Leonardo
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 Feb22 asked Does there exist any elements which I can add to a Relation such that the Relation remains Sym/Anti, Trans, and Reflexive? Feb19 accepted Prove that $f(n) = n^2 - 1$ is not injective and not surjective. Am I doing it right? 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. Feb19 asked Prove that $f(n) = n^2 - 1$ is not injective and not surjective. Am I doing it right? 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 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 $(s_1 + s_2 + . . . + s_n) / n$ 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 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 Feb5 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 $(s_1 + s_2 + . . . + s_n) / n$ Feb5 accepted Advice? homework: $\forall x,y \in \mathbb{R}, x \in \mathbb{Q} \land y \notin \mathbb{Q} \implies (x + y) \notin \mathbb{Q}$ 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. Feb5 asked Advice? homework: $\forall x,y \in \mathbb{R}, x \in \mathbb{Q} \land y \notin \mathbb{Q} \implies (x + y) \notin \mathbb{Q}$ Jan26 accepted How would one prove that $\sqrt{n}$ is the largest divisor that needs to be checked to determine if $n$ is prime? 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 $s \leq r$ Jan26 asked How would one prove that $\sqrt{n}$ is the largest divisor that needs to be checked to determine if $n$ is prime? Jan25 accepted If I am checking for $s$ divides $n$ on the interval $S = [3, n-x]$, how large can I make $x$ to ensure I have verified $n$ is prime? Jan25 comment If I am checking for $s$ divides $n$ on the interval $S = [3, n-x]$, how large can I make $x$ to ensure I have verified $n$ is prime? Actually, I have a feeling my first statement is false, let x = 3.