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So you're asking if there exists a number c such that [math]2(11+12c)^n + 1[/math] is not prime for any n? I'm confused, since as far as I can see, [math]2(11+12c)^n + 1[/math] will always be divisible by three, since it is of the form [math]2k+1[/math]. So it is never prime.

It's infinite.

No.

If this was part of a calculation I did I would probably apply the rule of turning the denominator upside down and multiplying , giving 0/1 , then finishing thinking that 0 divided by an integer is 0 .

 

0

No. The identity [math]\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}[/math] holds only for [math]b, c, d \neq 0[/math]. It is for this reason that, when simplifying an algebraic expression, the variable conditions you write out pertain to the original expression, not the one you end up with after simplification. Ie. [math]\frac{(x-2)(x-3)}{(x-4)(x-3)} = \frac{x-2}{x-4}, x \neq 4 \wedge x \neq 3[/math]. What you're saying by this is that [math]\frac{(x-2)(x-3)}{(x-4)(x-3)}[/math] is the same as [math]\frac{x-2}{x-4}[/math], but ONLY IF x isn't four or three. Because if it were equal to either three or four, the former expression wouldn't be the same as the latter. For example, if x=3, the former expression turns out to be [math]\frac{0}{0}[/math], while the latter -1. And I think we can agree that [math] \frac{0}{0} \neq -1[/math]. And it's the same with [math]\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}[/math]. The two are the same ONLY IF b, c and d aren't zero. If they are, the identity doesn't apply.

1/(1/0) is the same as 1*(0/1).

That gave me an idea, if [math]\frac{1}{(\frac{1}{0})} = 1 \times \frac{0}{1} = 1 \times 0 = 0[/math],

 

then we have [math]\frac{1}{(\frac{1}{0})} = {(\frac{1}{0})}^{-1} = (\frac{1}{0})^{-r} = 0[/math], besides [math](\frac{1}{0})^{0} = \frac{1^0}{0^0} = \frac{1}{?} = ?[/math]

 

we still get [math]\frac{1}{0} = {(\frac{1}{0})}^{+r} = {((\frac{1}{0})^{-1})}^{-1} = {0}^{-1} = \frac{1}{0} = ?[/math]

 

for any [math]-r \in \mathcal{R}^{-}[/math], and [math]+r \in \mathcal{R}^{+}[/math] ...

No. See above.

The denominator of the main fraction is undefined and and the answer should be zero, as 1/undefined = 0

No. Dividing one, or any number for that matter, by an undefined quantity results in yet another undefined quantity. Also, it's not just dividing; doing anything with an undefined quantity gives an undefined result.

[math]\frac{1}{\frac{1}{0}}=\frac{1}{?}=?[/math]. Division by zero is always undefined.

What I'm puzzling over are the values of b for which there is no "small" n.

As in there is no "small" n for which 2*b^n + 1 is prime?

For C++ I would definitely recommend http://www.learncpp.com/

While I'm not sure it directly answers your question, you should think about reading Where Mathematics Comes From by Lakoff and Núñez. While controversial, it does shed interesting light on mathematics as a whole.

OpenGL/DirectX would be my choice, but I don't know what the deal is with them and Java. Google, google, google.

Heh, I completely forgot about this topic, and seeing as I never solved my problem I stopped uploading the videos. I'm pretty sure I already tried Screen Vidshot, but I'll give it a whirl nonetheless.

@Greg Boyles: No, they're not. JS is based on Java, but it's a completely different language. You have specific variable types in Java, and you can't assign one to the other (unless they can be implicitly converted). This can however be easily achieved using generic programming.

Comparisons of Java and C++ are easily available using a simple Google search; try "C++ vs. Java". The Wikipedia article on this topic should give you all the necessary information: http://en.wikipedia.org/wiki/Comparison_of_Java_and_C%2B%2B

I would say expressing the formula for [math]\sum_{n=1}^{m}\frac{2^{n-1}}{3^{2^{n-1}}+1}[/math].

Oops. Right *blush*

For [math]x_1 = x_2 = 1[/math] it does not.

I'm no expert in these matters, but I'm next to positive you'll be needing the Lambert W Function.

You should first learn how to express a line in three dimensions; have a look here. For this specific problem, I would recommend the parametric expression, but to get a better grip on the subject it might not be a bad idea to try and solve this problem using all three expressions.

Hint: The point C lies on the line passing through A and B. How would one go about writing down the equation for this line?

Brilliant! Thanks

Take a union of closed intervals. The uperbound of the union will be the least upperbound, if it exists, of the intervals in question. That least upper bound may or may not be in the union.

This makes no sense to me. If I understand the concept of upper bound correctly, the upper bound of the union will be the larger upper bound of the two, ie. if we have [a, b] and [c, d] with b < d, b > c (so overlapping intervals), the union of the two will be [a, d]; d is not the least upper bound.

Furthermore, a union of two or more sets is the set containing every element which is contained in at least one of the former sets. If the intervals are closed then the bounds are included in the interval, and therefore will be included in the union.

And regardless, all of the intervals considered here are symmetric, ie; one interval is always a subset of the other, ie. the union will always result in the larger of the two as demonstrated in the OP. I don't understand how a union of closed intervals can result in an open interval, let alone how a union of symmetric intervals can result in an open interval.

The open interval (-r,r) can be realized as a union of intervals [-q_n,q_n] with each q_n rational by choosing q_n converging to r, where r vcould be any real number, rational or irrational.

 

If you like let r be pi and q_n the decimal approximation of pi to n decimal places.

I understand what you're saying, but I still don't see the connection to whether or not Tx is a topology. How does this prove that, let's say T3, isn't closed under arbitrary union or finite intersection?

Thanks for your help DrRocket, I appreciate it. However, I'm having trouble understanding the following

The others are ruled out because unions of closed intervals can be open intervals and because irrationals can be arbitrarily approximated by rationals and vice versa.

Do you think you could go into more detail, ideally give a specific example? Please understand, this is relatively new to me, I've yet to become comfortable with this type of thinking. How can unions of closed intervals be open intervals? And how does approximating irrational numbers have any bearing on the problem?

Okay, this confuses me a little; why would someone who knows enough about topology to write a book make such an obvious mistake. Also, while it's true that unions of intervals do not have to result in intervals, this is only true if the intervals have an empty intersection (except the case [math](a, b) U [b, c)[/math]), which if my logic is correct is only the case in T1. Let's be specific; could you give me an example of two intervals in T2 (ie. of the form (-r, r), r is real) who's intersection or union is not in T2? Or even better a hint so that I might find them myself?

I've started reading Sid Morris's Topology Without Tears and I'm stuck. Here's the exercise:

Let R be the set of all real numbers. Precisely three of the following ten collections of subsets of R are topologies. Identify these and justify your answer.

T1 = R, {} and every interval (a, b), with a, b any real numbers.

T2 = R, {} and every interval (-r, r) with r any positive real number.

T3 = R, {} and every interval (-r, r) with r any positive rational number.

T4 = R, {} and every interval [-r, r] with r any positive rational number.

T5 = R, {} and every interval (-r, r) with r any positive irrational number.

T6 = R, {} and every interval [-r, r] with r any positive irrational number.

T7 = R, {} and every interval [-r, r) with r any positive real number.

T8 = R, {} and every interval (-r, r] with r any positive real number.

T9 = R, {}, every interval (-r, r) and every interval [-r, r] with r any positive real number.

T10 = R, {}, every interval [-n, n] and every interval (-r, r) with r any positive real number.

Now, T1 is easy; not a topology. T2 through T10 though, that's a different story. Take T2 for example. I make the (possibly incorrect) assumption that if the union of any two sets from T2 is in T2 then every finite or infinite union of any number of sets from T2 is also in T2. The same with intersections.

I consider three different scenarios with two intervals (-r1, r1) and (-r2, r2); a) r1 < r2, b) r1 = r2, c) r2 > r1:

a) The intersection of (-r1, r1) and (-r2, r2) with r1 < r2 is always the interval (-r1, r1), which is in T2.

The union of (-r1, r1) and (-r2, r2) with r1 < r2 will always be the interval (-r2, r2), which is in T2.

b) Both the intersection and union of (-r1, r1) and (-r2, r2) with r1 = r2 is always the interval (-r1, r1) = (-r2, r2), which is in T2.

c) The intersection of (-r1, r1) and (-r2, r2) with r1 > r2 is always the interval (-r2, r2), which is in T2.

The union of (-r1, r1) and (-r2, r2) with r1 > r2 will always be the interval (-r1, r1), which is in T2.

Therefor, I conclude that T2 is a topology on R. Unfortunately, the exact same reasoning can be applied to all the collections T2 through T10, which leads to the result that all collections T2-10 are topologies, but only 3 of them are. What am I doing wrong?

At a second glance, I think I misunderstood the former excerpt. I understood cutting and gluing in topology to be the same as division by zero in algebra (or anywhere else). Thanks dave.

I've been meaning to ask this for a while. From a layman's point of view (me being the layman), Perelman's solution breaks the most basic rules of topology.

Topology [...] is a major area of mathematics concerned with properties that are preserved under continuous deformations of objects, such as deformations that involve stretching, but no tearing or gluing.

Completing the proof, Perelman takes any compact, simply connected, three-dimensional manifold without boundary and starts to run the Ricci flow. This deforms the manifold into round pieces with strands running between them. He cuts the strands and continues deforming the manifold until eventually he is left with a collection of round three-dimensional spheres. Then he rebuilds the original manifold by connecting the spheres together with three-dimensional cylinders, morphs them into a round shape and sees that, despite all the initial confusion, the manifold was in fact homeomorphic to a sphere.

So...what's the deal?

Or another puzzle:

 

Let 0.999... = x

 

Subtracting equals from equals:

 

9.999... = 10x

- .999... = -1x

 

9.000... = 9x

 

Dividing through by 9

 

1.000... = x

 

But we started with x = 0.999...?!

 

No wonder limits seemed so troubling to Leibniz and Newton in the 17th century.

http://lmgtfy.com/?q=0.999+%3D+1

There are many ways to solve simultaneous equations; my favorite is the substitution method.

Hey all,

I read this article on the Riemann hypothesis:

http://plus.maths.org/content/music-primes

It's the first article I've read so far that actually showed me the implications the Riemann hypothesis has on prime's. Unfortunately, after reading the article I googled "Riemann harmonics" and didn't find anything. I'm not that surprised that they made the name up and I would like to read more about them; what are they called?

Thanks.