Jump to content

the tree

Senior Members
  • Posts

  • Joined

  • Last visited

Everything posted by the tree

  1. Of a set S of size |S|=n. It's clear that |SxS|=n2, so n3 binary operations mapping SxS to S. The amount that would make a monoid would be less than that (I feel it's getting a little late for combinatronic questions right now). I don't think there would be a means of listing them all other than brute force.
  2. The definition of a monoid requires a binary operation mapping from SxS to S. Brute force I guess.
  3. That depends on what the binary operation is, in each case, that makes them more than just sets. If you can think of a binary operation that maps {(4,4),(4,9),...} to {4,9}, that meets the conditions to make it a monoid - and another for {(8,8),...} then there may or may not be a homomorphism between the monoids that you will have created. To be clear on that, different binary operations can make for different monoids - even with the same set. For instance, let [imath]S=\{ a,b \}[/imath] and take two binary operations, [imath]\circ_1[/imath] and [imath]\circ_2[/imath] that are defined thusly: [math]\begin{array}{c|cc} \circ_1 & a & b \\ \hline a&a&b\\ b&b&a\\ \end{array}[/math] [math]\begin{array}{c|cc} \circ_2 & a & b \\ \hline a&a&b\\ b&b&b\\ \end{array}[/math] Both [imath](S , \circ_1 )[/imath] and [imath](S , \circ_2 )[/imath] are monoids, but they are non-equivalent because no homomorphism exists between them. Now let [imath]T=\{ c , d \}[/imath] and define [imath]\circ_3[/imath] thusly: [math]\begin{array}{c|cc} \circ_3 & c & d \\ \hline c&c&d\\ d&d&c\\ \end{array}[/math] There exists a homomorphism between [imath]( T , \circ_3 )[/imath] and [imath]( S , \circ_1 )[/imath], making them equivalent monoids.
  4. Notation wise, that's just horrible. And it's quite understandable that you'd be confused. What it's trying to do is to give a more distinct name to the operation. (M,*M) and (N,*N) would be two monoids, consisting of a set each and an operation each - the point of calling the operations *M and *N is to make it clear that the two operations needn't be the same. A homomorphism is a function that preserves structure. You'll see a list of conditions in your text book, but what they amount to is that while each monoid will consist of elements with different names, once you strip that away - if a homomorphism exists between them then they will be equivalent to one another. If you're familiar with the meaning of symmetry, then this is a very similar concept. If not, then you'll just have to work through a lot of examples.
  5. Oh heavens no. The SxS is what * maps from and S is what * maps to. A Cartesian product is technically an example of a binary operation, but for the sake of this discussion - it's not the one that you're talking about. For solid examples of binary operations, you'll find that finite monoids make easily drawn tables. The common vernacular is 'over' the set, but yes, sort of. A monoid is both the set, and the operation. That would be something, but it would not be called a monoid.
  6. Strictly speaking, the Mandelbrot set doesn't have to be generated using imaginary numbers, you can get the same set using geometrical terms but it's messier and there is really no point. But it's pretty unclear what you're asking for. What is it that you'd actually like to know/understand?
  7. Say S were the set of integers, then addition over the integers would be a mapping from SxS to S. Addition is useful, so it is rumored. If you were to draw up a multiplication table then you'd see a fairly obvious example of mapping ZxZ to Z. Other than what? No. Not at all. Lists allow for repetition and have an inherent order. Sets do not have repeated elements and a set written in a different order is the same set. That's the name of the operation in question. e.g. addition over the integers would be written (Z,+).
  8. No. It's about 0.04% off. And for that matter, we've already covered that pi is an irrational number - so the answer there should have more than a little obvious.
  9. "A binary operation on a set S [...] maps elements of the Cartesian product S×S to S" -- WikiP The most obvious example of a binary operation would be addition, or possibly multiplication. Although it should cover any function that takes two inputs and gives one output, all from the same set.
  10. Practically speaking, a hundred sine waves will look an awful lot like an infinite amount. If you were to accurately describe a wave that was truly not a smooth function - then it wouldn't be able to accurately describe the behavior of a speaker's diaphragm, so it would be useless for a synthesizer.
  11. Do you fully understand the concept of a set? Do you fully understand the concept of a binary operation?
  12. No need to patronise the OP, the privilege of education doesn't make you any better than them - and just because they're a little misdirected that doesn't give you the license to be a dick about it.
  13. That is what I meant, I was sort of deliberately avoiding technical language there.
  14. Shoddy copy pasting on my part, sorry. Double the last digit, then subtract it from the rest.
  15. If you take a look at the wikipedia article I linked, you'll see that by taking determinants of the matrix in the middle and of a minor of that matrix, you can see what kind of conic you are looking at.
  16. Yeah but then you need an existing list of prime numbers, which weakens the point of the exercise.
  17. Yeah, no, pretty much every pattern that comes up in modular arithmetic - even if it isn't very interesting, will have some kind of aesthetic appeal to it. Which is why I'm so certain that you should try actually learning some of this stuff rather than just speculating about it. Start learning about the Chinese Remainder Theorem or Fermat's Little Theorem, start devising your own proofs for how that damned divisibility check works (and probably, some point along the line, learn what a mathematical proof is), seriously, just, go learn something.
  18. To check if n is prime, you only really need to check if it's divisible by numbers between 2 and sqrt(n). And even then you can be wasting a fair amount of time by checking every single potential divisor in that range.
  19. That would almost certainly be the 9 times table, or the 32 times table. But let's forgive that. The differences between the digits start off hinting a sort of oscillating pattern indeed but urm... that doesn't really continue unfortunately. 9 7 5 3 1 1 3 5 7 9 0 2 4 6 8 I think that's what we'd call a good attempt. But you're not going to get further until you admit to yourself that this has more to do with the behavior of the number 3 in base 10, than it does any other prime or any general case.
  20. There's no rush. It's a forum not a chatroom. Maybe read up modular arithmetic or something.
  21. Clearly yes, you are getting over excited about the second of the divisibility tests mentioned, that is all.
  22. Okay I have to ask, where on earth did you make the leap between "not divisible by 3" and "probably a prime"? If we pick a couple of numbers, we can run them through any amount of divisibility tests. Try 1921 and 2311 for utterly contrived examples. The last digit in both cases is odd. They've failed the divisibility test for 2 - does that make them both prime? The sum of all digits in both cases is not divisible by 3. They've failed the (your?) divisibility test for 3 - does that make them both prime? The last digit in both cases is not 0 or 5. They've failed the divisibility test for 5 - does that make them both prime? The last digit, subtracted from the rest of the number, is not divisible by 7 in either case. They've failed the divisibility test for 7 - does that make them both prime? The sum of every second digit minus the sum of the remaining digits, is not divisible by 11. They've failed the divisibility test for 11 - does that make them both prime? As it happens, one is prime and one is not, but they both ran through the same filters with the same results. While all these divisibility tests are plenty neat, none of them give results which are a sufficient condition for a number to be prime.
  23. The thing is, it's a really neat way of checking divisibility by 3. But getting over excited about the revelation that primes aren't divisible by three, is just going to make you look stupid especially if you're going to pick a composite number as your example.
  24. That's known as a digital sum. The word vedic refers to pretty much any ancient hindi tradition but not a specific operation. Except that 3398101=73×9907, that is to say that it is not prime. All the digital sum has told you is that 3398101(mod3)=1. Well it's because 10n(mod3)=1 for all n but I think somehow you've still missed what pattern we're talking about. It's also true of EVERY OTHER NUMBER THAT IS NOT DIVISIBLE BY THREE. We're talking about two thirds of the integers that just happens to have the primes as a subset.
  25. Okay really now, if something is transcendental then it has to be irrational. The transcendentals are a subset of the irrationals. Not wanting to scare you too much, but the irrationals are infinitely more numerous than the rationals - numbers don't fit into our systems as neatly as we'd presumed they would.
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.