MHB Proving Binary Relations: R & S

[andrew1]

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?

 

[Evgeny.Makarov]

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?

 

[andrew1]

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?

 

[Evgeny.Makarov]

Sorry about the latex commands thing, not sure how it works

Even after I said that you can surround formulas with dollar signs?

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.

 

[andrew1]

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

 

[Evgeny.Makarov]

reflexivity is just aRa

Yes. More accurately speaking, $R$ is reflexive if $aRa$ holds for all $a$.

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.

 

[andrew1]

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?

 

[Evgeny.Makarov]

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.

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.

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.

 

[Deveno]

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 $aRb$?