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Let $A = \mathbb Z \times ( \mathbb Z\setminus {0} )$.

A binary relation $R$ on $A$ is defined as follows: For all $(a,b),(c,d) \in A$ $$(a,b) \,R\,(c,d) \iff ad = bc$$

now how do I find if $R$ is a reflexive, symmetric, or anti-symmetric relation?

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How do you find these things? You check that the definitions hold. One by one. Case by case. –  Asaf Karagila Dec 4 '13 at 17:42
    

2 Answers 2

Reflexivity:

We need to be able to confirm that for all $(a, b) \in A$, it is true that $(a,b) \,R\,(a, b)$. Certainly we have that $ab = ba$ for all $(a, b) \in A$. $\;\large \color{green}{\checkmark}$.

Symmetry:

We need to be able to confirm that for all $(a, b), (c, d) \in A$, if $(a, b) \,R\,(c, d)$, then $(c, d) \,R\,(a, b)$. So suppose $(a, b), (c, d) \in A$, and suppose $(a, b) R (c, d).$ Then $ad = bc$, by definition. And by commutativity of multiplication, it follows that $da = cb$, and so $cb = da$. But this means precisely that $(c, d) \,R\, (a, b)$. Hence, our relation is symmetric. $\color{green}{\large \checkmark}$

Anti-symmetry:

We need to be able to confirm that for all $(a, b), (c, d) \in A$, if $(a, b) \,R\,(c, d)$ AND $(c, d) \,R\,(a, b)$, then $(a, b) = (c, d)$. Again, use the definition of the relation to show that antisymmetry fails. We need to find just one counterexample of two ordered pairs in $(a, b), (c, d) \in A$ for which $(a,b) \,R\,(c, d)$ but $(a, b) \neq (c, d)$. Let's choose $(1, 1), (3, 3).\;$ $(1, 1)\,R\,(3, 3)$ since $1\cdot 3 = 1\cdot 3$, and we already know $R$ is symmetric, so $(3, 3) \,R\,(1, 1),\,$ but clearly, $(1, 1) \neq (3, 3).\;$ Hence, since there exist two related elements which are symmetrically related, but which are not identical to one another, we know that anti-symmetry fails. $\large \require{cancel} \color{red}{\cancel \checkmark}$

Transitivity:

If you also need to prove transitivity, can you show that for all $(a,b),(c,d),(e,f)\in A$ if $(a,b)\:R\:(c,d)$ and $(c,d)\:R\:(e,f),$ then it must follow that $(a,b) \,R\,(e, f)$?

Suppose we have three arbitrary pairs: $(a,b),(c,d),(e,f)\in A$, such that $(a,b)\:R\:(c,d)$ and $(c,d)\:R\:(e,f).$ Now we need to determine whether this necessarily implies $(a, b) \,R\,(e, f).$

Since $(a,b) \,R\,(c, d),\; ad = bc \implies adf = bcf\,$

Since $\,(c, d)\,R\,(e,f),\; cf = de.$

$$adf = bcf,\; cf = de \implies adf= bde \implies af = be \implies (a, b)\,R\,(e, f)$$ It does indeed hold, and thus the relation is transitive. $\large \color{green}{\checkmark}$


In the end, you'll find that your relation is an equivalence relation, because it is reflexive, symmetric, and transitive. But it does not define a partial order, which requires that it be reflexive AND antisymmetric AND transitive.

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Wow, needs a TU just for the typing! +1 –  Amzoti Dec 5 '13 at 1:35
    
Reflexive, symmetric and antisymmetric: identity. –  egreg Dec 6 '13 at 18:22

Reflexivity and symmetry are completely straightforward and short proofs. Since it is symmetric, then it cannot be anti-symmetric unless it is the equality relation (which it is not, as you should be able to confirm).

I assume that you wish to verify that it is a transitive relation, instead (as you look like you're about to build the rational numbers). Suppose that $(a,b),(c,d),(e,f)\in A$ such that $(a,b)\:R\:(c,d)$ and $(c,d)\:R\:(e,f).$ That means that $a,c,e$ are integers and $b,d,f$ are non-zero integers, where $ad=bc$ and $cf=de.$ To show that $(a,b)\:R\:(e,f),$ we must demonstrate that $af=be.$ Well, since $ad=bc,$ then $adf=bcf,$ and since $cf=de,$ then $adf=bde.$ What can we then conclude, given that $d\ne 0?$

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