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I wish to plan a progressive dinner with 16 couples. If we have 4 courses and 4 hosts for each course, how can I plan a chart for where no one eats together twice?
Each couple would eat with 12 other couples.
What kind of math class are you in? Have you ever heard of a balanced incomplete block design?
Figuring out why a method works is interesting, just remember that nothing works as fast as Newton's method. (Actually I'm using Newton's method right now to calculate the initial conditions for a pde that models heat/gas mixture in an engine)
I found this to be a rather interesting proof. Use Baire's category theorem to prove the following:
If
is a collection of open dense sets in a complete metric space , then is dense.Give a counter example to show that "open" is a required property, without using the axiom of choice.
Modular arithmetic is not an operation, I don't believe you can talk about "it's inverse".
We can not stop the user, unless the user runs out of names, which is only possible if he is that small minded.
Actually no, it's not possible. If he were to get banned 10 times a day with different user names, he could keep that going for over 40 billion years. And that's only considering user names of length exactly 10 letters long.
ARB has learned how to use a proxy. We'll probably be seeing a lot more of him.
finite, your best bet is to visit your professor during office hours. As bobbym said as well, use our forums. Emails would be much slower and probably less helpful.
In problems such as this, always write out x in terms of y (or vice-versa). Then write an equation involving x and y, in this case the sum of squares equation. Make a substitution and solve.
You're certainly on the right track. But when you have that the intersection contains all of P, and hence all of Q, you actually get that P and Q are the same exact two subgroups. Since we started out with two different subgroups, obviously this can't happen.
In response... to bobbym.
Due to the fact that we are forced to act with incomplete information we should be aware that our rationalizations could be erroneous. You mean, we don't know of anything better.
Yes, we are always open to error. Scientists often never state it because it is so commonly know. Imagine a scientist saying:
"There are two asteroids coming toward the planet Earth. And just to remind you, two is the number that comes after one."
Of course this is an exaggeration, but the idea behind it is the same.
Funny you should mention that, originally my training was in Chemistry and Biochemistry and I do not think that Homeopathy is bunk. But not being a Doctor, know one cares what I think.
Your right. And if you were a doctor who spoke out on Homeopathy, most people (who bought into Homeopathy) also wouldn't care what you think. But this isn't the way it should be. How much you listen to a person should be proportional to how much reason and evidence they give for their positions. Just because society doesn't act like this doesn't mean you can't.
So if ridiculing strange theories does not prove they are right and it doesn't prove they are wrong, then why do it?
Cause it's fun? Seriously though, if the evidence can overcome ridicule, it will be accepted. Ridicule serves a purpose: It's a filter. The things that don't stand up to ridicule are filtered out, the ones that do stand up to it are allowed in.
The ridicule of the eyewitnesses by experts has prevented mainstream science from investigating them.
No, the lack of evidence has prevented mainstream science from investigating them. Many things would also completely overturn well-known science that we have a ton of real-life experience with, and have verified over and over again that it works. For example, cold-fusion or homeopathy would completely revolutionize the way chemistry and physics works. This isn't bad in itself, but when there is also no evidence for them, it is a fatal flaw.
Of course, testimony is evidence
No it's not! Testimony is not, can not, and will never be considered evidence. We can not have such low standards for evidence that a person could just make something up and it be used to support a hypothesis. And there are many other reasons I have provided before as to why testimony is not evidence.
It may be enough for you to trust zoologists and you may be right. I don't know. The question is, is it really enough. You have stated experts do not have all the answers. Trusting the opinions of zoologists may be an error. Margaret Geller once stated that in science the weight of authority has no meaning. They have been wrong before. This is not a math proof. we have no certainty here.
What you're failing to realize is it isn't just that they say it, but their reasons for saying it are what's important. And their reasons for excluding cryptozoology as a field of science are entirely sound.
I guess he means that, because we assumed that p,p+2 where the largest pair there does not exist any numbers with factors p+2k,p+2(k+1).
As I posted before, this statement is false. (p+2k)*(p+2(k+1)) has precisely those numbers as factors.
I'm having a lot of trouble understanding what it is you're trying to go for.
If we suppose that there exist a larger pair of the form p+2k and p+2(k+1) Than the set of numbers that have as factors p+2k and p+2(k+1) must be equal to 0 (since they do not exist).
We are supposing that there exists a larger pair where p+2k and p+2(k+1) are both prime? This doesn't make sense, we've already supposed that p and p+2 were the largest pair with this property.
And you say that the number of integers that have p+2k and p+2(k+1) as factors must be zero. This is false, just look at the number:
(p+2k)*(p+2(k+1))
This number has both those as a factor.
Ack, another ARB sock puppet.
(At least, I think so...)
Moderation has been discussing this and our (hmm, perhaps my) best guess is no, it is not.
To be fair, RickIsAnIdiot has a point. To say "0.999..." has a limit is wrong. The increasing sequence of adding 9's:
0.9, 0.99, 0.999, 0.9999, ...
This sequence has a limit. And of course, this is what you meant (as context in your post shows).
And I have no idea what you were trying to do here:
limit as x goes to infinity (1-x(.1+.01+.001+.0001...))=0
limit as x goes to infinity (x(.1+.01+.001+.0001...))=1
The limit as x goes to infinity of 1 - x * 0.111... = 0?
Well, the question doesn't state that my contribution of $15 was me paying my third. It could very well be that I was cheap and the candles totaled over $45. Is this a correct interpretation Ganesh?
You would think Anthony R. Brown would at least try to hide his posting style.
Largest Time Shifting of a Human
This is the correct phrasing. What happens is that time passes at different rates relative to how fast you're traveling. So if I leave Earth really fast and my twin stays behind, time for me slows down where as for earth it stays the same. If I return to earth, I will be younger than my twin.
You actually don't need to go fast, just being caught in a gravity well will do the trick. If you sit outside a black hole and are close enough to it, what seems like 1 year for you could be like 100 years for the people on Earth. You come back an all the people you used to know are gone. Or perhaps if you had a terminal disease, you could wait until they find a cure.
Better estimate. This number should be optimal (save actually trying to cut the darn things).
171 legos (each) for the left and right, but in order to get this the legos must be oriented so that they are taller than they are wide.
90 legos for the top, they must be oriented so that they are deeper than they are wide.
Total legos: 432.
I just need to know how many of the 1x2 legos I would need to cover: the right side, the left side, and the top
The back,front, and bottom need to be excluded, as well as the 12x12 section of light bright.
You contradicted yourself here. At first you say you just want the number to cover the right, left and top. Then you seem to imply you want to cover part of the bottom to (specifically, the bottom minus the 12x12). Assuming you don't want to cover the bottom:
Area of the right and left sides: D * H = 323.75
Area of the top: D * W = 151.7
Total area: 799.2
Area of a lego: 2
Number of legos needed: 799.2 / 2 = 399.6 ~ 400. I would buy 500 to be safe since legos will not fit exactly.
Jane,
is the cyclic group of order n.In a group G of order 56, the Sylow 7-subgroups have order 7. Now let's assume we have two Sylow subgroups, P and Q. Their intersection, call it I, has to be not only a subgroup of G, but quite clearly a subgroup of P as well. By Lagrange's theorem, the order of I must divide the order of P. But P's order is 7. So either we have that the order of I is 1 (the identity) or 7 (all of P).
I'll leave it to you to explain why the order of I can't be 7.
This is where
is the integers modulo n. Jane, why the notation with C? I typically see this in applications to physics and chemistry, but normally the group theoretic:
Or the number theoretic:
notations are used by mathematicians. Just curious where you got it from.
Another sock puppet of Anthony R. Brown.
Perhaps a more aesthetic name would be, "The locus of xyz=0".
I too agree that such an object is most likely unnamed. After a few seconds thought, you should realize that indeed most geometric shapes are unnamed.
For the general case, you want to do things... generally. You have two operations, and from your example you can see exactly where your new sets are coming from. The hope is that this is what always happens, that you will be able to prove that applying any more of the operations will give you something you already have.
Why is this? What would be a "good" topology question?
The standard topology of the real line should be used as a place where ideas come from (for example, the idea that a union of two opens sets is open). On the other hand, problems in topology should show you the topological reasons for what things do hold on the real line, why things don't hold without certain assumptions (i.e. Hausdorff), or just investigations of weird spaces.
Again, this is just me and my personal opinion. The more general question is better, and perhaps the example is necessary to find the general solution, but the problem is still rather tedious.
Ok, I think I got it, but if not, I'm close enough. By combining your previous example, and mathsyperson's example (I started with just rationals instead of just irrationals), I got 10 different sets.
I think the thing here is to just be stupid and keep adding on different sets like (4, 5) and [6, 7].
Points that are not limit points also need to be not interior points, right?
Correct.
And just to inject my personal beliefs, this is not a very good topology question. Interesting, sure, and perhaps a good math question in general, but it does not help you with topology much...