Leonardo
Reputation
500
Top tag
Next privilege 1,000 Rep.
Create new tags
 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. Feb 19 asked Prove that $f(n) = n^2 - 1$ is not injective and not surjective. Am I doing it right? Feb 5 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. Feb 5 accepted How does one select some $i$ to prove $\exists i$ s.t. $s_i \leq A$ where $A$ is the average of real numbers $(s_1 + s_2 + . . . + s_n) / n$ Feb 5 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? Feb 5 revised How does one select some $i$ to prove $\exists i$ s.t. $s_i \leq A$ where $A$ is the average of real numbers $(s_1 + s_2 + . . . + s_n) / n$ edited body Feb 5 asked How does one select some $i$ to prove $\exists i$ s.t. $s_i \leq A$ where $A$ is the average of real numbers $(s_1 + s_2 + . . . + s_n) / n$ Feb 5 accepted Prove $x \in \mathbb{Q}$ and $y \notin \mathbb{Q} \implies (x + y) \notin \mathbb{Q}$ Feb 5 comment Prove $x \in \mathbb{Q}$ and $y \notin \mathbb{Q} \implies (x + y) \notin \mathbb{Q}$ 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. Feb 5 asked Prove $x \in \mathbb{Q}$ and $y \notin \mathbb{Q} \implies (x + y) \notin \mathbb{Q}$ Jan 26 accepted How would one prove that $\sqrt{n}$ is the largest divisor that needs to be checked to determine if $n$ is prime? Jan 26 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 $s \leq \sqrt{n} \implies (r \land \lnot r)$ Jan 26 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 $s \leq r$