hi my real analysis book says 'Let $A \in \mathbf{Q}$ be a set that is bounded above such that whenever $x \in A$ and $t \in \mathbf{Q}$ with $t < x$, then $t \in A$. Suppose sup(A) is not in A. There exists a $y \in \mathbf{R}$ such that $A = \{ x \in \mathbf{Q}: x < y \}$'. I know that usually the rationals dont have the least upper bound property, but that they do when embedded in the reals. [...]
[...] but i dont get why, in this notation, it's not $y \in \mathbf{Q}$ because it says $x \in A$, which is a subset of $\mathbf{Q}$, so i feel this set $A$ is actually some rational number and every rational number less than it ?
it says in the passage i inserted that A is bounded above such that whenever x is in A and t is in Q with t less than x, t is in A, so this set choosing rationals, then including everything beneath them
sqrt 2 would never be an x value because it cannot be chosen from a subset of rationals, right?
it is crucial to the argument being made that the y they refer to is within reals, meaning it can be rational or irrational. under the notation they provided for A, y should only be rational
yes i know that rationals are reals but that's not the point of confusion
let me clarify something: based on the passage i wrote, does this mean take a subset of the rationals, now for any x in this set, every rational below it is also in this set?
i might just be misreading because based on what youre saying, this is not what the passage is saying & then this would be my problem
referring to this: " hi my real analysis book says 'Let $A \in \mathbf{Q}$ be a set that is bounded above such that whenever $x \in A$ and $t \in \mathbf{Q}$ with $t < x$, then $t \in A$. Suppose sup(A) is not in A. There exists a $y \in \mathbf{R}$ such that $A = \{ x \in \mathbf{Q}: x < y \}$'. I know that usually the rationals dont have the least upper bound property, but that they do when embedded in the reals. [...]"
well, the $\sup$ can be rational or not. Note that both $(-\infty,2) \cap \mathbb{Q}$ and $(-\infty,2] \cap \mathbb{Q}$ seem to be valid examples of $A$. But the first does not contain its $\sup$.
@TedShifrin Hi, I think could solve the problem. The key idea was to see $H_{f\circ g}(a)=J_{g}^{\top}(a) H_{f}(b)J_{g}(a)$ using the fact $b$ is a critical point of $f$. Then since $J_{g}(a)$ is inversible and the Hessians they are symmetric so the rank of $H_{f}(b)$ is the same of the rank of the Hessian of $H_{f\circ g}(a)$.
Is it possible to make text collapsible in posts? I know we can use spoilers (>!), but I'm looking to collapse multiple lines and would preferably like something toggleable, e.g. meta.stackoverflow.com/a/352631/10841085
dT/dt is a continuous function that is positive in a neighborhood of 0 by some reasoning that it sounds like you already know. the constant function T(t) = Tair is another solution to the differential equation, so taking for granted that the graphs od distinct solutions never cross, the graph of your T(t) doesn't cross the line y = Tair
333333331 is not prime; it is divisible by 17. This does not require a computer. Euler did calculations like this all the time.
What's more, in your sequence 31, 331, 3331, 33331, …, every 15th number is divisible by 31.
Proof:
An noted in lab bhattacharjee's answer,
the sequence has the form
...
That code can go a lot higher, but it will get slow for huge terms because it's using a provable prime test. However, Sage can also use a probababilistic test which is quite fast.
@Ajay Fair enough. But it's nice to have a computed list to check your hand calculations against.
Sage is fun. It's free, and I can use it on my phone without installing anything. But you do need to learn the basics of Python before starting on Sage.
Well I sorta lied, I did learn modular arithmetic and stuff like the Chinese remainder theorem. But that was more than a year ago and I practically don't remember anything.
I know how to do stuff like 10 mod 4, but I don't think that's enough to tackle this problem.
The above answer is only useful to prove some of those numbers composite (if one finds a small factor), if we want to find it out without electronic help. To prove a 7-digit number prime by hand is already cumbersome and very time consuming , even if you would apply modern tests. The largest prime found by hand was the $39$ - digit number $2^{127}-1$
You are most likely referring to the 1903 presentation by American mathematician Frank Cole. The original false conjecture was that the 67-th Mersenne number $M_{67}:=2^{67}-1$ is prime, and it goes back to the preface to Mersenne's own Cogitata Physica-Mathematica (1644). However, Cole was alrea...
I heard from a stroy (can anyone say whether it is true or not ?) where a student checked the primality of the rep-units (again) and discovered a probable prime (not discovered before!) that turned out to be prime , namely $R_{317}$.
If you want to explore this stuff, start with something a bit simpler. It can be shown that if n has no factors of 2 or 5 then the decimal of 1/n is purely periodic. That is, it doesn't have an initial non-periodic part. And by the pigeon-hole principle, the length of the period must be less than n. If n is prime, then the period divides n-1.
Eg, 1/13 = 0.076923076923... and 1/7 = 0.142857142857... So 7 and 13 must divide 999999
The length of the period of 1/7 is 6. 1000000×1/7 - 1/7 = 142857. So (1000000-1)/7 = 142857
If you do long division of 1/7, there are 6 possible remainders. Zero can't be a remainder, because that would give a terminating decimal. And as soon as we get a remainder that we got earlier, then the cycle of quotient digits repeats.
For a purely periodic reciprocal, the first remainder that repeats is 1. Play with some small examples on paper.
Sometimes that happens because the OP doesn't realise they can only accept one answer.
What's even more irritating is when they get 3 or 4 decent answers, but instead of accepting one of them they write their own answer that badly mashes together the separate answers in a way that demonstrates that they don't really understand those answers. And then they accept their own answer.
@PM2Ring This is called "rep-hunting". Self-answers are very rarely the proper way , only if the author has made progress making further efforts obsolete.
I don't mind self-answers if the OP has done a good job of summarising / synthesising the other answers. Or if it's a question & answer pair specifically created as a canonical question. But otherwise, self-answers tend to be very low quality, and pure rep-hunting, which rightly attract copious downvotes.
I think it works something like this: Since 1/14 = 0.07142857142 has a length of 6, we can do a 6 digit value, but since there is a zero beforehand, we go down by 1, so instead of 999999, we do 888888.
idk if that logic even makes sense but it seems to work somehow.
The Details:
Since definitions vary:
A topological space $(X,\tau)$ is a set $\tau$ of subsets of $X$, called closed subsets, such that
$\varnothing, X\in\tau$,
The intersection $$\bigcap_{i\in I}X_i$$ of any closed subsets $(X_i)_{i\in I}$ is closed, where $I$ is arbitrary, and
The union of fi...
As a general matter of style, it takes you a long time to get to the question. I would typically put the question right up at the top, and then explain the details.
I'm sure some votes are accidental. It's very easy to do on a phone, especially when scrolling one-handed.
A few months ago, when I finished reading a good answer, I scrolled up to upvote it and was horrified to see that I'd already downvoted. Luckily, I was able to reverse it in time. And now I try to be more careful while scrolling.
$\mathcal V(I)$ is usually called the variety associated to $I$. The game is to go back and forth between algebra (polynomial ideals) and geometry (subvarieties of $k^n$).
i think there should be varieties of downvotes: the 'i'm having a bad day' downvote, the 'you annoyed me so i am getting back at you' downvote, the 'there's something wrong with the answer' downvote, etc. that would be so much better.
I wish the system made it impossible for you to leave a comment if you downvote, and vice versa. Then people would be more likely to leave comments on downvoted posts, since the OP would know that the commenter wasn't the downvoter, so the OP wouldn't have a reason to revenge-downvote them.
i rarely downvote, but when i do it's usually old posts by ted shifrin, in retaliation for something unrelated or imagined that took place on this chat
I don't downvote much on Math.SE. I'm fairly ruthless downvoting bad answers on Physics.SE.
On Physics.SE we get answers by people attempting to promote crackpot theories, usually their own pet theory, but not always.
We don't seem to get many cranks on Math.SE. Sure, there's the occasional person who's baffled by stuff related to limits, and can't accept derivatives or 1 = 0.999... But we don't get many people with purported proofs of classic things like squaring the circle or duplicating the cube.
yes, and the cranks that do appear tend to be obsessed with very individual, specific things, and not 'famous' problems or anything that would attract a lot of casual attention
Compared to the Riemann hypothesis, it's pretty easy to understand the Collatz conjecture. And if you know a little number theory you might be hopeful that you can stumble across a proof. But the fact that experts haven't made much progress on it should be a clue...
At least with twin primes & Goldbach's there are various avenues of attack that people have tried. With Collatz, it's much harder, due to the problem Leslie mentioned.
FWIW, a decade or so ago I constructed a pattern in Conway's Game of Life that produces Collatz sequences in binary, using streams of gliders for the bit strings. :)
Ok so coming back to what I was writing previously, $(\ker T)^\perp = \overline{T^*(Y^*)}^{w^*}$ and so $((\ker T)^\perp)^\perp = (T^*(Y^*))^\perp = \ker T^{*}$ and the image being finite dimensional is not necessary for this to hold (this is because $Z^\perp = (\overline{Z}^{w^})^\perp$)
I'm not even sure why it doesn't render. Perhaps overload of asterisks
Now if you imagine x^3, you need to first stretch and only then shift downwards, right?
or else it will be 0,5(x³ -5)
But you could also "first shift" by making the coordinate origin (0, -5), then from there go left and right and imagine a new coordinate system where thats the origin in that case, you would first shift, right? Why are you allowed to do that?
koro i'm seeing a number of used copies on bookfinder.com in the $30-50 US range. it looks like some are duplicate listings of the same book through multiple sellers, and i dunno about shipping or the reliability of any individual seller. i've had some luck with bookfinder before.
i'm generally surprised though, it does looks like most used copies are... not priced like used books.
It's important because measure theory has this dumb thing to itself, that you can't prove obvious things unless you use something like monotone class theorem
The situation doesn't appear just once, it will keep on repeating itself. And you'll have to use the theorem again and again if you want to be theoretically correct
Suppose that we have a set X, semi algebra S on X and a measure $\mu$ on S. Then, 1) $\mu$ is extended naturally to A(S) (the algebra generated by S). 2) To extend $\mu$ to F(A(S)) (the sigma algebra generated by S), we do the following steps: 2a) Define outer-measure $\mu^*$ on P(X), 2b) Define M:=the set of all measurable sets, 2c) Show that M is a sigma algebra, 2d) Show that restriction of $\mu^{*}$ to M is actually a measure. 3) The restriction of measure obtained in 2c) to F(A(S)) turns out to be an extension of $\mu$.
I want to ask: why is monotone class theorem required at all here? 4) can be proven without it.
i wouldn't say it's "required," it's a technical tool that people like using because it is applicable frequently enough that it saves time in the long run to use it instead of ad hoc approaches
it unifies a lot of annoying technical details into one annoying technical detail
Leslie, I asked this question because in my class the teacher said seemed to imply that "it's very very difficult to prove 4) without knowing Monotone class theorem."
all of that stuff was known before halmos (or whoever the MCT guy was), the foundational papers don't use it or other machinery and often have accessible proofs
But then there is a theorem (2.36 if I recall correctly) in the book that does use monotone class theorem, which is introduced right before the theorem.
i never had to learn or teach from it, but i liked terence tao's book when i looked at it
generally though diving too deep into measure theory is a waste of time, it's just tons of annoying little tricks and constructions that are only useful in measure theory
read the hypotheses and the theorems but not the proofs
maybe pop out is too strong a word, but i get sin(z) = (-1)^n i sinh(y) and cos(z) = (-1)^n i cosh(y), and the quotient is tanh(y), and |tanh(y)| <= |y| from the power series for tan being alternating with decreasing positive coefficients, so |tan(z)| <= |y| <= |z|
333333331 is not prime; it is divisible by 17. This does not require a computer. Euler did calculations like this all the time.
What's more, in your sequence 31, 331, 3331, 33331, …, every 15th number is divisible by 31.
Proof:
An noted in lab bhattacharjee's answer,
the sequence has the form
...
