470 reputation
112
bio website
location
age
visits member for 1 year, 11 months
seen Apr 15 at 12:36

Feb
26
comment Is this a correct proof for this relation?
@dtldarek Thanks for the advice, it seems to be fairly clear now thanks to Stahl's edit.
Feb
26
reviewed Approve suggested edit on Is this a correct proof for this relation?
Feb
26
comment Is this a correct proof for this relation?
@Jim I was actually just using that, and that is where I found \lightning but alas it did not work.
Feb
26
comment Is this a correct proof for this relation?
@Jim Thanks, I also cant find a good contradiction symbol.
Feb
26
asked Is this a correct proof for this relation?
Feb
26
accepted Is the relation on the positive integers defined by $(x,y) \in R$ if $x = y^2$ only antisymmetric?
Feb
26
comment Is the relation on the positive integers defined by $(x,y) \in R$ if $x = y^2$ only antisymmetric?
@ThomasAndrews Ah thanks, I don't know why I am having trouble thinking such a simple example! Probably just in the mind-set that $y = x$ is the law of the land for some reason.
Feb
26
comment Is the relation on the positive integers defined by $(x,y) \in R$ if $x = y^2$ only antisymmetric?
@ThomasAndrews Wait, what elements? I need an example please. You are saying that there is a pair for $x, y \in \mathbb{Z}^+$ s.t. $x = y^2$ and $(x,y) \neq (1,1)$ ?
Feb
26
comment Is the relation on the positive integers defined by $(x,y) \in R$ if $x = y^2$ only antisymmetric?
Ah I see where I have been mistaken now, it is for all $x,y \in X$ not $(x,y) \in R$, that was bad mistake! I understand it now.
Feb
26
comment Is the relation on the positive integers defined by $(x,y) \in R$ if $x = y^2$ only antisymmetric?
@ThomasAndrews It is defined on the positive integers. We can use $x = 1$ and $y = 1$ if we want.
Feb
26
comment Is the relation on the positive integers defined by $(x,y) \in R$ if $x = y^2$ only antisymmetric?
I think I see what you mean, because I was looking at it as 1 related to 1 as the only member in the relation. But you are looking at it from the perspective of the rule. Unfortunately in my opinion since 1 is the only number that I can think of besides 0 which equals its square, $x = y^2$ and $y = z^2$ does indeed imply that $x = z^2$ since 1 is still the only option for $x, y,$ and $z$. I can not think of a counter-example to transitivity or symmetry.
Feb
26
asked Is the relation on the positive integers defined by $(x,y) \in R$ if $x = y^2$ only antisymmetric?
Feb
23
comment Does there exist any elements which I can add to a Relation such that the Relation remains Sym/Anti, Trans, and Reflexive?
Right now I am introducing "rules" that define the relation, along with the ability to randomly generate a relation with specified attributes.
Feb
23
accepted Does there exist any elements which I can add to a Relation such that the Relation remains Sym/Anti, Trans, and Reflexive?
Feb
22
comment Does there exist any elements which I can add to a Relation such that the Relation remains Sym/Anti, Trans, and Reflexive?
@alancalvitti I understand, so if I added an element $(1,5)$ to the relation would it no longer be a relation, of X?
Feb
22
asked Does there exist any elements which I can add to a Relation such that the Relation remains Sym/Anti, Trans, and Reflexive?
Feb
19
accepted Prove that $f(n) = n^2 - 1$ is not injective and not surjective. Am I doing it right?
Feb
19
comment Prove that $f(n) = n^2 - 1$ is not injective and not surjective. Am I doing it right?
Thanks, it helps to know I am not leaving anything out.
Feb
19
comment Prove that $f(n) = n^2 - 1$ is not injective and not surjective. Am I doing it right?
So what should I say "choose" $f(n) = 3$ instead?
Feb
19
comment Prove that $f(n) = n^2 - 1$ is not injective and not surjective. Am I doing it right?
Do I have to prove it is a function first? It seems like a dumb question, but thought I would ask anyways.