Proving Binary Relations: R & S

In summary: No, we cannot have $aRb$, because $aRb$ would mean that $a\le b$, which is false.Since $1R9$ does not exist, it seems that $R$ is not symmetric.No, we cannot have $aRb$, because $aRb$ would mean that $a\le b$, which is false.Since $1R9$ does not exist, it seems that $R$ is not symmetric.So $R$ is not symmetric.Since $1R9$ does not exist, it seems that $R$ is not transitive.It is not the case that
  • #1
andrew1
20
0
Hi,

I'm currently stuck on a few questions regarding binary relations as I'm unsure on how to prove their properties.

R is defined on N by aRb if and only if a <= b and b <= a+5

Is R reflexive, symmetric, antisymmetric, transitive?

S is defined on Z (union) {x + 1/2 : x is an element of all integers} by aSb if and only if a - b is an even integer.

Is S reflexive, symmetric, antisymmetric, transitive?
 
Physics news on Phys.org
  • #2
Is $\le$ reflexive? Why? What can you say about reflexivity of $R$?

Also why do you use LaTeX commands but do not surround formulas with dollar signs or [math] tags?
 
  • #3
Sorry about the latex commands thing, not sure how it works :S

In regards to R, would a + 5 be treated as just a? therefore it isn't reflexive?
 
  • #4
andrew said:
Sorry about the latex commands thing, not sure how it works
Even after I said that you can surround formulas with dollar signs?

andrew said:
In regards to R, would a + 5 be treated as just a? therefore it isn't reflexive?
$a+5$ is almost never treated as $a$; these are two different numbers. I suspect you don't know what "reflexive" means. Please review your textbook or Wikipedia and feel free to ask questions if you don't understand the explanation.
 
  • #5
reflexivity is just aRa so I guess it would be reflexive since if I use any natural number for a it will be related to itself
 
  • #6
andrew said:
reflexivity is just aRa
Yes. More accurately speaking, $R$ is reflexive if $aRa$ holds for all $a$.

andrew said:
so I guess it would be reflexive since if I use any natural number for a it will be related to itself
Yes. How to explain this in more detail? If $a\in\Bbb N$, then by definition of $R$, $aRa$ means $a\le a\le a+5$, which is true for any $a$. Now try proving reflexivity of $S$.

Concerning symmetry of $R$, what happens when $a<b$?

By the way, you can click the "Reply With Quote" button to see how formulas are typed.
 
  • #7
Evgeny.Makarov said:
Yes. More accurately speaking, $R$ is reflexive if $aRa$ holds for all $a$.

Yes. How to explain this in more detail? If $a\in\Bbb N$, then by definition of $R$, $aRa$ means $a\le a\le a+5$, which is true for any $a$. Now try proving reflexivity of $S$.

Concerning symmetry of $R$, what happens when $a<b$?

By the way, you can click the "Reply With Quote" button to see how formulas are typed.

Well when $a<b$ won't be related to a since 4 < 5 for example doesn't mean 5 < 4, but now the anti-symmetric property part of the question confuses me because if I take a as 4 again and b as 5, 4 $\ne$ 5. Would it also be sufficient to say that the relation isn't transitive as well since 1R5 for example and 5R11 but 1 isn't related to 11?
 
  • #8
andrew said:
Well when $a<b$ won't be related to a since 4 < 5 for example doesn't mean 5 < 4
Let's say it precisely. The expression $a<b$ cannot be related to $a$ because $a<b$ is not a number, and $R$ is a relation on numbers. Only a number can be related to another number via $R$. Further, 4 < 5 does not indeed imply 5 < 4, but so what? What does it have to do with $R$, which involved two inequalities (and both of them non-strict)? You have the right idea, but you need to say it precisely.

andrew said:
but now the anti-symmetric property part of the question confuses me because if I take a as 4 again and b as 5, 4 $\ne$ 5.
But is it the case that $4R5$ and $5R4$? If not, then this example does not violate antisymmetry.

andrew said:
Would it also be sufficient to say that the relation isn't transitive as well since 1R5 for example and 5R11 but 1 isn't related to 11?
It is not the case that $5R11$, but you are right that $R$ is not transitive.
 
  • #9
This is what I suggest:

Choose $a$ and $b$ from the set $\{1,2,3,4,5,6,7,8,9\}$.

This isn't all of $\Bbb N$, but it will give you an idea of which number-pairs are related.

For example, do we have $1R9$? Let's check:

Is $1 \leq 9$? Yes, so we passed the first test.

Is $9 \leq 1+5$? no, so we failed the second test, and thus the pair $(1,9)$ is not in $R$.

You might try to eliminate certain "kinds" of pairs at one fell swoop. For example, if $a > b$, can we have $aRb$?
 

Related to Proving Binary Relations: R & S

1. What is a binary relation?

A binary relation is a mathematical concept that represents the relationship or connection between two sets of objects or elements. It is typically denoted by R or S and can be thought of as a set of ordered pairs (x,y) where x is related to y in some way.

2. How can we prove that a relation R is reflexive?

To prove that a relation R is reflexive, we must show that for every element x in the set, (x,x) is a part of the relation. In other words, every element in the set must be related to itself. This can be shown by using a logical argument or by providing specific examples.

3. What is the difference between a symmetric and an antisymmetric relation?

A symmetric relation is one where if (x,y) is a part of the relation, then (y,x) is also a part of the relation. In other words, the order of the elements does not matter. On the other hand, an antisymmetric relation is one where if (x,y) is a part of the relation, then (y,x) cannot be a part of the relation unless x and y are the same element.

4. How do we prove that a relation R is transitive?

To prove that a relation R is transitive, we must show that if (x,y) and (y,z) are both part of the relation, then (x,z) must also be a part of the relation. In simpler terms, if x is related to y and y is related to z, then x must also be related to z. This can again be shown through logical arguments or by providing examples.

5. Can a relation be both symmetric and antisymmetric?

No, a relation cannot be both symmetric and antisymmetric at the same time. This is because in a symmetric relation, the order of the elements does not matter, while in an antisymmetric relation, the order does matter. Therefore, a relation can only be one or the other, not both.

Similar threads

MHB Give a set and a relation that satisfies the properties
    • Set Theory, Logic, Probability, Statistics
Replies
10
Views
2K
I Supremum of a set, relations and order
    • Set Theory, Logic, Probability, Statistics
2
Replies
35
Views
1K
MHB Satisfying S∘S=S: What Are The Conditions?
    • Set Theory, Logic, Probability, Statistics
Replies
7
Views
1K
MHB Antisymmetry & Partial Orderings - H&J Ch.2 Section 5 | Peter's Help
    • Set Theory, Logic, Probability, Statistics
Replies
3
Views
2K
MHB Q: Let S = {x, y, z } and R is a relation defined on S such that?
    • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
MHB Understanding Binary Relations: Reflexive, Symmetric, Antisymmetric & Transitive
    • Set Theory, Logic, Probability, Statistics
Replies
6
Views
2K
MHB Problem understanding relations syntax.
    • Set Theory, Logic, Probability, Statistics
Replies
1
Views
921
MHB Show that ~ is an equivalence relation
    • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
MHB Proving Reflexive, Symmetric and Transitive Properties of Relation R on P(U)
    • Set Theory, Logic, Probability, Statistics
Replies
3
Views
2K
MHB Can you explain how the law of logic was used to reach this conclusion?
    • Set Theory, Logic, Probability, Statistics
Replies
3
Views
2K

Hot Threads

Recent Insights

Back
Top