Recent content by Deveno

In my mind, mathematics is the study of logical structures divorced from context. Number is one of those structures.

For part (iii), show for any $a \in \Bbb R^{\ast}$, that there exists $A \in T$ with $f(A) = a$. This shows $f$ is surjective, and then you can use the Fundamental Isomorphism Theorem.

I like your spanning definition, as it dovetails nicely with the notion of generation by a set. I don't follow how $0$ is the value of the empty linear combination *by associativity*, could you explain?

It's a tough nut to crack. On the one hand, if we want a basis to define the *dimension* of a vector space, we ought to choose $\emptyset$ as the basis. On the other, if we want a basis to be a minimal *spanning set*, we ought to choose $\{0\}$ (since it is the only element we can form spanning...

Real polynomials are precisely those complex polynomials that equal their own conjugate-polynomial. Here is an example: $x^2 + x + 1$ is a REAL polynomial. ONE complex root is the complex number: $\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}$, since:$\omega^2 + \omega + 1 =...

$R$ is a ring, $S$ is a ring, and $M$ is an $(R,S)$-bimodule. A typical element of $A$ is: $\begin{pmatrix}r&m\\0&s\end{pmatrix}$ with $r\in R, m\in M$, and $s\in S$. We have an isomorphism (of abelian groups): $A \to R \oplus M \oplus S$ given by: $\begin{pmatrix}r&m\\0&s\end{pmatrix}...

We have the abelian group isomorphism: $\begin{pmatrix}a&b\\0&c\end{pmatrix} \mapsto (a,0,b,c)$ Now let's say $A$ is an abelian subgroup of $(R,+)$, where we consider $R$ as $\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$. Let $H' = \{(a,0,0,0) \in A\}$. By closure, we have $(a,0,0,0) +...

The Fundamental Theorem of Algebra states that every polynomial in $\Bbb C[x]$ splits into linear factors of the form $(x - z_i)$ for roots $z_i \in \Bbb C$. In other words, the roots of a complex polynomial are complex numbers. Since complex-conjugation $z \mapsto \overline{z}$ is a field...

We have: $\begin{pmatrix}0&m\\0&0\end{pmatrix}\begin{pmatrix}r&0\\0&0\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}$

Here is a start: Note that as an abelian group, our ring is (isomorphic) to: $\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$. So, first, convince yourself that any right (or left) ideal must be of the form: $H \oplus \{0\} \oplus K_1 \oplus K_2$, where $H$ is an additive subgroup of $\Bbb...

You're using the word "all" far too loosely. Ask yourself: how do you know any irrational number, such as say, $\sqrt{2}-1$ is reachable on your odometer? Yes, you've created a way of generating a countable number of real numbers, and in fact, I believe it can be proven that all of them are...

Your odometer is an example-we have a finite number of digits (0-9) and we are arranging them in an infinite sequence. The fallacy lies in you imagining that you will create *every* irrational number. In the language of cardinal arithmetic, if we call the cardinality of the natural numbers $N$...

The initial "seed value" of the odometer (its first setting) doesn't matter. If it truly goes through all the permutations, we may as well imagine it starts at $0000\dots$. But then it can be seen that it merely counts, just as the natural numbers do. The odometer *will* reach every...

Yes, that's the one. For definiteness- Suppose your odometer readings are: $a_1 = d_{11}d_{12}d_{13}\dots$ $a_2 = d_{21}d_{22}d_{23}\dots$ $a_3 = d_{31}d_{32}d_{33}\dots$ $\ \ \vdots$ So that, in particular, for any natural number $k$, we have: $a_k = d_{k1}d_{k2}d_{k3}\dots$, where $d_{kj}$...

Apply Cantor's second diagonal method to your list-what is your conclusion? How will you verify that $e$ (for example) is on your list? Surely you can find a series of numbers on your list that grow ever closer to $e$, but that is not quite the same thing. For $e$ to be on your list, the digits...