reference request - Dummit and Foote as a First Text in Abstract Algebra

I'm wondering how Dummit and Foote (3rd ed.) would fair as a first text in Abstract Algebra. I've researched this question on this site, and found a few opinions, which conflicted. Some people said it is better as a reference text, or something to read after one has a fair deal of exposure to the main ideas of abstract algebra, while others have said it is fine for a beginner. Is a text such as Herstein's Topics in Algebra, Artin's Algebra, or Fraleigh's A First Course in Algebra a better choice?

Here's a summary of the parts of my mathematical background that I presume are relevant. I've covered most of Spivak's famed Calculus text (in particular the section on fields, constructing $\mathbf{R}$ from $\mathbf{Q}$, and showing the uniqueness of $\mathbf{R}$ which is probably the most relevant to abstract algebra) so I am totally comfortable with rigorous proofs. I also have a solid knowledge of elementary number theory; the parts that I guess are most relevant to abstract algebra are that I have done some work with modular arithmetic (up to proving fundamental results like Euler's Theorem and the Law of Quadratic Reciprocity), the multiplicative group $(\mathbf{Z}/n\mathbf{Z})^{\times}$ (e.g. which ones are cyclic), polynomial rings such as $\mathbf{Z}[x],$ and studying the division algorithm and unique factorization in $\mathbf{Z}[\sqrt{d}]$ (for $d \in \mathbf{Z}$). I have only a little bit of experience with linear algebra (about the first 30 pages or so of Halmos' Finite Dimensional Vector Spaces and a little bit of computational knowledge with matrices) though.

With this said, I don't have much exposure to actual abstract algebra. I know what a group, ring, field, and vector space are but I haven't worked much with these structures (i.e. I can give a definition, but I have little intuition and only small lists of examples). I have no doubt that Dummit and Foote is comprehensive enough for my purposes (I hope to use it mostly for the sections on group theory, ring theory, and Galois Theory), but is it a good text for building intuition and lists of examples in abstract algebra for someone who has basically none of this? Will I, more than just learning theorems and basic techniques, develop a more abstract and intuitive understanding of the fundamental structures (groups, rings, modules, etc.)? It is a very large and supposedly dense text, so will the grand "picture" of group theory, for example, be lost? I've heard it is a book for people who have some basic intuition in group and ring theory, and I hesitate to put myself in this category given my description of my relevant knowledge in the paragraph above. Do you think the text is right for me, or would I be more successful with one of the three texts I mentioned in the first paragraph?

Thanks for reading this (lengthy) question. I look forward to your advice!

Is this function bijective, surjective and injective?

$\lfloor\cdot\rfloor: Q \rightarrow\mathbb Z$ with $\lfloor x\rfloor :=$ floor of $x$.

I know a function is injective by using $f(x_1)=f(x_2) \Rightarrow x_1=x_2$
and a function is surjective if each element of the codomain, $y\in Y$, is the image of some element in the domain $x\in X$,
and bijective if the function is both injective and surjective.

I don't know what floor of $x$ is.

calculus - Negation of the Definition of Limit of a Function

Question

Suppose we are dealing with real valued functions $f(x)$ of one real variable $x$. We say that the limit of $f(x)$ at point $a$ exists and equals to $L$ if and only if

$$\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| < \varepsilon } \right)} \right)$$

and show this by the symbolism

$$\mathop {\lim f(x)}\limits_{x \to a} = L$$

Now, what do we say when we want to state that the limit of function $f(x)$ does not exist at $a$? In fact, what is the negation of the above statement? I am interested to obtain the negation with a step by step approach using tautologies in logic. To be specific, I want to start from

$$\neg \left[ {\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| < \varepsilon } \right)} \right)} \right]$$

and then go through it to get the final form of negation (See the example below).

---

My Thought

I just wrote down the two following negations without going through a step by step approach.

$$\forall L,\exists \varepsilon > 0:\left( {\forall \delta > 0,\exists x:0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| \ge \varepsilon } \right)$$

or

$$\nexists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| < \varepsilon } \right)} \right)$$

I want to know weather these are true or not.

---

Example

I will give an example of what I mean by a step by step approach. Consider the following statement

$${P}\implies R$$

I want to take a step by step approach to obtain the negation of the above statement. Here is what they usually do in logic

\begin{align}
\, \neg \left( {P \implies R} \right) &\iff \neg \left( {\neg P \vee R} \right) & \text{Conditional Disjunction} \\
\qquad \qquad \quad &\iff \neg \neg P \wedge \neg R & \text{Demorgan's Law} \\

\qquad \qquad \quad &\iff P \wedge \neg R & \text{Double Negation}
\end{align}

Answer

The first negation is almost completely right. You forgot to negate the implication at the end.

Remember that the negation of an "if, then" statement is not an "if, then" statement. $A \implies B$ has negation "$A \land \neg B$" (read: $A$ and not $B$).

So, "if the sky is blue, then I love cheese" has negation "the sky is blue and I do not love cheese."

We say $\lim \limits_{x \to a} f(x) = L$ if $$\forall \epsilon > 0\text{, }\exists \delta > 0 \text{ such that }\forall x\text{, }|x - a| < \delta \implies |f(x) - L| < \epsilon.$$ Then the negation of this is: $$\exists \epsilon > 0\text{ such that }\forall \delta > 0\text{, }\exists x\text{ such that }|x - a| < \delta \text{ **and** }|f(x) - L| \geq \epsilon.$$

---

UPDATE Here is how to negate the following statement step by step

Negation of "$\lim \limits_{x \to a} f(x)$ exists", i.e., $$\exists L\forall \epsilon > 0\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| \geq \epsilon).$$

We say $\lim \limits_{x \to a} f(x)$ does not exist if:

$\neg[\exists L\forall \epsilon > 0\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$

$\forall L \neg[\forall\epsilon > 0\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$

$\forall L\exists \epsilon > 0\neg[\exists \delta > 0\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$

$\forall L\exists \epsilon > 0\forall \delta > 0\neg[\forall x:(|x - a| < \delta \implies |f(x) - L| < \epsilon)]$

$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: \neg[|x - a| < \delta \implies |f(x) - L| < \epsilon]$

$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: |x - a| < \delta \land \neg(|f(x) - L| < \epsilon)$ (Negation of implication)

$\forall L\exists \epsilon > 0 \forall \delta > 0\exists x: |x - a| < \delta \land |f(x) - L| \geq \epsilon$

soft question - 'Obvious' theorems that are actually false

It's one of my real analysis professor's favourite sayings that "being obvious does not imply that it's true".

Now, I know a fair few examples of things that are obviously true and that can be proved to be true (like the Jordan curve theorem).

But what are some theorems (preferably short ones) which, when put into layman's terms, the average person would claim to be true, but, which, actually, are false
(i.e. counter-intuitively-false theorems)?

The only ones that spring to my mind are the Monty Hall problem and the divergence of $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ (counter-intuitive for me, at least, since $\frac{1}{n} \to 0$
).

I suppose, also, that $$\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e=\sum\limits_{n=0}^{\infty}\frac{1}{n!}$$ is not obvious, since one 'expects' that $\left(1+\frac{1}{n}\right)^n \to (1+0)^n=1$.

I'm looking just for theorems and not their (dis)proof -- I'm happy to research that myself.

Thanks!

Answer

Theorem (false):

One can arbitrarily rearrange the terms in a convergent series without changing its value.

geometry - How to get center coordinates of circles on edge bigger circle?

I want to draw 8 smaller circles on the edge of a big circle. I know the distance between all small circles should be 20. How can I find the center coordinates of the circles? I already know the center coordinates of the first small circle.

What I know:

• Center coordinates of big circle is (100,100)


• Big circle radius is 200


• There are 8 small circles


• Distance between edges of small circles is 20


• Center coorindates of first small circle


• Size of all small circles is equal

What I want to know:

• Center coordinates of small circles

Sketch problem:

Answer

You already have a formula for finding the center of one of the
small circles:

x = xBigCircle + Math.round(200 * Math.cos(phi));
y = yBigCircle + Math.round(200 * Math.sin(phi));

Since you want the small circles all to be the same size and each one

is the same distance from each of its neighbors, they will be evenly
spaced around the circle. Since one full turn around the circle is
the angle 2 * Math.PI, you want one eighth of that, which is
0.25 * Math.PI. Stepping by that angle around the circle eight
times, starting at the first circle, gets you back to the first circle
while finding seven other equally-spaced points.

The centers of the small circles should be at

x = xBigCircle + Math.round(200 * Math.cos(phi + n * 0.25 * Math.PI));

y = yBigCircle + Math.round(200 * Math.sin(phi + n * 0.25 * Math.PI));

where n ranges from $0$ through $7$, inclusive
($0 \leq n < 8$).
The value $n=0$ is just the center of the first small circle,
which you already know.

To make a "gap" of size $20$ between each pair of small circles,
just set the radius of the small circles accordingly.

The distance between centers is 200 * 2 * Math.sin(Math.PI/8),
subtract $20$ from that for the desired gap, then divide by $2$
to get the desired radius.

Can one eigenvalue have two different eigenvectors?

I think the answer is no, but to be precise, is it correct to assume that if we have one eigenvalue that is the same, then the eigenvectors for these have to be the same too?

For example,

$$\begin{gathered}
T(1,0,0) = (0, - 2,0) \hfill \\
T(0,1,0) = (0,0.5,0) \hfill \\
T(0,0,1) = (0,1.5,0) \hfill \\
\end{gathered}$$

The transformation matrix $T$ has two eigenvalues that are zero, but this cannot be the case? The other one is $0.5$.

number theory - Solving $f(x)f(y) = f(x + y)$

I am a little lost trying to derive what form $f(x)$ must have if we know $f(x)f(y) = f(x + y)$ for real inputs $x, y$.

My attempt so far:

Set $y=0$ and we have $f(x)f(0) = f(x)$ meaning either $f(x) = 0$ or $f(0) = 1$. Not sure what to do with this.

What about setting $y=x$? Then $f(x)^2 = f(2x)$. Multiply both sides by $f(x)$ and then $f(x)^3 = f(2x)f(x) = f(2x + x) = f(3x)$ and so on, so $f(x)^n = f(nx)$ for some integer $n \geq 2$. But it's also true for $n=1$ because $f(x)^1 = f(1 \cdot x) = f(x)$ and it's also true for $n=0$ (if we assume $f(0) = 1$) since $f(x)^0 = f(0 \cdot x) = f(0) = 1$, so $f(x)^n = f(nx)$ holds for integer $n \geq 0$.

For $n > 0$: raise both sides to $1/n$ and we get

$f(x) = f(nx)^{1/n}$

I don't really know where I am going with this or if it's even the right track. Am I even allowed to do that in the first place? Am I supposed to be assuming $f(x)$ is real? Or complex? Or positive? Or something? Should I be assuming $x$ and $y$ are complex? I don't really know what assumptions to make exactly. I'm just trying to prove/show that this all implies $f(x)$ has some exponential form but pretending I don't know that yet.

Could use any corrections or a push in the right direction.

Answer

You're doing fine. So far you've managed to identify that either

1. $f$ is everywhere zero, or





2. $f(0) = 1$, in which case $f(nx) = f(x)^n$ for every positive integer $n$.

You can probably also manage to show that for every positive integer $k$, you have $f(x/k) = f(x)^{1/k}$, and then combine these to conclude that for any rational number $r$, $f(rx) = f(x)^r$.

A good next place to look is to say "let's say $f(1) = A$." Then we can work out $f(2), f(3), \ldots$ and $f(1/2), f(1/3), \ldots$, and maybe even $f(r)$ for every rational number $r$ with a little cleverness.

But what about irrationals? To say anything useful there, I believe you need an added assumption like "$f$ is continuous".

Post-comment additions

For things like this problem, it can be really helpful to write down everything in detail, rather than just as notes. You could, for instance, say this:

I'm studying the functional equation
$$
f(x + y) = f(x)f(y), \tag{1}
$$
which I'll assume is defined for $x$ a real number, and that the values taken by $f$ are also real, i.e., that I have
$$
f: \Bbb R \to \Bbb R : x \mapsto f(x)

$$

Lemma 1: If $f(0) = 0$, then $f(x) = 0$ for all $x \in \Bbb R$.

Proof: From equation 1, we have $f(x) = f(x + 0) = f(x) f(0) = f(x)\cdot 0 = 0.

Lemma 2: Assuming $c = f(0) \ne 0$, we have $f(0) = 1$.
Proof: $f(0) = f(0 + 0) = f(0)^2$, so $c = c^2$, hence $c - c^2 = c(1-c) = 0$, when $c = 0$ or $c = 1$. We've assumed $c \ne 0$, hence $c = 1$. QED.

Henceforth we'll assume $f(0) = 1$ and ignore the always-zero solution.

Lemma 3: For any $x\in \Bbb R$, $f(2x) = f(x)^2; f(3x) = f(x)^3$.

Proof: $f(2x) = f(x + x) = f(x) f(x)$ by equation 1. Similarly, breaking up $f(3x) = f(2x) + f(x)$ establishes the second claim.

Lemma 4: For any positive integer $n$, $f(nx) = f(x)^n$.
Proof, by induction: Let $P(m)$ be the statement that for the positive integer $m$, and for every real number $x$, $f(mx) = f(x)^m$. We know that for any real $x$, $f(1x) = f(x) = f(x)^1$, so $P(1)$ is true. Suppose that for some integer $k$, we know $f(kx) = f(x)^k$ (this is our induction hypothesis $P(k)$). Then let's examine $f((k+1) x)$:
\begin{align}
f((k+1)x)
&= f(kx + x) \\

&= f(kx)f(x) & \text{By equation 1} \\
&= f(x)^kf(x) & \text{By the induction hypothesis}\\ &= f(x)^{k+1}.
\end{align}
We see that $P(k)$ implies $P(k+1)$; combining this with the fact that $P(1)$ is true, we find (by induction) that $P(n)$ is true for all positive integers $n$.

...and you continue in this vein. It really helps to know what assumptions you're making in each step.