Results 1 to 3 of 3

Thread: Relations

  1. #1
    Junior Member
    Joined
    Nov 2007
    Posts
    29

    Relations

    Question:

    The relation R on {1,2,3,....} where aRb means a|b.

    Is it reflexive, is it symmetric, is it antisymmetric, is it transitive?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by oldguy View Post
    Question:

    The relation R on {1,2,3,....} where aRb means a|b.

    Is it reflexive,
    obviously, for all $\displaystyle a \in \mathbb{N}$ we have $\displaystyle a|a$, thus the relation is reflexive

    is it symmetric
    for $\displaystyle a,b \in \mathbb{N}$, if $\displaystyle a|b$ does it mean that $\displaystyle b|a$?

    is it antisymmetric
    for $\displaystyle a,b \in \mathbb{N}$, if $\displaystyle a|b$ and $\displaystyle b|a$ does it mean that $\displaystyle a = b$?

    is it transitive?
    for $\displaystyle a,b,c \in \mathbb{N}$. if $\displaystyle a|b$ and $\displaystyle b|c$, does it mean that $\displaystyle a|c$?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, oldguy!

    The relation $\displaystyle R$ on $\displaystyle \{1,2,3, \cdots\}$ where $\displaystyle a\text{R}b$ means $\displaystyle a|b.$

    Is it reflexive? .Symmetric? .Antisymmetric? .Transitive?

    For any natural numnber $\displaystyle a,\;a \div a \:=\:1$
    . . That is: .$\displaystyle a|a$
    Hence, $\displaystyle \text{R}$ is reflexive.

    If $\displaystyle a$ divides $\displaystyle b$, it does not follow that $\displaystyle b$ divides $\displaystyle a.$
    Hence, it is not symmetric.

    If $\displaystyle a|b$ and $\displaystyle b|a$, then $\displaystyle a = b.$
    Hence, $\displaystyle \text{R}$ is antisymmetric.

    If $\displaystyle a|b$ and $\displaystyle b|c$, then $\displaystyle a|c.$ . **
    Hence, $\displaystyle \text{r}$ is transitive.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    ** Proof of transitivity

    If $\displaystyle a|b$, then: .$\displaystyle b \:=\:ma$ for some integer $\displaystyle m.$

    If $\displaystyle b|c$, then: .$\displaystyle c \:=\:nb$ for some integer $\displaystyle n.$

    Then: .$\displaystyle c \:=\:nb \:=\:n(ma) \:=\:(mn)a$

    Therefore: .$\displaystyle a|c.$

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Relations and Functions - Inverse Relations Question
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Nov 13th 2011, 12:20 PM
  2. Replies: 1
    Last Post: Sep 19th 2011, 01:09 PM
  3. [SOLVED] Relations on A
    Posted in the Discrete Math Forum
    Replies: 10
    Last Post: Nov 21st 2010, 11:25 AM
  4. Relations in a set
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Sep 5th 2010, 10:03 PM
  5. relations help (3)
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Apr 18th 2010, 04:49 AM

Search Tags


/mathhelpforum @mathhelpforum