# Relation, reflexive, symmetric and trasitive

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• September 27th 2009, 08:14 AM
Madspeter
Relation, reflexive, symmetric and trasitive
The given relation S on real numbers, R, is given by

xSy <=> x*y > 0

Determine wether S is reflexive, symmetric and transitive.

How can I solve this assignment by proof or contradiction?
• September 27th 2009, 08:30 AM
Swlabr
Quote:

Originally Posted by Madspeter
The given relation S on real numbers, R, is given by

xSy <=> x*y > 0

Determine whether S is reflexive, symmetric and transitive.

How can I solve this assignment by proof or contradiction?

Note that reflexivity fails. Can you find the single element which this does not hold for? This is your contradiction.

However, the other two do hold. Symmetry is pretty simple (it comes from the commutativity of * over $\mathbb{R}$). For transitivity, we can use the fact that $x*y>0 \Leftrightarrow x, y > 0 \text{ or } x, y <0$. Two cases:

(1) $xSy, ySz, x<0$ $\Rightarrow y<0 \Rightarrow z<0 \Rightarrow x*z>0 \Rightarrow xSz$.

(2) is identical, but we take $x>0$.
• September 27th 2009, 08:53 AM
Madspeter
The element which you speak of must be the fact that either x or y can be chosen as 0 - giving us a false statement. 0 is not bigger than 0.
Is this how you mean reflexitvity fails?

suppose in your 1) that x = 0 instead of x < 0 - wouldnt this give us a non-transitive relation?
• September 27th 2009, 09:06 AM
Plato
Quote:

Originally Posted by Madspeter
The element which you speak of must be the fact that either x or y can be chosen as 0 - giving us a false statement. 0 is not bigger than 0.
Is this how you mean reflexitvity fails?

wouldnt this give us a non-transitive relation?

That is correct.

Rethink the transitive part. If $x~\&~y$ are related that is saying that they are not zero and both have the same sign.
• September 27th 2009, 09:28 AM
Madspeter
Quote:

Originally Posted by Plato
That is correct.

Rethink the transitive part. If $x~\&~y$ are related that is saying that they are not zero and both have the same sign.

of course

if xSy <=> x*y > 0

then either x or y can't be 0, cause they have to > 0, because they are related.

----

this, however, does not hold for the reflexivity, where x can be 0.
right?

Do you think you have timefor one more task of relation? if not, have a nice evening and of course thanks for your help with xSy, it brought a bit more insight to relations.

Here goes.

The relation T on positive R+ = {x E R | x > 0 } is given by:

xTy <=> [x E Q and x*y E Q]

Determine reflexivity, symmetry and transivity. Use proof and contradiction.

Im not sure wether to be confused by the extra amount of structere in this relation or to recognize this as very similar to the prior assignment.

I'd go for not reflexive since either x or y can be 0

The relation symmetric (commutitave rule, (x*y)=(y*x)

though for transitive relation, we can't use your 1) and have to proof xRz by having x>0

am I far off?
• September 27th 2009, 10:29 AM
Swlabr
Quote:

Originally Posted by Madspeter
this, however, does not hold for the reflexivity, where x can be 0.
right?

No - the relation is both symmetric and transitive because 0 is related to nothing. If we have $xSy$ then neither $x$ nor $y$ is 0 (as $x*0=0 \forall x \in \mathbb{R}$).

As for your other problem, you should note that for $x, y \in \mathbb{Q} \Rightarrow x*y \in \mathbb{Q}$, and also that $0 \in \mathbb{Q}$. You can use this result to show that your relation is, in fact, an equivalence relation.
• September 27th 2009, 10:56 AM
Madspeter
I need to clarify a couple of things.

quote
No - the relation is both reflexive and transitive because 0 is related to nothing. If we have http://www.mathhelpforum.com/math-he...c24432c4-1.gif then neither http://www.mathhelpforum.com/math-he...155c67a6-1.gif nor http://www.mathhelpforum.com/math-he...904f345d-1.gif is 0 (as http://www.mathhelpforum.com/math-he...e6aa0532-1.gif).
quote

I thought you meant xSy = x*y > 0 failed to be reflexive?
• September 27th 2009, 10:59 AM
Plato
I think the way you defined $S$ is confusing.
I read it to be a relation defined on $\mathbb{R}$ as $xSy$ if and only $x\cdot y>0$.
In that case, the relation $S$ is not reflexive, $(0,0)\not\in S$ because $0\cdot 0 \not >0$.
• September 27th 2009, 11:13 AM
Swlabr
Quote:

Originally Posted by Madspeter
I need to clarify a couple of things.

quote
No - the relation is both reflexive and transitive because 0 is related to nothing. If we have http://www.mathhelpforum.com/math-he...c24432c4-1.gif then neither http://www.mathhelpforum.com/math-he...155c67a6-1.gif nor http://www.mathhelpforum.com/math-he...904f345d-1.gif is 0 (as http://www.mathhelpforum.com/math-he...e6aa0532-1.gif).
quote

I thought you meant xSy = x*y > 0 failed to be reflexive?

That should read "is both symmetric and transitive", not reflexive. I've edited my previous post. I hope that stops any confusion...?
• September 27th 2009, 11:16 AM
Madspeter
Quote:

Originally Posted by Plato
Deleted

this is very clear!

as goes for relation T:

quote
The relation T on positive R+ = {x E R | x > 0 } is given by:

xTy <=> [x E Q and x*y E Q]
quote

Since x has to be bigger than 0, then my guess is that T is reflexive.
x is related to x since any x*x > 0.
• September 27th 2009, 11:19 AM
Madspeter
Quote:

Originally Posted by Swlabr
That should read "is both symmetric and transitive", not reflexive. I've edited my previous post. I hope that stops any confusion...?

I see now (Nod), thanks.