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visits member for 1 year, 10 months
seen Apr 15 at 12:36

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 Advice? homework: $\forall x,y \in \mathbb{R}, x \in \mathbb{Q} \land y \notin \mathbb{Q} \implies (x + y) \notin \mathbb{Q}$
Feb
5
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.
Feb
5
asked Advice? homework: $\forall x,y \in \mathbb{R}, x \in \mathbb{Q} \land y \notin \mathbb{Q} \implies (x + y) \notin \mathbb{Q}$
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 $s \leq r$
Jan
26
asked How would one prove that $\sqrt{n}$ is the largest divisor that needs to be checked to determine if $n$ is prime?
Jan
25
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?
Jan
25
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.
Jan
25
asked 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?
Jan
22
revised How would one prove this in predicate logic
markup changes
Jan
22
suggested suggested edit on How would one prove this in predicate logic
Jan
22
accepted Have I justified that $\forall x \in \mathbb{R}$, $x > 1 \rightarrow x^2 > x$
Jan
22
awarded  Custodian