Is Relation S Reflexive, Symmetric, and Transitive?

  • Thread starter Thread starter kingstar
  • Start date Start date
  • Tags Tags
    Set
kingstar
Messages
38
Reaction score
0

Homework Statement



a) Consider the relation S de fiend on the set {t : t is a person} such that xSy holds exactly if
person x is taller than y. Determine if the relation S is reflexive, symmetric and transitive.
Is the relation S an equivalence relation?

Homework Equations


Recall that a relation R de ned on a set A is reflexive if for all x 2 A xRx.
Recall that a relation R de ned on a set A is symmetric if for all x 2 A and y 2 A the xRy implies
yRx.
Recall that a relation R e ned on a set A is transitive if for all x; y; z in A, both xRy and yRz
holds, then xRz holds as well.
Finally recall that a relation R is an equivalence relation if its reflexive, symmetric and transitive.


The Attempt at a Solution



As far as i can see the set is not symmetric or reflexive but I'm not 100% on transitive...
It would be transitive if x > y and y > z then x > z holds...but we aren't given any information on y > z?

So would this set be transitive?
 
Physics news on Phys.org
Yes the relationship is not symmetric or reflexive, but it IS transitive.

If ##xRy## holds, then ##x > y##, if ##yRz## holds, then ##y > z##. So person ##x## is taller than person ##y## and person ##y## is taller than person ##z##.

It must be the case that person ##x## must be taller than person ##z##.
 
Zondrina said:
Yes the relationship is not symmetric or reflexive, but it IS transitive.

If ##xRy## holds, then ##x > y##, if ##yRz## holds, then ##y > z##. So person ##x## is taller than person ##y## and person ##y## is taller than person ##z##.

It must be the case that person ##x## must be taller than person ##z##.

Yes, i agree with what you said...but the question does not mention that y > z
 
kingstar said:
Yes, i agree with what you said...but the question does not mention that y > z

According to your definition (with some slight corrections):

Recall that a relation ##R## defined on a set ##A## is transitive if ##\forall x, y, z \in A##, if both ##xRy## and ##yRz## hold, then ##xRz## holds as well.

Notice I added the word 'if'. If you assume they hold, does the conclusion still hold?
 
Yeah then it holds. The definition is provided as part of the information the exam question but i'll just assume it in the exam as well. Thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top