Need help with equivalence relation question

[tacit isle]

I am having trouble understanding what the ~ symbol means in this context and (m, n) ∼ (p, q) iff m + q = n + p. Also part b.

[charred basin]

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[left trench]

Well it’s just a notation for a binary relation when two elements a and b are in relation we write a~b

Here you have to prove that it’s an equivalence relation so it’s so it’s a reflexive symmetrical and transitive relation

[tacit isle]

no but i think i can do that by myself but I still dont understand the (m, n) ∼ (p, q) iff m + q = n + p part

[keen spire]

that's how the relation is defined

e.g (1,2) ~ (2,3)

[left trench]

Well you can look at it as a definition (a,b) is in relation with (e,f) if and only if a+f=b+e

[keen spire]

what is there to understand, is it the ~ symbol that throws you off? the iff?

the pairs (m,n) and (p,q) are in relation if and only if m+q = n+p

the condition on the right is the defining property

[tacit isle]

ok so i think i understand it now

[keen spire]

good, now verify reflexivity, transitivity and symmetricity

[tacit isle]

reflexive: (m, n) ∼ (m, n)

symmetry: (m, n) ∼ (p, q) implies (p, q) ∼ (m, n)

transitive: (m, n) ∼ (p, q) and (p, q) ∼ (r, s), then m + q = n + p and p + s = q + r

[keen spire]

what's the condition to check here?

[keen spire]

similarly, here

[tacit isle]

commutative

[keen spire]

that's the reason why equality holds

but which equality?

what's the equality when you replace p,q with m,n

[tacit isle]

m + n = m + n

[keen spire]

m+n = n+m, keep the order, easier to follow that way

[tacit isle]

ok. I am still lost at part b tho

[keen spire]

by definition the equivalence class of (0,0) contains all pairs (x,y) for which (0,0) ~ (x,y)

what condition does this set on x,y?

@tacit isle

[tacit isle]

x = y?

[keen spire]

yes

so, what are the elements of the equivalence class of (0,0)?

[tacit isle]

x = y

[keen spire]

give a proper answer

[tacit isle]

im not sure why x = y doesnt qualify as a proper answer

[keen spire]

an equivalence class is a set

x=y is not a set

what are the elements of [(0,0)]

dont try this sh* with your grader either

one-way ticket to F-land

[tacit isle]

so something along the lines of {x| x =y}

[keen spire]

a direct product is a set of ordered pairs

[tacit isle]

(X x Y)?

[keen spire]

revise definitions, please

the relation consists of pairs such as this

$$ [(0,0)] = {(m,n) \mid (m,n)\sim (0,0)} = {(m,n) \mid m=n} $$

[native fiber]

aL

[keen spire]

this is nonsense

[keen spire]

$$ \Rightarrow [(0,0)] = {(m,m) \mid m\in\omega} $$

[native fiber]

aL

[tacit isle]

ah ok i get it

Thank you @keen spire

[tired bloom]

@tacit isle has given 1 rep to @keen spire

[tacit isle]

+close