Equivalence Relation with multiples

Question

How can I prove the equivalence of this relation, and how can I calculate the equivalence class of (4,8)?

On the set the relation R is definded by (a,b)R(c,d) ⇔ ad=bc. Find out if this is an equivalence relation and, if so, calculate the equivalence class of the element (4,8).

I tried to prove the Reflexive property first. I say a = d, then $d^2$ = bc , second member doesn't change, I think this is a not reflexive expression. Is right my assumption? If it is right, the relation has not equivalence!!

Answer 1

In order to show that $R$ is an equivalence relation (which it happens to be), you need to show that it is reflexive, symmetric, and transitive.

Answer 2

If division is allowed, then this is easy: $ad=bc$ if and only if $a/b=c/d$, and going on from there it's easy.

But in a way that's cheating, because it assumes that the rules of arithmetic with fractions make sense, and the point of proving that this is an equivalence relation would be to use that fact in order to prove that the rules of arithmetic with fractions make sense.

You've got $\dfrac a b = \dfrac c d$ and $\dfrac c d = \dfrac e f$ but you want to avoid the easy way using fractions since that would defeat the purpose. So first do it working with fractions and then modify that.

Since $\dfrac a b = \dfrac c d$, use a common denominator, getting $\dfrac{ad}{bd}=\dfrac{bc}{bd}$, and thus $ad=bc$, and similarly $\dfrac{cf}{df}=\dfrac{de}{df}$, so $cf = de$. This should all suggest thinking about a denominator common to all three. That would be $bdf$.

So you have $ad=bc$ and $cf=de$.

Multiply both sides by $f$ in the first equality and by $b$ in the second.

Then you have $adf=bcf$ and $bcf=bde$.

Consequently $adf=bde$. If $d\ne0$, this implies $af=be$, so that $(a,b)\sim(e,f)$.

I'll leave you with two exercises:

• Figure out what to do in the case where $d=0$.
• Think about whether this can be done without considering separately the case where $d=0$ and the case where $d\ne 0$.

Answer 3

We want to show that this is an equivalence relation if possible

Reflexive Property: We see that $(a,b)\sim(a,b)$ where $a,b\in\mathbb{R}$ since $ab=ba$. This follows from the commutative property for real numbers.

Symmetric Property: We want to show if $(a,b)\sim(c,d)$. then $(c,d)\sim(a,b)$ where $a,b,c,d\in\mathbb{R}$. From this assumption we see that $ad=bc$. By the commutative property we get $cb=da$. Thus $(c,d)\sim(a,b)$.

Transitive Property: We want to show if $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$ where $c,d\neq 0$ then $(a,b)\sim(e,f)$. By assumption we get $ad=bc$ and $cf=de$. We want to show $af=be$. We see that $adcf=bcde\implies af=be$. If $c=0$ or $d=0$ then equality follows automatically.

Thus this is an equivalence relation.

And for the last part you need to describe the equivalence relation in terms of a set(after all these are sets). That is $[4,8]\equiv\{(a,b)| (a,b)\sim(4,8):8a=4b, a,b\in\mathbb{R}\}$. This is just the outline. I leave it up to you to advance this idea further.