I know what mod is but i don\'t understand this notation it says x =_ y mod 6 do
ID: 3141899 • Letter: I
Question
I know what mod is but i don't understand this notation it says x =_ y mod 6 does it mean x - y = n and n is divisible by 6 without remainder?
To be reflexive means (x,x) so here it says its reflexive cause x - x = 0 = 6 * 0 how does that prove reflexiveness?
To be symmetric you have (x,y) and (y,x) for all in the set of integers so how does x - y = 6k prove its symmetric? what is k? and why wasn't it used in the first explanation of reflexive
To be transitive (a,b) and (b,c) and (a,c) are also related i don't understand the proof so now they introduced j ?
The equivalence class of [1] is {x e R | xR1} so that means all pairs which have 1 as their 2nd number in the relation so what does the relation look like(what is the a and b inside it)? I'm confused as to what goes in there from the x _=mod 6
Show that the relation R on the set of integers R (x,y)lx ymod6) is an equivalence relation on the set of integers. State the equivalence classes of R.Explanation / Answer
I know what mod is but i don't understand this notation it says x =_ y mod 6 does it mean x - y = n and n is divisible by 6 without remainder?
yes ,it means x - y = 6k , where k is integer
To be reflexive means (x,x) so here it says its reflexive cause x - x = 0 = 6 * 0 how does that prove reflexiveness?
for reflexive (x,x) should satisfy given relation
here ( x -x = 0 = 6*k, here k = 0
hence (x,x) satisfy relation ,hence relation is reflexive
To be symmetric you have (x,y) and (y,x) for all in the set of integers so how does x - y = 6k prove its symmetric? what is k? and why wasn't it used in the first explanation of reflexive
if (xRy) then x -y = 6k
which means y -x = -6k = 6*(-k) = 6 c , which mean (y,x) satisfy relation
hence if (xRy) then (yRx)
To be transitive (a,b) and (b,c) and (a,c) are also related i don't understand the proof so now they introduced j ?
if aRb mean a -b = 6*x
bRc means b - c = 6*y
now a -c = (a-b)+(b-c) = 6x + 6y =6(x+Y)
which means (a,c) satisfy relation
so if (a,b) & (b,c) satisfy relation then (a,c) also satisfy relation
The equivalence class of [1] is {x e R | xR1} so that means all pairs which have 1 as their 2nd number in the relation so what does the relation look like(what is the a and b inside it)? I'm confused as to what goes in there from the x _=mod 6
let equivalence class of [1] is y
1 - y = 6*k
hence y = 6k+ 1
hence
equivalence class of [1] is 6k+1
clearly 1 -(6k+1)= -6k ,