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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

user153519
  • 11

1 Answers1

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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.

TZakrevskiy
  • 22,980
  • Are you sure about transitive. ---- Transitivity: xRy and yRz => xRz
    xRy <=> k ∈ R , x-y=3k+1
    yRz <=> s ∈ R , y-z=3s+1
    x-y +y -z=3k +3s+2
    x-z=3k+3s+2 is Transitivity.
    because x-z=3k+3s+2 is possible.
    I'm right or not. –     – user153519 May 26 '14 at 21:33
  • @user153519 no, in fact, $xRy$ and $yRz $ imply $zRx$, but not $xRz $. – TZakrevskiy May 27 '14 at 21:52