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On the set of integers, let be related to precisely when x ≠ y

  1. Is this Reflexive?
  2. Is this Symmetric?
  3. Is this Transitive?

I'm also wondering if it can be multiple? I assume it can maybe be two but maybe not all 3.

To my understanding:

  • Reflexive is when each element is related to itself, I am not sure how to apply that to x ≠ y? (Edit: If x = 3 and y = 3, then x ≠ y, so it can't be reflexive as it would be an incorrect statement, so for not equals to it can never be reflexive from what I studied going back over notes)

  • Symmetric is when x is related to y, it implies that y is related to x (which may be fitting here as x is related to y when they don't equal each other?)

  • Transitive: When x is related to y, and y is related to z, then x is related to z (Not applicable here? Unsure)

I'm not sure if it's reflex as x ∈ Z and y ∈ Z (both are related to the set of integers), it could be symmetric as they are related when x ≠ y is the same as being related when y ≠ x, then I'm not sure of transitive.

CantCodeForShit
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  • $x\ne x$ does not hold. Hence the relation is not reflexive. It is neither transitive since $x\ne y$ and $y\ne z$ does not imply $x\ne z$. But it is symmetric. – Peter Oct 13 '21 at 12:25
  • @Peter Thankyou for your response! I literally just refreshed after editing the reflexive part and hopefully I got the right understanding of why it can't be reflexive (as if y is the same value as x, then its incorrect) and also what you said x not being equal to x makes no sense either. – CantCodeForShit Oct 13 '21 at 12:28
  • $(1)$ The relation is symmetric because $x\ne y$ implies $y\ne x$. $(2)$ For the reflexive part, you verify $x$~$x$ (here $x\ne x)$ – Peter Oct 13 '21 at 12:30
  • @Peter Did you want to make a longer answer incorporating your two comments + a little bit more info so I can accept it as an answer? Your help has been amazing! :) – CantCodeForShit Oct 13 '21 at 12:35
  • As an aside, not only is the given relation not reflexive (meaning there is at least one example of an element who is not related to itself)... an even stronger statement can be made here... that it is irreflexive (meaning that all elements are examples of elements who are not related to themselves). – JMoravitz Oct 13 '21 at 13:13
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    "I'm also wondering if it can be multiple? I assume it can maybe be two but maybe not all 3." Relations may be any combination of these three properties, including all three. Those relations who are all three (reflexive, symmetric, and transitive) are of particular importance in maths and are called "Equivalence Relations." – JMoravitz Oct 13 '21 at 13:17
  • Is the question "Can a relation satisfy any combination of these three properties?" If so, this is a duplicate. On the other hand, if you are specifically interested in the relation ${(x,y) : x\ne y}$, that is a distinct question... – Xander Henderson Oct 13 '21 at 14:01

1 Answers1

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You are correct, the relation is not relfexive.

Now, it is time to formally prove that.

To prove that it is not, you must prove that the statement "$\neq$ is a reflexive relation" is false.

First we use the definition of reflexivity to rewrite the above statement into:

$$\forall x\in Z: x\neq x$$

Now, we must prove the above statement is false. Since the statement is of the type "$\forall x\in X: P(x)$", it is enough to find one value of $x$ such that $P(x)$ is not true (this value is then called the *counterexample). In your case, taking $x=3$ is perfectly OK, because $3\neq 3$ is false.

Alternatively, you could just prove the negation of the statement. The negation is

$$\exists x\in Z: x=x$$

this statement can be proven, because $x=3$ satisfies the relation $x=x$.

For symmetry, you are correct that the relation is symmetric.

You can do this by proving the statement:

$$\forall x,y\in Z: x\neq y \implies y\neq y$$

Formally, can prove any statement of the type $\forall x,y\in A: P(x,y)\implies Q(x,y)$ by:

  • Taking any two values $x,y\in A$
  • Assuming $P(x,y)$ is true
  • From that, proving $Q(x,y)$ must also be true.

Now, try it on your case!

For transitivity, if you think the relation is not transitive, then try to find some couterexample. Once you do, follow the steps I outlined above to formally prove the relation is not transitive.

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5xum
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