check if relation r is reflexivity, transitivity, symmetry. r is a binary relation in the set of natural numbers such that x r y (x mod 3) = (y+1 mod 3).
x-y-1≡(3 mod) <=>x-y-1=3k, for some k ∈ R.
1). Reflexivity:
xRx <=> x-x-1 =3k => -1= 3k *-1 => relation is Reflexivity
2). Symmetry:
xRy => yRx
if xRy <=> x-y-1=3k, k ∈ R
yRx <=> y+1 -x = 3d , d ∈ R
x-y-1= - (y+1 + x) => k = -d relation is Symmetry
3). Transitivity:
xRy and yRz => xRz
xRy <=> k ∈ R , x-y-1=3k
yRz <=> s ∈ R , y-z-1=3s
x-y-1 + y -z-1=3k +3s
x-z-2=3(k+s) is not Transitivity
can you tell me if this is correct
The relation is not reflexive, because $$xRx\iff 0=x-x=3k+1,$$which is impossible for natural $k$.
The relation is not symetric. $$xRy\iff x-y=3k+1,$$ $$yRx\iff y-x=3l+1,$$ for some naturals $k$, $l$, which leads to (take a sum) $3k+3l+2=0$, which is impossible.
In the same spirit, this relation is not transitive.