Leonardo
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 Feb26 comment Is this a correct proof for this relation? @Mohan Nice, I should add that and then I think it is a complete proof. So the part by contradiction is good enough for anti-symmetry? Feb26 comment Is this a correct proof for this relation? @Mohan It seemed like my proof for transitivity and antisymmetry was so similar that I stuck them together. Can I separate them and say something like "similarly R is transitive?" Feb26 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. Feb26 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. Feb26 comment Is this a correct proof for this relation? @Jim Thanks, I also cant find a good contradiction symbol. Feb26 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. Feb26 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 $(x,y) \neq (1,1)$ ? Feb26 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. Feb26 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. Feb26 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. Feb23 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. Feb22 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? Feb19 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. Feb19 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? Feb19 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. Feb5 comment How does one select some $i$ to prove $\exists i$ s.t. $s_i \leq A$ where $A$ is the average of real numbers $(s_1 + s_2 + . . . + s_n) / n$ Yes thank you, the more I stare at the answer the more obvious it seems, its like one of those hidden paintings or something. Feb5 comment How does one select some $i$ to prove $\exists i$ s.t. $s_i \leq A$ where $A$ is the average of real numbers $(s_1 + s_2 + . . . + s_n) / n$ So how might I write this then. To yield a contradiction, assume for all $i$, $i > A$. This implies the average of such a series is actually equal to, and not greater than, $A$. And this is the contradiction? Very much as you have written? Do I need to declare $i$ be a natural or something like that? Feb5 comment Advice? homework: $\forall x,y \in \mathbb{R}, x \in \mathbb{Q} \land y \notin \mathbb{Q} \implies (x + y) \notin \mathbb{Q}$ I guess if a rational plus a rational is a rational, then assuming x + y is rational means (x + y) - x is also rational. Therefore leading to the conclusion that y is also rational, contradicting the hypothesis. Jan26 comment How would one prove that $\sqrt{n}$ is the largest divisor that needs to be checked to determine if $n$ is prime? I am still on line 3, all I see is that you took the square root of $s^2 = n$ but I still do not see how $s \leq \sqrt{n} \implies (r \land \lnot r)$ Jan26 comment How would one prove that $\sqrt{n}$ is the largest divisor that needs to be checked to determine if $n$ is prime? So $s^2$ is less than or equal to $rs$ because $s \leq r$