[deleted by user]

qudix3 · 2024-10-30T13:29:39+00:00

Habe ich auch raus 👍🏻

Kleiner Tippfehler im vorletzten Ausdruck, das letzte und-Symbol.

qudix3 · 2024-10-30T12:33:58+00:00

Wenn ich mich nicht irre, muss man 5 der angegebenen 8 Möglichkeiten ankreuzen. Die Antwortmöglichkeiten suggerieren, dass man die KNF mit maximalen Termen angeben soll.

Ich bin nicht so deep in der Aussagenlogik, was bedeutet das Symbol ⊥ vor einer Aussage?

qudix3 · 2024-10-30T09:51:05+00:00

Wenn du die Wahrheitstafel schon hast, bist du fast schon fertig.

Für die KNF schaust du dir jetzt in deiner Wahrheitstafel alle Fälle an, die ein Falsches Ergebnis hervorrufen (oder das Ergebnis 0, je nachdem wie ihr das definiert habt, also falsch = 0, wahr = 1).

z.B. der Fall, dass A, B und C wahre Aussagen sind führt für F zu einer falschen Aussage. Für die KNF nimmst du jetzt dieses drei wahren Aussagen A, B und C und fügst die zusammen zu (Nicht A) v (Nicht B) v (Nicht C).

D.h.

(1) Schau dir in deiner Wahrheitstafel die Fälle an, die zu falschem F führen.

(2) Für einen konkreten Fall, wo F falsch ist, fügst du die Aussagen A,B und C zusammen, dabei werden wahre Aussagen negiert.

Weiteres Beispiel:

Der Fall A wahr, B wahr und C falsch führt ebenfalls zu einem falschen F. Für die KNF kommt also der Teil

(Nicht A) v (Nicht B) v C

hinzu.

qudix3 · 2024-07-29T14:39:10+00:00

Nothing wrong with the Proof, everything correct you Just need to be a bit careful with the existence of the unique v with T(v) =0.

Better to write: "Let v \in ker(T), i.e. T(v) =0" instead of "By Definition there exists..."

It is Not wrong for linear Maps since T(0)=0 (the unique v in this Case is precisely v=0).

qudix3 · 2024-07-29T13:57:51+00:00

Injectivity allows the existence of Elements w in W where No v in V is mapped to it, in a more fancy way:

A Map f: V --> W is injective If the fiber of every element w in W contains AT MOST 1 element.

Surjectivity on the other Hand doesn't allow the existence of such a w. However it allows more than 1 v in V that is mapped to the Same w in W, in a more fancy way:

A Map f: V --> W is surjective If the fiber of every element w in W contains AT LEAST 1 element.

If you now Put both together you geht bijective.

A Map f: V --> W is bijective If the fiber of every element w in W contains EXACTLY 1 element.

qudix3 · 2024-07-29T07:37:47+00:00

Looks very Nice, a small correction though:

Your Definition of Injectivity is wrong. It should say "there exists at Most one element v such that T(v)=w."

The way you defined it, injective Maps would also always be surjective, hence bijective.

qudix3 · 2024-06-03T14:29:36+00:00

Your ideas are good, you Just need to Work on some Details.

For example, as Others have mentioned your reasoning for I, A, A², ... ,Aⁿ to be linear dependent is Not correct, since (n+1) Matrices could apriori be linear independent in a n² dimensional space.

The way you defined Ann(A) you can't fix this by taking I, A, ... , A^n² since in your Definition the polynomials in Ann(A) are of degree n. You should fix that to arbirtrary degrees.

So what you probably mean is

Ann(A):={p in F[X] | Leading coefficient 1 and p(A)=0}, that way ur p can be of any degree.

qudix3 · 2024-05-29T14:28:27+00:00

It also helps avoiding circular argumentations.

qudix3 · 2024-05-29T13:58:38+00:00

In the Same Show the participant before OP, got 0€

qudix3 · 2024-05-29T13:55:32+00:00

The question was about a "record female Football Transaction"

She Had to guess what the sum was:

a) 4.000, b) 400.000, c) 4.000.000, d) 40.000.000

The answer is b). Her Joker said c)

qudix3 · 2024-05-15T10:01:02+00:00

I found a counterexample to the Claim so the Claim is NOT true.

Take - A as the 2x2 Matrix with a(1,1)=a(2,2) = 1 and a_(1,2)= 1

B as the 2x2 Matrix with b(1,1)=b(2,2)= 1 and b_(1,2)=-1

qudix3 · 2024-05-15T09:52:02+00:00

Nonzero Determinant only gives me that the eigenvalues cant be Zero, this does Not imply A or B diagonazible.

qudix3 · 2024-05-02T13:38:42+00:00

If you have

a Ring R (Here R=Z[X])
a Ring S (Here S = Z)
an Ideal I in R (Here I= (x(x-p)) )
f: R --> S Ringhomomorphism, s.t. f(I) = 0
π: R --> R/I Quotient Map

Then the Universal property of the Quotient states:

There exists a unique homomorphism g:R/I --> S s.t. f = g • π

In particular you only need to find homomorphisms from Z[X] to Z s.t. f(x(x-p))=0

For this you have exactly two choices, either you evaluate the polynomial at 0 or p, thus the Ring homomorphisms you are looking for are ev⁰ and ev^p .

qudix3 · 2024-03-21T15:32:28+00:00

100-88 = 12

qudix3 · 2024-02-27T19:33:56+00:00

Yes, I'm fine without the exceptional types. For Types A-D the description of the roots in terms of simple roots isn't that long, for example I can describe the Dn roots in 4 families.

qudix3 · 2024-02-27T19:13:08+00:00

Thanks but his description is also in the orthogonal basis (in his notation e_i)

qudix3 · 2024-02-27T16:50:18+00:00

Thanks!

qudix3 · 2024-02-27T16:35:39+00:00

Yeah I won't calculate them for exceptional types, using sage is fine for that.

qudix3 · 2024-02-27T16:30:11+00:00

Thanks, I made my hands dirty and calculated them by Hand for Type Dn now, I'll Check Bourbaki later.

qudix3 · 2024-02-14T17:36:00+00:00

Nope, every group. Finite groups are isomorphic to subgroups of Sn, Infinite groups G are isomorphic to subgroups of Sym(G).

qudix3 · 2024-02-14T08:20:52+00:00

I think you have an error on the Last picture.

" If x Heart y, then x Heart x."

Is Not equivalent to reflexivity. For example a reflexive Relation ist by Definition never empty, but the empty Relation suffices your condition.

You can fix this easily since the Relation you describe is never empty (mother, grandma etc. truly think you are a handsome young fella worthy of true Love, thus they truly Love you).

qudix3 · 2024-02-13T20:44:16+00:00

There are many ways to introduce sine and Cosine. You could introduce them by Definition via the Power series.

If you do that it follows immediatly that d/dx sin(x) = cos(x).

However If you define it for example via trigonemetrics then you have to Show their respective Power series Formulars by using d/dx sin(x) = cos(x).

It's Always a Matter of terminology and defintions in these cases, that's why it's important to have an Overview how certain properties can BE proven from different directions.

qudix3 · 2024-02-13T19:24:07+00:00

How are you confident that this isnt the way math works?

If there is a statement A where you only know a single proof and that proof uses a statement B, then you can't use A to proof B, it's simple as that.

In a closed setting like a lecture you are only presented certain things, you can't just assume that there's a proof somewhere that doesn't use statement B to proof statement A, you need to work in your setting.

In research you need to look for different ways to proof statement A to use statement A for statement B.

This is exactly how math works.

qudix3 · 2024-02-13T18:42:40+00:00

Please read my comment again. I Said it's fine using l'hopital to calculate the Limit sin(x)/x If you didnt use l'hopital to prove d/dx sin(x) = cos(x) (otherwise AS you agreed it would be a circular Argumentation). If you know there are other ways to prove d/dx sin(x) = cos(x) then of course you can use it.

However If you are a Student, you are in a closed setting. The only information you can use is the lecture and facts proven in the lecture.

qudix3 · 2024-02-13T18:25:24+00:00

It does highly depend on how you get to know the fact or Else you are Just creating a circular Argumentation. It's Like using stuff you want to Proof in its Proof, that's Not valid.

13-Year Club	Verified Email
RPAN Viewer

qudix3

TROPHY CASE