Question

Define the relation ~ on \(Z \times (Z \backslash {0})\) by \((a,b)~(c,d) \leftrightarrow ad=bc\). Prove that ~ is an equivalence relation. (Note that elements of this relation are ordered pairs of ordered pairs)

Answer 1

hold up getting my definitions in a bit

Answer 2

looks a equivalence class for a fraction

Answer 3

looks *like*...

Answer 4

for (a,b) ~ (a,b) because ab = ab

Answer 5

wanna try symmetry?

Answer 6

hold on

Answer 7

Definition 6.2.3 states that \(R\) is an equivalence relation if \(R\) is reflexive, symmetric, and transitive. \\

R is reflexive if \((\forall x \in S)(x,x) \in R]\)\\

R is symmetric if \((\forall x,y \in S)[(x,y) \in R \rightarrow (y,x) \in R]\)\\

R is transitive if \((\forall x, y, z \in S)[((x,y) \in R \land (y,z) \in R) \rightarrow (x,z) \in R]\)\\

Answer 8

so if we let x = (a,b) then for reflexive we have (\forall (a,b) \in S)((a,b),(a,b)) \in R]

Answer 9

UGH! os why!

Answer 10

here is the real question
what are these things? look familiar?

Answer 11

no I'm trying to refresh my memory..my prof lost my homework and I accidentally overwritten the file so now I'm scrambling through my notes

Answer 12

does \(ad=bc\) ring a bell?

Answer 13

demand an A+ for losing your homework XD

Answer 14

how the heck ... really... I have notes from 2/23 and 2/24 how did he lose it? anyway that's like matrix stuff like ad-bc

Answer 15

so let me try attempt symmetric

Answer 16

R is symmetric if \((\forall (a,b),(c,d) \in S)[((a,b),(c,d)) \in R \rightarrow ((c,d),(a,b)) \in R]\)\\

Answer 17

no no think third grade

Answer 18

the \(Z \times (Z \backslash {0})\) should be a big hint
why the restriction about the zero?

Answer 19

we can't have a 0 in the denominator

Answer 20

zactly
you are building fractions

Answer 21

symmetry comes from the fact that if \(ad=bc\) then since equality is symmetric you have \(bc=ad\)