@AndrewSalmon I was working rather hastily, but unless I slipped up, the answer is yes, and you can prove it by element-chasing, showing that each side is a subset of the other.
I hope, I am able to tell this in english, I sometimes have the feeling, that in higher math there is a lot more heuristical arguments than in the low level stuff
@DominicMichaelis If you mean that there’s more use of heuristics to gain insight, that may well be true. If you mean that heuristics actually replace rigorous argument, it’s not.
@OrangeHarvester According to this, I’ve lost $52086$ rep to the cap. I can’t say that it distresses me, but I do find it amusing that Arturo and I could ‘fund’ five $20$K users from our combined lost rep!
Is there a way to show that the only solution of
$$\sin(x)=y$$
is $x=y=0$ with $x,y\in \mathbb{Q}$.
I am seaching a way to proof it with the stuff you learn in linear algebra and analysis 1+2 (the stuff you learn in the first and second term).
@DominicMichaelis Some flag types are disputed and some flag types are declined. I don't think the same flag type has the option of being declined if it can be disputed, and vice versa.
@DominicMichaelis it will be if it is a complete duplicate, give it some time. Also the part b of the questions is a little different, so it may not be marked as duplicate.
Is this proof sufficient: Let $|G| = 8$ Prove that $G$ has an element of order 2. Let $a \in G$. It must be true that $a^{|G|} = e$. Assume that $|a| \neq 2$ then, $a^8 \neq 8$.
For instance, one can use Lagrange's Theorem to show that the order of the cyclic subgroup generated by $a\ne e$ can only be one of $2,4,8$. Thus, either $a^1,a^2,a^4$ has order 2.
@Eric: The easiest argument, I think, is to start with any non-identity element $a\in G$. Then the order of $a$ is $2,4$, or $8$, by Lagrange’s theorem. If it’s $2$, you’re done. If it’s $4$; use $a^2$. And if it’s $8$, use $a^4$.
@BrianM.Scott ok that makes much more sense but how does $a^4$ prove there is an element of order 2? Oh wait, because we can then use the previous idea that if it is 4 then take $a^2$?
@BrianM.Scott What do you tend to do more: explain your work or just write a line's answer? The former is definitely very fancy, but not understandable.
@skullpatrol As an example, I wrote my doctoral dissertation over a Christmas break. I was able to do this because I’d already written up very complete proofs of my results as I discovered them, so it was really just a matter of organizing what I wanted to say and polishing the exposition. (And writing it out longhand $-$ this was $1974$.)
@Ethereal I’d say that I prefer whatever I think is most helpful, which most of the time is either a pretty thorough explanation or a very carefully directed hint. But I do like to explain things!
@Ethereal Answers that strike people as elegant or usefully general often get a lot of upvotes from people who appreciate the sophistication, even if they’re less useful to the OP.
@Ethereal Yes, I was; I was just saying that liking to explain things was undoubtedly one of the reasons that I wanted to be one in the first place.
@Ethereal Precisely. But it’s a fairly general phenomenon: I often have an accepted answer with fewer upvotes than one or more of the other answers. Sometimes that’s just because I said something a bit better, but sometimes it’s because I judged the OP’s level better.
But sometimes, when I explain everything, I end up with a zero-votes answer with a short answer accepted. That is a painful feeling when you didn't really need to type a lot: you did it for the sake of others.
@BrianM.Scott Populist!
No wonder why you-know-who got two Populist badges.
I think that I’d emphasize that there are really two parts to a direct argument, showing that the sequence converges, and then finding the limit, and that it’s the second that’s the easy part.
@Ethereal It’s the graph of a function with the desired properties. Put the ends of the W at $x=1$ and $x=5$, the feet at $x=2$ and $x=4$, and the central peak at $x=3$.
@Ethereal No idea at all. Even if I were familiar with and capable of judging all of the answers, I couldn’t pick a single one: there are too many different ways for an answer to be good.
Suppose $u=\frac{z(1+i)-i}{z+1}$ as $z\in \mathbb{C} \setminus{-i}$
What is the set of points $M(z)$ for which $u$ is a real number?
What is the set of points $M(z)$ for which $u$ is pure imaginary number?
What is the set of points $M(z)$ for which $|u|=\sqrt{2}$?
@Ethereal You can do it inline like that for a short sequence of equations, or you can use \begin{align}...\end{align} or simply separate the lines of the sequence by `\\`
@Ethereal That depends on whether the steps are reversible or not
@OrangeHarvester: major part of them are created by my brother, and some of them are still unanswered. He has thousands of problems in calculus. As regards this question, I need to ask him. Of course, some problems come from other sources like some local math contests. (see some of my last questions)
@Chris'ssisterandpals Ahh, that explains it. :-) In high school, I would look out for math problems from Romanian School Contests on AoPS. They were quite famous.
Now for the point: the question is pretty well-researched and I have posted a very simple proof, but I do not think it's OK because as they say, "If math is too easy, you're doing it wrong."
If $n > m$, then $m + k = n$ where $k$ is a natural number using the closure property under addition. Write $m + k = n$ as $k = n - m$ but we already know that $k$ is a natural number...
"The Standard Model is wrong! Booo to the standard model." "But I though you need $3 billion dollars to build a machine to find a missing piece in the Standard Model..." "The standard model is great! Long live the standard model!"
I am reading a research paper and in one step they have done $S = B \pmod{19}$ where $S$ and $B$ are both matrices. What does it mean? How do I calculate this?
Is there a way to show that the only solution of
$$\sin(x)=y$$
is $x=y=0$ with $x,y\in \mathbb{Q}$.
I am seaching a way to proof it with the stuff you learn in linear algebra and analysis 1+2 (the stuff you learn in the first and second term).
@DominicMichaelis I simply mean that for instance, $A=\eft( \matrix{1&1\\ 0& 1}\right)$ is not diagonalizable. In the case where there are repeated eigenvalues.
@DominicMichaelis Yes. If I remember correctly, I said it was true that the matrix is automatically diagonalizable when the diagonal elements are pairwise distinct. But the OP also asked whether there was a clasification in general for upper triangular matrices. And I just said that when there are repeated eigenvalues, both cases can occur: diagonalizable and not diagonalizable.
user19161
6:03 PM
@peoplepower Surely, a great book would be MacLane's Categories for the working mathematician.
@DominicMichaelis Okay. Langranigan multipliers are not constant, rather they are variables. And time being the independent variable in physics, the multiplier can have dependence on time.
Is there a way to show that the only solution of
$$\sin(x)=y$$
is $x=y=0$ with $x,y\in \mathbb{Q}$.
I am seaching a way to proof it with the stuff you learn in linear algebra and analysis 1+2 (the stuff you learn in the first and second term).
Suppose $u=\frac{z(1+i)-i}{z+1}$ as $z\in \mathbb{C} \setminus{-i}$
What is the set of points $M(z)$ for which $u$ is a real number?
What is the set of points $M(z)$ for which $u$ is pure imaginary number?
What is the set of points $M(z)$ for which $|u|=\sqrt{2}$?
If $n > m$, then $m + k = n$ where $k$ is a natural number using the closure property under addition. Write $m + k = n$ as $k = n - m$ but we already know that $k$ is a natural number...
I am reading a research paper and in one step they have done $S = B \pmod{19}$ where $S$ and $B$ are both matrices. What does it mean? How do I calculate this?
