Yes, removing all unnecessary notation, the result your are looking for is the following:
Let k be a real number. For real numbers n and x, define n\circx=nx+k(x-1). Then (n\circx)\circy=n\circ(xy).
The proof of this is trivial, and your result is the special case of k=-1/8, and where...
Hi Robert,
Your notion of "width" is of course not well defined.
Your idea of distinguishing between numbers with finite and infinite "width" is not a new one. For example, the notion of a definable real number, does perhaps capture your idea, and more.
It is an elementary fact that...
Hi Robert,
My first question about this is: What is the significance of prime numbers here? Could you just as well use any increasing sequence of natural numbers for which the sum of the reciprocals diverges?
You should consider also informing us about the relevant equivalence relation, as the truth or falsity of your statement heavily depends on that information.
For example, if you for your eq.rel. use existence of a bijection, the statement is true, but other relations, like having same parity...
Hi Aija, your statement above is just wrong. From those 3 equations, you should immeditately observe the solution a=b=c=M, where M is any invertible matrix, and the "problem" is to determine the remaining solutions, if any.
You are asking about the number of derangments Sn of a set with n elements. If you are looking for an exact answer, you can either use the recurrence relation Sn+1=n(Sn+Sn-1), or compute the alternating sum Ʃ(-1)in!/i! where i goes from 0 to n.
A solution to An=I is obviously attained if A is a suitable diagonal or rotation matrix, and also for all similar matrices PAP-1, where P is invertible.
Hi, msmath, are you serious about any of the above? I find it all completely incoherent. Would you mind explaining in complete sentences what is going on, and what your motivation for this is?
Hi there, would you mind informing us of your proposed betting procedure, an in particular how you arrive at your numbers?
And again, since your coin is fair, and I suppose you are given fair odds on each bet, all bets, both individually and collectively, will have zero EV. That is a fact...
Hi there, you have observed some cases of powers of 2 being almost in the middle of consecutive squares, and you ask about the next occurence. A quick computer search reveals no such cases with exponent less than 1000. So unless either I or my computer made a mistake, which may very well happen...
The answer is no. The only value for x satisfying the above requirement is x=c, and whether or not f(c) is defined is irrelevant for the existence of the limit.