Please do try the questions by yourself before resorting to these hints!
- {(2,z), (3,x), (3,z)}.
- R ∪ {(3,3)}.
R ∪ {(2,4)}.
2R3 & 3R2 but 2 ≠ 3.
R ∪ {(3,3), (4,3)}.
- 2km and 2kmn.
- 23.3 = 512 and 5.
- (y,x) ∈ (R∩S)-1
⇔ (x,y) ∈ R∩S
⇔ (x,y) ∈ R ∧ (x,y) ∈ S ⇔ (y,x) ∈ R-1 ∧ (y,x) ∈ S-1
⇔ (y,x) ∈ R-1∩S-1.
Similarly.
- A = {1, 2} and R = {(1,2), (2,1)}, suitably extended for general n.
A×A.
- R is reflexive and R ⊆ R ∪ S ⊆ t(R∪S), so that is reflexive too.
(x,y) ∈ t(R∪S) ⇒ ∃ x0 = x, x1, x2, ... xn = y with (xi,xi+1) ∈ R∪S for 0 ≤ i < n.
If (xi,xi+1) ∈ R then (xi+1,xi) ∈ R and if (xi,xi+1) ∈ S then (xi+1,xi) ∈ S, so (xi+1,xi) ∈ R∪S.Hence (y,x) ∈ t(R∪S), so that is symmetric.
Clearly t(R∪S) is transitive, so it is an equivalence relation.
Moreover, any equivalence relation containing R∪S must contain t(R∪S) so that is the smallest such.
- r(R) - treat 1 as a prime.
s(R) - x is a multiple or divisor of y by a prime amount. t(R) - x is a strict factor of y.
- Yes, yes, no (t(s(R)) is reflexive for any elements related to anything at all, but s(t(R)) need not be).
Yes, yes, no.
- t(s(r(R))).
Divisibility order. No - symmetry precludes anti-symmetry in general.
- Diagonal order and lexicographic order.
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