definition of symmetry and antisymmetry
If {(a, b) belongs to R}, then implies {(B, a), belongs to R}. This is the definition of symmetry
If {(a, b) belongs to R} and {(BA), belongs to R}, then implies a = B
According to the definition of symmetry, R1 should be reflexive and symmetric
But according to the definition of antisymmetry, {(a, b) belongs to R} and {(B, a), belongs to R} then implies a = B. then R1 is reflexive, symmetric and antisymmetric. Is there such a relationship?
If R1 is antisymmetric, why is R2 symmetric? Can there be a symmetric antisymmetric relationship in the set?

Yes, there are both symmetric and antisymmetric relations. Your conclusions are all right
R1 satisfies reflexivity, symmetry and antisymmetry (R1 also satisfies transitivity)
R2 satisfies symmetry (R2 also satisfies transitivity)
R3 satisfies anti symmetry (R1 also satisfies anti reflexivity and transitivity)

Discrete mathematics -- transitive relation
S. R ∈ a, S-Transitive, r-transitive, s ∪ r-transitive (judgment, need to prove)

S ∪ R is not necessarily transitive, e.g
S = {(1,2)}, r = {(2,3)} are transitive relations of a = {1,2,3}, but s ∪ r = {(1,2), (2,3)} are not transitive

The nature of relation -- transmission
I don't quite understand. Ask for guidance
For example, the example in the book: x = {1,2,3}, R1 = {}, R2 = {}, R3 = {}. The answer here is that R1 and R2 are transitive, R3 is not transitive,

In R1, if it is transitive, it must be. It conforms to the definition of transitivity, so it is transitive
There is in R3, but there is no, there is no, it does not meet the requirements of definition, so it is not delivered
R2 is special, because the definition requires "every time XRY and yrz, there is xrz". There is only one order pair, so it can't be judged by the definition. Here, R.R (compound operation of relation R) can be used to judge. If R.R is a subset of R, then R is transitive, otherwise it is not transitive. Here, r2.r2 is an empty set, which is a subset of R2, so it is transitive

For the transfer relation in Discrete Mathematics

For example, the relation {} on the set a = {a, B} satisfies transitivity

What does the term binary operation mean in discrete mathematics

Binary operation is actually an operation with two variables,
Addition, subtraction, multiplication, division and or are binary operations, and the non binary operation is taken
There is also a function of one variable f (x) and a function of two variables f (x, y)

What is the meaning of binary relation in discrete mathematics? Is there only two elements?

Let s be a nonempty set and R be a condition about the elements of S. if any pair of ordered elements in S (a, b), we can always determine whether a and B satisfy the condition R, then R is a relation of S. if a and B satisfy the condition R, then a and B satisfy the condition R, then a and B have a relation R

The binary relation of Discrete Mathematics
If R is a transitive relation on a, then R2 is also a transitive relation on set a
Right
No, take a counterexample

Let R be transitive over a, that is, if XRY and yrz, then there is xrz. Now if there is XRY & # 178; y and yr & # 178; Z, then there exists u, V ∈ a, such that xru, ury and YRV, vrz, then there is XRY and yrz, that is, XRY & # 178; Z, that is, R & # 178; which is also transitive on a

Reflexive and anti reflexive, symmetric and antisymmetric problems in discrete mathematics! A = {1,2,3} let R1 = {,}; R2 = {} Why is R1 neither reflexive nor anti reflexive Why is R2 symmetric or antisymmetric