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John

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Everything posted by John

  1. You've now given, by my count, four different processes here. 1. Sum 2(R/(6p)) for 3 < p <= p_2, where p is prime. 2. Sum 2(R/(6p)) for 3 < p <= p_1, where p is prime. 3. Sum 2(R/(6p)) for 3 < p <= sqrt(p_1), where p is prime. 4. Sum 2(R/(6p)) for p = 5 and p = 7. The first three may all just be the result of mistakes on either of our parts, but the last one seems to be cherry picking values to make the math work. In any case, if we only include p = 5 and p = 7, then (again using imatfaal's example) we get [408/6 - 1] - [408/15+408/21] = 67 - 46.63 = 20.37, whereas imatfaal only found 11 in the range. Is LL supposed to be the lower limit for twin prime pairs (as I'm assuming here) or the lower limit for primes that are members of twin prime pairs (in which case the subtraction step seems off)?
  2. Again, the multiplication by 2 makes sense. I'm asking about the division by 6p. Also, 2(72/30) and 72/15 are the same number, so there's no difference in the results they provide. I only included up to 11 in the first example, because your earlier comments led me to believe we needed to total R/(3p) for p greater than 3 up to and including p_2. If we only include primes greater than 3 and less than p_2 (i.e., primes from 5 to p_1), then using imatfaal's example above, we end up with LL being [408/6 - 1] - [408/15+408/21+408/33+408/39+408/51+408/57+408/69+408/87+408/93] = 67 - 99.6 = -32.6 (rounding to the nearest tenth), which again is a negative number.
  3. That's what I get for replying in a hurry. You're still making that assertion without justification. You divide R by 6, and take that to be the number of multiples of 6 in R. I can see that. Why does dividing (twice) the number of multiples of 6 by some other number clear out composite multiples of that number? If R = 72, for instance, then yes, there are 13 multiples of 6 in that range (I guess we only count 11, though, since 0 and 72 are the end points). Why does dividing 11 by 15 "clear out" multiples of 5? The total idea's been clear for a few posts. It's just those two steps that seem rather odd. But assuming it's all valid, using 7 and 11, and knowing the process now, let's see. R then equals 72; R/6 - 1 = 11; the sum of R/(3p) for p = 5, 7, 11 = 72/15 + 72/21 + 72/33 = 4008/385 = ~10.41, thus LL = 11 - 10.41 = 0.59 Alright. Now, for 11 and 13. R then equals 48; R/6 - 1 = 7; the sum is then 48/15+48/21+48/33+48/39 = 40896/5005 = ~8.17, thus LL = 7 - 8.17 = -1.17 This is the problem that I pointed out a few posts ago, though the fraction is larger this time because I didn't factor the R/6 out of it. Since the lower limit is negative, then (using only this reasoning) the possibility exists that zero pairs of twin primes exist in the range.
  4. I understand the concept, just don't necessarily agree that it works as you claim. Multiplying by 2 is obvious. It's the R/(6p) bit that throws me. For instance, if p were 7, then 2R/(6p) = R/(3p), which in this case I believe would be 62/21. But there are more "relevant products" (with respect to 7--do we add this to R/(3(5)) = 62/15?) in this range than 62/21. I guess you could round, but even then, you must show that this process is correct for all ranges up to infinity. Response to this edited out, as it was redundant. This makes it seem like you're basically restating the twin prime conjecture in a more complicated way. The fact that, assuming a pair of twin primes in some range, there must be at least one pair of twin primes in a larger range, is trivial. And while the number of twin primes in an increasing range increases (at least to a certain point), the conjecture is that it increases to infinity, which you haven't shown. I could still be missing something obvious, of course. Maybe someone else can shed some light, or I'll try to later when I have more time.
  5. You should justify the second sentence for me. I don't see how that sum will give you the number of composite n in R such that n = 6x +/- 1. But, assuming it is valid, the third statement strikes me as incorrect, or at least useless. If I'm understanding you correctly, letting [math]L_l[/math] be the lower limit, and letting your [math]P_2[/math] be the [math]n[/math]th prime, what you end up with is [math]L_l = \frac{R}{6} - \sum_{i=3}^n \frac{2R}{6p_i} = \frac{R}{6}\left(1-\sum_{i=3}^n\frac{2}{p_i}\right)[/math] where [math]p_i[/math] is the [math]i[/math]th prime, e.g. [math]p_3 = 5[/math], [math]p_4 = 7[/math], etc. However, for [math]n \geq 6[/math], that fancy sum is greater than 1 (2/5+2/7+2/11+2/13 = 5112/5005), which means [math]L_l[/math] becomes negative, which doesn't tell us anything about whether there are infinitely many primes in some arbitrary interval, and also further supports the notion that the second statement is incorrect.
  6. I see that Lu is the upper limit for twin primes in R, containing all numbers in R that can be expressed at 6n +/- 1. Pr is the number of composite numbers in R that can be expressed as 6n +/- 1, having no prime factors less than 5. LL is the lowest possible number of twin primes in R. I still don't see how LL equals 2(R/(6p)), though, especially in light of the value given for LL in example 2, which is zero, meaning R must be zero, which is contradictory since R is 48.
  7. I follow you until this sentence. I'm assuming R/6p is supposed to be R/(6p), but why is LL equal to twice this value? And what exactly do you mean when you say that LL is equal to that value for all primes up to P_2? What is L? What is P? I won't comment on the argument itself, since (as my previous comments show) I'm unclear as to what exactly the argument is--or rather, I see the argument, but I don't see the justification for certain steps building to it. Maybe I'm a derp. Regardless, clumsy is alright, though elegance is, of course, preferred. That said, if you provide even a kludgy proof of the twin prime conjecture, as long as it's valid, you'll be famous.
  8. This is all correct (I'm assuming we're assuming amy's already figured out her answers, since we're freely discussing the specific solution). More completely, we know one card has to be the ace of spades. We can either conceptualize that as the equivalent of removing the ace of spades from the deck and calculating the number of four-card hands we can get from the remaining cards, which is C(51,4), or be fancy and multiply the number of ways to choose the ace of spades by the number of ways to choose the other cards, i.e. C(1,1)C(51,4). The latter method is especially useful for number 5 Yeah, I was going to mention in my earlier post that starting with number 3 might be easier, but decided to relegate that to a later post if the OP didn't mention it herself. It's sad how many times I've gotten caught up trying to directly calculate some hairy combinatorics or probability result in situations where simply subtracting a previous result from the total (or 1) would give the correct answer much more easily. But that comes with being a derp, I suppose. :[
  9. I'm assuming you have to make the choice soon, and can't alter it later, else you could just wait and see how the self-study goes before deciding. If so, then I would recommend the third option, for the extra calculus education and especially for the calculus-based physics course. Take what I say here with a grain of salt, as I've not taken algebra-based physics; but I think if you're truly interested in learning the material, and especially if you're planning to focus on a scientific field in college, then algebra-based physics will probably not be very satisfying. As for the forum, the Science Education forum might be a better place, but if it's a problem, then one of the mods will move the thread as needed. No harm, no foul. Best of luck to you, in any case.
  10. Since the ace of spades must be in the hand, the number of cards left to deal is one smaller, as is the number of cards left in the deck.
  11. I may be giving too much away, but remember that, when we factor [math]x^2+2gx+c=0[/math], we'll arrive at [math](x+a)(x+b)=0[/math], where the following must be true: [math]a+b=2g[/math] and [math]ab=c[/math].
  12. They're both valid within their respective domains of applicability. I suppose something is missing with respect to general relativity in the sense that we haven't yet united it with quantum mechanics. As for Euclidean geometry, as far as I know nothing's missing. It's a mathematical system based on intuitive axioms, and I believe it's complete and probably consistent. A consequence of general relativity is that real space isn't Euclidean, but Euclidean geometry is still a good approximation given certain conditions (namely, I think, in regions where gravitational effects are fairly weak). No worries, my response here was mostly tongue in cheek. However, though I currently intend to go into applied mathematics, I don't agree that pure mathematics and its results are worthless or need not exist. If the decimal representation of pi were of finite length, then one could just multiply pi by a sufficient power of 10 to get an integer P, which could then be written as P/1, a ratio of two integers. Since pi is irrational, it cannot be expressed as a ratio of two integers, which means its decimal representation must be infinitely long. Based on some of what you've said so far, you might find finitism somewhat pleasing. It's not exactly the same as what you're talking about here, but you might enjoy looking into some of its ideas.
  13. Though I'm not sure whether you mean GR is wrong or Euclidean geometry is wrong, this is probably incorrect in either case, depending on what exactly you mean by "wrong." Don't let pure mathematicians hear you say that. Also, mathematics that seems to have no application now may find applications in the future. For an example, see the history of number theory. I won't speak for D H, but I will say "perfect" still strikes me as odd terminology for what seemingly amounts to manipulating the natural numbers. I'm not sure what exactly you're asking here. Maybe I'm just too tired. If no one else answers, perhaps you could clarify your question a bit. A circle is simply the set of all points at some distance from a given central point. It's not a physical object, just a geometric shape. It fits into reality insofar as certain objects or regions are approximately (but not perfectly) circular. I'm not sure what you're asking with regards to the "infinite" question. If you're asking about pi's decimal representation, then it's infinitely long because pi is an irrational number, one proof of which can be found here.
  14. Can you provide an example of a problem you've attempted to solve that gave you trouble, along with what steps you took?
  15. Correct. This seems an odd definition of perfection for a mathematical system. I'm not sure how a lack of infinity renders pi meaningless. Assuming that by "getting rid of infinity," you mean nothing can be infinite (including the number of digits after a decimal point), the ratio of a circle's circumference to its diameter would still be approximately pi. You'd need to ask someone better educated in physics to be more sure, but my understanding is that it would be impossible for us to verify that anything physical is in fact a perfect circle, due to limitations in measuring equipment at small scales and eventually due to the uncertainty principle at quantum scales. The degree is defined as 1/360 of a full rotation, therefore a full rotation (i.e. a circle) is 360 degrees. Dropping 90 degrees would result in three quarters of a circle. You could redefine the degree to be 1/270 of a rotation, I suppose, but then we'd just say a right angle is 67.5 degrees.
  16. I'm not sure what you mean by the "nature" of a circle. The ratio of a circle's diameter to its circumference is pi, which is irrational, and the area of a circle is equal to its radius multiplied by pi squared, so in that sense, I guess it is. As for circles existing in reality, you'd be hard-pressed to find anything perfectly circular, especially since we can't measure things like length with infinite precision. I'm also not certain what you mean by "perfect mathematical system," but assuming such a thing exists, irrational numbers aren't necessarily excluded. While we usually think of "irrational" as meaning "unreasonable," in mathematics the "rational numbers" are simply those that can be expressed as a ratio of two integers. Irrational numbers cannot be written as the ratio of two integers, hence their name. As for the last bit, keep in mind that the sphere (a 3D object) is similar in that finding its volume and surface area, based on its radius, also involves pi.
  17. Given C/D = pi, the fact that pi is irrational means C or D is irrational. If C and D are both rational, then since rational numbers are closed under division, C/D cannot equal pi. Just to be clear, when we say a set is closed under some operation, we mean the result of applying that operation to members of the set results in another member of the set. Thus, what I'm saying is that given any two rational numbers, dividing one by the other yields another rational number. Therefore, pi, being irrational, cannot be the result of dividing one rational number by another.
  18. The rational numbers are closed under addition, multiplication, subtraction and division, so no finite combination of rational numbers using those operations will give you an exact result for pi, which is not a rational number.
  19. I'm not sure I follow. If we have an [math]n \times k[/math] matrix with [math]n=k^2[/math] and each element set to [math]\frac{1}{k}[/math], then each column will consist of [math]k^2[/math] elements of [math]\frac{1}{k}[/math] each. Since condition 2 deals with the sum of the squares along each column, we end up with [math]k^2 \left(\frac{1}{k^2}\right) =1[/math], as condition 2 requires.
  20. It's a bit cheesy, but if you let [math]n=k^2[/math] and set every element to [math]\frac{1}{k}[/math], then for sufficiently large [math]k[/math] you'll have such a matrix.
  21. There is actually a flaw in my little proof, which is that it kind of assumes [math]\frac{0}{0} = 1[/math], which is its own related can of worms. This can be avoided by taking a few extra steps, though, and going from there. Regardless, you can manipulate the equation to get into all kinds of silliness, which reinforces the notion that the equation is wrong. Your errors here are common, and I can't really fault you for them (though a quick Google search would help a lot). You're treating infinity as if it were just some very large number (even though you give lip service to the idea that it isn't), and you're misunderstanding the notion of the limit. When we take derivatives or integrals, we're assuming some value (let's call it [math]\Delta x[/math]) is getting closer and closer to (but never actually reaching) zero. [math]\Delta x[/math] gets arbitrarily small, and we call it an infinitesimal in recognition of that fact, but it's never actually zero. So while [math]\lim_{x \to \infty} \frac{1}{x} = 0[/math], the fact remains that [math]\frac{1}{\infty} \neq 0[/math].
  22. This discussion comes up more often than it probably should. Let's assume that [math] \frac{1}{0} = \infty [/math] is valid and true. This means [math] 1 = 0 \infty [/math]. But what happens if we multiply by a real number [math]n \neq 1[/math]? Then we have [math](1)(n) = (0)(\infty)(n) \implies n = (0)(\infty) = 1[/math] But [math]n \neq 1[/math]. Thus [math]\frac{1}{0}\neq\infty[/math]. If we just went with the concept of "putting nothing into something forever," it seems to me that would result in [math]\frac{1}{0} > \infty[/math], which contradicts the definition of infinity as being greater than any value.
  23. The idea is that f(x) is some curve such that at some point on the curve, the tangent is parallel to the (otherwise unrelated) line 3x - 4y = 1 => y = (3/4)x - 1/4. As imatfaal has directly and mathematic has indirectly pointed out, there are infinitely many curves that have this property. Therefore, the answer to the OP's question depends on what, exactly, f(x) is.
  24. As an example (this is encountered early on in courses dealing with proof techniques, and isn't difficult to follow), we can prove that [math]\sqrt{2}[/math] is irrational using proof by contradiction, as follows: Suppose [math]\sqrt{2}[/math] is rational. Then it can be written as the ratio of two relatively prime integers, i.e. [math]\sqrt{2} = \frac{p}{q}[/math] where [math]p[/math] and [math]q[/math] are integers sharing no factor greater than 1. Since [math]\sqrt{2} = \frac{p}{q}[/math], then [math]2 = \frac{p^2}{q^2} \implies 2q^2 = p^2[/math]. Since [math]q[/math] is an integer, then [math]q^2[/math] must be an integer, which means [math]p^2[/math] is equal to 2 times an integer and is therefore even. Since [math]p^2[/math] is even, [math]p[/math] must be even, which means it's also equal to 2 times an integer, say [math]k[/math]. Now, [math]p = 2k \implies p^2 = 4k^2[/math], which, from our earlier calculations, means [math]4k^2 = 2q^2 \implies 2k^2 = q^2[/math]. By the same reasoning we used earlier, this means [math]q[/math] must be even. So we have that [math]p[/math] and [math]q[/math] are both even. But this contradicts our earlier assumption that they're relatively prime, and so [math]\sqrt{2}[/math] must be irrational. Proofs that pi is irrational are a bit harder to follow, but a variety of such proofs do exist. They can easily be found using Google if you'd like to take a gander.
  25. An observer can distinguish between those values, but an observer isn't required. Consider, for instance, gravitation. The force due to gravitation between two objects is, among other things, proportional to the objects' masses. More massive objects will attract each other more strongly than will less massive objects, other things being equal, whether anyone's around to measure that force or not. Much of the rest of what you said boils down to the difference between something and how we describe that something. And yes, beyond a certain point, axioms come into play. If you're looking to ponder truth without resorting to any axioms at all, then I suppose you'd need to talk to a philosopher (for instance, the guy you've most recently been arguing about logic with in this thread ), which is something I'm not, about whether such a thing is worthwhile or even possible. You might this this article vaguely interesting: http://en.wikipedia.org/wiki/M%C3%BCnchhausen_Trilemma As for why 1 + 1 = 2, there are formal proofs to be found. Edit: You added to your post while I was typing, and rendered my last sentence redundant. Also, I'm not sure anyone's claiming that the entirety of mathematics is independent of human thought. My entire point rests with the notion that while the labels we assign to various quantities are constructs, the variations in quantity themselves are not. I take our number systems to be labels assigned to quantities that exist independent of labels, if that makes sense.
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