# Math Help - Abstract Algebra - Equivalnce Relation

1. ## Abstract Algebra - Equivalnce Relation

Let R be an integral domain.

Let Q = { (a, b) | a, b ∈ R and b 0 }

Deﬁne a relation ∼ on Q by (a, b) ∼ (c, d) if and only if ad = bc.

Prove that ∼ is an equivalence relation (that is, ∼ is reﬂexive, symmetric, and transitive).

So for reﬂexivity, for instance, you must prove that (a, b) ∼ (a, b).

Would this be correct:

1.) For all (a, b) ∈ Q, we have (a, b) ~ (a, b) since ab = ba. So ~ is reflexive.

2.) If (a, b), (c, d) ∈ Q and (a, b) ~ (c, d), then

da = cb
cb = da
(c, d) ~ (a, b)
So, ~ is symmetric

3.) If (a, b), (c, d), (e, f) ∈ Q and (a, b) ~ (c, d), (c, d) ~ (e, f), then

ad = bc and cf = de

adf = bcf and bcf = bde

daf = dbe

d(af - be) = 0

From (c, d) ∈ Q, we know that d is not equal to 0, since R has no zero divisors, we obtain af - be = 0. Then, af = be, so (a, b) ~ (e, f) and ~ is transitive

So ~ is an equivalnce relation on Q

2. ## Re: Abstract Algebra - Equivalnce Relation

probably a good start would be looking up what reflexive, symmetric, and transitive relations are

3. ## Re: Abstract Algebra - Equivalnce Relation

Originally Posted by romsek
probably a good start would be looking up what reflexive, symmetric, and transitive relations are

4. ## Re: Abstract Algebra - Equivalnce Relation

Originally Posted by AwesomeHedgehog
Let R be an integral domain.

Let Q = { (a, b) | a, b ∈ R and b 0 }

Deﬁne a relation ∼ on Q by (a, b) ∼ (c, d) if and only if ad = bc.
Prove that ∼ is an equivalence relation (that is, ∼ is reﬂexive, symmetric, and transitive).
So for reﬂexivity, for instance, you must prove that (a, b) ∼ (a, b).

1.) For all (a, b) ∈ Q, we have (a, b) ~ (a, b) since ab = ba. So ~ is reflexive.
correct

2.) If (a, b), (c, d) ∈ Q and (a, b) ~ (c, d), then

da = cb
cb = da
(c, d) ~ (a, b)
So, ~ is symmetric
not quite.

$(a,b)\sim (c,d) \Rightarrow ac=bd$

show $(c,d)\sim (a,b) \Rightarrow ac=db$ is True

3.) If (a, b), (c, d), (e, f) ∈ Q and (a, b) ~ (c, d), (c, d) ~ (e, f), then

ad = bc and cf = de
no. look at your relation again