Because whenever you answer a stack question at all, you have to make sure that this person has proven the requisite linear algebra results and whatnot
My point is that he's probably already done that proof in class. Many classes do ask when they do a computation to manually verify on a particular example some previously proven general theorem
I mean, they might not ask you to prove it on a homework assignment necessarily, but they either did the proof in class, have it in the book and are expected to read it, or the class just takes the statement as granted. In any of those three cases, let it be
@Daminark No, you don't get it. The class I was in literally never mentioned the issue of the uniqueness of A. I'm cranking my gears thinking back and AFAIK it was never once brought up (vectors were, but not matrices). So now I'm curious if I can even construct an equation with infinite A.
it might be that I'm asking him to prove something as trivial as 4k= 8 having a unique k. Obviously it does. It's just... assumed.
I mean, take an elementary row operation, you can undo it. So once you show that an elementary operation has a matrix representation (do it to the identity), you know that is invertible
Anyway, I've got a lecture to prep so I should tab out now. Though last thing, what new row operations would you be defining then? And I guess, what would having those operations get you?
I wonder if its possible to make two curves meet at one point (but not cross) and have different tangents at that point. My guess is its not possible (at least if the curve is smooth and differentiable).
@AkivaWeinberger wait... I only need the statement: "If a + bx > c + dx, then a > c or b > d". I am pretty sure that is all I need. Can it be proven in general? If you can prove it in general then by all means do so!
on another note, I think I can prove it when the "b" in x^2 + ax + b = 0 is negative. However, that is just a thought.
@Daminark Your Very Secret Number Theory Study Group is slowly getting to 14 days of inactivity (last post here 11d ago) - at 14 days it might be frozen.
I've already been proving this stuff. I've been challenging Akiva.
XD
@shaihorowitz It's not. It's just i wasn't asking for help. Nor did i need it. I was sharing a proposition I had already proved to see if akiva could do it. I didn't know you were trying to help me.
@shaihorowitz that's really just a computation issue. Brute force is the easiest method there. By definition something is either prime, composite, or a unit. There's only two integer units. So... I'd use google to find the largest unchecked number.
@Daminark Is it ok to joint that room even if I do not plan to actively read the book, only occasionally look at how the reading is going, perhaps with posting a comment here and there?
Because I think it is not good that only a room owner can create an event for the room, I have created the room, Discussing Specific Topics . In this room, people can discuss the topics they have specified before. Each topic lasts in this room for at most one day; It depends on its popularity among others. Its duration also can be specified before beginning.
And who knows, maybe the reason it couldn't fit was that he had to build up all of algebraic geometry to do so and just didn't have space in the margin!
@Typhon well, he allegedly had it, though the proof that people think he had in mind is false
user84215
4:20 AM
But there were not enough space to write it in his notebook.
So we don't know if Fermat had a legit proof or not on basis of the stuff. There's a suspect which is convincing but false, and unless he's devilishly clever to either come up with an elementary argument or develop the theory of modular elliptic curves single-handedly (only the former would've been remotely possible :P), there's not yet good reason to believe his proof worked
there's a hell of a lot of daylight between "Fermat's last theorem is provably true" and "Fermat could have proven it using the mathematics of his era"
Also, though, this line of reasoning is under the assumption that he had an argument which could convince others. Is there any historical evidence of such?
if a man convinces 100000 mathematicians that they've proven the Riemann Hypothesis and nobody can debate it, is there any reason to not rename it the Riemann Theorem?
@Semiclassical I know I read it somewhere reputable once. Don't recall where. I've always taken it as a "Library of Alexandria" type situation. The proof was lost.
Unrelated to the issue I was giving akiva earlier, here is another proof you guys can help me with. Suppose x is a negative irrational number. Prove that for integers a, b, c, and d that if a + bx > c + dx, then a > c or b > d.
The issue here is that I've tried proof by contradiction and gotten to where I have a <= c and b <= d. The issue is that multiplying by the latter gives bx <= dx.... and I don't think that can be used to form a contradiction.
Suppose not. Then, a <= c and b >= d. Multiplying the latter equation by x we get bx <= dx. Then, adding the equations together we get a + bx <= c + dx. This is a contradiction. Therefore the statement is true. []
@Semiclassical basically I proved that if for Z[x] where x is a positive real irrational and there exists a unit y that is neither 1 nor -1, then there exists a unit w such that all units are powers of w or the negatives of powers of w.
the thing I'm doing now is to build up for the case that x is negative.
If b<d, then the conclusion is true. If b=d, then a>c follows immediately. If b>d, then the inequality becomes x>(c-a)/(b-d). But x<0, so this is only true when a>c.
technically I handwaved away the proof that there exists a y^n and y^n+1 such that the potential unit is between them
but.... the professor I had the original course said "it was quite clever" and "needed real analysis to patch an error". I didn't understand the real analysis so I just take that as granted for the sake of sanity.
About the only mathematical 'discovery' I can ever claim to have made in number theory is a rule to find remainders mod 7, and that's a silly little thing.
@Semiclassical It was an introductory abstract proofing course that happened to be number theory themed in one of the homeworks for an advanced extra project for special credit. So when the professor said, "you need real analysis" he was saying it more in the sense of: "how in the world did you think of this?"
x^2 + ax + b = 0 for integers a and b. Both values of x are negative.
Definition of norm/given info:
N(c + dx) = c^2 - cda + bd^2 = 1 N(e + fx) = 1 c + dx > e + fx c, d, e, and f are positive integers.
Prove:
c > e and d < f
Proof:
Suppose c <= e. Then d < f. Multiplying together and squaring we get that c^2 <= e^2 and cd < ef. Since both values of x are negative, we know that a is a positive number and b is negative number as if x and x' are the two solutions of the polynomial we can say that a = -x - x' and xx' = b. Therefore, cda < efa and c^2 <= e^2 and bd^2 < bf^2.
hi. can sb help me w this? there are 4 people we want to group them into groups of 4, we do this twice, means each is assigned to a group twice. so how many ways can we assign these people?
actually lets say there are 4 teachers to be assigned to teach in 4 schools and each school needs 2 teachers. now that makes sense !
I'd say we have $ 4 ! $ conditions for the first time, and then once again we do it and there is another $ 4! $ so as the answer I'd get $ 4! ^2 $ is that right?
Suppose that $Z[x]$ is composed of all numbers of the form $c + dx$ where $c$ and d are integers and $x$ is an irrational positive solution to the polynomial $x^2 + b = 0$ where $b$ is an integer.
I seek to show that there exist any elements that are units aside from $1$ and $-1$. I know that...
@s.harp @TimTheEnchanter you mean what I'm doing is ONE condition among n conditions it could be? like, alice twice to 2nd sch and petre to 2nd sch, but no other choices is considered..? is that so ?
edit :
alice to 2nd school and petre to 2nd school or alice twice to 2nd school (only one of the conditions)
Some time ago I think at least I read that the geometric series converges everywhere on it's radius of convergence except for "+1". Is there a question about it? Could find anything...
hi guys, let $\omega_j$ a sequence such that $$ \sum_{j=0}^{+\infty} \omega_j \le \infty $$ and for all $k$ we have $$ \omega_k \leq \sum_{j=k+1}^{+\infty} \omega_j $$
and assume (but you can prove it is true) that for al $x$ in a certain interval $[-A,A]$ we have
my question is it true that if I pick $x,y$ in such interval with $x < y$ then there's a $j_0$ such that $d_{x,j} = d_{y,j}$ for $j < j_0$ and $d_{x,j} < d_{y,j}$ for $j = j_0$
namely
if given $x$ and is expansion
and I get the sequence $d_j$
is the order preserved?
my guess would be yes, because the binary expansion is a specific case of this one
user84215
9:25 AM
Why don't you suggest your ideas for starting a new event in Discussing Specific Topics in order to discuss a specific math subject?