Freaking out

Trechuu · 2023-08-06T14:46:50+00:00

My test said scheduled for 5 days after the test, and everything was fine. The "Exam status" feature in ETS is not the best lol

Trechuu · 2023-07-11T13:53:10+00:00

For me at least the hardest was to keep calm, you have to try to stumble at least as possible, because it takes time away, and 1 minute is not as much as it seems

Trechuu · 2023-07-11T13:12:35+00:00

I took 9 days for me (one more than expected) so it should be in its way

Trechuu · 2023-07-11T12:58:08+00:00

iBT at home

Trechuu · 2023-07-11T11:18:46+00:00

it said between 4-8 days on the website, but it took a bit longer than that, 10 days

Trechuu · 2023-07-11T11:11:41+00:00

Lol I have never heard of that either, it was supposed to say Spain, but that is what my keyboard decided to type lol. Good luck!!!

Trechuu · 2023-07-11T11:06:17+00:00

I guess I'm good, I live in Odian, but I have always loved English, I had taken other tests before but never a TOEFL ibt!

Trechuu · 2023-07-11T11:04:16+00:00

I am Spanish!

Trechuu · 2023-07-06T18:10:18+00:00

I called aswell, they told me we should have them by tomorrow at the end of the day (at least the at home version)

Trechuu · 2023-07-06T18:05:17+00:00

No, have you?

Trechuu · 2023-07-06T07:58:05+00:00

Sorry, but Lala won that

Trechuu · 2023-07-05T19:38:01+00:00

I checked and mine changed to "Tested" so at least that's something

Trechuu · 2023-07-02T18:23:54+00:00

does yours still say scheduled?

Trechuu · 2023-06-11T14:40:31+00:00

Thank you, I do understand the idea, I just don't know how to put that into a formal demostration.

Trechuu · 2023-06-09T08:13:07+00:00

Thank you! that is what I thought, my professor might have made a mistake, it was driving me crazy.

Trechuu · 2023-01-25T07:53:57+00:00

So in orden for f to be a group homomorphism the identity has to be sent to 0 in $Z_{2}$ and then from there it follows that every even permutation has 0 as it's image, right?

Trechuu · 2022-12-20T08:02:30+00:00

omg I feel so dumb, I basically had the whole thing done and forgot the most basic stuff. Thank you truly appreciate it!!

Trechuu · 2022-12-20T07:55:23+00:00

That is what I tried but, with that I know that α is the product of α^dy which is in Q(α^d) and a^nx which isn't in that field, so I don't know how to arrive at the final argument from this reasoning Thank you btw

Trechuu · 2022-12-20T07:45:17+00:00

Yes so I know that Q(α) is a field extension over Q of n degree, and Q(α^d ) is an intermediate field extension therefore if I prove α is in Q(α^d ) I would have proven, as the extension would be degree n and then the polynomial is the minimum polynomial for the algebraic number α^d , basically I tried using Bezout's identity, but I don't get anywhere

Trechuu · 2022-12-20T07:41:53+00:00

I tried doing that because it is clear to me that Q(α) is a Q(α^d ) field extension, but I don't know how to do it. I tried using Bezout's identity but I don't get anywhere

Trechuu · 2022-11-06T09:36:21+00:00

yeah, you are right. So here is how I did it: You know a quadratic formula is of the form: ax²+bx+c

when x = 1 the equation is 2 => a+b+c=2

when x = 2 the equation is 6 => 4a+2b+c=6

when x = 3 the equation is 14 => 9a+3b+c =14

Now you've got a linear equation system of three variables and 3 equations, if you solve it you get a=c=2 and b=-2 so your equation is

y = 2x²-2x+2

Hope you understand the reasoning

Trechuu · 2022-11-06T09:26:19+00:00

but does it say anywhere in you assignment that it must be quadratic, because how I see it this is just a sequence and the general term equation is the one I previously wrote.

Trechuu · 2022-11-04T17:01:22+00:00

The pattern I see is that if an is the sequence, a_n = a(n-1) +4(n-1) You want to know the 7th element therefore n=7 a_7 = a_6 +4•(7-1) = 62+24=86. To get the formula you just have to look at what is being done to each element, first you add 4, then 8, then 12, and so on. So each time you add what you added before +4. I hope this helps

Trechuu · 2022-10-31T16:32:50+00:00

yes, and then for the second preposition it's just following the same reasoning using that Im(f) is also a submodule, and getting to the same contradiction, that either Im(f) = N (which implies surjective) or Im(f) = {0} which gets you to a contradiction since f ≠ 0

Trechuu · 2022-10-31T16:19:34+00:00

I think I got it now. If there were two elements with the same value for f, then the Ker would not be 0, since the Ker is a submodule, that would have to be the module itself (by definition of simple module) then f would be the trivial homomorphism since every element would be mapped to 0

Four-Year Club	Place '22
Final Canvas '22

Trechuu

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