Let $R$ be a unique factorization ring and $d$ a nonzero element of $R$. There are only a finite number of distinct principal ideals that contain the ideal $(d)$.
Suppose that $d \neq 0$ is not a unit. Then $d = d_1 ... d_n$ for irreducible/prime elements $d_i$; and suppose that $(d) \subse...
@LeakyNun that polynomial has Galois group $S_4$ over $\Bbb Q$, so it contains a lot of quadratic subfields. Also note that the expression of a number in terms of radicals is non-unique for example, the roots of $x^4-10x^2+1$ may be written as $\pm \sqrt{2} \pm \sqrt{3}$ or as $\pm \sqrt{5\pm 2\sqrt{6}}$
That equation is an example of something I am bad at. For me, I often just conclude it is one equation in two unknowns, and then get stuck. I am bad at special cases like Ax=Bx
Some set theory comments: 1. I need to find a model of ZF-P to mimic the ability for powersets to generate uncountable sets in a more controlled manner order to bypass the problem that no uncountable sets can be explicitly well ordered 2. $\aleph_1$-dedekind finite sets should still survive under ZF+DC I think... since DC only allow you to pick a countable sequence from any arbitrary set with some relation...
@LeakyNun the only explanation I know is that this is similar to the casus irreducibilis in some sense. $\sqrt{78}$ is contained in the expression of the roots, but it's not contained in the field itself, just like the cubic root of unity is not contained in the field in the casus irreducibilis
Some definition: A $\aleph_{\alpha}$-dedekind finite set is a generalisation of a dedekind finite set in that all self injections are bijective and it does not contain a subset of cardinality $\aleph_{\alpha}$
Let $X$ be a set and consider the collection $\mathcal{A}(X)$ of countable or cocountable subsets of $X$, that is, $E \in \mathcal{A}(X)$ if $E$ is countable or $X-E$ is countable. If $X$ is countable, then $\mathcal{A}(X)$ coincides with the power set $\mathcal{P}(X)$ of $X$. Now suppose that $X...
@Daminark let $x_1, \cdots, x_n$ be the roots of an $n$-degree polynomial. $\delta = \prod_{j<k}^n (x_j - x_k)$ is not fixed by $\Bbb S_n$, but $\delta^2$ is. The discriminant is $\delta^2$
@Daminark since $\delta^2$ is a symmetric polynomial, by the fundamental theorem of symmetric polynomials, it is a polynomial in the coefficient of the $n$-degree polynomials
Changing up notation isn't exactly cool/dank so much as confusing... but okay so wouldn't you need to multiply it by $a^2$ then? Like the discriminant of a quadratic is classically $b^2 - 4ac$, I thought
@gian when you first wrote $\mathbb{B}^n$ I was like, okay that's just disgusting
I think I can demonstrate one example. Take an infinite dedekind finite set $D$. Pick out 5 elements $\{x\}_i$ (a choice function always exists in ZF for a finite family) and compute $A=D-\{x\}_i$. Now take $D$ again, but this time, pick out a disjoint set of 5 elements $\{y\}_i$ and compute $B=D-\{y\}_i$. Then it should follow that $|A|=|B|$ since we can demonstrate a bijection by pairing D with D, and the 5 element sets with the 5 element set
I will be surprised if it is consistent with ZF that there can exists sets whoose cardinality of it subsets depends on the nature of the elements being picked out and not just the number
Definition: A filtered space $X$ of formal dimension $n$ is locally cone-like if for all $i$, $0 \le i \le n$, and for each $x \in X^i - X^{i-1} = X_i$ there is an open neighborhood $U$ of $x$ in $X_i$, a neighborhood $N$ of $x$ in $X$, a compact filtered space $L$, and a homeomorphism $h:U \time...
@Adeek lol, I remember just a few months back you were all for the algebra. Ted fears that algebra is replacing geometry in this room but maybe it's the other way around
I've found myself to be more into the topology side than the geometry side, at least the classical diffgeo on curves/surfaces just wasn't my style when we did it over the summer, and I didn't absorb much
Of what I've seen, I've been happy with number theory, combinatorics, algebraic topology, complex analysis, to a lesser degree functional, had a bit of fun with differential topology
Yeah I mean the only subjects I've so far found myself disliking to some degree were the very crunchy side of analysis, and stuff like calculus/differential geometry
And it's not heavy dislike as much as I was moderately interested in the lectures and didn't find the psets that enjoyable. Somehow I don't think I'm as much of a visual thinker, or at least not good at visualizing rigid objects
There may be a chance that I'll be able to process things visually in topology more, my experience in AT so far has been very formal so maybe if I ever have a prof like DCal who believes very strongly in pictures, it'll response better
If I can visualise everything in the foundation of mathematics (because they are often some of the most counterintuitive things) then I will be able to visualise any field of mathematics
Despite that, there are many situations where pictures are more misleading than just working directly with the formalism
Depends on what your visualization is meant to contain.
But yes, I suppose being able to visualize everything in the foundations of mathematics is probably sufficient to visualize literally anything. However, I believe that the premise cannot hold.
Yeah, it is quite apparent when one tries to work with results related to e.g. inaccessible cardinals, or some theorems tied to axiom of choice, (and many of the theorems that I don't know about yet given my novice knowledge on the topic) then pictures will either become too cluttered or is downright misleading
One example I like to quote about how pictures can be misleading is when one tries to plot any dense linearly ordered set such as the rationals and irrationals: You will get a line that looks no different from the real number line (i.e. the drawing will have no holes) and you will get a lot of topology of these sets wrong by just relying on those pictures
Consider the prolem where we need to both minimize and maximize $f(x,y) = x^2+y^2$ subject to $x = y+1$. (So $g(x,y) = x-y=1.)$
Setting up the Lagrange Multiplier equation $\nabla f(x,y) = \lambda \nabla g(x,y)$, we obtain
$$ \langle 2x,2y \rangle = \lambda \langle 1, -1\rangle $$
which implies...
Hey!! Why does it hold that every triangle is contained in some plane? Also we have that a plane is a surface with zero volume, right? Does it then hold also that a triangle is a surface with zero volume?
Akiva, your $\Bbb{Q}\to \Bbb{Q}\setminus \{0\}$ order preserving bijection exercise you gave me the other day allow me to realise one thing about the rationals: It appears that the "global structure" of the rationals is completely determined by the harmonic sequence and the naturals, i.e. Start with the sequence $\pm(...,4,3,2,1,0,\frac{1}{2},\frac{1}{3},\frac{1}{4},...)$. Then the map $\frac{p}{q}\mapsto \frac{p+n}{q+n}$ for denominators $q \in s$, numerators $p \in s$, $n \in \Bbb{Z}$ will
generate all rationals (after omitting nonsensical terms such as division by zero expressions)
So in a sense, the rationals does have a repeating unit
Let $A=[a_{ij}]$ be an $n \times n$ matrix such that $a_{ij}$ is an integer $\forall i,j$. Let $AB=I$ with $B=[b_{ij}]$. Which of the following is true?
(A) If $\det A=1$ then $\det B=1$
(B) A sufficient condition for each $b_{ij}$ to be an integer is that $\det A$ is an integer
(C) $B$ is always an integer matrix.
(D) Necessary condition for each $b_{ij}$ to be an integer is $\det A \in \{1,-1\}$
LOL: Prolonged starvation can leave a body so fragile that eating food can, in itself, be fatal. Filling the stomach only causes blood flow to shift to the digestive tract, denying it to other organs too weak to endure the loss. Only intravenous nutrition and tiny mouthfuls of liquids can ease such a person back from the brink; if neither option is possible, they're Already Dead. Additionally, food has a thermogenic effect; it requires energy to process food before the food can be used as energy. Overfeeding a starving person whose body has shut down a lot of digestive functions puts an eno…
I want to calculate the area of the triangle with vertices (1,1,0), (2,1,2), (2,3,3).
We can parametrize the triangle using the function $\Sigma (x,y)=\left (x, y, 2x+\frac{y}{2}-\frac{5}{2}\right )$, right?
For the boundaries of x and y I have done the following:
From the verices we see that $1\leq x\leq 2$ and $0\leq z\leq 3\Rightarrow 0\leq 2x+\frac{y}{2}-\frac{5}{2}\leq 3\Rightarrow \frac{5}{2}-2x\leq \frac{y}{2}\leq 3+\frac{5}{2}-2x\Rightarrow \frac{5}{2}-2x\leq \frac{y}{2}\leq \frac{11}{2}-2x\Rightarrow 5-4x\leq y\leq 11-4x$. Since from the vertices the smaller value of $y$ is $1$ …
Let $S$ be a set. Then $f(S)$ is a set such that $f(S)$ does not inject into $S$ while $S$ injects into $f(S)$. Find all possible $f$
Here $f$ is not just an ordinary map, but actually a map which take models as domains and spit out models. Therefore there exists a formula $\phi$ such that $\phi(f)$ is true.
Simplest way to enforce an injection is to have $S \subsetneq f(S)$
Therefore, what remains is to determine the internal structure of the set $f(S) - S$
Let $R$ be a unique factorization ring and $d$ a nonzero element of $R$. There are only a finite number of distinct principal ideals that contain the ideal $(d)$.
Suppose that $d \neq 0$ is not a unit. Then $d = d_1 ... d_n$ for irreducible/prime elements $d_i$; and suppose that $(d) \subse...
Consider the prolem where we need to both minimize and maximize $f(x,y) = x^2+y^2$ subject to $x = y+1$. (So $g(x,y) = x-y=1.)$
Setting up the Lagrange Multiplier equation $\nabla f(x,y) = \lambda \nabla g(x,y)$, we obtain
$$ \langle 2x,2y \rangle = \lambda \langle 1, -1\rangle $$
which implies...
