Q: Is the following relation reflexive, symmetric, or transitive?
R, where (x,y)R(z,w) iff x+z ? y+w, on the set RxR.
I am not sure how to treat the relation of two ordered pairs.
Here is what I did.
Reflexive: Suppose (x,y)?R. Then x+y?x+y since x+y=x+y. This, (x,y)R(x,y). So, R is reflexive.
Symmetric: Suppose (x,y)R(z,w) and (z,w)R(x,y). Then, x+z ? y+w ? y+w ? x+z so, x+z ? y+w ? x+z which implies x+z=y+w. Therefore, R is symmetric. (I am pretty sure this is wrong, because I don't think I can just stick the relation "?" in the middle of the two relations. If not, I cannot state R is symmetric.)
Transitive: Suppose (x,y)R(z,w) and (z,w)R(u,v) where (u,v) are arbitrary coordinates in R. Then, x+z ? y+w and z+u ? w+v. so, we can rewrite this as x+z+u ? y+w+v. So, R is transitive.
Thats me attempt at it.
Thank you
--Dan
R, where (x,y)R(z,w) iff x+z ? y+w, on the set RxR.
I am not sure how to treat the relation of two ordered pairs.
Here is what I did.
Reflexive: Suppose (x,y)?R. Then x+y?x+y since x+y=x+y. This, (x,y)R(x,y). So, R is reflexive.
Symmetric: Suppose (x,y)R(z,w) and (z,w)R(x,y). Then, x+z ? y+w ? y+w ? x+z so, x+z ? y+w ? x+z which implies x+z=y+w. Therefore, R is symmetric. (I am pretty sure this is wrong, because I don't think I can just stick the relation "?" in the middle of the two relations. If not, I cannot state R is symmetric.)
Transitive: Suppose (x,y)R(z,w) and (z,w)R(u,v) where (u,v) are arbitrary coordinates in R. Then, x+z ? y+w and z+u ? w+v. so, we can rewrite this as x+z+u ? y+w+v. So, R is transitive.
Thats me attempt at it.
Thank you
--Dan