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comment A question about Quadratic residue
I don't see how this helps. Let $q = 4np+2p-1$ be prime, then there exists an $a$ so that $-1 \equiv a^2 \mod q$, which means $a^2+1 = x(4np+2p-1)$ for some $x \in \mathbb Z$. Then $a^2+1 \equiv -x \mod p$, so... ??? Was this what you were suggesting, or did you have something else in mind?
2d
revised A question about Quadratic residue
edited body
2d
answered A question about Quadratic residue
May
25
comment Can I approximate a measurable set with an open set for integration purposes?
$X$ may not have an open subset; do you want to allow $f$ to be extended to a larger domain? If so, and $X$ is measurable then the answer to your question taken literally is yes, because you can extend $f$ to $\mathbb R$ so that $\int_{\mathbb R} f > \int_X f$, and then use intermediate value theorem on the family of open intervals $(-a, a)$ for $a \in \mathbb R^{\ge 0}$ (if $\mathbb R-X$ has measure zero, just take $X^O=\mathbb R$). But morally, I'm thinking it's likely what you want is more like an open set $X^O$ that differs from $X$ on a set of measure zero, in which case the answer is no.
May
22
reviewed Approve Proving an entire function is constant
May
22
comment Proving an entire function is constant
Per @DanielFischer's comment, assume the equation $\frac{(f(z))^3}{z^2} + f(z)=0$ is satisfied for all nonzero values of $z$, and see what $f$ can be.
May
22
revised Proving an entire function is constant
added 2 characters in body
May
22
answered economist puzzle for
May
15
comment Why $a^n - b^n$ is divisible by $a-b$?
I'm saying you should make that step more explicit. Literally, include the line $1-(b/a)^n = (1-(b/a))(1+(b/a)+\ldots+(b/a)^{n-1})$, and then multiply by $a^n$.
May
14
comment Isn't the modus ponens just the definition of what 'if' means?
Your confusion about the infinite regress is actually exactly why we need to distinguish between axioms and rules of inference: axioms are merely statements about a system, wheras rules of inference tell you what kinds of steps are valid in a proof deriving new statements from previous statements. We simply chop off the regress at the point of rules of inference.
May
14
comment Why $a^n - b^n$ is divisible by $a-b$?
There's a good proof along these lines, but you didn't give it; it appears like you just pulled the second line out of the blue (the one with the red font). How do you get from the first line to the second line?
May
14
comment Isn't the modus ponens just the definition of what 'if' means?
A small technicality: modus ponens is not typically treated as an axiom, but rather as a rule of inference. In answer to your question, the modus ponens rule is precisely how the "semantics of implication" is treated in formal reasoning; in some sense, the rule of inference in this case defines the meaning of implication.
May
13
reviewed Reject binomial-distribution tag wiki
May
12
comment Why is it true that any polynomial $f(x)$ with coefficients in a finite field $F_q$ is separable over that field?
Ah. ${}{}{}{}{}$
May
12
comment Why is it true that any polynomial $f(x)$ with coefficients in a finite field $F_q$ is separable over that field?
Any irreducible polynomial with coefficients in $\mathbb F_q$ is separable. If you drop the irreducibility condition, it's never true, over any field, because you can always consider $f(x) = x^2$, which has $0$ as a repeated root.
May
12
answered logic puzzle (truth-tellers / liars)
May
12
comment logic puzzle (truth-tellers / liars)
Oh, I see, you have to incorporate "da" or "ja" into your questions somehow. My analysis assumes you haven't done so.
May
12
comment logic puzzle (truth-tellers / liars)
There is a typo in my first comment, btw; I meant to say "I haven't figured out a way to prove it completely impossible." What makes you say the puzzle has a solution? I just proved it impossible by a straightforward application of the pigeonhole principle.
May
12
comment logic puzzle (truth-tellers / liars)
Actually, I can prove it impossible: After asking your first question, every permutation of the three people remains possible, no matter how the first person responded (as a consequence of the fact that you don't know what "da" and "ja" mean). So there are 6 possible outcomes that you need to account for, but you only get $4$ remaining possible sequences of answers from the $3$ people.
May
12
comment logic puzzle (truth-tellers / liars)
If there is a solution to this problem, it has to involve not always being able to figure out what "da" and "ja" mean within the allotted number of questions, which strikes me as mildly ludicrous, but I haven't figured out a way to prove it completely possible. If you were asked to figure out what "da" and "ja" mean on top of figuring out who's who, then there would be $2 \cdot 3! = 12$ possible answers that you need to be able to give, but only $2^3=8$ possible sequences of answers to the $3$ questions.