Equivalence - Proofs (University)

Question

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Comments

Could you provide a higher resolution photo please?

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is this better

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Much better thank you.

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hi you were working on an answer what happened

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When I continued working on your problem I realised I didn’t know enough about the subject to continue, and I wasn’t sure that the answers I had provided already were good enough or relevant to be valid. So I deleted my answer to avoid confusion. However, I did find something interesting which may help for part (2). See the next comment.

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(2) Let A={a∈ℤ} (A is the set of all integers), f(a)=a⁶, and f:A→B. That is, a∈A maps to b∈B (such that b=f(a)). The mapping ensures that for every element in A there is one and only one element in B, but two different elements of A can map to the same element in B (for example, -2 and 2 both map to 64). ᴮ~ is ∀b∈B, b modulo 9 ∈{0,1}, that is, f(a₁) ᴮ~ f(a₂). The equivalence relation in B does not require any equivalence relation in A, so that any equivalence relation in A is independent of that in B. It is the function that permits the equivalence relation in B.

Answer 1

(1)

a ¹~ b⇒

(a) a ¹~ a (reflexivity)

(b) a ¹~ b and b ¹~ a (symmetry)

(c) a ¹~ b and b ¹~ c⇒a ¹~ c (transitivity);

a ²~ b⇒

(a) a ²~ a (reflexivity)

(b) a ²~ b and b ²~ a (symmetry)

(c) a ²~ b and b ²~ c⇒a ²~ c (transitivity).

³~ = ¹~ ∩²~, ⁴~ = ¹~ ∪²~.

(a) a ³~ a⇒

     (a ¹~ a)∩(a ²~ a)⇒TRUE∩TRUE=TRUE (reflexivity)

(b) a ³~ b and b ³~ a⇒

     ((a ¹~ b)∩(a ²~ b))∩((b ¹~ a)∩(b ²~ a))=

     (TRUE∩TRUE)∩(TRUE∩TRUE)=TRUE∩TRUE=TRUE (symmetry)

(c) [a ³~ b and b ³~ c⇒a ³~ c]⇒

      [((a ¹~ b)∩(a ²~ b))∩((b ¹~ c)∩(b ²~ c))⇒((a ¹~ c)∩(a ²~ c))];

      [(TRUE∩TRUE)∩(TRUE∩TRUE)⇒(TRUE∩TRUE)]⇒

      [(TRUE∩TRUE)⇒(TRUE∩TRUE)].

Since the statement and its implication are identical, there is no inconsistency, so ³~ is transitive. All three requirements have been met, so ³~ is an equivalence relation.

(a) a ⁴~ a⇒

     (a ¹~ a)∪(a ²~ a)⇒TRUE∪TRUE=TRUE (reflexivity)

(b) a ⁴~ b and b ⁴~ a⇒

     ((a ¹~ b)∪(a ²~ b))∩((b ¹~ a)∪(b ²~ a))=

     (TRUE∪TRUE)∩(TRUE∪TRUE)=TRUE∩TRUE=TRUE (symmetry)

(c) [a ⁴~ b and b ⁴~ c⇒a ⁴~ c]⇒

      [((a ¹~ b)∪(a ²~ b))∩((b ¹~ c)∪(b ²~ c))⇒((a ¹~ c)∪(a ²~ c))];

      [(TRUE∪TRUE)∩(TRUE∪TRUE)⇒(TRUE∩TRUE)]⇒

      [(TRUE∩TRUE)⇒(TRUE∩TRUE)].

Since the statement and its implication are identical, there is no inconsistency, so ⁴~ is transitive. All three requirements have been met, so ⁴~ is an equivalence relation.

(2)

Let B={1,8,27,...,k³}, b=f(a)=a³.

b ᴮ~ b⇒[(b mod 9)∈{0,1}]=[(b mod 9)∈{0,1}] (reflexive).

When k=3n, k³=27n³ (∀n∈ℤ) b mod 9=0;

when k=3n±1, k³=(3n)³±3(3n)²+3(3n)±1=27n³±27n²+9n±1.

This can be written 9n(3n²±3n+1)±1 or 9P±1. (9P±1)mod 9=1 or 9-1=8. Hence ∀b∈B, ᴮ~ is (b mod 9)∈{0,1,8}.

b₁ ᴮ~ b₂ and b₂ ᴮ~ b₁ for (b₁,b₂)∈B,  and for (b₁,b₂,b₃)∈B:

b₁ ᴮ~ b₂ and b₂ ᴮ~ b₃⇒b₁ ᴮ~ b₃ (symmetry and transitivity).

Let A={-k,...,-2,-1,0,1,2,3,...,k}. But ∀a∈A is not required to obey an equivalence relationship dependent on that in B, so ᴬ~ does not need to be an equivalence relation.

(3)

Let A be the set of all married women in a town and B be the set of all married man. ᴬ~ relates the elements of A (all married female) and ᴮ~ relates those of B (all married male). The product X=A×B is the set of married couples (a,b) ∀a∈A and ∀b∈B. But of all the possible pairs (a,b) only one represents a couple married to one another. So, if ˣ~ represents marriages, this equivalence relation cannot apply to all pairs in X. Marriages would be a subset of X and such an equivalence relation would apply.