Recent content by Erland

The three classical geometry problems: Doubling the cube, Trisecting the angle, and Squaring the circle, have no solutions. The proofs of that use field extensions, which includes using linear algebra. So. linear algebra did contribute to the solutions of these famous problems. Of course, the...

Yes of course. Thank you and sorry for the mistake!

Actually, a stronger result than your original can be proved. The Borsuk-Ulam Theorem says that if ##f: S^n\to \Bbb R^n## (where ##S^n## is the ##n##-sphere) is a continuous map, then there exists a point ##x\in S^n## such that ##f(-x)=f(x)##. In the case ##n=2##, an example is that at any given...

So, let us look at two specific examples and use pasmith's method: Example 1. $$y''-5y'+6y=0$$ The characterstic equation ##r^2-5r+6=0## has two distinct roots: ##r_1=2## and ##r_2=3##. Therefore, the characteristic polynomial can be factorized ##(r-2)(r-3)##. Now, we can calculate with...

How does Bloch define ##\le## for natural numbers? I noticed that he used his theorem to prove ##1\le n## for all ##n \in \Bbb N##, but by any reasonable definition of ##\le##, this should be trivial to prove, and then Theorem 1.2.10 would be trivial.

You can also view it this way: ##3\times 2=6## ##3.1\times 2.7=8.37## ##3.14\times 2.71=8.5094## ##3.141\times 2.718=8.527238## ##3.1415\times 2.7182=8.5392253## etc. The limit of this sequence is ##\pi\times e = 8.539734222673567\dots## We can also do the same from above, getting...

For this to work, the functions defined by the power series must be continuous at ##0##, as you pointed out in a later post. This continuity needs to be proved, and this is done by using the property that a power series converges uniformly on compact subsets inside its circle of convergence (or...

We can also view it this way: 1. ##A\rightarrow B##. (Hypothesis) 2. ##A##. (Hypothesis) 3. ##B##. (1, 2: Modus ponens) 4. ##\neg A \lor B##. (3: Introduction of disjunction) Thus: ##A\rightarrow B, \ A\vdash\neg A \lor B##. 1. ##A\rightarrow B##. (Hypothesis) 2. ##\neg A##. (Hypothesis) 3...

It is not clear to me what the OP means by "taking random points that corresponds to hyperreal numbers on the hyperreal line". In the literal sense, we need a probability distribution to do this, and no one is given. I think that what the OP wants is an explicit example of a (positive)...

Smallest infinitesimal?? There is no such thing. More precisely, there is no smallest positive infinitesimal. If there was one, say ##\varepsilon##, then ##\varepsilon/2## would be a smaller positive infinitesimal. And of course there are non-real hyperreals between the positive infinitesimal...

I think the simplest way to do this is to prove that an affine transformation (linear tranformation + translation) changes the measure of by constant factor, namely the absolute value of the determinant of the matrix that gives the linear transformation. Prove this for elementary linear...

It's somewhat unclear to me what you mean by "pulling back". A real or complex valued function on a product space is measurable (w.r.t. the measurable space with the product ##\sigma##-algebra), if the inverse image of open sets are measurable (w.r.t this space). So, no difference here from how...

You are on the right track, I think. The product ##\sigma##-algebra in ##\Bbb R^2## generated by measurable rectangles (that is, cartesian products of one dimensional Lebesgue measurable sets) is not the same as the ##\sigma##-algebra of Lebesgue measurable sets in ##\Bbb R^2##. Your example...

Of course, if ##\varepsilon## is a positive infinitesimal, then ##-\varepsilon## is a negative infinitesimal.

So, putting together the pieces in this thread, the probability for winning a game is $$p^4 + 4p^4(1-p) + 10p^4(1-p)^2 + \frac {20p^5(1-p)^3}{1-2p+2p^2},$$ if the probabilty for winning one point is ##p##.