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

ad = bc

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

Re: Abstract Algebra - Equivalnce Relation

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

Re: Abstract Algebra - Equivalnce Relation

Quote:

Originally Posted by

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

I posted my answer above.

Re: Abstract Algebra - Equivalnce Relation

Quote:

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

Quote:

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

ad = bc

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

Quote:

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