If you want a job, go to a community college, take a year or two of welding classes, and go forth.
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Fun conversation with one of our welding instructors last week: "Yeah, the college certainly pays a reasonable wage, but I have students who, with only a year of classes under their belts, are making more than I do. Most of my students end up making four or five times what I do."
"Grinding" in a game usually refers to repetitive and unrewarding gaming which is done in order to level up so that one can actually beat the next boss or whatnot.
@YourLordJoyBoy Sure. I'm just pointing out that having a college degree is not a guarantee of getting a job. You should go to college because there is something you are interested in and want to study. The fact that you are, perhaps, qualified for certain jobs is a perk, not the point.
@XanderHenderson Let me rephrase what I was saying before. I'm getting a job once I graduate. But that doesn't necessarily require me to have this degree. My point is that I can't have a job and be getting my degree too, it's too much for me.
But, in my area, there is a profound demand both for skilled workers and nurses. Train up in one of those areas, and (a) you are basically guaranteed to find work and (b) it is going to pay quite well.
@Jakobian I've welded. It is a lot of fun. I am terrible at it, but it is something that I would like to get back to at some point. I have ideas for building fractal-ish objects out of steel.
I really want to play with the CNC machine on campus.
@XanderHenderson wielding is precise enough for fractals? I'd assume you need something like this plastic machine that got popular.. I forgot the name of it
@Jakobian Do it correctly I mean. Obviously I'd need a printer and the needed files.
And 3d printers require 3d printer money :P
And since someone mentioned games, I have a confession. I am a huge Shadow The Hedgehog fan. Yet, I never got through Sonic Adventure 2, nor did I play Sonic Heroes or finish Shadow The Hedgehog or even play Sonic 06.
@Jakobian Depends on what you want to do, and how big you want it to be.
@YourLordJoyBoy Sega made a total of three Sonic games: Sonic the Hedgehog, Sonic the Hedgehog 2, and Sonic the Hedgehog 3 + Knuckles. There are no others.
@YourLordJoyBoy I don't know what you are trying to type there, but you can either write it as a fraction, using \frac{p}{q}, or use inline division with \div.
@SoumikMukherjee "Zoomer" is what I call a person from Gen Z (I think others use this term, too). I believe that this cohort is the group of people born (roughly) between 1998 and 2010. They are characterized by access to cell phones and social media from a young age, and by being in school / college during the COVID pandemic (thus they took many classes via Zoom; they are Zoomers).
how is $g:\mathbb{R}^{*}\to \mathbb{Z}_2$ defined by $g(x)=0$ if $x>0$ and $g(x) = 1$ if $x<1$ a homomorphism? $g(3\cdot -1) = g(-3) = 1$ but $g(3)g(-1) = 0$
@XanderHenderson no I'd say there are times when someone is too young and that's when they haven't developed their critical thinking skills and so on. A young person absorbs concepts like a sponge and that's bad for politics because of how much garbage it contains. That's what I think
@Jakobian You might be interested in some of the work on moral development done by folk like Piaget, Kohlberg, and the like. People's relationship with authority develops only by experimenting with those relationships.
You can't really have political opinions or thoughts on politics if you don't experiment with those thoughts and ideas. Kids (even young ones) need to be given the space to expression political ideas, and to have people around them push back and test those ideas.
@Obliv I just act like myself when I meet new people. And funny enough, I've been in class with a friend for months but only met him for the first time like a week or three ago.
That's like the most common bad advice is to just "act like yourself" whatever that means. I get the gist is to just be calm and not overthink but yeah I overthink
for surjectivity does this make sense: Let $h: \mathbb{Z}_{16}\to\mathbb{Z}_4$ defined by $h([a]_{16})=[3a]_4$. $h$ is surjective since for any $a \in \mathbb{Z}_4$ we have $a\mapsto 3a\pmod 4$
@Obliv this is composition of the quotient map $\mathbb{Z}_{16}\to\mathbb{Z}_4$ given by $[a]_{16}\to [a]_4$ and multiplication by $3$ in $\mathbb{Z}_4$
@LuckyChouhan It's going. At the moment I'm currently doing both the study guide alongside some homework. Figured the homework should be done before the school site goes on maintenance.
yeah nvm it doesn't even matter. for any $a \in \mathbb{Z}_4$ we have $a \mapsto 3a\pmod 4$. $h([a]_{16}[b]_{16})=h([a+b]_{16})=[3(a+b)]_{4} = [3a+3b]_{4} = [3a]_4+[3b]_4=h([a]_{16})h([b]_{16})$. I guess it does technically matter
The kernel is $a \in \mathbb{Z}_{16}$ s.t. $3a \equiv 0 \pmod 4$ so $a \in \{0,4,8,12\}$, why does what I'm reading say it's the cycle group $\mathbb{Z}_4$ generated by $[4]_{16}$
why would you call that "the cycle group $\mathbb{Z}_4$ generated by $[4]_{16}$" and not just the cycle group generated by $[4]_{16}$ which is isomorphic to $\mathbb{Z}_4$
I mean it's not an official solution so I could just be picking a fight with some random person on the internet
OH
Part of the question was identify the cycle group which the kernel is isomorphic to. I should probably read the whole question
@Obliv when doing group theory its common to not specify we are actually thinking about an isomorphic copy of $\mathbb{Z}_4$. Similarly when doing topology you might be thinking of an isomorphic copy of say, $[0, 1]$
Now that I think about it, it's maybe better to start with abstractions and then once people get used to it, only then give specific examples. It may take time, but the understanding will be more solid than to first look at examples and then go for abstractions
@SoumikMukherjee It takes time regardless, we don't think abstractly until later in life and even then it's only because we focus on it i.e, go to school.
I know it's isomorphic to $\mathbb{Z}$ because it's in the problem statement, but isn't $a \in \mathbb{Z}\times\mathbb{Z}/\langle(1,1)\rangle$ given by $(x\pmod 1,y\pmod 1)$
How do you think about $\mathbb{Z}\times \mathbb{Z}/\langle (1, 1)\rangle$? You identify two elements if they differ by a multiple of $(1, 1)$ i.e. if they differ by $(n, n)$ for some integer $n$
obliv: the subgroup generated by g always contains the inverse of g, more or less by definition. it is, among other things, a subgroup
this might be a thing about mixing additive and multiplicative notation. if you write elements of ZxZ with multiplicative notation, "(1,1)^{-1}" is (-1,-1)
in additive notation the statement would be that the subgroup generated by g in G is {ng: n in Z} where ng has an "iterated addition of g" intepretation when n is positive and "iterated addition of the inverse of g" in interpretation when n is negative
obliv: and g^{-1} might not be some iterated sum of a nonnegative number of elements of g (as for example, in the integers, -1 is not of the form 1 + 1 + 1 + ... + 1 (n times) for any nonnegative n, and only becomes that by some kind of abuse of notation of what it will mean to add something to itself a negative number of times)
Why did it take me so long to find this smh $$W_{n}\left(e^{-\operatorname{floor}\left(\ln x\right)}xF\left(\ln x\right)\right)=F\left(\ln x\right)=W_{n}\left(y\ln y\right)$$
so then if $\langle (1,1)\rangle = \{(n,n)\mid n \in \mathbb{Z}\}$ then we have the set $\{(1,2),(2,3),...\}$ which is isomorphic to $\mathbb{Z}$ *forgot the negatives again lol
obliv: i'd ask the same question about "throwing away the pairs," but maybe a reset of sorts would help. often when showing that two things A and B are isomorphic, one exhibits (or somehow proves the existence of) a function from A to B. it might help to write down what A is here, and what B is here, whatever is playing the role of the function here (i.e. specifically saying, given one element of A, what the corresponding element of B is)
if you are into isomorphism theorems, there is a version of this where you prove a quotient object A = C/D is isomorphic to B by exhibiting in the first instance not a function not from A to B, but from the "non-quotiented" object C to B, and then proving stuff about that function
john: i don't see any problem with it, at least if the linked post is not needed to understand the question. i wouldn't rely on an external link to provide any necessary context for the question, but for true "background" material, i would almost certainly prefer an external link to someone just pasting it into the OP
@Obliv my question is as such: If a real analytic function $f$ is involutive i.e. $f(f(x))=x$ and its Mellin transform can be taken, does this imply that $\tilde f$ inherits a functional equation as well?
