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?

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- Sep 27th 2009, 08:14 AMMadspeterRelation, 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? - Sep 27th 2009, 08:30 AMSwlabr
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 $\displaystyle \mathbb{R}$). For transitivity, we can use the fact that $\displaystyle x*y>0 \Leftrightarrow x, y > 0 \text{ or } x, y <0$. Two cases:

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

(2) is identical, but we take $\displaystyle x>0$. - Sep 27th 2009, 08:53 AMMadspeter
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? - Sep 27th 2009, 09:06 AMPlato
- Sep 27th 2009, 09:28 AMMadspeter
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.

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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? - Sep 27th 2009, 10:29 AMSwlabr
No - the relation is both symmetric and transitive because 0 is related to nothing. If we have $\displaystyle xSy$ then neither $\displaystyle x$ nor $\displaystyle y$ is 0 (as $\displaystyle x*0=0 \forall x \in \mathbb{R}$).

As for your other problem, you should note that for $\displaystyle x, y \in \mathbb{Q} \Rightarrow x*y \in \mathbb{Q}$, and also that $\displaystyle 0 \in \mathbb{Q}$. You can use this result to show that your relation is, in fact, an equivalence relation. - Sep 27th 2009, 10:56 AMMadspeter
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? - Sep 27th 2009, 10:59 AMPlato
I think the way you defined $\displaystyle S$ is confusing.

I read it to be a relation defined on $\displaystyle \mathbb{R}$ as $\displaystyle xSy$ if and only $\displaystyle x\cdot y>0$.

In that case, the relation $\displaystyle S$ is not reflexive, $\displaystyle (0,0)\not\in S$ because $\displaystyle 0\cdot 0 \not >0$. - Sep 27th 2009, 11:13 AMSwlabr
- Sep 27th 2009, 11:16 AMMadspeter
- Sep 27th 2009, 11:19 AMMadspeter