Recent content by ThirstyDog

I would agree with you that part (ii) of the proposition is incorrect (unless matrices are not acting on vectors from the left). If you look at bijection part http://en.wikipedia.org/wiki/Bijection,_injection_and_surjection it reads: "If g o f is a bijection, then it can only be concluded...

I think the easiest way to do it is consider the two E's as a single letter thus your words are built from D I S C R EE T - thus you have 7 distinct "letters". As for have the ordering between the first and last letter, you can ignore this and calculate all possible words then divide by 2 to...

I know this is slightly off topic but I remember an interesting talk in which it was stated that there an infinite number of primes, p, such that the is a prime, q, with the property p < q < p + 16. They proved this in an attempt to prove the twin prime conjecture.

If na = 0 (mod m) then na = km for some k. To find all a consider d = gcd(n,m), if we let a = m/d * l for l any positive integer then a is guaranteed to be an integer as d divides m and as d divides n we must have na = m * (n*l/d) which implies na is a integer multiple of m. I assume that a =...

I was under the impression that it was more sum S_{p} = \sum_{n=1}^{\infty} \frac{1}{n^{p}} was only convergent for p greater than one. But if you take its analytic continuation of the function then you obtain the result [tex] S_{-1} = \frac{-1}{12}. [/itex] So I think it reduces to a question...

Yes it is true. Say F maps from A to B then: 1) As F is onto every element in B corresponds to at least one element in A i.e. for all b \in B there is at least one a \in A such that f(a) = b . 2) As F is 1-1 every element in B can correspond to at most one element in A ie. f(a) = f(a')...

The only thing you will need is f \circ g = g \circ f \hspace{0.3cm} \Rightarrow f(g(v)) = g(f(v)). If the linear maps are matrices, A and B, then \circ is just the usual matrix multiplication which gives AB = BA. This is exactly what you had. I don't think that there is anything else...

I hope this answers in part your questions: If you have two linear maps f: V \rightarrow W \mbox{ and } g: V' \rightarrow W' if you want the two maps to commute then what you are saying is you want the composition of maps to satisfy f \circ g = g \circ f \hspace{0.3cm} \Rightarrow...

Ok. I interpreted what you said as multiplication by a group element is injective and surjective (hence invertible) thus proving 1 must lay in the set you defined, rather than you proving it is injective and hence invertible.

Matt provides a nice statement which is conceptually useful in terms of groups but it doesn't answer the actual question. It assumes that the binary operation (along with the associated set) satisfy the group axioms. In which case it must be that every element x has an inverse. It appeared to...

I am going on that it depends if you keep looking under cups once you have found a six. The logic defying component might just be dependent on this.

To understand this try considering Linear Diophantine equations modulo p. Note: gcd(x,p) = 1 for all 0 < x < p.

The point I was making was that although through the rearranging you can get the right answer in the end in the middle you made statement which were not entirely true. This is does not matter significantly if you are just looking for a sketch of the proof or aren't required to be completely...

If you are referring to a system of coupled ODE's then 'strong' eigenvalues correspond to the dominate eigenvectors. When I say dominate I mean ones that when time tends to infinity that the system follows a straight line given by the eigenvector. Suppose you have two coupled 1st order ODE's...

To go from parametric form x = sin(t) \mbox{ and } y =cos(t), to the form x^{2} + y^{2} = 1 you do not use the rearrangement as you have given. What you do is consider x^{2} + y^{2} = sin^{2}(t) + cos^{2}(t) = 1. The last step is using a trigonometric identity. The...