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Posts posted by John


If I'm understanding you correctly, then the short answer is "yes," the point of intersection will be "the same."
The longer answer is that while R^{2} is a subset of R^{3}, vectors in R^{2} aren't really the same objects as vectors of the form <x, y, 0> in R^{3}. However, R^{2} is isomorphic to (i.e. has the same structure as) the subspace of R^{3} containing these vectors, so results in one will "carry over" to the other.
I hope that explanation isn't too muddy.0 
When you ask about "irregular curves" on the main curve, are you just talking about singular points? In any case, with the caveat that analysis isn't exactly my strong point, I don't know if there's a slick general way to determine whether any given function is smooth, and I wouldn't expect there to be such a method.
In some cases, smoothness or lack thereof is apparent, e.g. in the case of polynomials, sine and cosine, absolute value, the greatest integer function, etc. Simply taking derivatives may yield an obvious pattern and make smoothness (or not) obvious. Along these lines, I would imagine mathematical induction is useful at least some of the time as well, i.e. showing the function in question is continuous, and then showing that if the n^{th} derivative is continuous, then the (n + 1)^{th} derivative must also be continuous.0 
While the given solution to the first problem is probably what's intended, I'm going to be "that guy" and note that from the diagram, we can't assume we're dealing with right angles at all, and thus we can't be sure we simply have a rectangle with one corner "flipped."
In a geometry class, just based on the illustration given, the correct answer (though it might make a teacher roll his eyes) is that there isn't enough information to solve the problem.
Edit: As a fun example, and the one my geometry textbook uses as part of its section on reasoning from diagrams, consider the fallacy of the isosceles triangle.0 
It may just be noting that, given our a and p, the first p  1 positive multiples of a will, mod p, give us the first p  1 natural numbers, though not necessarily in order.
As an example, consider a = 8 and p = 5. Then the first p  1 = 4 multiples of a are 8, 16, 24 and 32. These are congruent (mod 5) to 3, 1, 4 and 2, respectively.
2 
Simply put, Euclidean geometry satisfies Euclid's five postulates. Notably, in Euclidean geometry, the Euclidean parallel postulate holds. While stated by Euclid in terms of transversals and interior angles, this postulate equivalently says that given a line m and a point P not on m, there is at most one line through P parallel to m.
"NonEuclidean geometry" is generally taken to mean any geometry in which the Euclidean parallel postulate (or anything logically equivalent to it) is not assumed. Perhaps the bestknown examples of this are hyperbolic geometry (in which, given m and P as before, there are at least two lines through P parallel to m) and elliptic geometry (in which parallel lines do not exist).
To avoid possible confusion when reading about or discussing this, note that Euclid's first four postulates are enough to prove that parallel lines exist (and thus you may note that elliptic geometry requires throwing out more than just the fifth postulate). However, the statement that "there is at most one parallel to a given line through any point not on the line" cannot be proven from the other four axioms.0 
If a theorem is known to be valid under some set of axioms, and then a valid counterexample is also given under the same set of axioms, then this would simply indicate that the set of axioms is logically inconsistent.
As noted earlier in the thread, there is no complete axiomatization of mathematics. However, we do have axiomatizations for large parts of mathematics, and some of these are strong enough to be subject to Gödel's incompleteness theorems. In this case, the second theorem is of interest. It says that no consistent sufficiently strong axiomatic system can prove its own consistency. Here "sufficiently strong" relates to the ability to make certain statements about arithmetic (more technically, the ability to interpret Robinson arithmetic).
ZFC, which is probably the closest thing we have to an axiomatization of mathematics in general, falls under this umbrella. While we believe ZFC is consistent, we cannot prove it, and thus it's not inconceivable that some inconsistency may eventually be found. If such an inconsistency is found, then we'll be in some trouble, since an inconsistent axiomatic system can be used to derive any statement, thus rendering theorems built on ZFC rather suspect. In that case, we'll need to find some new system on which to rigorously build mathematics, but again we won't be certain the new system is consistent either.0 
I'll assume that by IANAM, you mean 'I Am Not A Moron'.
M probably stands for "mathematician," and I'm not one either, though I may be someday.
I've read from a book about how to write scientific reports that it's always better to provide too much information than too little. I may have over explained things a bit, but I just wanted to be on the safe side.
For the part of (((a+b)/2)^2ab)^0.5x2=ab not being true, I don't get why you're saying that it isn't (more overexplanation, incoming!):
11+13=24 10007+9887=19894
24/2=12 19894/2=9947
12^2=144 9947^2=98942809
11*13=143 10007*9887=98939209
144143=1 9894280998939209=3600
1^0.5=1 3600^0.5=60
1*2=2 60*2=120
1113=1311=2 100079887=988710007=120
Two examples aren't enough to establish the equality, but it can be shown true (though at first glance I thought it was false too). We have
[math]2\sqrt{\left(\frac{a+b}{2}\right)^{2}  ab} = 2\sqrt{\frac{a^{2} + 2ab + b^{2}  4ab}{4}} = \sqrt{a^{2}  2ab + b^{2}} = a  b[/math].
Since addition and multiplication are commutative, it's easy to see that b  a is also a square root here (and in fact is the one we want if b > a, since we're looking for the principal square root), and we can codify this by using the absolute value as you've done.
It may just be associative rule, but the method I used to reach the step 9!=5*24*21*16*9 is what (I believe) makes this algorithm faster than normal.
Please keep in mind that I'm just a high school math enthusiast, and not a legitimate mathematician. Thanks.
I'm not a computer scientist, and I don't know whether the extra required comparisons in your method outweigh any potential savings in terms of multiplication operations, but if nothing else, this does yield some insight into the mechanics of the factorial, even if (as imatfaal noted) it's ultimately an application of the associative (and, maybe more importantly, commutative) properties of multiplication.
Edit: I retract some of this, because I don't actually think any comparisons are added to what's required originally, and this method may require fewer. It is a nice little algorithm, in any case.
1 
We're talking here about primes relative to our initial set. If the initial set is {2, 3}, then 25 is such a relative prime, since neither 2 nor 3 divides 25.
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In post #42, I actually made a mistake in my paragraph about adding just 2 to a set of primes greater than 3. I split the cases where our product is even and where it is odd, but I defined n earlier in the thread to be the product of all the primes in the set, so the product will definitely be even. The rest of the reasoning still applies, though, as far as I can tell.
As for a finite set of primes including both 2 and 3, my point is that I'm not sure this changes the question much versus just talking about the entire set of primes, and I don't know (though admittedly I haven't thought about it a whole lot) how to go about finding successive twin prime pairs. Including 2 and 3 already gives us two of the major properties regarding consecutive twin prime pairs, namely that no even integer is prime and every third odd after 3 is not prime.If we look at the set containing only 2 and 3, then we already see these restrictions in practice. The first prime relative to the set that isn't already a prime regardless is 25, making the first set of successive relative twin prime pairs {(23, 25), (29, 31)}. I don't immediately see an easy way to guarantee that we can always find successive pairs of twin primes relative to a set containing 2, 3 and other primes, but perhaps someone better versed in number theory could easily do it.
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It's not an objection. I'm just saying that finding infinitely consecutive twin prime pairs relative to a finite set of primes seems easily doable, and the question doesn't appear to become difficult until we start with the set of all primes, which yields something related to the actual twin prime conjecture.
What I showed above is that given any set of primes greater than 3 (really greater than 2), we can immediately find consecutive twin prime pairs relative to the set. Similarly if we start with 2 and all primes greater than 3. It's only when we include 2 and 3 that things get interesting, and at that point (since all primes greater than 3 are similar in the sense of being one of two numbers mod 6) we're basically talking about the set of all primes.
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I suppose it could be useful. However, I think we'd ultimately end up just talking about actual consecutive twin prime pairs.
Without being very rigorous, the fact that all primes greater than 3 are equal to 1 or 5 mod 6 means any product involving any subset of such primes will also be equal to 1 or 5 mod 6. Clearly we have plenty of gaps between multiples in which to find our relative twin prime pairs in this case. In particular, any set of four successive even integers will constitute a pair of twin prime pairs relative to the set of primes greater than 3.Adding just 3 to this set means any product of a subset will be equal to 1, 3 or 5 mod 6 (not 0, since any multiple of 3 landing on a multiple of 6 must be an even multiple, and we're not including 2 in our initial set of primes yet). Again, any set of four successive even integers gives us what we want.
Similarly, adding just 2 to the set means any product of a subset will be equal to 1, 2, 4 or 5 mod 6. We're still able to find consecutive twin prime pairs relative to this set. If our product n is even, then we still have that n  1 and n + 1 are prime relative to our set. Each of these will be odd, and the closest any two odd "composites relative to the set" can be in this case is 6, i.e. if c is an odd composite relative to the set, then the closest possible odd composites relative to the set will be c  6 and c + 6. There are a few cases to try here, but I'm fairly certain we can find consecutive relative twin prime pairs for each case. Similar reasoning applies if our product n is odd in the first place.
Finally, adding 2 and 3 both to our set ultimately just yields the question of consecutive twin prime pairs in the first place.
Of course, there's always the chance I've made some error in reasoning or arithmetic that yields the above invalid.0 
Is this a homework question, or is it something you need to know to answer a homework question? It's hard to give hints here without just giving away the answer immediately, but I can try:
If we write a number in base n, then we're using a system of n digits (usually from 0 to n  1), with each position representing a power of n. For instance, in our usual base 10, we write the number 243 to represent 2 * 10^{2} + 4 * 10^{1} + 3 * 10^{0}.So you have this list of numbers written in "base of triangular numbers." Comparing the lists and in light of what we mean by numbers in base n, you might notice the pattern.
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My reading of what you're asking now is whether we can show that the statement "there are only finitely many sets of consecutive twin prime pairs," or even the statement that "there are only finitely many twin prime pairs," is not provable in whatever axiom system (I guess ZF or ZFC). This is not trivial (as far as I can see anyway).
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That's the clarification I'm asking for: are we talking about consecutive pairs of actual twin primes, or are we talking about consecutive pairs of twin "primes relative to the set," i.e. consecutive pairs of numbers not having one of the primes in our original set as a factor?
If the former, then I don't see multiples of the primes in our initial set having any impact one way or the other. If the latter, then we can immediately simplify the question as hinted at before.
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Since the product of any number of primes is by definition not prime, I don't see how generating multiples from a list of primes would have any effect on the infinitude or lack thereof of consecutive twin prime pairs.
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But in your OP you didn't seem to be asking about primes, but rather numbers not having any of a given finite set of primes as a factor. Are you wanting to know about actual prime quadruplets now? Or just prime quadruplets relative to the original set of primes? The latter won't necessarily be of the form 6n +/ 1.
And no, two primes p and p + 4 are not considered twin primes.
It might help if you restate precisely what you're wanting proved.
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Okay. Perhaps I should have said 'prime quadruplets'.
My thought is this. On the face if it, it is possible that there is only a finite quantity of prime quadruplets. This is regardless of the truth of the TPC. It could well be that the first 10,000 primes are enough to ensure that there is a highest quadruplet. A proof that it would not be possible to prove that this is the case seemed like it might be useful, or would at least be a curiosity. But maybe not. I just wondered.
Am I be misreading your proof above? It appears to relevant only to single twin primes, and to more or less restate Euclid's argument.
It is possible, as no proof to the contrary has been found. What I'm reading now is you're looking for sets of prime quadruplets relative to a finite set of primes, i.e. four numbers p, p + 2, p + 6, p + 8 such that no prime in our set divides any of these four numbers. Is that correct? Or is it p, p + 2, p + n, p + n + 2 for some integer n, such that there are no relative primes between p + 2 and p + n?
My proof above uses similar reasoning to Euclid's proof of the infinitude of primes. What my proof says is that, given any finite set of prime numbers, we can find a twin prime pair relative to the set, i.e. numbers p, p + 2 such that no prime in the set divides either number. Moreover, there are infinitely many such "relative twin prime pairs." This is what I understood you to be asking in your OP.
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I think there may be some miscommunication going on here.
A twin prime is a prime with another prime two units away from it. The prime and its twin are referred to as a twin prime pair. So when you say "two twin primes," I'm taking you to mean a single twin prime pair, but you may be meaning something else.Since every third odd number greater than three is divisible by three, the closest two twin prime pairs can be to each other is three units apart. So when we talk about "prime quadruplets," for example, we mean four primes of the form p, p + 2, p + 6, and p + 8. It's an open question whether there are infinitely many sets of prime quadruplets, and if there are, then this immediately implies that the twin prime conjecture is true. The proof of any similar statement involving the infinitude of twin prime pairs with some property also implies that the twin prime conjecture is true.
In your OP, my understanding is that you asked for a proof of the infinitude of twin prime pairs relative to a given set of primes, i.e. pairs of numbers n, n + 2 such that for any prime p in our original set, p does not divide n and p does not divide n + 2. This is what my earlier proof was about.
You seem to be asking for something else now, but I'm still not sure what.1 
My understanding was you were asking for pairs of integers (n, n + 2) such that no prime in our set was a factor of n or n + 2. Are you asking for something else?
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If we have some finite set S of primes, then take the product of those primes. Call this product n. Now consider kn where k is some positive integer. Since each prime in S divides n, each prime in S must also divide kn. Now, it's a fact that consecutive integers are always coprime. Thus, for any kn, we have that neither kn  1 nor kn + 1 is a multiple of any prime in S, and so they must form a pair of "primes relative to the set." Since there are infinitely many positive integers, it follows that there must be infinitely many such pairs.
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While the integral is defined in this context as a limit in terms of increasing numbers of increasingly small units of ndimensional volume, each unit of volume never actually becomes zero. Rather, we ultimately "arrive at" infinitesimally small units of volume, but each of these infinitesimal units still has a particular shape that changes depending on what coordinate system we're using. The Jacobian serves as the ratio of these infinitesimal volumes in one coordinate system to those in another.
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If I'm reading your question correctly, then yes. The sample space is the set of all possible outcomes for a given experiment. Every individual outcome is a subset of the sample space.
For our die roll example, the sample space is {1, 2, 3, 4, 5, 6}. The outcome that the result is odd is the subset {1, 3, 5}, the outcome that the result is 1 or 2 is {1, 2}, the outcome that the result is greater than 5 and odd is the empty set, etc.
1 
The notation for probabilities sort of depends on what exactly you're trying to say, how precise you're trying to be, etc.
As an example, take the case of rolling one sixsided die. Then our sample space is [math]\Omega = \{1, 2, 3, 4, 5, 6\}[/math]. Now say we're interested in the probability that the result of a single die roll (denoted [math]X[/math]) will be both less than five and even. As a shorthand, we might write [math]P(X \textnormal{ is less than 5 and } X \textnormal{ is even})[/math] or [math]P(X < 5 \wedge 2  X)[/math]. If we were writing a paper or something, then we'd probably want to take [math]P[/math] as the function I described in your recent thread, and letting [math]A = \{1, 2, 3, 4\}[/math] and [math]B = \{2, 4, 6\}[/math], we'd write [math]P(A \cap B)[/math]. Of course, it's easy enough to see that [math]A = \{x \in \Omega \mid x < 5\}[/math] and [math]B = \{x \in \Omega \mid 2  x\}[/math], so writing [math]P(A \wedge B)[/math] probably wouldn't get you lynched by the mathematical community, but it'd look a bit strange.
1 
If you take a look at the laws of Boolean algebra and set algebra, you'll notice the logical and set operators share quite a few properties. This perhaps becomes clearer if you consider set builder notation, where we say things like [math]A \cup B = \{ x \mid x \in A \vee x \in B \}[/math] or [math]A^{C} = \{ x \mid \neg ( x \in A) \}[/math] (though for the latter, we'd usually say [math]x \not\in A[/math] instead). That is to say, we define our set operations in terms of combinations of sentences connected or modified by logical operators.
The difference, then, lies in what objects are used with each set of operators. The logical operators are used for combining or modifying logical statements, while the set operators are used for denoting various collections of elements from sets or their complements.
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Determining smoothness of a function
in Mathematics
Posted · Edited by John
Yes, by smooth, I'm taking you to mean "has derivatives of all orders." Do you mean "continuous" or something instead?