1. Let - be a relation over Zx (Z\ {0}) defined by (a,b)-(c,d) iff ad = bc. Prove that this is an equivalence relation and describe the resulting quotient set
Need help with this Foundations of Mathematics homework problem. I received feedback from my professor for this problem and he said Quotient set description needs considerably more detail. Hint: Can you relate the quotient set to a well-known number system? Below is my work for this problem. The image uploaded is the homework problem. Make sure handwriting is neat and readable.
- K = Ζ x (Ζ ∖ {0}) is the provided set which a relation ∼ is determined by (a,b) ∼ (c,d) iff ad = bc.
We will prove that ∼ is an equivalence relation. If you want to prove that ∼ is an equivalence relation, we are required to prove that ∼ is reflexive , symmetric, and transitive.
(i) Reflexive: Let (a,b)∈K, we are required to prove that (a,b) ∼ (a,b).
Now since the commutative under multiplication is Ζ, we will be able to write
ab = ba (where a,b ∈ Ζ∖{0})
⇒ (a,b) ∼ (b,a)
For this reason, this proves that ∼ is reflexive.
(ii) Symmetric: Let (a,b) ∼ (c,d), where (a,b) and (c,d) are elements of K, we are required to prove that (c,d) ∼ (a,b).
Now, (a,b) ∼ (c,d)
⇒ ad = bc
With Z being the commutative under multiplication, we can therefore write it as
da = cb (since ad = da and bc = cb) or cb = da
⇒ (c,d) ∼ (a,b)
For this reason, this proves that ∼ is symmetric.
(iii) Transitive: Let (a,b), (c,d) and (e,f) be in K such that
(a,b) ∼ (c,d) and (c,d) ∼ (e,f)
Then we are required to prove that
(a,b) ∼ (e,f)
Now we have
(a,b) ∼ (c,d) ⇒ ad = bc
and (c,d) ∼ (e,f) ⇒ cf = de
Here, ef = de gives us c = de/f (As f ≠ 0)
Using this in
ad = bc
⇒ ad = b(de/f)
⇒ a = be/f (cancel d on both sides and note that d ≠ 0)
⇒ af = be
⇒ (a,b) ∼ (f,e)
Earlier in the problem, ∼ is reflexive, which allows us to now write
(a,b) ∼ (f,e) ∼ (e,f)
that is, (a,b) ∼ (e,f)
⇒ ∼ is transitive.
For this reason, it follows that ∼ is an equivalence relation.
Now we are required to explain the subsequent quotient set. Relation ∼ that is on set K is the quotient set, which is described as the set of all equivalence classes of relation ∼.
Ex: x = (a,b)∈K
We define
[x] = [(a,b)] = {y∈K ∣ y∼K}
= {y = (c,d)∈K ∣ (c,d) ∼ (a,b)}
= {y = (c,d)∈K ∣ (a,b) ∼ (c,d)}
= {y = (c,d)∈K ∣ ad = bc}
Then the quotient set is defined as
K∕∼ = {[x] ∣ x∈K}
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Need help with this Foundations of Mathematics homework problem. I received feedback from my professor for this problem and he said Quotient set description needs considerably more detail. Hint: Can you relate the quotient set to a well-known number system? Below is my work for this problem. The image uploaded is the homework problem. Make sure handwriting is neat and readable.
- K = Ζ x (Ζ ∖ {0}) is the provided set which a relation ∼ is determined by (a,b) ∼ (c,d) iff ad = bc.
We will prove that ∼ is an equivalence relation. If you want to prove that ∼ is an equivalence relation, we are required to prove that ∼ is reflexive , symmetric, and transitive.
(i) Reflexive: Let (a,b)∈K, we are required to prove that (a,b) ∼ (a,b).
Now since the commutative under multiplication is Ζ, we will be able to write
ab = ba (where a,b ∈ Ζ∖{0})
⇒ (a,b) ∼ (b,a)
For this reason, this proves that ∼ is reflexive.
(ii) Symmetric: Let (a,b) ∼ (c,d), where (a,b) and (c,d) are elements of K, we are required to prove that (c,d) ∼ (a,b).
Now, (a,b) ∼ (c,d)
⇒ ad = bc
With Z being the commutative under multiplication, we can therefore write it as
da = cb (since ad = da and bc = cb) or cb = da
⇒ (c,d) ∼ (a,b)
For this reason, this proves that ∼ is symmetric.
(iii) Transitive: Let (a,b), (c,d) and (e,f) be in K such that
(a,b) ∼ (c,d) and (c,d) ∼ (e,f)
Then we are required to prove that
(a,b) ∼ (e,f)
Now we have
(a,b) ∼ (c,d) ⇒ ad = bc
and (c,d) ∼ (e,f) ⇒ cf = de
Here, ef = de gives us c = de/f (As f ≠ 0)
Using this in
ad = bc
⇒ ad = b(de/f)
⇒ a = be/f (cancel d on both sides and note that d ≠ 0)
⇒ af = be
⇒ (a,b) ∼ (f,e)
Earlier in the problem, ∼ is reflexive, which allows us to now write
(a,b) ∼ (f,e) ∼ (e,f)
that is, (a,b) ∼ (e,f)
⇒ ∼ is transitive.
For this reason, it follows that ∼ is an equivalence relation.
Now we are required to explain the subsequent quotient set. Relation ∼ that is on set K is the quotient set, which is described as the set of all equivalence classes of relation ∼.
Ex: x = (a,b)∈K
We define
[x] = [(a,b)] = {y∈K ∣ y∼K}
= {y = (c,d)∈K ∣ (c,d) ∼ (a,b)}
= {y = (c,d)∈K ∣ (a,b) ∼ (c,d)}
= {y = (c,d)∈K ∣ ad = bc}
Then the quotient set is defined as
K∕∼ = {[x] ∣ x∈K}
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