symmetry

symmetry on an operation R over Z is defined thus,

xRy => yRx

e.g. if x=y then y=x, if 3|(x-y) then 3|(y-x)

anti-symmetry on R over Z is defined thus,

if xRy and yRx then x=y,

e.g. x(< =)y and y(< =)x then x=y (< = is less than or equal to)

Question: is the operation ' < ' symmetric, anti symmetric, neither or both?

[Blank posting]

Originally posted by genius
symmetry on an operation R over Z is defined thus,

xRy => yRx

e.g. if x=y then y=x, if 3|(x-y) then 3|(y-x)

anti-symmetry on R over Z is defined thus,

if xRy and yRx then x=y,

e.g. x(< =)y and y(< =)x then x=y (< = is less than or equal to)

Question: is the operation ' < ' symmetric, anti symmetric, neither or both?

Originally posted by ilywrin
Maybe relation is more appropriate than operation.
Depends what meaning you impose on that symbol > but if we assume the default it is neither symmetric, nor anti-symmetric.

you're discussing relations and antisymetric is if xRY then NOT yRx as a relation can be either reflexive (for all x xRx) or anti-reflexive (for all x NOT xRx)

Originally posted by aginis
you're discussing relations and antisymetric is if xRY then NOT yRx as a relation can be either reflexive (for all x xRx) or anti-reflexive (for all x NOT xRx)

Originally posted by Skinn13
no no, antisymmetric really is this by definition: if xRy and yRx, then x=y.
It is not the opposite of symmetric

Originally posted by Skinn13
not sure, but it seems like it is anti-symmetric to me, because a>b and b>a cannot happen both, and therefore the statement: if a>b and b>a , then a=b is true

Originally posted by genius
yes-i meant relation not operation 😛

i think you've got it, but could you give a better explanation?

The relation is not symmetric, and it is anti-symmetric.